Wavefunctions as gravitational waves

This is the paper I always wanted to write. It is there now, and I think it is good – and that‘s an understatement. 🙂 It is probably best to download it as a pdf-file from the viXra.org site because this was a rather fast ‘copy and paste’ job from the Word version of the paper, so there may be issues with boldface notation (vector notation), italics and, most importantly, with formulas – which I, sadly, have to ‘snip’ into this WordPress blog, as they don’t have an easy copy function for mathematical formulas.

It’s great stuff. If you have been following my blog – and many of you have – you will want to digest this. 🙂

Abstract : This paper explores the implications of associating the components of the wavefunction with a physical dimension: force per unit mass – which is, of course, the dimension of acceleration (m/s2) and gravitational fields. The classical electromagnetic field equations for energy densities, the Poynting vector and spin angular momentum are then re-derived by substituting the electromagnetic N/C unit of field strength (mass per unit charge) by the new N/kg = m/s2 dimension.

The results are elegant and insightful. For example, the energy densities are proportional to the square of the absolute value of the wavefunction and, hence, to the probabilities, which establishes a physical normalization condition. Also, Schrödinger’s wave equation may then, effectively, be interpreted as a diffusion equation for energy, and the wavefunction itself can be interpreted as a propagating gravitational wave. Finally, as an added bonus, concepts such as the Compton scattering radius for a particle, spin angular momentum, and the boson-fermion dichotomy, can also be explained more intuitively.

While the approach offers a physical interpretation of the wavefunction, the author argues that the core of the Copenhagen interpretations revolves around the complementarity principle, which remains unchallenged because the interpretation of amplitude waves as traveling fields does not explain the particle nature of matter.

Introduction

This is not another introduction to quantum mechanics. We assume the reader is already familiar with the key principles and, importantly, with the basic math. We offer an interpretation of wave mechanics. As such, we do not challenge the complementarity principle: the physical interpretation of the wavefunction that is offered here explains the wave nature of matter only. It explains diffraction and interference of amplitudes but it does not explain why a particle will hit the detector not as a wave but as a particle. Hence, the Copenhagen interpretation of the wavefunction remains relevant: we just push its boundaries.

The basic ideas in this paper stem from a simple observation: the geometric similarity between the quantum-mechanical wavefunctions and electromagnetic waves is remarkably similar. The components of both waves are orthogonal to the direction of propagation and to each other. Only the relative phase differs : the electric and magnetic field vectors (E and B) have the same phase. In contrast, the phase of the real and imaginary part of the (elementary) wavefunction (ψ = a·ei∙θ = a∙cosθ – a∙sinθ) differ by 90 degrees (π/2).[1] Pursuing the analogy, we explore the following question: if the oscillating electric and magnetic field vectors of an electromagnetic wave carry the energy that one associates with the wave, can we analyze the real and imaginary part of the wavefunction in a similar way?

We show the answer is positive and remarkably straightforward.  If the physical dimension of the electromagnetic field is expressed in newton per coulomb (force per unit charge), then the physical dimension of the components of the wavefunction may be associated with force per unit mass (newton per kg).[2] Of course, force over some distance is energy. The question then becomes: what is the energy concept here? Kinetic? Potential? Both?

The similarity between the energy of a (one-dimensional) linear oscillator (E = m·a2·ω2/2) and Einstein’s relativistic energy equation E = m∙c2 inspires us to interpret the energy as a two-dimensional oscillation of mass. To assist the reader, we construct a two-piston engine metaphor.[3] We then adapt the formula for the electromagnetic energy density to calculate the energy densities for the wave function. The results are elegant and intuitive: the energy densities are proportional to the square of the absolute value of the wavefunction and, hence, to the probabilities. Schrödinger’s wave equation may then, effectively, be interpreted as a diffusion equation for energy itself.

As an added bonus, concepts such as the Compton scattering radius for a particle and spin angular, as well as the boson-fermion dichotomy can be explained in a fully intuitive way.[4]

Of course, such interpretation is also an interpretation of the wavefunction itself, and the immediate reaction of the reader is predictable: the electric and magnetic field vectors are, somehow, to be looked at as real vectors. In contrast, the real and imaginary components of the wavefunction are not. However, this objection needs to be phrased more carefully. First, it may be noted that, in a classical analysis, the magnetic force is a pseudovector itself.[5] Second, a suitable choice of coordinates may make quantum-mechanical rotation matrices irrelevant.[6]

Therefore, the author is of the opinion that this little paper may provide some fresh perspective on the question, thereby further exploring Einstein’s basic sentiment in regard to quantum mechanics, which may be summarized as follows: there must be some physical explanation for the calculated probabilities.[7]

We will, therefore, start with Einstein’s relativistic energy equation (E = mc2) and wonder what it could possibly tell us. 

I. Energy as a two-dimensional oscillation of mass

The structural similarity between the relativistic energy formula, the formula for the total energy of an oscillator, and the kinetic energy of a moving body, is striking:

  1. E = mc2
  2. E = mω2/2
  3. E = mv2/2

In these formulas, ω, v and c all describe some velocity.[8] Of course, there is the 1/2 factor in the E = mω2/2 formula[9], but that is exactly the point we are going to explore here: can we think of an oscillation in two dimensions, so it stores an amount of energy that is equal to E = 2·m·ω2/2 = m·ω2?

That is easy enough. Think, for example, of a V-2 engine with the pistons at a 90-degree angle, as illustrated below. The 90° angle makes it possible to perfectly balance the counterweight and the pistons, thereby ensuring smooth travel at all times. With permanently closed valves, the air inside the cylinder compresses and decompresses as the pistons move up and down and provides, therefore, a restoring force. As such, it will store potential energy, just like a spring, and the motion of the pistons will also reflect that of a mass on a spring. Hence, we can describe it by a sinusoidal function, with the zero point at the center of each cylinder. We can, therefore, think of the moving pistons as harmonic oscillators, just like mechanical springs.

Figure 1: Oscillations in two dimensionsV-2 engine

If we assume there is no friction, we have a perpetuum mobile here. The compressed air and the rotating counterweight (which, combined with the crankshaft, acts as a flywheel[10]) store the potential energy. The moving masses of the pistons store the kinetic energy of the system.[11]

At this point, it is probably good to quickly review the relevant math. If the magnitude of the oscillation is equal to a, then the motion of the piston (or the mass on a spring) will be described by x = a·cos(ω·t + Δ).[12] Needless to say, Δ is just a phase factor which defines our t = 0 point, and ω is the natural angular frequency of our oscillator. Because of the 90° angle between the two cylinders, Δ would be 0 for one oscillator, and –π/2 for the other. Hence, the motion of one piston is given by x = a·cos(ω·t), while the motion of the other is given by x = a·cos(ω·t–π/2) = a·sin(ω·t).

The kinetic and potential energy of one oscillator (think of one piston or one spring only) can then be calculated as:

  1. K.E. = T = m·v2/2 = (1/2)·m·ω2·a2·sin2(ω·t + Δ)
  2. P.E. = U = k·x2/2 = (1/2)·k·a2·cos2(ω·t + Δ)

The coefficient k in the potential energy formula characterizes the restoring force: F = −k·x. From the dynamics involved, it is obvious that k must be equal to m·ω2. Hence, the total energy is equal to:

E = T + U = (1/2)· m·ω2·a2·[sin2(ω·t + Δ) + cos2(ω·t + Δ)] = m·a2·ω2/2

To facilitate the calculations, we will briefly assume k = m·ω2 and a are equal to 1. The motion of our first oscillator is given by the cos(ω·t) = cosθ function (θ = ω·t), and its kinetic energy will be equal to sin2θ. Hence, the (instantaneous) change in kinetic energy at any point in time will be equal to:

d(sin2θ)/dθ = 2∙sinθ∙d(sinθ)/dθ = 2∙sinθ∙cosθ

Let us look at the second oscillator now. Just think of the second piston going up and down in the V-2 engine. Its motion is given by the sinθ function, which is equal to cos(θ−π /2). Hence, its kinetic energy is equal to sin2(θ−π /2), and how it changes – as a function of θ – will be equal to:

2∙sin(θ−π /2)∙cos(θ−π /2) = = −2∙cosθ∙sinθ = −2∙sinθ∙cosθ

We have our perpetuum mobile! While transferring kinetic energy from one piston to the other, the crankshaft will rotate with a constant angular velocity: linear motion becomes circular motion, and vice versa, and the total energy that is stored in the system is T + U = ma2ω2.

We have a great metaphor here. Somehow, in this beautiful interplay between linear and circular motion, energy is borrowed from one place and then returns to the other, cycle after cycle. We know the wavefunction consist of a sine and a cosine: the cosine is the real component, and the sine is the imaginary component. Could they be equally real? Could each represent half of the total energy of our particle? Should we think of the c in our E = mc2 formula as an angular velocity?

These are sensible questions. Let us explore them. 

II. The wavefunction as a two-dimensional oscillation

The elementary wavefunction is written as:

ψ = a·ei[E·t − px]/ħa·ei[E·t − px]/ħ = a·cos(px E∙t/ħ) + i·a·sin(px E∙t/ħ)

When considering a particle at rest (p = 0) this reduces to:

ψ = a·ei∙E·t/ħ = a·cos(E∙t/ħ) + i·a·sin(E∙t/ħ) = a·cos(E∙t/ħ) i·a·sin(E∙t/ħ) 

Let us remind ourselves of the geometry involved, which is illustrated below. Note that the argument of the wavefunction rotates clockwise with time, while the mathematical convention for measuring the phase angle (ϕ) is counter-clockwise.

Figure 2: Euler’s formula760px-eulers_formula

If we assume the momentum p is all in the x-direction, then the p and x vectors will have the same direction, and px/ħ reduces to p∙x/ħ. Most illustrations – such as the one below – will either freeze x or, else, t. Alternatively, one can google web animations varying both. The point is: we also have a two-dimensional oscillation here. These two dimensions are perpendicular to the direction of propagation of the wavefunction. For example, if the wavefunction propagates in the x-direction, then the oscillations are along the y– and z-axis, which we may refer to as the real and imaginary axis. Note how the phase difference between the cosine and the sine  – the real and imaginary part of our wavefunction – appear to give some spin to the whole. I will come back to this.

Figure 3: Geometric representation of the wavefunction5d_euler_f

Hence, if we would say these oscillations carry half of the total energy of the particle, then we may refer to the real and imaginary energy of the particle respectively, and the interplay between the real and the imaginary part of the wavefunction may then describe how energy propagates through space over time.

Let us consider, once again, a particle at rest. Hence, p = 0 and the (elementary) wavefunction reduces to ψ = a·ei∙E·t/ħ. Hence, the angular velocity of both oscillations, at some point x, is given by ω = -E/ħ. Now, the energy of our particle includes all of the energy – kinetic, potential and rest energy – and is, therefore, equal to E = mc2.

Can we, somehow, relate this to the m·a2·ω2 energy formula for our V-2 perpetuum mobile? Our wavefunction has an amplitude too. Now, if the oscillations of the real and imaginary wavefunction store the energy of our particle, then their amplitude will surely matter. In fact, the energy of an oscillation is, in general, proportional to the square of the amplitude: E µ a2. We may, therefore, think that the a2 factor in the E = m·a2·ω2 energy will surely be relevant as well.

However, here is a complication: an actual particle is localized in space and can, therefore, not be represented by the elementary wavefunction. We must build a wave packet for that: a sum of wavefunctions, each with their own amplitude ak, and their own ωi = -Ei/ħ. Each of these wavefunctions will contribute some energy to the total energy of the wave packet. To calculate the contribution of each wave to the total, both ai as well as Ei will matter.

What is Ei? Ei varies around some average E, which we can associate with some average mass m: m = E/c2. The Uncertainty Principle kicks in here. The analysis becomes more complicated, but a formula such as the one below might make sense:F1We can re-write this as:F2What is the meaning of this equation? We may look at it as some sort of physical normalization condition when building up the Fourier sum. Of course, we should relate this to the mathematical normalization condition for the wavefunction. Our intuition tells us that the probabilities must be related to the energy densities, but how exactly? We will come back to this question in a moment. Let us first think some more about the enigma: what is mass?

Before we do so, let us quickly calculate the value of c2ħ2: it is about 1´1051 N2∙m4. Let us also do a dimensional analysis: the physical dimensions of the E = m·a2·ω2 equation make sense if we express m in kg, a in m, and ω in rad/s. We then get: [E] = kg∙m2/s2 = (N∙s2/m)∙m2/s2 = N∙m = J. The dimensions of the left- and right-hand side of the physical normalization condition is N3∙m5. 

III. What is mass?

We came up, playfully, with a meaningful interpretation for energy: it is a two-dimensional oscillation of mass. But what is mass? A new aether theory is, of course, not an option, but then what is it that is oscillating? To understand the physics behind equations, it is always good to do an analysis of the physical dimensions in the equation. Let us start with Einstein’s energy equation once again. If we want to look at mass, we should re-write it as m = E/c2:

[m] = [E/c2] = J/(m/s)2 = N·m∙s2/m2 = N·s2/m = kg

This is not very helpful. It only reminds us of Newton’s definition of a mass: mass is that what gets accelerated by a force. At this point, we may want to think of the physical significance of the absolute nature of the speed of light. Einstein’s E = mc2 equation implies we can write the ratio between the energy and the mass of any particle is always the same, so we can write, for example:F3This reminds us of the ω2= C1/L or ω2 = k/m of harmonic oscillators once again.[13] The key difference is that the ω2= C1/L and ω2 = k/m formulas introduce two or more degrees of freedom.[14] In contrast, c2= E/m for any particle, always. However, that is exactly the point: we can modulate the resistance, inductance and capacitance of electric circuits, and the stiffness of springs and the masses we put on them, but we live in one physical space only: our spacetime. Hence, the speed of light c emerges here as the defining property of spacetime – the resonant frequency, so to speak. We have no further degrees of freedom here.

 

The Planck-Einstein relation (for photons) and the de Broglie equation (for matter-particles) have an interesting feature: both imply that the energy of the oscillation is proportional to the frequency, with Planck’s constant as the constant of proportionality. Now, for one-dimensional oscillations – think of a guitar string, for example – we know the energy will be proportional to the square of the frequency. It is a remarkable observation: the two-dimensional matter-wave, or the electromagnetic wave, gives us two waves for the price of one, so to speak, each carrying half of the total energy of the oscillation but, as a result, we get a proportionality between E and f instead of between E and f2.

