Wave functions and equations: a summary

Post scriptum note added on 11 July 2016: This is one of the more speculative posts which led to my e-publication analyzing the wavefunction as an energy propagation. With the benefit of hindsight, I would recommend you to immediately the more recent exposé on the matter that is being presented here, which you can find by clicking on the provided link. In fact, I actually made some (small) mistakes when writing the post below.

Original post:

Schrödinger’s wave equation for spin-zero, spin-1/2, and spin-one particles in free space differ from each other by a factor two:

  1. For particles with zero spin, we write: ∂ψ/∂t = i·(ħ/m)·∇2ψ. We get this by multiplying the ħ/(2m) factor in Schrödinger’s original wave equation – which applies to spin-1/2 particles (e.g. electrons) only – by two. Hence, the correction that needs to be made is very straightforward.
  2. For fermions (spin-1/2 particles), Schrödinger’s equation is what it is: ∂ψ/∂t = i·[ħ/(2m)]·∇2ψ.
  3. For spin-1 particles (photons), we have ∂ψ/∂t = i·(2ħ/m)·∇2ψ, so here we multiply the ħ/m factor in Schrödinger’s wave equation for spin-zero particles by two, which amounts to multiplying Schrödinger’s original coefficient by four.

Look at the coefficients carefully. It’s a strange succession:

  1. The ħ/m factor (which is just the reciprocal of the mass measured in units of ħ) works for spin-0 particles.
  2. For spin-1/2 particles, we take only half that factor: ħ/(2m) = (1/2)·(ħ/m).
  3. For spin-1 particles, we double that factor: 2ħ/m = 2·(ħ/m).

I describe the detail on my Deep Blue page, so please go there for more detail. What I did there, can be summarized as follows:

  • The spin-one particle is the photon, and we derived the photon wavefunction from Maxwell’s equations in free space, and found that it solves the ∂ψ/∂t = i·(2ħ/m)·∇2ψ equation, not the ∂ψ/∂t = i·(ħ/m)·∇2ψ or ∂ψ/∂t = i·[ħ/(2m)]·∇2ψ equations.
  • As for the spin-zero particles, we simplified the analysis by assuming our particle had zero rest mass, and we found that we were basically modeling an energy flow.
  • The analysis for spin-1/2 particles is just the standard analysis you’ll find in textbooks.

We can speculate how things would look like for spin-3/2 particles, or for spin-2 particles, but let’s not do that here. In any case, we will come back to this. Let’s first focus on the more familiar terrain, i.e. the wave equation for spin-1/2 particles, such as protons or electrons. [A proton is not elementary – as it consists of quarks – but it is a spin-1/2 particle, i.e. a fermion.]

The phase and group velocity of the wavefunction for spin-1/2 particles (fermions)

We’ll start with the very beginning of it all, i.e. the two equations that the young Comte Louis de Broglie presented in his 1924 PhD thesis, which give us the temporal and spatial frequency of the wavefunction, i.e. the ω and k in the θ = ω·t − k·t argument  of the a·ei·θ wavefunction:

  1. ω = E/ħ
  2. k = p/ħ

This allows to calculate the phase velocity of the wavefunction:

vp = ω/k = (E/ħ)/(p/ħ) = E/p

This is an elementary wavefunction, several of which we would add with appropriate coefficients, with uncertainty in the energy and momentum ensuring our component waves have different frequencies, and, therefore, the concept of a group velocity does not apply. In effect, the a·ei·θ wavefunction does not describe a localized particle: the probability to find it somewhere is the same everywhere. We may want to think of our wavefunction being confined to some narrow band in space, with us having no prior information about the probability density function, and, therefore, we assume a uniform distribution. Assuming our box in space is defined by Δx = x2 − x1, and imposing the normalization condition (all probabilities have to add up to one), we find that the following logic should hold:

(Δx)·a2 = (x2−x1a= 1 ⇔ Δx = 1/a2


However, we are, of course, interested in the group velocity, as the group velocity should correspond to the classical velocity of the particle. The group velocity of a composite wave is given by the vg = ∂ω/∂k formula. Of course, that formula assumes an unambiguous relation between the temporal and spatial frequency of the component waves, which we may want to denote as ωn and kn, with n = 1, 2, 3,… However, we will not use the index as the context makes it quite clear what we are talking about.