However, such reflections do not answer the fundamental question we started out with: what is mass? At this point, it is hard to go beyond the circular definition that is implied by Einstein’s formula: energy is a two-dimensional oscillation of mass, and mass packs energy, and c emerges us as the property of spacetime that defines how exactly.

When everything is said and done, this does not go beyond stating that mass is some scalar field. Now, a scalar field is, quite simply, some real number that we associate with a position in spacetime. The Higgs field is a scalar field but, of course, the theory behind it goes much beyond stating that we should think of mass as some scalar field. The fundamental question is: why and how does energy, or matter, condense into elementary particles? That is what the Higgs mechanism is about but, as this paper is exploratory only, we cannot even start explaining the basics of it.

What we can do, however, is look at the wave equation again (Schrödinger’s equation), as we can now analyze it as an energy diffusion equation. 

IV. Schrödinger’s equation as an energy diffusion equation

The interpretation of Schrödinger’s equation as a diffusion equation is straightforward. Feynman (Lectures, III-16-1) briefly summarizes it as follows:

“We can think of Schrödinger’s equation as describing the diffusion of the probability amplitude from one point to the next. […] But the imaginary coefficient in front of the derivative makes the behavior completely different from the ordinary diffusion such as you would have for a gas spreading out along a thin tube. Ordinary diffusion gives rise to real exponential solutions, whereas the solutions of Schrödinger’s equation are complex waves.”[17]

Let us review the basic math. For a particle moving in free space – with no external force fields acting on it – there is no potential (U = 0) and, therefore, the Uψ term disappears. Therefore, Schrödinger’s equation reduces to:

∂ψ(x, t)/∂t = i·(1/2)·(ħ/meff)·∇2ψ(x, t)

The ubiquitous diffusion equation in physics is:

∂φ(x, t)/∂t = D·∇2φ(x, t)

The structural similarity is obvious. The key difference between both equations is that the wave equation gives us two equations for the price of one. Indeed, because ψ is a complex-valued function, with a real and an imaginary part, we get the following equations[18]:

  1. Re(∂ψ/∂t) = −(1/2)·(ħ/meffIm(∇2ψ)
  2. Im(∂ψ/∂t) = (1/2)·(ħ/meffRe(∇2ψ)

These equations make us think of the equations for an electromagnetic wave in free space (no stationary charges or currents):

  1. B/∂t = –∇×E
  2. E/∂t = c2∇×B

The above equations effectively describe a propagation mechanism in spacetime, as illustrated below.

Figure 4: Propagation mechanismspropagation

The Laplacian operator (∇2), when operating on a scalar quantity, gives us a flux density, i.e. something expressed per square meter (1/m2). In this case, it is operating on ψ(x, t), so what is the dimension of our wavefunction ψ(x, t)? To answer that question, we should analyze the diffusion constant in Schrödinger’s equation, i.e. the (1/2)·(ħ/meff) factor:

  1. As a mathematical constant of proportionality, it will quantify the relationship between both derivatives (i.e. the time derivative and the Laplacian);
  2. As a physical constant, it will ensure the physical dimensions on both sides of the equation are compatible.

Now, the ħ/meff factor is expressed in (N·m·s)/(N· s2/m) = m2/s. Hence, it does ensure the dimensions on both sides of the equation are, effectively, the same: ∂ψ/∂t is a time derivative and, therefore, its dimension is s1 while, as mentioned above, the dimension of ∇2ψ is m2. However, this does not solve our basic question: what is the dimension of the real and imaginary part of our wavefunction?

At this point, mainstream physicists will say: it does not have a physical dimension, and there is no geometric interpretation of Schrödinger’s equation. One may argue, effectively, that its argument, (px – E∙t)/ħ, is just a number and, therefore, that the real and imaginary part of ψ is also just some number.

To this, we may object that ħ may be looked as a mathematical scaling constant only. If we do that, then the argument of ψ will, effectively, be expressed in action units, i.e. in N·m·s. It then does make sense to also associate a physical dimension with the real and imaginary part of ψ. What could it be?

We may have a closer look at Maxwell’s equations for inspiration here. The electric field vector is expressed in newton (the unit of force) per unit of charge (coulomb). Now, there is something interesting here. The physical dimension of the magnetic field is N/C divided by m/s.[19] We may write B as the following vector cross-product: B = (1/c)∙ex×E, with ex the unit vector pointing in the x-direction (i.e. the direction of propagation of the wave). Hence, we may associate the (1/c)∙ex× operator, which amounts to a rotation by 90 degrees, with the s/m dimension. Now, multiplication by i also amounts to a rotation by 90° degrees. Hence, we may boldly write: B = (1/c)∙ex×E = (1/c)∙iE. This allows us to also geometrically interpret Schrödinger’s equation in the way we interpreted it above (see Figure 3).[20]

Still, we have not answered the question as to what the physical dimension of the real and imaginary part of our wavefunction should be. At this point, we may be inspired by the structural similarity between Newton’s and Coulomb’s force laws:F4Hence, if the electric field vector E is expressed in force per unit charge (N/C), then we may want to think of associating the real part of our wavefunction with a force per unit mass (N/kg). We can, of course, do a substitution here, because the mass unit (1 kg) is equivalent to 1 N·s2/m. Hence, our N/kg dimension becomes:

N/kg = N/(N·s2/m)= m/s2

What is this: m/s2? Is that the dimension of the a·cosθ term in the a·eiθ a·cosθ − i·a·sinθ wavefunction?

My answer is: why not? Think of it: m/s2 is the physical dimension of acceleration: the increase or decrease in velocity (m/s) per second. It ensures the wavefunction for any particle – matter-particles or particles with zero rest mass (photons) – and the associated wave equation (which has to be the same for all, as the spacetime we live in is one) are mutually consistent.

In this regard, we should think of how we would model a gravitational wave. The physical dimension would surely be the same: force per mass unit. It all makes sense: wavefunctions may, perhaps, be interpreted as traveling distortions of spacetime, i.e. as tiny gravitational waves.

V. Energy densities and flows

Pursuing the geometric equivalence between the equations for an electromagnetic wave and Schrödinger’s equation, we can now, perhaps, see if there is an equivalent for the energy density. For an electromagnetic wave, we know that the energy density is given by the following formula:F5E and B are the electric and magnetic field vector respectively. The Poynting vector will give us the directional energy flux, i.e. the energy flow per unit area per unit time. We write:F6Needless to say, the ∙ operator is the divergence and, therefore, gives us the magnitude of a (vector) field’s source or sink at a given point. To be precise, the divergence gives us the volume density of the outward flux of a vector field from an infinitesimal volume around a given point. In this case, it gives us the volume density of the flux of S.

We can analyze the dimensions of the equation for the energy density as follows:

  1. E is measured in newton per coulomb, so [EE] = [E2] = N2/C2.
  2. B is measured in (N/C)/(m/s), so we get [BB] = [B2] = (N2/C2)·(s2/m2). However, the dimension of our c2 factor is (m2/s2) and so we’re also left with N2/C2.
  3. The ϵ0 is the electric constant, aka as the vacuum permittivity. As a physical constant, it should ensure the dimensions on both sides of the equation work out, and they do: [ε0] = C2/(N·m2) and, therefore, if we multiply that with N2/C2, we find that is expressed in J/m3.[21]

Replacing the newton per coulomb unit (N/C) by the newton per kg unit (N/kg) in the formulas above should give us the equivalent of the energy density for the wavefunction. We just need to substitute ϵ0 for an equivalent constant. We may to give it a try. If the energy densities can be calculated – which are also mass densities, obviously – then the probabilities should be proportional to them.

Let us first see what we get for a photon, assuming the electromagnetic wave represents its wavefunction. Substituting B for (1/c)∙iE or for −(1/c)∙iE gives us the following result:F7Zero!? An unexpected result! Or not? We have no stationary charges and no currents: only an electromagnetic wave in free space. Hence, the local energy conservation principle needs to be respected at all points in space and in time. The geometry makes sense of the result: for an electromagnetic wave, the magnitudes of E and B reach their maximum, minimum and zero point simultaneously, as shown below.[22] This is because their phase is the same.

Figure 5: Electromagnetic wave: E and BEM field

Should we expect a similar result for the energy densities that we would associate with the real and imaginary part of the matter-wave? For the matter-wave, we have a phase difference between a·cosθ and a·sinθ, which gives a different picture of the propagation of the wave (see Figure 3).[23] In fact, the geometry of the suggestion suggests some inherent spin, which is interesting. I will come back to this. Let us first guess those densities. Making abstraction of any scaling constants, we may write:F8We get what we hoped to get: the absolute square of our amplitude is, effectively, an energy density !

|ψ|2  = |a·ei∙E·t/ħ|2 = a2 = u

This is very deep. A photon has no rest mass, so it borrows and returns energy from empty space as it travels through it. In contrast, a matter-wave carries energy and, therefore, has some (rest) mass. It is therefore associated with an energy density, and this energy density gives us the probabilities. Of course, we need to fine-tune the analysis to account for the fact that we have a wave packet rather than a single wave, but that should be feasible.

As mentioned, the phase difference between the real and imaginary part of our wavefunction (a cosine and a sine function) appear to give some spin to our particle. We do not have this particularity for a photon. Of course, photons are bosons, i.e. spin-zero particles, while elementary matter-particles are fermions with spin-1/2. Hence, our geometric interpretation of the wavefunction suggests that, after all, there may be some more intuitive explanation of the fundamental dichotomy between bosons and fermions, which puzzled even Feynman:

“Why is it that particles with half-integral spin are Fermi particles, whereas particles with integral spin are Bose particles? We apologize for the fact that we cannot give you an elementary explanation. An explanation has been worked out by Pauli from complicated arguments of quantum field theory and relativity. He has shown that the two must necessarily go together, but we have not been able to find a way of reproducing his arguments on an elementary level. It appears to be one of the few places in physics where there is a rule which can be stated very simply, but for which no one has found a simple and easy explanation. The explanation is deep down in relativistic quantum mechanics. This probably means that we do not have a complete understanding of the fundamental principle involved.” (Feynman, Lectures, III-4-1)

The physical interpretation of the wavefunction, as presented here, may provide some better understanding of ‘the fundamental principle involved’: the physical dimension of the oscillation is just very different. That is all: it is force per unit charge for photons, and force per unit mass for matter-particles. We will examine the question of spin somewhat more carefully in section VII. Let us first examine the matter-wave some more. 

VI. Group and phase velocity of the matter-wave

The geometric representation of the matter-wave (see Figure 3) suggests a traveling wave and, yes, of course: the matter-wave effectively travels through space and time. But what is traveling, exactly? It is the pulse – or the signal – only: the phase velocity of the wave is just a mathematical concept and, even in our physical interpretation of the wavefunction, the same is true for the group velocity of our wave packet. The oscillation is two-dimensional, but perpendicular to the direction of travel of the wave. Hence, nothing actually moves with our particle.

Here, we should also reiterate that we did not answer the question as to what is oscillating up and down and/or sideways: we only associated a physical dimension with the components of the wavefunction – newton per kg (force per unit mass), to be precise. We were inspired to do so because of the physical dimension of the electric and magnetic field vectors (newton per coulomb, i.e. force per unit charge) we associate with electromagnetic waves which, for all practical purposes, we currently treat as the wavefunction for a photon. This made it possible to calculate the associated energy densities and a Poynting vector for energy dissipation. In addition, we showed that Schrödinger’s equation itself then becomes a diffusion equation for energy. However, let us now focus some more on the asymmetry which is introduced by the phase difference between the real and the imaginary part of the wavefunction. Look at the mathematical shape of the elementary wavefunction once again:

ψ = a·ei[E·t − px]/ħa·ei[E·t − px]/ħ = a·cos(px/ħ − E∙t/ħ) + i·a·sin(px/ħ − E∙t/ħ)

The minus sign in the argument of our sine and cosine function defines the direction of travel: an F(x−v∙t) wavefunction will always describe some wave that is traveling in the positive x-direction (with the wave velocity), while an F(x+v∙t) wavefunction will travel in the negative x-direction. For a geometric interpretation of the wavefunction in three dimensions, we need to agree on how to define i or, what amounts to the same, a convention on how to define clockwise and counterclockwise directions: if we look at a clock from the back, then its hand will be moving counterclockwise. So we need to establish the equivalent of the right-hand rule. However, let us not worry about that now. Let us focus on the interpretation. To ease the analysis, we’ll assume we’re looking at a particle at rest. Hence, p = 0, and the wavefunction reduces to:

ψ = a·ei∙E·t/ħ = a·cos(−E∙t/ħ) + i·a·sin(−E0∙t/ħ) = a·cos(E0∙t/ħ) − i·a·sin(E0∙t/ħ)

E0 is, of course, the rest mass of our particle and, now that we are here, we should probably wonder whose time we are talking about: is it our time, or is the proper time of our particle? Well… In this situation, we are both at rest so it does not matter: t is, effectively, the proper time so perhaps we should write it as t0. It does not matter. You can see what we expect to see: E0/ħ pops up as the natural frequency of our matter-particle: (E0/ħ)∙t = ω∙t. Remembering the ω = 2π·f = 2π/T and T = 1/formulas, we can associate a period and a frequency with this wave, using the ω = 2π·f = 2π/T. Noting that ħ = h/2π, we find the following:

T = 2π·(ħ/E0) = h/E0 ⇔ = E0/h = m0c2/h

This is interesting, because we can look at the period as a natural unit of time for our particle. What about the wavelength? That is tricky because we need to distinguish between group and phase velocity here. The group velocity (vg) should be zero here, because we assume our particle does not move. In contrast, the phase velocity is given by vp = λ·= (2π/k)·(ω/2π) = ω/k. In fact, we’ve got something funny here: the wavenumber k = p/ħ is zero, because we assume the particle is at rest, so p = 0. So we have a division by zero here, which is rather strange. What do we get assuming the particle is not at rest? We write:

vp = ω/k = (E/ħ)/(p/ħ) = E/p = E/(m·vg) = (m·c2)/(m·vg) = c2/vg

This is interesting: it establishes a reciprocal relation between the phase and the group velocity, with as a simple scaling constant. Indeed, the graph below shows the shape of the function does not change with the value of c, and we may also re-write the relation above as:

vp/= βp = c/vp = 1/βg = 1/(c/vp)

Figure 6: Reciprocal relation between phase and group velocitygraph

We can also write the mentioned relationship as vp·vg = c2, which reminds us of the relationship between the electric and magnetic constant (1/ε0)·(1/μ0) = c2. This is interesting in light of the fact we can re-write this as (c·ε0)·(c·μ0) = 1, which shows electricity and magnetism are just two sides of the same coin, so to speak.[24]

Interesting, but how do we interpret the math? What about the implications of the zero value for wavenumber k = p/ħ. We would probably like to think it implies the elementary wavefunction should always be associated with some momentum, because the concept of zero momentum clearly leads to weird math: something times zero cannot be equal to c2! Such interpretation is also consistent with the Uncertainty Principle: if Δx·Δp ≥ ħ, then neither Δx nor Δp can be zero. In other words, the Uncertainty Principle tells us that the idea of a pointlike particle actually being at some specific point in time and in space does not make sense: it has to move. It tells us that our concept of dimensionless points in time and space are mathematical notions only. Actual particles – including photons – are always a bit spread out, so to speak, and – importantly – they have to move.

For a photon, this is self-evident. It has no rest mass, no rest energy, and, therefore, it is going to move at the speed of light itself. We write: p = m·c = m·c2/= E/c. Using the relationship above, we get:

vp = ω/k = (E/ħ)/(p/ħ) = E/p = c ⇒ vg = c2/vp = c2/c = c

This is good: we started out with some reflections on the matter-wave, but here we get an interpretation of the electromagnetic wave as a wavefunction for the photon. But let us get back to our matter-wave. In regard to our interpretation of a particle having to move, we should remind ourselves, once again, of the fact that an actual particle is always localized in space and that it can, therefore, not be represented by the elementary wavefunction ψ = a·ei[E·t − px]/ħ or, for a particle at rest, the ψ = a·ei∙E·t/ħ function. We must build a wave packet for that: a sum of wavefunctions, each with their own amplitude ai, and their own ωi = −Ei/ħ. Indeed, in section II, we showed that each of these wavefunctions will contribute some energy to the total energy of the wave packet and that, to calculate the contribution of each wave to the total, both ai as well as Ei matter. This may or may not resolve the apparent paradox. Let us look at the group velocity.

To calculate a meaningful group velocity, we must assume the vg = ∂ωi/∂ki = ∂(Ei/ħ)/∂(pi/ħ) = ∂(Ei)/∂(pi) exists. So we must have some dispersion relation. How do we calculate it? We need to calculate ωi as a function of ki here, or Ei as a function of pi. How do we do that? Well… There are a few ways to go about it but one interesting way of doing it is to re-write Schrödinger’s equation as we did, i.e. by distinguishing the real and imaginary parts of the ∂ψ/∂t =i·[ħ/(2m)]·∇2ψ wave equation and, hence, re-write it as the following pair of two equations:

  1. Re(∂ψ/∂t) = −[ħ/(2meff)]·Im(∇2ψ) ⇔ ω·cos(kx − ωt) = k2·[ħ/(2meff)]·cos(kx − ωt)
  2. Im(∂ψ/∂t) = [ħ/(2meff)]·Re(∇2ψ) ⇔ ω·sin(kx − ωt) = k2·[ħ/(2meff)]·sin(kx − ωt)

Both equations imply the following dispersion relation:

ω = ħ·k2/(2meff)

Of course, we need to think about the subscripts now: we have ωi, ki, but… What about meff or, dropping the subscript, m? Do we write it as mi? If so, what is it? Well… It is the equivalent mass of Ei obviously, and so we get it from the mass-energy equivalence relation: mi = Ei/c2. It is a fine point, but one most people forget about: they usually just write m. However, if there is uncertainty in the energy, then Einstein’s mass-energy relation tells us we must have some uncertainty in the (equivalent) mass too. Here, I should refer back to Section II: Ei varies around some average energy E and, therefore, the Uncertainty Principle kicks in. 

VII. Explaining spin

The elementary wavefunction vector – i.e. the vector sum of the real and imaginary component – rotates around the x-axis, which gives us the direction of propagation of the wave (see Figure 3). Its magnitude remains constant. In contrast, the magnitude of the electromagnetic vector – defined as the vector sum of the electric and magnetic field vectors – oscillates between zero and some maximum (see Figure 5).

We already mentioned that the rotation of the wavefunction vector appears to give some spin to the particle. Of course, a circularly polarized wave would also appear to have spin (think of the E and B vectors rotating around the direction of propagation – as opposed to oscillating up and down or sideways only). In fact, a circularly polarized light does carry angular momentum, as the equivalent mass of its energy may be thought of as rotating as well. But so here we are looking at a matter-wave.

The basic idea is the following: if we look at ψ = a·ei∙E·t/ħ as some real vector – as a two-dimensional oscillation of mass, to be precise – then we may associate its rotation around the direction of propagation with some torque. The illustration below reminds of the math here.

Figure 7: Torque and angular momentum vectorsTorque_animation

A torque on some mass about a fixed axis gives it angular momentum, which we can write as the vector cross-product L = r×p or, perhaps easier for our purposes here as the product of an angular velocity (ω) and rotational inertia (I), aka as the moment of inertia or the angular mass. We write:

L = I·ω

Note we can write L and ω in boldface here because they are (axial) vectors. If we consider their magnitudes only, we write L = I·ω (no boldface). We can now do some calculations. Let us start with the angular velocity. In our previous posts, we showed that the period of the matter-wave is equal to T = 2π·(ħ/E0). Hence, the angular velocity must be equal to:

ω = 2π/[2π·(ħ/E0)] = E0

We also know the distance r, so that is the magnitude of r in the Lr×p vector cross-product: it is just a, so that is the magnitude of ψ = a·ei∙E·t/ħ. Now, the momentum (p) is the product of a linear velocity (v) – in this case, the tangential velocity – and some mass (m): p = m·v. If we switch to scalar instead of vector quantities, then the (tangential) velocity is given by v = r·ω. So now we only need to think about what we should use for m or, if we want to work with the angular velocity (ω), the angular mass (I). Here we need to make some assumption about the mass (or energy) distribution. Now, it may or may not sense to assume the energy in the oscillation – and, therefore, the mass – is distributed uniformly. In that case, we may use the formula for the angular mass of a solid cylinder: I = m·r2/2. If we keep the analysis non-relativistic, then m = m0. Of course, the energy-mass equivalence tells us that m0 = E0/c2. Hence, this is what we get:

L = I·ω = (m0·r2/2)·(E0/ħ) = (1/2)·a2·(E0/c2)·(E0/ħ) = a2·E02/(2·ħ·c2)

Does it make sense? Maybe. Maybe not. Let us do a dimensional analysis: that won’t check our logic, but it makes sure we made no mistakes when mapping mathematical and physical spaces. We have m2·J2 = m2·N2·m2 in the numerator and N·m·s·m2/s2 in the denominator. Hence, the dimensions work out: we get N·m·s as the dimension for L, which is, effectively, the physical dimension of angular momentum. It is also the action dimension, of course, and that cannot be a coincidence. Also note that the E = mc2 equation allows us to re-write it as:

L = a2·E02/(2·ħ·c2)

Of course, in quantum mechanics, we associate spin with the magnetic moment of a charged particle, not with its mass as such. Is there way to link the formula above to the one we have for the quantum-mechanical angular momentum, which is also measured in N·m·s units, and which can only take on one of two possible values: J = +ħ/2 and −ħ/2? It looks like a long shot, right? How do we go from (1/2)·a2·m02/ħ to ± (1/2)∙ħ? Let us do a numerical example. The energy of an electron is typically 0.510 MeV » 8.1871×10−14 N∙m, and a… What value should we take for a?

We have an obvious trio of candidates here: the Bohr radius, the classical electron radius (aka the Thompon scattering length), and the Compton scattering radius.

Let us start with the Bohr radius, so that is about 0.×10−10 N∙m. We get L = a2·E02/(2·ħ·c2) = 9.9×10−31 N∙m∙s. Now that is about 1.88×104 times ħ/2. That is a huge factor. The Bohr radius cannot be right: we are not looking at an electron in an orbital here. To show it does not make sense, we may want to double-check the analysis by doing the calculation in another way. We said each oscillation will always pack 6.626070040(81)×10−34 joule in energy. So our electron should pack about 1.24×10−20 oscillations. The angular momentum (L) we get when using the Bohr radius for a and the value of 6.626×10−34 joule for E0 and the Bohr radius is equal to 6.49×10−59 N∙m∙s. So that is the angular momentum per oscillation. When we multiply this with the number of oscillations (1.24×10−20), we get about 8.01×10−51 N∙m∙s, so that is a totally different number.

The classical electron radius is about 2.818×10−15 m. We get an L that is equal to about 2.81×10−39 N∙m∙s, so now it is a tiny fraction of ħ/2! Hence, this leads us nowhere. Let us go for our last chance to get a meaningful result! Let us use the Compton scattering length, so that is about 2.42631×10−12 m.

This gives us an L of 2.08×10−33 N∙m∙s, which is only 20 times ħ. This is not so bad, but it is good enough? Let us calculate it the other way around: what value should we take for a so as to ensure L = a2·E02/(2·ħ·c2) = ħ/2? Let us write it out:F9

In fact, this is the formula for the so-called reduced Compton wavelength. This is perfect. We found what we wanted to find. Substituting this value for a (you can calculate it: it is about 3.8616×10−33 m), we get what we should find:F10

This is a rather spectacular result, and one that would – a priori – support the interpretation of the wavefunction that is being suggested in this paper. 

VIII. The boson-fermion dichotomy

Let us do some more thinking on the boson-fermion dichotomy. Again, we should remind ourselves that an actual particle is localized in space and that it can, therefore, not be represented by the elementary wavefunction ψ = a·ei[E·t − px]/ħ or, for a particle at rest, the ψ = a·ei∙E·t/ħ function. We must build a wave packet for that: a sum of wavefunctions, each with their own amplitude ai, and their own ωi = −Ei/ħ. Each of these wavefunctions will contribute some energy to the total energy of the wave packet. Now, we can have another wild but logical theory about this.

Think of the apparent right-handedness of the elementary wavefunction: surely, Nature can’t be bothered about our convention of measuring phase angles clockwise or counterclockwise. Also, the angular momentum can be positive or negative: J = +ħ/2 or −ħ/2. Hence, we would probably like to think that an actual particle – think of an electron, or whatever other particle you’d think of – may consist of right-handed as well as left-handed elementary waves. To be precise, we may think they either consist of (elementary) right-handed waves or, else, of (elementary) left-handed waves. An elementary right-handed wave would be written as:

ψ(θi= ai·(cosθi + i·sinθi)

In contrast, an elementary left-handed wave would be written as:

ψ(θi= ai·(cosθii·sinθi)

How does that work out with the E0·t argument of our wavefunction? Position is position, and direction is direction, but time? Time has only one direction, but Nature surely does not care how we count time: counting like 1, 2, 3, etcetera or like −1, −2, −3, etcetera is just the same. If we count like 1, 2, 3, etcetera, then we write our wavefunction like:

ψ = a·cos(E0∙t/ħ) − i·a·sin(E0∙t/ħ)

If we count time like −1, −2, −3, etcetera then we write it as:

 ψ = a·cos(E0∙t/ħ) − i·a·sin(E0∙t/ħ)= a·cos(E0∙t/ħ) + i·a·sin(E0∙t/ħ)

Hence, it is just like the left- or right-handed circular polarization of an electromagnetic wave: we can have both for the matter-wave too! This, then, should explain why we can have either positive or negative quantum-mechanical spin (+ħ/2 or −ħ/2). It is the usual thing: we have two mathematical possibilities here, and so we must have two physical situations that correspond to it.

It is only natural. If we have left- and right-handed photons – or, generalizing, left- and right-handed bosons – then we should also have left- and right-handed fermions (electrons, protons, etcetera). Back to the dichotomy. The textbook analysis of the dichotomy between bosons and fermions may be epitomized by Richard Feynman’s Lecture on it (Feynman, III-4), which is confusing and – I would dare to say – even inconsistent: how are photons or electrons supposed to know that they need to interfere with a positive or a negative sign? They are not supposed to know anything: knowledge is part of our interpretation of whatever it is that is going on there.

Hence, it is probably best to keep it simple, and think of the dichotomy in terms of the different physical dimensions of the oscillation: newton per kg versus newton per coulomb. And then, of course, we should also note that matter-particles have a rest mass and, therefore, actually carry charge. Photons do not. But both are two-dimensional oscillations, and the point is: the so-called vacuum – and the rest mass of our particle (which is zero for the photon and non-zero for everything else) – give us the natural frequency for both oscillations, which is beautifully summed up in that remarkable equation for the group and phase velocity of the wavefunction, which applies to photons as well as matter-particles:

(vphase·c)·(vgroup·c) = 1 ⇔ vp·vg = c2

The final question then is: why are photons spin-zero particles? Well… We should first remind ourselves of the fact that they do have spin when circularly polarized.[25] Here we may think of the rotation of the equivalent mass of their energy. However, if they are linearly polarized, then there is no spin. Even for circularly polarized waves, the spin angular momentum of photons is a weird concept. If photons have no (rest) mass, then they cannot carry any charge. They should, therefore, not have any magnetic moment. Indeed, what I wrote above shows an explanation of quantum-mechanical spin requires both mass as well as charge.[26] 

IX. Concluding remarks

There are, of course, other ways to look at the matter – literally. For example, we can imagine two-dimensional oscillations as circular rather than linear oscillations. Think of a tiny ball, whose center of mass stays where it is, as depicted below. Any rotation – around any axis – will be some combination of a rotation around the two other axes. Hence, we may want to think of a two-dimensional oscillation as an oscillation of a polar and azimuthal angle.

Figure 8: Two-dimensional circular movementoscillation-of-a-ball

The point of this paper is not to make any definite statements. That would be foolish. Its objective is just to challenge the simplistic mainstream viewpoint on the reality of the wavefunction. Stating that it is a mathematical construct only without physical significance amounts to saying it has no meaning at all. That is, clearly, a non-sustainable proposition.