The relation between ωn and kn is known as the dispersion relation, and one particularly nice way to calculate ω as a function of k is to distinguish the real and imaginary parts of the ∂ψ/∂t =i·[ħ/(2m)]·∇2ψ wave equation and, hence, re-write it as a pair of two equations:

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

Both equations imply the following dispersion relation:

ω = ħ·k2/(2m)

We can now calculate vg = ∂ω/∂k as:

vg = ∂ω/∂k = ∂[ħ·k2/(2m)]/∂k = 2ħk/(2m) = ħ·(p/ħ)/m = p/m = m·v/m = v

That’s nice, because it’s what we wanted to find. If the group velocity would not equal the classical velocity of our particle, then our model would not make sense.

We used the classical momentum formula in our calculation above: p = m·v. To calculate the phase velocity of our wavefunction, we need to calculate that E/p ratio and, hence, we need an energy formula. Here we have a lot of choice, as energy can be defined in many ways: is it rest energy, potential energy, or kinetic energy? At this point, I need to remind you of the basic concepts.

The argument of the wavefunction as the proper time

It is obvious that the energy concept that is to be used in the ω = E/ħ is the total energy. Louis de Broglie himself noted that the energy of a particle consisted of three parts:

  1. The particle’s rest energy m0c2, which de Broglie referred to as internal energy (Eint): it includes the rest mass of the ‘internal pieces’, as de Broglie put it (now we call those ‘internal pieces’ quarks), as well as their binding energy (i.e. the quarks’ interaction energy);
  2. Any potential energy (V) it may have because of some field (so de Broglie was not assuming the particle was traveling in free space): the field(s) can be anything—gravitational, electromagnetic—you name it: whatever changes the energy because of the position of the particle;
  3. The particle’s kinetic energy, which he wrote in terms of its momentum p: K.E. = m·v2/2 = m2·v2/(2m) = (m·v)2/(2m) = p2/(2m).

So the wavefunction, as de Broglie wrote it, can be written as follows:

ψ(θ) = ψ(x, t) = a·eiθ = a·e−i[(Eint + p2/(2m) + V)·t − p∙x]/ħ 

This formula allows us to analyze interesting phenomena such as the tunneling effect and, hence, you may want to stop here and start playing with it. However, you should note that the kinetic energy formula that is used here is non-relativistic. The relativistically correct energy formula is E = mvc, and the relativistically correct formula for the kinetic energy is the difference between the total energy and the rest energy:

K.E. = E − E0 = mv·c2 − m0·c2 = m0·γ·c2 − m0·c2 = m0·c2·(γ − 1), with γ the Lorentz factor.

At this point, we should simplify our calculations by adopting natural units, so as to ensure the numerical value of = 1, and likewise for ħ. Hence, we assume all is described in Planck units, but please note that the physical dimensions of our variables do not change when adopting natural units: time is time, energy is energy, etcetera. But when using natural units, the E = mvc2 reduces to E = mv. As for our formula for the momentum, this formula remains p = mv·v, but is now some relative velocity, i.e. a fraction between 0 and 1. We can now re-write θ = (E/ħ)·t – (p/ħ)·x as:

θ = E·t – p·x = E·t − p·v·t = mv·t − mv·v·v·t = mv·(1 − v2)·t

We can also write this as:

ψ(x, t) = a·ei·(mv·t − p∙x) = a·ei·[(m0/√(1−v2))·t − (m0·v/√(1−v2)∙x) = a·ei·m0·(t − v∙x)/√(1−v2)

The (t − v∙x)/√(1−v2) factor in the argument is the proper time of the particle as given by  the formulas for the Lorentz transformation of spacetime:


However, both the θ = mv·(1 − v2)·t and θ = m0·t’ = m0·(t − v∙x)/√(1−v2) are relativistically correct. Note that the rest mass of the particle (m0) acts as a scaling factor as we multiply it with the proper time: a higher m0 gives the wavefunction a higher density, in time as well as in space.

Let’s go back to our vp = E/p formula. Using natural units, it becomes:

vp = E/p = mv/mv·v = 1/v

Interesting! The phase velocity is the reciprocal of the classical velocity! This implies it is always superluminal, ranging from vp = ∞ to vp= 1 for going from 0 to 1 = c, as illustrated in the simple graph below.

phase velocity

Let me note something here, as you may also want to use the dispersion relation, i.e. ω = ħ·k2/(2m), to calculate the phase velocity. You’d write:

vp = ω/k = [ħ·k2/(2m)]/k = ħ·k/(2m) = ħ·(p/ħ)/(2m) = m·v/(2m) = v/2

That’s a nonsensical result. Why do we get it? Because we are mixing two different mass concepts here: the mass that’s associated with the component wave, and the mass that’s associated with the composite wave. Think of it. That’s where Schrödinger’s equation is different from all of the other diffusion equations you’ve seen: the mass factor in the ∂ψ/∂t = i·[ħ/(2m)]·∇2ψ equation is the mass of the particle that’s being represented by the wavefunction that solves the equation. Hence, the diffusion constant ħ/(2m) is not a property of the medium. In that sense, it’s different from the κ/k factor in the ∂T/∂t = (κ/k)·∇2T heat diffusion, for example. We don’t have a medium here and, therefore, Schrödinger’s equation and the associated wavefunction are intimately connected.