The interpretation that is offered here looks at amplitude waves as traveling fields. Their physical dimension may be expressed in force per mass unit, as opposed to electromagnetic waves, whose amplitudes are expressed in force per (electric) charge unit. Also, the amplitudes of matter-waves incorporate a phase factor, but this may actually explain the rather enigmatic dichotomy between fermions and bosons and is, therefore, an added bonus.

The interpretation that is offered here has some advantages over other explanations, as it explains the how of diffraction and interference. However, while it offers a great explanation of the wave nature of matter, it does not explain its particle nature: while we think of the energy as being spread out, we will still observe electrons and photons as pointlike particles once they hit the detector. Why is it that a detector can sort of ‘hook’ the whole blob of energy, so to speak?

The interpretation of the wavefunction that is offered here does not explain this. Hence, the complementarity principle of the Copenhagen interpretation of the wavefunction surely remains relevant.

Appendix 1: The de Broglie relations and energy

The 1/2 factor in Schrödinger’s equation is related to the concept of the effective mass (meff). It is easy to make the wrong calculations. For example, when playing with the famous de Broglie relations – aka as the matter-wave equations – one may be tempted to derive the following energy concept:

  1. E = h·f and p = h/λ. Therefore, f = E/h and λ = p/h.
  2. v = λ = (E/h)∙(p/h) = E/p
  3. p = m·v. Therefore, E = v·p = m·v2

E = m·v2? This resembles the E = mc2 equation and, therefore, one may be enthused by the discovery, especially because the m·v2 also pops up when working with the Least Action Principle in classical mechanics, which states that the path that is followed by a particle will minimize the following integral:F11Now, we can choose any reference point for the potential energy but, to reflect the energy conservation law, we can select a reference point that ensures the sum of the kinetic and the potential energy is zero throughout the time interval. If the force field is uniform, then the integrand will, effectively, be equal to KE − PE = m·v2.[27]

However, that is classical mechanics and, therefore, not so relevant in the context of the de Broglie equations, and the apparent paradox should be solved by distinguishing between the group and the phase velocity of the matter wave.

Appendix 2: The concept of the effective mass

The effective mass – as used in Schrödinger’s equation – is a rather enigmatic concept. To make sure we are making the right analysis here, I should start by noting you will usually see Schrödinger’s equation written as:F12This formulation includes a term with the potential energy (U). In free space (no potential), this term disappears, and the equation can be re-written as:

∂ψ(x, t)/∂t = i·(1/2)·(ħ/meff)·∇2ψ(x, t)

We just moved the i·ħ coefficient to the other side, noting that 1/i = –i. Now, in one-dimensional space, and assuming ψ is just the elementary wavefunction (so we substitute a·ei∙[E·t − p∙x]/ħ for ψ), this implies the following:

a·i·(E/ħ)·ei∙[E·t − p∙x]/ħ = −i·(ħ/2meffa·(p22 ei∙[E·t − p∙x]/ħ 

⇔ E = p2/(2meff) ⇔ meff = m∙(v/c)2/2 = m∙β2/2

It is an ugly formula: it resembles the kinetic energy formula (K.E. = m∙v2/2) but it is, in fact, something completely different. The β2/2 factor ensures the effective mass is always a fraction of the mass itself. To get rid of the ugly 1/2 factor, we may re-define meff as two times the old meff (hence, meffNEW = 2∙meffOLD), as a result of which the formula will look somewhat better:

meff = m∙(v/c)2 = m∙β2

We know β varies between 0 and 1 and, therefore, meff will vary between 0 and m. Feynman drops the subscript, and just writes meff as m in his textbook (see Feynman, III-19). On the other hand, the electron mass as used is also the electron mass that is used to calculate the size of an atom (see Feynman, III-2-4). As such, the two mass concepts are, effectively, mutually compatible. It is confusing because the same mass is often defined as the mass of a stationary electron (see, for example, the article on it in the online Wikipedia encyclopedia[28]).

In the context of the derivation of the electron orbitals, we do have the potential energy term – which is the equivalent of a source term in a diffusion equation – and that may explain why the above-mentioned meff = m∙(v/c)2 = m∙β2 formula does not apply.

References

This paper discusses general principles in physics only. Hence, references can be limited to references to physics textbooks only. For ease of reading, any reference to additional material has been limited to a more popular undergrad textbook that can be consulted online: Feynman’s Lectures on Physics (http://www.feynmanlectures.caltech.edu). References are per volume, per chapter and per section. For example, Feynman III-19-3 refers to Volume III, Chapter 19, Section 3.

Notes

[1] Of course, an actual particle is localized in space and can, therefore, not be represented by the elementary wavefunction ψ = a·ei∙θa·ei[E·t − px]/ħ = a·(cosθ i·a·sinθ). We must build a wave packet for that: a sum of wavefunctions, each with its own amplitude ak and its own argument θk = (Ek∙t – pkx)/ħ. This is dealt with in this paper as part of the discussion on the mathematical and physical interpretation of the normalization condition.

[2] The N/kg dimension immediately, and naturally, reduces to the dimension of acceleration (m/s2), thereby facilitating a direct interpretation in terms of Newton’s force law.

[3] In physics, a two-spring metaphor is more common. Hence, the pistons in the author’s perpetuum mobile may be replaced by springs.

[4] The author re-derives the equation for the Compton scattering radius in section VII of the paper.

[5] The magnetic force can be analyzed as a relativistic effect (see Feynman II-13-6). The dichotomy between the electric force as a polar vector and the magnetic force as an axial vector disappears in the relativistic four-vector representation of electromagnetism.

[6] For example, when using Schrödinger’s equation in a central field (think of the electron around a proton), the use of polar coordinates is recommended, as it ensures the symmetry of the Hamiltonian under all rotations (see Feynman III-19-3)

[7] This sentiment is usually summed up in the apocryphal quote: “God does not play dice.”The actual quote comes out of one of Einstein’s private letters to Cornelius Lanczos, another scientist who had also emigrated to the US. The full quote is as follows: “You are the only person I know who has the same attitude towards physics as I have: belief in the comprehension of reality through something basically simple and unified… It seems hard to sneak a look at God’s cards. But that He plays dice and uses ‘telepathic’ methods… is something that I cannot believe for a single moment.” (Helen Dukas and Banesh Hoffman, Albert Einstein, the Human Side: New Glimpses from His Archives, 1979)

[8] Of course, both are different velocities: ω is an angular velocity, while v is a linear velocity: ω is measured in radians per second, while v is measured in meter per second. However, the definition of a radian implies radians are measured in distance units. Hence, the physical dimensions are, effectively, the same. As for the formula for the total energy of an oscillator, we should actually write: E = m·a2∙ω2/2. The additional factor (a) is the (maximum) amplitude of the oscillator.

[9] We also have a 1/2 factor in the E = mv2/2 formula. Two remarks may be made here. First, it may be noted this is a non-relativistic formula and, more importantly, incorporates kinetic energy only. Using the Lorentz factor (γ), we can write the relativistically correct formula for the kinetic energy as K.E. = E − E0 = mvc2 − m0c2 = m0γc2 − m0c2 = m0c2(γ − 1). As for the exclusion of the potential energy, we may note that we may choose our reference point for the potential energy such that the kinetic and potential energy mirror each other. The energy concept that then emerges is the one that is used in the context of the Principle of Least Action: it equals E = mv2. Appendix 1 provides some notes on that.

[10] Instead of two cylinders with pistons, one may also think of connecting two springs with a crankshaft.

[11] It is interesting to note that we may look at the energy in the rotating flywheel as potential energy because it is energy that is associated with motion, albeit circular motion. In physics, one may associate a rotating object with kinetic energy using the rotational equivalent of mass and linear velocity, i.e. rotational inertia (I) and angular velocity ω. The kinetic energy of a rotating object is then given by K.E. = (1/2)·I·ω2.

[12] Because of the sideways motion of the connecting rods, the sinusoidal function will describe the linear motion only approximately, but you can easily imagine the idealized limit situation.

[13] The ω2= 1/LC formula gives us the natural or resonant frequency for a electric circuit consisting of a resistor (R), an inductor (L), and a capacitor (C). Writing the formula as ω2= C1/L introduces the concept of elastance, which is the equivalent of the mechanical stiffness (k) of a spring.

[14] The resistance in an electric circuit introduces a damping factor. When analyzing a mechanical spring, one may also want to introduce a drag coefficient. Both are usually defined as a fraction of the inertia, which is the mass for a spring and the inductance for an electric circuit. Hence, we would write the resistance for a spring as γm and as R = γL respectively.

[15] Photons are emitted by atomic oscillators: atoms going from one state (energy level) to another. Feynman (Lectures, I-33-3) shows us how to calculate the Q of these atomic oscillators: it is of the order of 108, which means the wave train will last about 10–8 seconds (to be precise, that is the time it takes for the radiation to die out by a factor 1/e). For example, for sodium light, the radiation will last about 3.2×10–8 seconds (this is the so-called decay time τ). Now, because the frequency of sodium light is some 500 THz (500×1012 oscillations per second), this makes for some 16 million oscillations. There is an interesting paradox here: the speed of light tells us that such wave train will have a length of about 9.6 m! How is that to be reconciled with the pointlike nature of a photon? The paradox can only be explained by relativistic length contraction: in an analysis like this, one need to distinguish the reference frame of the photon – riding along the wave as it is being emitted, so to speak – and our stationary reference frame, which is that of the emitting atom.

[16] This is a general result and is reflected in the K.E. = T = (1/2)·m·ω2·a2·sin2(ω·t + Δ) and the P.E. = U = k·x2/2 = (1/2)· m·ω2·a2·cos2(ω·t + Δ) formulas for the linear oscillator.

[17] Feynman further formalizes this in his Lecture on Superconductivity (Feynman, III-21-2), in which he refers to Schrödinger’s equation as the “equation for continuity of probabilities”. The analysis is centered on the local conservation of energy, which confirms the interpretation of Schrödinger’s equation as an energy diffusion equation.

[18] The meff is the effective mass of the particle, which depends on the medium. For example, an electron traveling in a solid (a transistor, for example) will have a different effective mass than in an atom. In free space, we can drop the subscript and just write meff = m. Appendix 2 provides some additional notes on the concept. As for the equations, they are easily derived from noting that two complex numbers a + i∙b and c + i∙d are equal if, and only if, their real and imaginary parts are the same. Now, the ∂ψ/∂t = i∙(ħ/meff)∙∇2ψ equation amounts to writing something like this: a + i∙b = i∙(c + i∙d). Now, remembering that i2 = −1, you can easily figure out that i∙(c + i∙d) = i∙c + i2∙d = − d + i∙c.

[19] The dimension of B is usually written as N/(m∙A), using the SI unit for current, i.e. the ampere (A). However, 1 C = 1 A∙s and, hence, 1 N/(m∙A) = 1 (N/C)/(m/s).     

[20] Of course, multiplication with i amounts to a counterclockwise rotation. Hence, multiplication by –i also amounts to a rotation by 90 degrees, but clockwise. Now, to uniquely identify the clockwise and counterclockwise directions, we need to establish the equivalent of the right-hand rule for a proper geometric interpretation of Schrödinger’s equation in three-dimensional space: if we look at a clock from the back, then its hand will be moving counterclockwise. When writing B = (1/c)∙iE, we assume we are looking in the negative x-direction. If we are looking in the positive x-direction, we should write: B = -(1/c)∙iE. Of course, Nature does not care about our conventions. Hence, both should give the same results in calculations. We will show in a moment they do.

[21] In fact, when multiplying C2/(N·m2) with N2/C2, we get N/m2, but we can multiply this with 1 = m/m to get the desired result. It is significant that an energy density (joule per unit volume) can also be measured in newton (force per unit area.

[22] The illustration shows a linearly polarized wave, but the obtained result is general.

[23] The sine and cosine are essentially the same functions, except for the difference in the phase: sinθ = cos(θ−π /2).

[24] I must thank a physics blogger for re-writing the 1/(ε0·μ0) = c2 equation like this. See: http://reciprocal.systems/phpBB3/viewtopic.php?t=236 (retrieved on 29 September 2017).

[25] A circularly polarized electromagnetic wave may be analyzed as consisting of two perpendicular electromagnetic plane waves of equal amplitude and 90° difference in phase.

[26] Of course, the reader will now wonder: what about neutrons? How to explain neutron spin? Neutrons are neutral. That is correct, but neutrons are not elementary: they consist of (charged) quarks. Hence, neutron spin can (or should) be explained by the spin of the underlying quarks.

[27] We detailed the mathematical framework and detailed calculations in the following online article: https://readingfeynman.org/2017/09/15/the-principle-of-least-action-re-visited.

[28] https://en.wikipedia.org/wiki/Electron_rest_mass (retrieved on 29 September 2017).

Entropy, energy and enthalpy

Phew! I am quite happy I got through Feynman’s chapters on thermodynamics. Now is a good time to review the math behind it. We thoroughly understand the gas equation now:

PV = NkT = (γ–1)U

The gamma (γ) in this equation is the specific heat ratio: it’s 5/3 for ideal gases (so that’s about 1.667) and, theoretically, 4/3 ≈ 1.333 or 9/7 ≈ 1.286 for diatomic gases, depending on the degrees of freedom we associate with diatomic molecules. More complicated molecules have even more degrees of freedom and, hence, can absorb even more energy, so γ gets closer to one—according to the kinetic gas theory, that is. While we know that the kinetic gas theory is not quite accurate – an approach involving molecular energy states is a better match for reality – that doesn’t matter here. As for the term (specific heat ratio), I’ll explain that later. [I promise. 🙂 You’ll see it’s quite logical.]

The point to note is that this body of gas (or whatever substance) stores an amount of energy U that is directly proportional to the temperature (T), and Nk/(γ–1) is the constant of proportionality. We can also phrase it the other way around: the temperature is directly proportional to the energy, with (γ–1)/Nk the constant of proportionality. It means temperature and energy are in a linear relationship. [Yes, direct proportionality implies linearity.] The graph below shows the T = [(γ–1)/Nk]·U relationship for three different values of γ, ranging from 5/3 (i.e. the maximum value, which characterizes monatomic noble gases such as helium, neon or krypton) to a value close to 1, which is characteristic of more complicated molecular arrangements indeed, such as heptane (γ = 1.06) or methyl butane ((γ = 1.08). The illustration shows that, unlike monatomic gas, more complicated molecular arrangements allow the gas to absorb a lot of (heat) energy with a relatively moderate rise in temperature only.

CaptureWe’ll soon encounter another variable, enthalpy (H), which is also linearly related to energy: H = γU. From a math point of view, these linear relationships don’t mean all that much: they just show these variables – temperature, energy and enthalphy – are all directly related and, hence, can be defined in terms of each other.

We can invent other variables, like the Gibbs energy, or the Helmholtz energy. In contrast, entropy, while often being mentioned as just some other state function, is something different altogether. In fact, the term ‘state function’ causes a lot of confusion: pressure and volume are state variables too. The term is used to distinguish these variables from so-called process functions, notably heat and work. Process functions describe how we go from one equilibrium state to another, as opposed to the state variables, which describe the equilibrium situation itself. Don’t worry too much about the distinction—for now, that is.

Let’s look at non-linear stuff. The PV = NkT = (γ–1)U says that pressure (P) and volume (V) are inversely proportional one to another, and so that’s a non-linear relationship. [Yes, inverse proportionality is non-linear.] To help you visualize things, I inserted a simple volume-pressure diagram below, which shows how pressure and volume are related for three different values of U (or, what amounts to the same, three different values of T).

graph 2

The curves are simple hyperbolas which have the x- and y-axis as horizontal and vertical asymptote respectively. If you’ve studied social sciences (like me!) – so if you know a tiny little bit of the ‘dismal science’, i.e. economics (like me!) – you’ll note they look like indifference curves. The x- and y-axis then represent the quantity of some good X and some good Y respectively, and the curves closer to the origin are associated with lower utility. How much X and Y we will buy then, depends on (a) their price and (b) our budget, which we represented by a linear budget line tangent to the curve we can reach with our budget, and then we are a little bit happy, very happy or extremely happy, depending on our budget. Hence, our budget determines our happiness. From a math point of view, however, we can also look at it the other way around: our happiness determines our budget. [Now that‘s a nice one, isn’t it? Think about it! 🙂 And, in the process, think about hyperbolas too: the y = 1/x function holds the key to understanding both infinity and nothingness. :-)]

U is a state function but, as mentioned above, we’ve got quite a few state variables in physics. Entropy, of course, denoted by S—and enthalpy too, denoted by H. Let me remind you of the basics of the entropy concept:

  1. The internal energy U changes because (a) we add or remove some heat from the system (ΔQ), (b) because some work is being done (by the gas on its surroundings or the other way around), or (c) because of both. Using the differential notation, we write: dU = dQ – dW, always. The (differential) work that’s being done is PdV. Hence, we have dU = dQ – PdV.
  2. When transferring heat to a system at a certain temperature, there’s a quantity we refer to as the entropy. Remember that illustration of Feynman’s in my post on entropy: we go from one point to another on the temperature-volume diagram, taking infinitesimally small steps along the curve, and, at each step, an infinitesimal amount of work dW is done, and an infinitesimal amount of entropy dS = dQ/T is being delivered.
  3. The total change in entropy, ΔS, is a line integral: ΔS = ∫dQ/T = ∫dS.

That’s somewhat tougher to understand than economics, and so that’s why it took me more time to come with terms with it. 🙂 Just go through Feynman’s Lecture on it, or through that post I referenced above. If you don’t want to do that, then just note that, while entropy is a very mysterious concept, it’s deceptively simple from a math point of view: ΔS = ΔQ/T, so the (infinitesimal) change in entropy is, quite simply, the ratio of (1) the (infinitesimal or incremental) amount of heat that is being added or removed as the system goes from one state to another through a reversible process and (2) the temperature at which the heat is being transferred. However, I am not writing this post to discuss entropy once again. I am writing it to give you an idea of the math behind the system.

So dS = dQ/T. Hence, we can re-write dU = dQ – dW as:

dU = TdS – PdV ⇔ dU + d(PV) = TdS – PdV + d(PV)

⇔ d(U + PV) = dH = TdS – PdV + PdV + VdP = TdS + VdP

The U + PV quantity on the left-hand side of the equation is the so-called enthalpy of the system, which I mentioned above. It’s denoted by H indeed, and it’s just another state variable, like energy: same-same but different, as they say in Asia. We encountered it in our previous post also, where we said that chemists prefer to analyze the behavior of substances using temperature and pressure as ‘independent variables’, rather than temperature and volume. Independent variables? What does that mean, exactly?

According to the PV = NkT equation, we only have two independent variables: if we assign some value to two variables, we’ve got a value for the third one. Indeed, remember that other equation we got when we took the total differential of U. We wrote U as U(V, T) and, taking the total differential, we got:

dU = (∂U/∂T)dT + (∂U/∂V)dV

We did not need to add a (∂U/∂P)dP term, because the pressure is determined by the volume and the temperature. We could also have written U = U(P, T) and, therefore, that dU = (∂U/∂T)dT + (∂U/∂P)dP. However, when working with temperature and pressure as the ‘independent’ variables, it’s easier to work with H rather than U. The point to note is that it’s all quite flexible really: we have two independent variables in the system only. The third one (and all of the other variables really, like energy or enthalpy or whatever) depend on the other two. In other words, from a math point of view, we only have two degrees of freedom in the system here: only two variables are actually free to vary. 🙂

Let’s look at that dH = TdS + VdP equation. That’s a differential equation in which not temperature and pressure, but entropy (S) and pressure (P) are ‘independent’ variables, so we write:

dH(S, P) = TdS + VdP

Now, it is not very likely that we will have some problem to solve with data on entropy and pressure. At our level of understanding, any problem that’s likely to come our way will probably come with data on more common variables, such as the heat, the pressure, the temperature, and/or the volume. So we could continue with the expression above but we don’t do that. It makes more sense to re-write the expression substituting TdS for dQ once again, so we get:

dH = dQ + VdP

That resembles our dU = dQ – PdV expression: it just substitutes V for –P. And, yes, you guessed it: it’s because the two expressions resemble each other that we like to work with H now. 🙂 Indeed, we’re talking the same system and the same infinitesimal changes and, therefore, we can use all the formulas we derived already by just substituting H for U, V for –P, and dP for dV. Huh? Yes. It’s a rather tricky substitution. If we switch V for –P (or vice versa) in a partial derivative involving T, we also need to include the minus sign. However, we do not need to include the minus sign when substituting dV and dP, and we also don’t need to change the sign of the partial derivatives of U and H when going from one expression to another! It’s a subtle and somewhat weird point, but a very important one! I’ll explain it in a moment. Just continue to read as for now. Let’s do the substitution using our rules:

dU = (∂Q/∂T)VdT + [T(∂P/∂T)V − P]dV becomes:

dH = (∂Q/∂T)PdT + (∂H/∂P)TdP = CPdT + [–T·(∂V/∂T)P + V]dP

Note that, just as we referred to (∂Q/∂T)as the specific heat capacity of a substance at constant volume, which we denoted by CV, we now refer to (∂Q/∂T)P as the specific heat capacity at constant pressure, which we’ll denote, logically, as CP. Dropping the subscripts of the partial derivatives, we re-write the expression above as:

dH = CPdT + [–T·(∂V/∂T) + V]dP

So we’ve got what we wanted: we switched from an expression involving derivatives assuming constant volume to an expression involving derivatives assuming constant pressure. [In case you wondered what we wanted, this is it: we wanted an equation that helps us to solve another type of problem—another formula for a problem involving a different set of data.]

As mentioned above, it’s good to use subscripts with the partial derivatives to emphasize what changes and what is constant when calculating those partial derivatives but, strictly speaking, it’s not necessary, and you will usually not find the subscripts when googling other texts. For example, in the Wikipedia article on enthalpy, you’ll find the expression written as:

dH = CPdT + V(1–αT)dP with α = (1/V)(∂V/∂T)

Just write it all out and you’ll find it’s the same thing, exactly. It just introduces another coefficient, α, i.e. the coefficient of (cubic) thermal expansion. If you find this formula is easier to remember, then please use this one. It doesn’t matter.

Now, let’s explain that funny business with the minus signs in the substitution. I’ll do so by going back to that infinitesimal analysis of the reversible cycle in my previous post, in which we had that formula involving ΔQ for the work done by the gas during an infinitesimally small reversible cycle: ΔW = ΔVΔP = ΔQ·(ΔT/T). Now, we can either write that as:

  1. ΔQ = T·(ΔP/ΔT)·ΔV = dQ = T·(∂P/∂T)V·dV – which is what we did for our analysis of (∂U/∂V)or, alternatively, as
  2. ΔQ = T·(ΔV/ΔT)·ΔP = dQ = T·(∂V/∂T)P·dP, which is what we’ve got to do here, for our analysis of (∂H/∂P)T.

Hence, dH = dQ + VdP becomes dH = T·(∂V/∂T)P·dP + V·dP, and dividing all by dP gives us what we want to get: dH/dP = (∂H/∂P)= T·(∂V/∂T)+ V.

[…] Well… NO! We don’t have the minus sign in front of T·(∂V/∂T)P, so we must have done something wrong or, else, that formula above is wrong.

The formula is right (it’s in Wikipedia, so it must be right :-)), so we are wrong. Indeed! The thing is: substituting dT, dV and dP for ΔT, ΔV and ΔP is somewhat tricky. The geometric analysis (illustrated below) makes sense but we need to watch the signs.

Carnot 2

We’ve got a volume increase, a temperature drop and, hence, also a pressure drop over the cycle: the volume goes from V to V+ΔV (and then back to V, of course), while the pressure and the temperature go from P to P–ΔP and T to T–ΔT respectively (and then back to P and T, of course). Hence, we should write: ΔV = dV, –ΔT = dT, and –ΔP = dP. Therefore, as we replace the ratio of the infinitesimal change of pressure and temperature, ΔP/ΔT, by a proper derivative (i.e. ∂P/∂T), we should add a minus sign: ΔP/ΔT = –∂P/∂T. Now that gives us what we want: dH/dP = (∂H/∂P)= –T·(∂V/∂T)+ V, and, therefore, we can, indeed, write what we wrote above:

dU = (∂Q/∂T)VdT + [T(∂P/∂T)V − P]dV becomes:

dH = (∂Q/∂T)PdT + [–T·(∂V/∂T)P + V]dP = CPdT + [–T·(∂V/∂T)P + V]dP

Now, in case you still wonder: what’s the use of all these different expressions stating the same? The answer is simple: it depends on the problem and what information we have. Indeed, note that all derivatives we use in our expression for dH expression assume constant pressure, so if we’ve got that kind of data, we’ll use the chemists’ representation of the system. If we’ve got data describing performance at constant volume, we’ll need the physicists’ formulas, which are given in terms of derivatives assuming constant volume. It all looks complicated but, in the end, it’s the same thing: the PV = NkT equation gives us two ‘independent’ variables and one ‘dependent’ variable. Which one is which will determine our approach.

Now, we left one thing unexplained. Why do we refer to γ as the specific heat ratio? The answer is: it is the ratio of the specific heat capacities indeed, so we can write:

γ = CP/CV

However, it is important to note that that’s valid for ideal gases only. In that case, we know that the (∂U/∂V)derivative in our dU = (∂U/∂T)VdT + (∂U/∂V)TdV expression is zero: we can change the volume, but if the temperature remains the same, the internal energy remains the same. Hence, dU = (∂U/∂T)VdT = CVdT, and dU/dT = CV. Likewise, the (∂H/∂P)T derivative in our dH = (∂H/∂T)PdT + (∂H/∂P)TdP expression is zero—for ideal gases, that is. Hence, dH = (∂H/∂T)PdT = CPdT, and dH/dT = CP. Hence,

CP/CV = (dH/dT)/(dU/dT) = dH/dU

Does that make sense? If dH/dU = γ, then H must be some linear function of U. More specifically, H must be some function H = γU + c, with c some constant (it’s the so-called constant of integration). Now, γ is supposed to be constant too, of course. That’s all perfectly fine: indeed, combining the definition of H (H = U + PV), and using the PV = (γ–1)U relation, we have H = U + (γ–1)U = γU (hence, c = 0). So, yes, dH/dU = γ, and γ = CP/CV.

Note the qualifier, however: we’re assuming γ is constant (which does not imply the gas has to be ideal, so the interpretation is less restrictive than you might think it is). If γ is not a constant, it’s a different ballgame. […] So… Is γ actually constant? The illustration below shows γ is not constant for common diatomic gases like hydrogen or (somewhat less common) oxygen. It’s the same for other gases: when mentioning γ, we need to state the temperate at which we measured it too. 😦  However, the illustration also shows the assumption of γ being constant holds fairly well if temperature varies only slightly (like plus or minus 100° C), so that’s OK. 🙂

Heat ratio

I told you so: the kinetic gas theory is not quite accurate. An approach involving molecular energy states works much better (and is actually correct, as it’s consistent with quantum theory). But so we are where we are and I’ll save the quantum-theoretical approach for later. 🙂

So… What’s left? Well… If you’d google the Wikipedia article on enthalphy in order to check if I am not writing nonsense, you’ll find it gives γ as the ratio of H and U itself: γ = H/U. That’s not wrong, obviously (γ = H/U = γU/U = γ), but that formula doesn’t really explain why γ is referred to as the specific heat ratio, which is what I wanted to do here.

OK. We’ve covered a lot of ground, but let’s reflect some more. We did not say a lot about entropy, and/or the relation between energy and entropy. Too bad… The relationship between entropy and energy is obviously not so simple as between enthalpy and energy. Indeed, because of that easy H = γU relationship, enthalpy emerges as just some auxiliary variable: some temporary variable we need to calculate something. Entropy is, obviously, something different. Unlike enthalpy, entropy involves very complicated thinking, involving (ir)reversibility and all that. So it’s quite deep, I’d say – but I’ll write more about that later. I think this post has gone as far as it should. 🙂

A post for my kids: on energy and potential

We’ve been juggling with a lot of advanced concepts in the previous post. Perhaps it’s time I write something that my kids can understand too. One of the things I struggled with when re-learning elementary physics is the concept of energy. What is energy really? I always felt my high school teachers did a poor job in trying to explain it. So let me try to do a better job here.