It’s an interesting point, because if we’re going to be measuring the mass as multiples of ħ/2 (as suggested by the ħ/(2m) = 1/[m/[ħ/2)] factor itself), then its possible values (for ħ = 1) will be 1/2, 1, 3/2, 2, 5/2,… Now that should remind you of a few things—things like harmonics, or allowable spin values, or… Well… So many things. 🙂

Let’s do the exercise for bosons now.

The phase and group velocity of the  wavefunction for spin-0 particles

My Deep Blue page explains why we need to drop the 1/2 factor in Schrödinger’s equation to make it fit the wavefunction for bosons. We distinguished two bosons: (1) the (theoretical) zero-mass particle (which has spin zero), and the (actual) photon (which has spin one). Let’s first do the analysis for the spin-zero particle.

  • A zero-mass particle (i.e. a particle with zero rest mass) should be traveling at the speed of light: both its phase as well as its group velocity should be equal to = 1. In fact, we’re not talking composite wavefunctions here, so there’s no such thing as a group velocity. We’re not adding waves: there is only one wavefunction. [Note that we don’t need to add waves with different frequencies in order to localize our particle, because quantum mechanics and relativity theory come together here in what might well be the most logical and absurd conclusion ever: as an outside observer, we’re going to see all those zero-mass particles as point objects whizzing by because of the relativistic length contraction. So their wavefunction is only all over spacetime in their proper space and time, but not in ours!]
  • Now, it’s easy to show that, if we choose our time and distance units such that c = 1, then the energy formula reduces to E = m∙c2 = m. Likewise, we find that p = m∙c = m. So we have this strange condition: E = p = m.
  • Now, this is not consistent with the ω = ħ·k2/(2m) we get out of the ∂ψ/∂t = i·[ħ/(2m)]·∇2ψ equation, because E/ħ = ħ·(p/ħ)2/(2m) ⇔ E = m2/(2m) = m/2. That does not fit the E = p = m condition. The only way out is to drop the 1/2 factor, i.e. to multiply Schrödinger’s coefficient with 2.

Let’s quickly check if it does the trick. We assume E, p and m will be multiples of ħ/2 (E = p = m = n·(ħ/2), so the wavefunction is ei∙[t − x]n·/2, Schrödinger’s constant becomes 2/n, and the derivatives for ∂ψ/∂t = i·(ħ/m)·∇2ψ are:

  • ∂ψ/∂t = −i·(n/2)·ei∙[t − x]·n/2
  • 2ψ = ∂2[ei∙[t − x]·n/2]/∂x= i·(n/2)·∂[ei∙[t − x]·n/2]/∂x = −(n2/4)·ei∙[t − x]·n/2

So the Schrödinger equation becomes:

i·(n/2)·ei∙[t − x]n·/2) = −i·(2/n)·(n2/4)·ei∙[t − x]·n/2 ⇔  n/2 = n/2 ⇔ 1 = 1

As Feynman would say, it works like a charm, and note that n does not have to be some integer to make this work.

So what makes spin-1/2 particles different? The answer is: they have both linear as well as angular momentum, and the equipartition theorem tells us the energy will be shared equally among both , so they will pick up linear and angular momentum. Hence, the associated condition is not E = p = m, but E = p = 2m. We’ll come back to this.

Let’s now summarize how it works for spin-one particles

The phase and group velocity of the  wavefunction for spin-1 particles (photons)

Because of the particularities that characterize an electromagnetic wave, the wavefunction packs two waves, capturing both the electric as well as the magnetic field vector (i.e. E and B). For the detail, I’ll refer you to the mentioned page, because the proof is rather lengthy (but easy to follow, so please do check it out). I will just briefly summarize the logic here.

1. For the spin-zero particle, we measured E, m and p in units of – or as multiples of – the ħ/2 factor. Hence, the elementary wavefunction (i.e. the wavefunction for E = p = m = 1) for the zero-mass particle is ei(x/2 − t/2).