A high-school level course usually introduces the topic using the gravitational force, i.e. Newton’s Third Law: F = GmM/r2. This law states that the force of attraction is proportional to the product of the masses m and M, and inversely proportional to the square of the distance r between those two masses. The factor of proportionality is equal to G, i.e. the so-called universal gravitational constant, aka the ‘big G’ (G ≈ 6.674×10-11 N(m/kg)2), as opposed to the ‘little g’, which is the gravity of Earth (g ≈ 9.80665 m/s2). As far as I am concerned, it is at this point where my high-school teacher failed.

Indeed, he would just go on and simplify Newton’s Third Law by writing F = mg, noting that g = GM/rand that, for all practical purposes, this g factor is constant, because we are talking small distances as compared to the radius of the Earth. Hence, we should just remember that the gravitational force is proportional to the mass only, and that one kilogram amounts to a weight of about 10 newton (9.80665 kg·m/s2 (N) to be precise). That simplification would then be followed by another simplification: if we are lifting an object with mass m, we are doing work against the gravitational force. How much work? Well, he’d say, work is – quite simply – the force times the distance in physics, and the work done against the force is the potential energy (usually denoted by U) of that object. So he would write U = Fh = mgh, with h the height of the object (as measured from the surface of the Earth), and he would draw a nice linear graph like the one below (I set m to 10 kg here, and h ranges from 0 to 100 m).

Potential energy uniform gravitation field

Note that the slope of this line is slightly less than 45 degrees (and also note, of course, that it’s only approximately 45 degrees because of our choice of scale: dU/dh is equal to 98.0665, so if the x and y axes would have the same scale, we’d have a line that’s almost vertical).

So what’s wrong with this graph? Nothing. It’s just that this graph sort of got stuck in my head, and it complicated a more accurate understanding of energy. Indeed, with examples like the one above, one tends to forget that:

  1. Such linear graphs are an approximation only. In reality, the gravitational field, and force fields in general, are not uniform and, hence, g is not a constant: the graph below shows how g varies with the height (but the height is expressed in kilometer this time, not in meter).
  2. Not only is potential energy usually not a linear function but – equally important – it is usually not a positive real number either. In fact, in physics, U will usually take on a negative value. Why? Because we’re indeed measuring and defining it by the work done against the force.

Erdgvarp

So what’s the more accurate view of things? Well… Let’s start by noting that potential energy is defined in relation to some reference point and, taking a more universal point of view, that reference point will usually be infinity when discussing the gravitational (or electromagnetic) force of attraction. Now, the potential energy of the point(s) at infinity – i.e. the reference point – will, usually, be equated with zero. Hence, the potential energy curve will then take the shape of the graph below (y = –1/x), so U will vary from zero (0) to minus infinity (–∞) , as we bring the two masses closer together. You can readily see that the graph below makes sense: its slope is positive and, hence, as such it does capture the same idea as that linear mgh graph above: moving a mass from point 1 to point 2 requires work and, hence, the potential energy at point 2 is higher than at point 1, even if both values U(2) and U(1) are negative numbers, unlike the values of that linear mgh curve.

Capture

How do you get a curve like that? Well… I should first note another convention which is essential for making the sign come out alright: if the force is gravity, then we should write F = –GmMr/r3. So we have a minus sign here. And please do note the boldface type: F and r are vectors, and vectors have both a direction and magnitude – and so that’s why they are denoted by a bold letter (r), as opposed to the scalar quantities G, m, M or r).

Back to the minus sign. Why do we have that here? Well… It has to do with the direction of the force, which, in case of attraction, will be opposite to the so-called radius vector r. Just look at the illustration below, which shows, first, the direction of the force between two opposite electric charges (top) and then (bottom), the force between two masses, let’s say the Earth and the Moon.

Force and radius vector

So it’s a matter of convention really.

Now, when we’re talking the electromagnetic force, you know that likes repel and opposites attract, so two charges with the same sign will repel each other, and two charges with opposite sign will attract each other. So F12, i.e. the force on q2 because of the presence of q1, will be equal to F12 = q1q2r/r3. Therefore, no minus sign is needed here because qand q2 are opposite and, hence, the sign of this product will be negative. Therefore, we know that the direction of F comes out alright: it’s opposite to the direction of the radius vector r. So the force on a charge q2 which is placed in an electric field produced by a charge q1 is equal to F12 = q1q2r/r3. In short, no minus sign needed here because we already have one. Of course, the original charge q1 will be subject to the very same force and so we should write F21 = –q1q2r/r3. So we’ve got that minus sign again now. In general, however, we’ll write Fij = qiqjr/r3 when dealing with the electromagnetic force, so that’s without a minus sign, because the convention is to draw the radius vector from charge i to charge j and, hence, the radius vector r in the formula F21 would point in the other direction and, hence, the minus sign is not needed.

In short, because of the way that the electromagnetic force works, the sign always come out right: there is no need for a minus sign in front. However, for gravity, there are no opposite charges: masses are always alike, and so likes actually attract when we’re talking gravity, and so that’s why we need the minus sign when dealing with the gravitational force: the force between a mass i and another mass j will always be written as Fij = –mimjr/r3, so here we do have to put the minus sign, because the direction of the force needs to be opposite to the direction of the radius vector and so the sign of the ‘charges’ (i.e. the masses in this case), in the case of gravity, does not take care of that.

One last remark here may be useful: always watch out to not double-count forces when considering a system with many charges or many masses: both charges (or masses) feel the same force, but with opposite direction. OK. Let’s move on. If you are confused, don’t worry. Just remember that (1) it’s very important to be consistent when drawing that radius vector (it goes from the charge (or mass) causing the force field to the other charge (or mass) that is being brought in), and (2) that the gravitational and electromagnetic forces have a lot in common in terms of ‘geometry’ – notably that inverse proportionality relation with the square of the distance between the two charges or masses – but that we need to put a minus sign when we’re dealing with the gravitational force because, with gravitation, likes do not repel but attract each other, as opposed to electric charges.

Now, let’s move on indeed and get back to our discussion of potential energy. Let me copy that potential energy curve again and let’s assume we’re talking electromagnetics here, and that we’re have two opposite charges, so the force is one of attraction.

Capture

Hence, if we move one charge away from the other, we are doing work against the force. Conversely, if we bring them closer to each other, we’re working with the force and, hence, its potential energy will go down – from zero (i.e. the reference point) to… Well… Some negative value. How much work is being done? Well… The force changes all the time, so it’s not constant and so we cannot just calculate the force times the distance (Fs). We need to do one of those infinite sums, i.e. an integral, and so, for point 1 in the graph above, we can write:

Capture

Why the minus sign? Well… As said, we’re not increasing potential energy: we’re decreasing it, from zero to some negative value. If we’d move the charge from point 1 to the reference point (infinity), then we’d be doing work against the force and we’d be increasing potential energy. So then we’d have a positive value. If this is difficult, just think it through for a while and you’ll get there.

Now, this integral is somewhat special because F and s are vectors, and the F·ds product above is a so-called dot product between two vectors. The integral itself is a so-called path integral and so you may not have learned how to solve this one. But let me explain the dot product at least: the dot product of two vectors is the product of the magnitudes of those two vectors (i.e. their length) times the cosine of the angle between the two vectors:

F·d=│F││ds│cosθ

Why that cosine? Well… To go from one point to another (from point 0 to point 1, for example), we can take any path really. [In fact, it is actually not so obvious that all paths will yield the same value for the potential energy: it is the case for so-called conservative forces only. But so gravity and the electromagnetic force are conservative forces and so, yes, we can take any path and we will find the same value.] Now, if the direction of the force and the direction of the displacement are the same, then that angle θ will be equal to zero and, hence, the dot product is just the product of the magnitudes (cos(0) = 1). However, if the direction of the force and the direction of the displacement are not the same, then it’s only the component of the force in the direction of the displacement that’s doing work, and the magnitude of that component is Fcosθ. So there you are: that explains why we need that cosine function.

Now, solving that ‘special’ integral is not so easy because the distance between the two charges at point 0 is zero and, hence, when we try to solve the integral by putting in the formula for F and finding the primitive and all that, you’ll find there’s a division by zero involved. Of course, there’s a way to solve the integral, but I won’t do it here. Just accept the general result here for U(r):

U(r) = q1q2/4πε0r

You can immediately see that, because we’re dealing with opposite charges, U(r) will always be negative, while the limit of this function for r going to infinity is equal to zero indeed. Conversely, its limit equals –∞ for r going to zero. As for the 4πεfactor in this formula, that factor plays the same role as the G-factor for gravity. Indeed, εis an ubiquitous electric constant: ε≈ 8.854×10-12 F/m, but it can be included in the value of the charges by choosing another unit and, hence, it’s often omitted – and that’s what I’ll also do here. Now, the same formula obviously applies to point 2 in the graph as well, and so now we can calculate the difference in potential energy between point 1 and point 2:

Capture

Does that make sense? Yes. We’re, once again, doing work against the force when moving the charge from point  1 to point 2. So that’s why we have a minus sign in front. As for the signs of qand q2, remember these are opposite. As for the value of the (r2 – r1) factor, that’s obviously positive because  r2 > r1. Hence, ΔU = U(1) – U(2) is negative. How do we interpret that? U(2) and U(1) are negative values, the difference between those two values, i.e. U(1) – U(2), is negative as well? Well… Just remember that ΔU is minus the work done to move the charge from point 1 to point 2. Hence, the change in potential energy (ΔU) is some negative value because the amount of work that needs to be done to move the charge from point 1 to point 2 is decidedly positive. Hence, yes, the charge has a higher energy level (albeit negative – but that’s just because of our convention which equates potential energy at infinity with zero) at point 2 as compared to point 1.

What about gravity? Well… That linear graph above is an approximation, we said, and it also takes r = h = 0 as the reference point but it assigns a value of zero for the potential energy there (as opposed to the –∞ value for the electromagnetic force above). So that graph is actually an linearization of a graph resembling the one below: we only start counting when we are on the Earth’s surface, so to say.

Capture

However, in a more advanced physics course, you will probably see the following potential energy function for gravity: U(r) = –GMm/r, and the graph of this function looks exactly the same as that graph we found for the potential energy between two opposite charges: the curve starts at point (0, –∞) and ends at point (∞, 0).

OK. Time to move on to another illustration or application: the covalent bond between two hydrogen atoms.

Application: the covalent bond between two hydrogen atoms

The graph below shows the potential energy as a function of the distance between two hydrogen atoms. Don’t worry about its exact mathematical shape: just try to understand it.

potential-energy-curve-H2-moleculecovalent_bond_hydrogen_3

 

 

Natural hydrogen comes in Hmolecules, so there is a bond between two hydrogen atoms as a result of mutual attraction. The force involved is a chemical bond: the two hydrogen atoms share their so-called valence electron, thereby forming a so-called covalent bond (which is a form of chemical bond indeed, as you should remember from your high-school courses). However, one cannot push two hydrogen atoms too close, because then the positively charged nuclei will start repelling each other, and so that’s what is depicted above: the potential energy goes up very rapidly because the two atoms will repel each other very strongly.

The right half of the graph shows how the force of attraction vanishes as the two atoms are separated. After a while, the potential energy does not increase any more and so then the two atoms are free.

Again, the reference point does not matter very much: in the graph above, the potential energy is assumed to be zero at infinity (i.e. the ‘free’ state) but we could have chosen another reference point: it would only shift the graph up or down. 

This brings us to another point: the law of energy conservation. For that, we need to introduce the concept of kinetic energy once again.

The formula for kinetic energy

In one of my previous posts, I defined the kinetic energy of an object as the excess energy over its rest energy:

K.E. = T = mc– m0c= γm0c– m0c= (γ–1)m0c2

γ is the Lorentz factor in this formula (γ = (1–v2/c2)-1/2), and I derived the T = mv2/2 formula for the kinetic energy from a Taylor expansion of the formula above, noting that K.E. = mv2/2 is actually an approximation for non-relativistic speeds only, i.e. speeds that are much less than c and, hence, have no impact on the mass of the object: so, non-relativistic means that, for all practical purposes, m = m0. Now, if m = m0, then mc– m0c2 is equal to zero ! So how do we derive the kinetic energy formula for non-relativistic speeds then? Well… We must apply another method, using Newton’s Law: the force equals the time rate of change of the momentum of an object. The momentum of an object is denoted by p (it’s a vector quantity) and is the product of its mass and its velocity (p = mv), so we can write

F = d(mv)/dt (again, all bold letters denote vectors).

When the speed is low (i.e. non-relativistic), then we can just treat m as a constant and so we can write F = mdv/dt = ma (the mass times the acceleration). If m would not be constant, then we would have to apply the product rule: d(mv) = (dm/dt)v + m(dv/dt), and so then we would have two terms instead of one. Treating m as a constant also allows us to derive the classical (Newtonian) formula for kinetic energy:

Capture

So if we assume that the velocity of the object at point O is equal to zero (so vo = 0), then ΔT will be equal to T and we get what we were looking for: the kinetic energy at point P will be equal to T = mv2/2.

Energy conservation

Now, the total energy – potential and kinetic – of an object (or a system) has to remain constant, so we have E = T + U = constant. As a consequence, the time derivative of the total energy must equal zero. So we have:

E = T + U = constant, and dE/dt = 0

Can we prove that with the formulas T = mv2/2 and U = q1q2/4πε0r? Yes, but the proof is a bit lengthy and so I won’t prove it here. [We need to take the derivatives ∂T/∂t and ∂U/∂t and show that these derivatives are equal except for the sign, which is opposite, and so the sum of those two derivatives equals zero. Note that ∂T/∂t = (dT/dv)(dv/dt) and that ∂U/∂t = (dU/dr)(dr/dt), so you have to use the chain rule for derivatives here.] So just take a mental note of that and accept the result:

(1) mv2/2 + q1q2/4πε0= constant when the electromagnetic force is involved (no minus sign, because the sign of the charges makes things come out alright), and
(2) mv2/2 – GMm/= constant when the gravitational force is involved (note the minus sign, for the reason mentioned above: when the gravitational force is involved, we need to reverse the sign).

We can also take another example: an oscillating spring. When you try to compress a (linear) spring, the spring will push back with a force equal to F = kx. Hence, the energy needed to compress a (linear) spring a distance x from its equilibrium position can be calculated from the same integral/infinite sum formula: you will get U = kx2/2 as a result. Indeed, this is an easy integral (not a path integral), and so let me quickly solve it:

Capture

While that U = kx2/2 formula looks similar to the kinetic energy formula, you should note that it’s a function of the position, not of velocity, and that the formula does not involve the mass of the object we’re attaching to the string. So it’s a different animal altogether. However, because of the energy conservation law, the graph of both the potential and kinetic energy will obviously reflect each other, just like the energy graphs of a swinging pendulum, as shown below. We have:

T + U = mv2/2 + kx2/2 = C

energy - pendulum - 2 energy - pendulum

Note: The graph above mentions an ‘ideal’ pendulum because, in reality, there will be an energy loss due to friction and, hence, the pendulum will slowly stop, as shown below. Hence, in reality, energy is conserved, but it leaks out of the system we are observing here: it gets lost as heat, which is another form of kinetic energy actually.

energy - pendulum - 3

Another application: estimating the radius of an atom

very nice application of the energy concepts introduced above is the so-called Bohr model of a hydrogen atom. Feynman introduces that model as an estimate of the size (or radius) of an atom (see Feynman’s Lectures, Vol. III, p. 2-6). The argument is the following.

The radius of an atom is more or less the spread (usually denoted by Δ or σ) in the position of the electron, so we can write that Δx = a. In words, the uncertainty about the position is the radius a. Now, we know that the uncertainty about the position (x) also determines the uncertainty about the momentum (p = mv) of the electron because of the Uncertainty Principle ΔxΔp ≥ ħ/2 (ħ ≈ 6.6×10-16 eV·s). The principle is illustrated below, and in a previous posts I proved the relationship. [Note that k in the left graph actually represents the wave number of the de Broglie wave, but wave number and momentum are related through the de Broglie relation p = ħk.]

example of wave packet

Hence, the order of magnitude of the momentum of the electron will – very roughly – be p ≈ ħ/a. [Note that Feynman doesn’t care about factors 2 or π or even 2π (h = 2πħ): the idea is just to get the order of magnitude (Feynman calls it a ‘dimensional analysis’), and that he actually equates p with p = h/a, so he doesn’t use the reduced Planck constant (ħ).]

Now, the electron’s potential energy will be given by that U(r) = q1q2/4πε0formula above, with q1= e (the charge of the proton) and q2= –e (i.e. the charge of the electron), so we can simplify this to –e2/a. 

The kinetic energy of the electron is given by the usual formula: T = mv2/2. This can be written as T = mv2/2 = m2v2/2m = p2/2m =  h2/2ma2. Hence, the total energy of the electron is given by

E = T + U = h2/2ma– e2/a

What does this say? It says that the potential energy becomes smaller as a gets smaller (that’s because of the minus sign: when we say ‘smaller’, we actually mean a larger negative value). However, as it gets closer to the nucleus, it kinetic energy increases. In fact, the shape of this function is similar to that graph depicting the potential energy of a covalent bond as a function of the distance, but you should note that the blue graph below is the total energy (so it’s not only potential energy but kinetic energy as well).

CaptureI guess you can now anticipate the rest of the story. The electron will be there where its total energy is minimized. Why? Well… We could call it the minimum energy principle, but that’s usually used in another context (thermodynamics). Let me just quote Feynman here, because I don’t have a better explanation: “We do not know what a is, but we know that the atom is going to arrange itself to make some kind of compromise so that the energy is as little as possible.”

He then calculates, as expected, the derivative dE/da, which equals dE/da = –h2/ma3 e2/a2. Setting dE/da equal to zero, we get the ‘optimal’ value for a: 

ah2/me=0.528×10-10 m = 0.528 Å (angstrom)

Note that this calculation depends on the value one uses for e: to be correct, we need to put the 4πε0 factor back in. You also need to ensure you use proper and compatible units for all factors. Just try a couple of times and you should find that 0.528 value.

Of course, the question is whether or not this back-of-the-envelope calculation resembles anything real? It does: this number is very close to the so-called Bohr radius, which is the most probable distance between the proton and and the electron in a hydrogen atom (in its ground state) indeed. The Bohr radius is an actual physical constant and has been measured to be about 0.529 angstrom. Hence, for all practical purposes, the above calculation corresponds with reality. [Of course, while Feynman started with writing that we shouldn’t trust our answer within factors like 2, π, etcetera, he concludes his calculation by noting that he used all constants in such a way that it happens to come out the right number. :-)]

The corresponding energy for this value for a can be found by putting the value aback into the total energy equation, and then we find:

E= –me4/2h= –13.6 eV

Again, this corresponds to reality, because this is the energy that is needed to kick an electron out of its orbit or, to use proper language, this is the energy that is needed to ionize a hydrogen atom (it’s referred to as a Rydberg of energy). By way of conclusion, let me quote Feynman on what this negative energy actually means: “[Negative energy] means that the electron has less energy when it is in the atom than when it is free. It means it is bound. It means it takes energy to kick the electron out.”

That being said, as we pointed out above, it is all a matter of choosing our reference point: we can add or subtract any constant C to the energy equation: E + C = T + U + C will still be constant and, hence, respect the energy conservation law. But so I’ll conclude here and – of course – check if my kids understand any of this.

And what about potential?

Oh – yes. I forgot. The title of this post suggests that I would also write something on what is referred to as ‘potential’, and it’s not the same as potential energy. So let me quickly do that.

By now, you are surely familiar with the idea of a force field. If we put a charge or a mass somewhere, then it will create a condition such that another charge or mass will feel a force. That ‘condition’ is referred to as the field, and one represents a field by field vectors. For a gravitational field, we can write:

F = mC

C is the field vector, and F is the force on the mass that we would ‘supply’ to the field for it to act on. Now, we can obviously re-write that integral for the potential energy as

U = –∫F·ds = –m∫C·ds = mΨ with Ψ (read: psi) = ∫C·d= the potential

So we can say that the potential Ψ is the potential energy of a unit charge or a unit mass that would be placed in the field. Both C (a vector) as well Ψ (a scalar quantity, i.e. a real number) obviously vary in space and in time and, hence, are a function of the space coordinates x, y and z as well as the time coordinate t. However, let’s leave time out for the moment, in order to not make things too complex. [And, of course, I should not say that this psi has nothing to do with the probability wave function we introduced in previous posts. Nothing at all. It just happens to be the same symbol.]

Now, U is an integral, and so it can be shown that, if we know the potential energy, we also know the force. Indeed, the x-, y and z-component of the force is equal to:

F= – ∂U/∂x, F= – ∂U/∂y, F= – ∂U/∂z or, using the grad (gradient) operator: F = –∇U  

Likewise, we can recover the field vectors C from the potential function Ψ:

C= – ∂Ψ/∂x, C= – ∂Ψ/∂y, C= – ∂Ψ/∂z, or C = –∇Ψ

That grad operator is nice: it makes a vector function out of a scalar function.

In the ‘electrical case’, we will write:

F = qE

 And, likewise,

U = –∫F·ds = –q∫E·ds = qΦ with Φ (read: phi) = ∫E·d= the electrical potential.

Unlike the ‘psi’ potential, the ‘phi’ potential is well known to us, if only because it’s expressed in volts. In fact, when we say that a battery or a capacitor is charged to a certain voltage, we actually mean the voltage difference between the parallel plates of which the capacitor or battery consists, so we are actually talking the difference in electrical potential ΔΦ = Φ– Φ2., which we also express in volts, just like the electrical potential itself.

Post scriptum:

The model of the atom that is implied in the above derivation is referred to as the so-called Bohr model. It is a rather primitive model (Wikipedia calls it a ‘first-order approximation’) but, despite its limitations, it’s a proper quantum-mechanical view of the hydrogen atom and, hence, Wikipedia notes that “it is still commonly taught to introduce students to quantum mechanics.” Indeed, that’s Feynman also uses it in one of his first Lectures on Quantum Mechanics (Vol. III, Chapter 2), before he moves on to more complex things.

Light

I started the two previous posts attempting to justify why we need all these mathematical formulas to understand stuff: because otherwise we just keep on repeating very simplistic but nonsensical things such as ‘matter behaves (sometimes) like light’, ‘light behaves (sometimes) like matter’ or, combining both, ‘light and matter behave like wavicles’. Indeed: what does ‘like‘ mean? Like the same but different? 🙂 However, I have not said much about light so far.

Light and matter are two very different things. For matter, we have quantum mechanics. For light, we have quantum electrodynamics (QED). However, QED is not only a quantum theory about light: as Feynman pointed out in his little but exquisite 1985 book on quantum electrodynamics (QED: The Strange Theory of Light and Matter), it is first and foremost a theory about how light interacts with matter. However, let’s limit ourselves here to light.

In classical physics, light is an electromagnetic wave: it just travels on and on and on because of that wonderful interaction between electric and magnetic fields. A changing electric field induces a magnetic field, the changing magnetic field then induces an electric field, and then the changing electric field induces a magnetic field, and… Well, you got the idea: it goes on and on and on. This wonderful machinery is summarized in Maxwell’s equations – and most beautifully so in the so-called Heaviside form of these equations, which assume a charge-free vacuum space (so there are no other charges lying around exerting a force on the electromagnetic wave or the (charged) particle whom’s behavior we want to study) and they also make abstraction of other complications such as electric currents (so there are no moving charges going around either).

I reproduced Heaviside’s Maxwell equations below as well as an animated gif which is supposed to illustrate the dynamics explained above. [In case you wonder who’s Heaviside? Well… Check it out: he was quite a character.] The animation is not all that great but OK enough. And don’t worry if you don’t understand the equations – just note the following:

  1. The electric and magnetic field E and B are represented by perpendicular oscillating vectors.
  2. The first and third equation (∇·E = 0 and ∇·B = 0) state that there are no static or moving charges around and, hence, they do not have any impact on (the flux of) E and B.
  3. The second and fourth equation are the ones that are essential. Note the time derivatives (∂/∂t): E and B oscillate and perpetuate each other by inducing new circulation of B and E.

Heaviside form of Maxwell's equations

The constants μ and ε in the fourth equation are the so-called permeability (μ) and permittivity (ε) of the medium, and μ0 and ε0 are the values for these constants in a vacuum space. Now, it is interesting to note that με equals 1/c2, so a changing electric field only produces a tiny change in the circulation of the magnetic field. That’s got something to do with magnetism being a ‘relativistic’ effect but I won’t explore that here – except for noting that the final Lorentz force on a (charged) particle F = q(E + v×B) will be the same regardless of the reference frame (moving or inertial): the reference frame will determine the mixture of E and B fields, but there is only one combined force on a charged particle in the end, regardless of the reference frame (inertial or moving at whatever speed – relativistic (i.e. close to c) or not). [The forces F, E and B on a moving (charged) particle are shown below the animation of the electromagnetic wave.] In other words, Maxwell’s equations are compatible with both special as well as general relativity. In fact, Einstein observed that these equations ensure that electromagnetic waves always travel at speed c (to use his own words: “Light is always propagated in empty space with a definite velocity c which is independent of the state of motion of the emitting body.”) and it’s this observation that led him to develop his special relativity theory.

Electromagneticwave3Dfromside

325px-Lorentz_force_particle

The other interesting thing to note is that there is energy in these oscillating fields and, hence, in the electromagnetic wave. Hence, if the wave hits an impenetrable barrier, such as a paper sheet, it exerts pressure on it – known as radiation pressure. [By the way, did you ever wonder why a light beam can travel through glass but not through paper? Check it out!] A very oft-quoted example is the following: if the effects of the sun’s radiation pressure on the Viking spacecraft had been ignored, the spacecraft would have missed its Mars orbit by about 15,000 kilometers. Another common example is more science fiction-oriented: the (theoretical) possibility of space ships using huge sails driven by sunlight (paper sails obviously – one should not use transparent plastic for that). 

I am mentioning radiation pressure because, although it is not that difficult to explain radiation pressure using classical electromagnetism (i.e. light as waves), the explanation provided by the ‘particle model’ of light is much more straightforward and, hence, a good starting point to discuss the particle nature of light:

  1. Electromagnetic radiation is quantized in particles called photons. We know that because of Max Planck’s work on black body radiation, which led to Planck’s relation: E = hν. Photons are bona fide particles in the so-called Standard Model of physics: they are defined as bosons with spin 1, but zero rest mass and no electric charge (as opposed to W bosons). They are denoted by the letter or symbol γ (gamma), so that’s the same symbol that’s used to denote gamma rays. [Gamma rays are high-energy electromagnetic radiation (i.e. ‘light’) that have a very definite particle character. Indeed, because of their very short wavelength – less than 10 picometer (10×10–12 m) and high energy (hundreds of KeV – as opposed to visible light, which has a wavelength between 380 and 750 nanometer (380-750×10–9 m) and typical energy of 2 to 3 eV only (so a few hundred thousand times less), they are capable of penetrating through thick layers of concrete, and the human body – where they might damage intracellular bodies and create cancer (lead is a more efficient barrier obviously: a shield of a few centimeter of lead will stop most of them. In case you are not sure about the relation between energy and penetration depth, see the Post Scriptum.]
  2. Although photons are considered to have zero rest mass, they have energy and, hence, an equivalent relativistic mass (m = E/c2) and, therefore, also momentum. Indeed, energy and momentum are related through the following (relativistic) formula: E = (p2c+ m02c4)1/2 (the non-relativistic version is simply E = p2/2m0 but – quite obviously – an approximation that cannot be used in this case – if only because the denominator would be zero). This simplifies to E = pc or p = E/c in this case. This basically says that the energy (E) and the momentum (p) of a photon are proportional, with c – the velocity of the wave – as the factor of proportionality.
  3. The generation of radiation pressure can then be directly related to the momentum property of photons, as shown in the diagram below – which shows how radiation force could – perhaps – propel a space sailing ship. [Nice idea, but I’d rather bet on nuclear-thermal rocket technology.]