2. For the spin-1 particle (the photon), one can show that we get two of these elementary wavefunctions (ψand ψB), and one can then prove that we can write the sum of the electric and magnetic field vector as:

E + BE + B = ψ+ ψ= E + i·E

= √2·ei(x/2 − t/2+ π/4) = √2·ei(π/4)·ei(x/2 − t/2) = √2·ei(π/4)·= √2·ei(π/4)·ei(x/2 − t/2)

Hence, the photon has a special wavefunction. Does it solve the Schrödinger equation? It does when we use the 2ħ/m diffusion constant, rather than the ħ/m or ħ/(2m) coefficient. Let us quickly check it. The derivatives are:

  • ∂ψ/∂t = −√2·ei(π/4)·ei∙[t − x]/2·(i/2)
  • 2ψ = ∂2[√2·ei(π/4)·ei∙[t − x]/2]/∂x= √2·ei(π/4)·∂[ei∙[t − x]/2·(i/2)]/∂x = −√2·ei(π/4)·ei∙[t − x]/2·(1/4)

Note, however, that we have two mass, energy and momentum concepts here: EE, pE, mE and EB, pB, and mB respectively. Hence, if E= p= mE = E= p= mB = 1/2, then E = E+ EB, p = p+ pB and m = m+ mare all equal to 1. Hence, because E = p = m = 1 and we measure in units of ħ, the 2ħ/m factor is equal to 2 and, therefore, the modified Schrödinger’s equation ∂ψ/∂t = i·(2ħ/m)·∇2ψ becomes:

i·√2·ei(π/4)·ei∙[t − x]/2·(1/2) = −i·√2·2·ei(π/4)·ei∙[t − x]/2·(1/4) ⇔ 1/2 = 2/4 = 1/2

It all works out. Let’s quickly check it for E, m and p being multiples of ħ, so we write: E = p = m = n·ħ = n, so the wavefunction is √2·ei(π/4)·ei∙[t − x]n·/2, Schrödinger’s 2ħ/m constant becomes 2ħ/m = 2/n, and the derivatives for ∂ψ/∂t = i·(ħ/m)·∇2ψ are:

  • ∂ψ/∂t = −i·(n/2)·√2·ei(π/4)·ei∙[t − x]·n/2
  • 2ψ = ∂2[ei∙[t − x]·n/2]/∂x= i·√2·ei(π/4)·(n/2)·∂[ei∙[t − x]·n/2]/∂x = −√2·(n2/4)·ei(π/4)·ei∙[t − x]·n/2

So the Schrödinger equation becomes:

i·√2·ei(π/4)·(n/2)·ei∙[t − x]·n/2) = −i·√2·ei(π/4)·(2/n)·(n2/4)·ei∙[t − x]·n/2 ⇔  n/2 = n/2 ⇔ 1 = 1

It works like a charm again. Note the subtlety of the difference between the ħ/(2m) and 2ħ/m factor: it depends on us measuring the mass (and, hence, the energy and momentum as well) in units of ħ/2 (for spin-0 particles) or, alternatively (for spin-1 particles), in units of ħ. This is very deep—but it does make sense in light of the En =n·ħ·ω = n·h·f formula that solves the black-body radiation problem, as illustrated below. [The formula next to the energy levels is the probability of an atomic oscillator occupying that energy level, which is given by Boltzmann’s Law. You can check things in my post on it.]

energy levels

It is now time to look at something else.

Schrödinger’s equation as an energy propagation mechanism

The Schrödinger equations above are not complete. The complete equation includes force fields, i.e. potential energy:

schrodinger 5

To write the equation like this, we need to move the on the right-hand side of our ∂ψ/∂t = i·(2ħ/m)·∇2ψ equation to the other side, and multiply both sides with −1. [Remember: 1/i = −i.] Now, it is very interesting to do a dimensional analysis of this equation. Let’s do the right-hand side first. The ħfactor in the ħ/(2m) is expressed in J2·s2. Now that doesn’t make much sense, but then that mass factor in the denominator makes everything come out alright. Indeed, we can use the mass-equivalence relation to express m in J/(m/s)2 units. So we get: (J2·s2)·[(m/s)2/J] = J·m2. But so we multiply that with some quantity (the Laplacian) that’s expressed per m2. So −(ħ2/2m)·∇2ψ is something expressed in joule, so it’s some amount of energy! Interesting, isn’t it? [Note that it works out fine with the addition Vψ term, which is also expressed in joule.] On the left-hand side, we have ħ, and its dimension is the action dimension: J·s, i.e. force times distance times time (N·m·s). So we multiply that with a time derivative and we get J once again, the unit of energy. So it works out: we have joule units both left and right. But what does it mean?

Well… The Laplacian on the right-hand side works just the same as for our heat diffusion equation: it gives us a flux density, i.e. something expressed per square meter (1/m2). Likewise, the time derivative on the left-hand side gives us a flow per second. But so what is it that is flowing here? Well… My interpretation is that it is energy, and it’s flowing between a real and an imaginary space—but don’t be fooled by the terms here: both spaces are equally real, as both have an actual physical dimension. Let me explain.