Sail-Force1

I said in my introduction to this post that light and matter are two very different things. They are, and the logic connecting matter waves and electromagnetic radiation is not straightforward – if there is any. Let’s look at the two equations that are supposed to relate the two – the de Broglie relation and the Planck relation:

  1. The de Broglie relation E = hassigns a de Broglie frequency (i.e. the frequency of a complex-valued probability amplitude function) to a particle with mass m through the mass-energy equivalence relation E = mc2. However, the concept of a matter wave is rather complicated (if you don’t think so: read the two previous posts): matter waves have little – if anything – in common with electromagnetic waves. Feynman calls electromagnetic waves ‘real’ waves (just like water waves, or sound waves, or whatever other wave) as opposed to… Well – he does stop short of calling matter waves unreal but it’s obvious they look ‘less real’ than ‘real waves’. Indeed, these complex-valued psi functions (Ψ) – for which we have to square the modulus to get the probability of something happening in space and time, or to measure the likely value of some observable property of the system – are obviously ‘something else’! [I tried to convey their ‘reality’ as well as I could in my previous post, but I am not sure I did a good job – not all really.]
  2. The Planck relation E = hν relates the energy of a photon – the so-called quantum of light (das Lichtquant as Einstein called it in 1905 – the term ‘photon’ was coined some 20 years later it is said) – to the frequency of the electromagnetic wave of which it is part. [That Greek symbol (ν) – it’s the letter nu (the ‘v’ in Greek is amalgamated with the ‘b’) – is quite confusing: it’s not the v for velocity.]

So, while the Planck relation (which goes back to 1905) obviously inspired Louis de Broglie (who introduced his theory on electron waves some 20 years later – in his PhD thesis of 1924 to be precise), their equations look the same but are different – and that’s probably the main reason why we keep two different symbols – f and ν – for the two frequencies.

Photons and electrons are obviously very different particles as well. Just to state the obvious:

  1. Photons have zero rest mass, travel at the speed of light, have no electric charge, are bosons, and so on and so on, and so they behave differently (see, for example, my post on Bose and Fermi, which explains why one cannot make proton beam lasers). [As for the boson qualification, bosons are force carriers: photons in particular mediate (or carry) the electromagnetic force.]
  2. Electrons do not weigh much and, hence, can attain speeds close to light (but it requires tremendous amounts of energy to accelerate them very near c) but so they do have some mass, they have electric charge (photons are electrically neutral), and they are fermions – which means they’re an entirely different ‘beast’ so to say when it comes to combining their probability amplitudes (so that’s why they’ll never get together in some kind of electron laser beam either – just like protons or neutrons – as I explain in my post on Bose and Fermi indeed).

That being said, there’s some connection of course (and that’s what’s being explored in QED):

  1. Accelerating electric charges cause electromagnetic radiation (so moving charges (the negatively charged electrons) cause the electromagnetic field oscillations, but it’s the (neutral) photons that carry it).
  2. Electrons absorb and emit photons as they gain/lose energy when going from one energy level to the other.
  3. Most important of all, individual photons – just like electrons – also have a probability amplitude function – so that’s a de Broglie or matter wave function if you prefer that term.

That means photons can also be described in terms of some kind of complex wave packet, just like that electron I kept analyzing in my previous posts – until I (and surely you) got tired of it. That means we’re presented with the same type of mathematics. For starters, we cannot be happy with assigning a unique frequency to our (complex-valued) de Broglie wave, because that would – once again – mean that we have no clue whatsoever where our photon actually is. So, while the shape of the wave function below might well describe the E and B of a bona fide electromagnetic wave, it cannot describe the (real or imaginary) part of the probability amplitude of the photons we would associate with that wave.

constant frequency waveSo that doesn’t work. We’re back at analyzing wave packets – and, by now, you know how complicated that can be: I am sure you don’t want me to mention Fourier transforms again! So let’s turn to Feynman once again – the greatest of all (physics) teachers – to get his take on it. Now, the surprising thing is that, in his 1985 Lectures on Quantum Electrodynamics (QED), he doesn’t really care about the amplitude of a photon to be at point x at time t. What he needs to know is:

  1. The amplitude of a photon to go from point A to B, and
  2. The amplitude of a photon to be absorbed/emitted by an electron (a photon-electron coupling as it’s called).

And then he needs only one more thing: the amplitude of an electron to go from point A to B. That’s all he needs to explain EVERYTHING – in quantum electrodynamics that is. So that’s partial reflection, diffraction, interference… Whatever! In Feynman’s own words: “Out of these three amplitudes, we can make the whole world, aside from what goes on in nuclei, and gravitation, as always!” So let’s have a look at it.

I’ve shown some of his illustrations already in the Bose and Fermi post I mentioned above. In Feynman’s analysis, photons get emitted by some source and, as soon as they do, they travel with some stopwatch, as illustrated below. The speed with which the hand of the stopwatch turns is the angular frequency of the phase of the probability amplitude, and it’s length is the modulus -which, you’ll remember, we need to square to get a probability of something, so for the illustration below we have a probability of 0.2×0.2 = 4%. Probability of what? Relax. Let’s go step by step.

Stopwatch

Let’s first relate this probability amplitude stopwatch to a theoretical wave packet, such as the one below – which is a nice Gaussian wave packet:

example of wave packet

This thing really fits the bill: it’s associated with a nice Gaussian probability distribution (aka as a normal distribution, because – despite its ideal shape (from a math point of view), it actually does describe many real-life phenomena), and we can easily relate the stopwatch’s angular frequency to the angular frequency of the phase of the wave. The only thing you’ll need to remember is that its amplitude is not constant in space and time: indeed, this photon is somewhere sometime, and that means it’s no longer there when it’s gone, and also that it’s not there when it hasn’t arrived yet. 🙂 So, as you long as you remember that, Feynman’s stopwatch is a great way to represent a photon (or any particle really). [Just think of a stopwatch in your hand with no hand, but then suddenly that hand grows from zero to 0.2 (or some other random value between 0 and 1) and then shrinks back from that random value to 0 as the photon whizzes by. […] Or find some other creative interpretation if you don’t like this one. :-)]

Now, of course we do not know at what time the photon leaves the source and so the hand of the stopwatch could be at 2 o’clock, 9 o’clock or whatever: so the phase could be shifted by any value really. However, the thing to note is that the stopwatch’s hand goes around and around at a steady (angular) speed.

That’s OK. We can’t know where the photon is because we’re obviously assuming a nice standardized light source emitting polarized light with a very specific color, i.e. all photons have the same frequency (so we don’t have to worry about spin and all that). Indeed, because we’re going to add and multiply amplitudes, we have to keep it simple (the complicated things should be left to clever people – or academics). More importantly, it’s OK because we don’t need to know the exact position of the hand of the stopwatch as the photon leaves the source in order to explain phenomena like the partial reflection of light on glass. What matters there is only how much the stopwatch hand turns in the short time it takes to go from the front surface of the glass to its back surface. That difference in phase is independent from the position of the stopwatch hand as it reaches the glass: it only depends on the angular frequency (i.e. the energy of the photon, or the frequency of the light beam) and the thickness of the glass sheet. The two cases below present two possibilities: a 5% chance of reflection and a 16% chance of reflection (16% is actually a maximum, as Feynman shows in that little book, but that doesn’t matter here).

partial reflection

But – Hey! – I am suddenly talking amplitudes for reflection here, and the probabilities that I am calculating (by adding amplitudes, not probabilities) are also (partial) reflection probabilities. Damn ! YOU ARE SMART! It’s true. But you get the idea, and I told you already that Feynman is not interested in the probability of a photon just being here or there or wherever. He’s interested in (1) the amplitude of it going from the source (i.e. some point A) to the glass surface (i.e. some other point B), and then (2) the amplitude of photon-electron couplings – which determine the above amplitudes for being reflected (i.e. being (back)scattered by an electron actually).

So what? Well… Nothing. That’s it. I just wanted you to give some sense of de Broglie waves for photons. The thing to note is that they’re like de Broglie waves for electrons. So they are as real or unreal as these electron waves, and they have close to nothing to do with the electromagnetic wave of which they are part. The only thing that relates them with that real wave so to say, is their energy level, and so that determines their de Broglie wavelength. So, it’s strange to say, but we have two frequencies for a photon: E= hν and E = hf. The first one is the Planck relation (E= hν): it associates the energy of a photon with the frequency of the real-life electromagnetic wave. The second is the de Broglie relation (E = hf): once we’ve calculated the energy of a photon using E= hν, we associate a de Broglie wavelength with the photon. So we imagine it as a traveling stopwatch with angular frequency ω = 2πf.

So that’s it (for now). End of story.

[…]

Now, you may want to know something more about these other amplitudes (that’s what I would want), i.e. the amplitude of a photon to go from A to B and this coupling amplitude and whatever else that may or may not be relevant. Right you are: it’s fascinating stuff. For example, you may or may not be surprised that photons have an amplitude to travel faster or slower than light from A to B, and that they actually have many amplitudes to go from A to B: one for each possible path. [Does that mean that the path does not have to be straight? Yep. Light can take strange paths – and it’s the interplay (i.e. the interference) between all these amplitudes that determines the most probable path – which, fortunately (otherwise our amplitude theory would be worthless), turns out to be the straight line.] We can summarize this in a really short and nice formula for the P(A to B) amplitude [note that the ‘P’ stands for photon, not for probability – Feynman uses an E for the related amplitude for an electron, so he writes E(A to B)].

However, I won’t make this any more complicated right now and so I’ll just reveal that P(A to B) depends on the so-called spacetime interval. This spacetime interval (I) is equal to I = (z2– z1)+ (y2– y1)+ (x2– x1)– (t2– t1)2, with the time and spatial distance being measured in equivalent units (so we’d use light-seconds for the unit of distance or, for the unit of time, the time it takes for light to travel one meter). I am sure you’ve heard about this interval. It’s used to explain the famous light cone – which determines what’s past and future in respect to the here and now in spacetime (or the past and present of some event in spacetime) in terms of

  1. What could possibly have impacted the here and now (taking into account nothing can travel faster than light – even if we’ve mentioned some exceptions to this already, such as the phase velocity of a matter wave – but so that’s not a ‘signal’ and, hence, not in contradiction with relativity)?
  2. What could possible be impacted by the here and now (again taking into account that nothing can travel faster than c)?

In short, the light cone defines the past, the here, and the future in spacetime in terms of (potential) causal relations. However, as this post has – once again – become too long already, I’ll need to write another post to discuss these other types of amplitudes – and how they are used in quantum electrodynamics. So my next post should probably say something about light-matter interaction, or on photons as the carriers of the electromagnetic force (both in light as well as in an atom – as it’s the electromagnetic force that keeps an electron in orbit around the (positively charged) nucleus). In case you wonder, yes, that’s Feynman diagrams – among other things.

Post scriptum: On frequency, wavelength and energy – and the particle- versus wave-like nature of electromagnetic waves

I wrote that gamma waves have a very definite particle character because of their very short wavelength. Indeed, most discussions of the electromagnetic spectrum will start by pointing out that higher frequencies or shorter wavelengths – higher frequency (f) implies shorter wavelength (λ) because the wavelength is the speed of the wave (c in this case) over the frequency: λ = c/f – will make the (electromagnetic) wave more particle-like. For example, I copied two illustrations from Feynman’s very first Lectures (Volume I, Lectures 2 and 5) in which he makes the point by showing

  1. The familiar table of the electromagnetic spectrum (we could easily add a column for the wavelength (just calculate λ = c/f) and the energy (E = hf) besides the frequency), and
  2. An illustration that shows how matter (a block of carbon of 1 cm thick in this case) looks like for an electromagnetic wave racing towards it. It does not look like Gruyère cheese, because Gruyère cheese is cheese with holes: matter is huge holes with just a tiny little bit of cheese ! Indeed, at the micro-level, matter looks like a lot of nothing with only a few tiny specks of matter sprinkled about!

And so then he goes on to describe how ‘hard’ rays (i.e. rays with short wavelengths) just plow right through and so on and so on.

  electromagnetic spectrumcarbon close-up view

Now it will probably sound very stupid to non-autodidacts but, for a very long time, I was vaguely intrigued that the amplitude of a wave doesn’t seem to matter when looking at the particle- versus wave-like character of electromagnetic waves. Electromagnetic waves are transverse waves so they oscillate up and down, perpendicular to the direction of travel (as opposed to longitudinal waves, such as sound waves or pressure waves for example: these oscillate back and forth – in the same direction of travel). And photon paths are represented by wiggly lines, so… Well, you may not believe it but that’s why I stupidly thought it’s the amplitude that should matter, not the wavelength.

Indeed, the illustration below – which could be an example of how E or B oscillates in space and time – would suggest that lower amplitudes (smaller A’s) are the key to ‘avoiding’ those specks of matter. And if one can’t do anything about amplitude, then one may be forgiven to think that longer wavelengths – not shorter ones – are the key to avoiding those little ‘obstacles’ presented by atoms or nuclei in some crystal or non-crystalline structure. [Just jot it down: more wiggly lines increase the chance of hitting something.] But… Both lower amplitudes as well as longer wavelengths imply less energy. Indeed, the energy of a wave is, in general, proportional to the square of its amplitude and electromagnetic waves are no exception in this regard. As for wavelength, we have Planck’s relation. So what’s wrong in my very childish reasoning?

Cosine wave concepts

As usual, the answer is easy for those who already know it: neither wavelength nor amplitude have anything to do with how much space this wave actually takes as it propagates. But of course! You didn’t know that? Well… Sorry. Now I do. The vertical y axis might measure E and B indeed, but the graph and the nice animation above should not make you think that these field vectors actually occupy some space. So you can think of electromagnetic waves as particle waves indeed: we’ve got ‘something’ that’s traveling in a straight line, and it’s traveling at the speed of light. That ‘something’ is a photon, and it can have high or low energy. If it’s low-energy, it’s like a speck of dust: even if it travels at the speed of light, it is easy to deflect (i.e. scatter), and the ’empty space’ in matter (which is, of course, not empty but full of all kinds of electromagnetic disturbances) may well feel like jelly to it: it will get stuck (read: it will be absorbed somewhere or not even get through the first layer of atoms at all). If it’s high-energy, then it’s a different story: then the photon is like a tiny but very powerful bullet – same size as the speck of dust, and same speed, but much and much heavier. Such ‘bullet’ (e.g. a gamma ray photon) will indeed have a tendency to plow through matter like it’s air: it won’t care about all these low-energy fields in it.

It is, most probably, a very trivial point to make, but I thought it’s worth doing so.

[When thinking about the above, also remember the trivial relationship between energy and momentum for photons: p = E/c, so more energy means more momentum: a heavy truck crashing into your house will create more damage than a Mini at the same speed because the truck has much more momentum. So just use the mass-energy equivalence (E = mc2) and think about high-energy photons as armored vehicles and low-energy photons as mom-and-pop cars.]