Things become somewhat more comprehensible when we remind ourselves that the Schrödinger equation is equivalent to the following pair of equations:

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

So what? Let me insert an illustration here. See what happens. The wavefunction acts as a link function between our physical spacetime and some other space whose dimensions – in my humble opinion – are also physical. We have those sines and cosines, which mirror the energy of the system at any point in time, as measured by the proper time of the system.


Let me more precise. The wavefunction, as a link function between two spaces here, associates every point in spacetime with some real as well as some imaginary energy here—but, as mentioned above, that imaginary energy is as real as the real energy. What it embodies really is the energy conservation law: at any point in time (as measured by the proper time) the sum of kinetic and potential energy must be equal to some constant, and so that’s what’s shown here. Indeed, you should note the phase shift between the sine and the cosine function: if one reaches the +1 or −1 value, then the other function reaches the zero point—and vice versa. It’s a beautiful structure.

Of course, the million-dollar question is: is it a physical structure, or a mathematical structure? The answer is: it’s a bit of both. It’s a mathematical structure but, at the same time, its dimension is physical: it’s an energy space. It’s that energy that explains why amplitudes interfere—which, as you know, is what they do. So these amplitudes are something real, and as the dimensional analysis of Schrödinger’s equation reveals their dimension is expressed in joule, then… Well… Then these physical equations say what they say, don’t they? And what they say, is something like the diagram below.

summary 2

Note that the diagram above does not show the phase difference between the two springs. The animation below does a better job here, although you need to realize the hand of the clock will move faster or slower as our object travels through force fields and accelerates or decelerates accordingly.


We may relate that picture above to the principle of least action, which ensures that the difference between the kinetic energy (KE) and potential energy (PE) in the integrand of the action integral, i.e.


is minimized along the path of travel.

The spring metaphor should also make us think of the energy formula for a harmonic oscillator, which tells us that the total energy – kinetic (i.e. the energy related to its momentum) plus potential (i.e. the energy stored in the spring) – is equal to T + U = m·ω02/2. The ωhere is the angular velocity, and we have two springs here, so the total energy would be the sum of both, i.e. m·ω02, without the 1/2 factor. Does that make sense? It’s like an E = m·vequation, so that’s twice the (non-relativistic) kinetic energy. Does that formula make any sense?

In the context of what we’re discussing here, it does. Think about the limit situation by trying to imagine a zero-mass particle here (I am talking a zero-mass spin-1/2 particle this time). It would have no rest energy, so it’s only energy is kinetic, which is equal to:

K.E. = E − E0 = mv·c2 − m0·c2 = mc·c2

Why is mequal to mc? Zero-mass particles must travel at the speed of light, as the slightest force on them gives them an infinite acceleration. So there we are: the m·ω02 equation makes sense! But what if we have a non-zero rest mass? In that case, look at that pair of equations again: they give us a dispersion relation, i.e. a relation between ω and k. Indeed, using natural units again, so the numerical value of ħ = 1, we can write:

ω = k2/(2m) = p2/(2m) = (m·v)2/(2m) = m·v2/2

This equation seems to represent the kinetic energy but m is not the rest mass here: it’s the relativistic mass, so that makes it different from the classical kinetic energy formula (K.E. = m0·v2/2). [It may be useful here to remind you of how we get that classical formula. We basically integrate force over distance, from some start to some final point of a path in spacetime. So we write: ∫ F·ds = ∫ (m·a)·ds = ∫ (m·a)·ds = ∫ [m·(dv/dt)]·ds = ∫ [m·(ds/dt)]·d= ∫ m·v·ds. So we can solve that using the m·v2/2 primitive but only if m does not vary, i.e. if m = m0. If velocity are high, we need the relativistic mass concept.]

So we have a new energy concept here: m·v2, and it’s split over those two springs. Hmm… The interpretation of all of this is not so easy, so I will need to re-visit this. As for now, however, it looks like the Universe can be represented by a V-twin engine! 🙂

V-Twin engine


Is it real?

You may still doubt whether that new ‘space’ has an actual energy dimension. It’s a figment of our mind, right? Well… Yes and no. Again, it’s a bit of a mixture between a mathematical and a physical space: it’s definitely not our physical space, as it’s not the spacetime we’re living in. But, having said that, I don’t think this energy space is just a figment of our mind. Let me give you some additional reasons, beside the dimensional analysis we did above.

For example, there is the fact that we need to to take the absolute square of the wavefunction to get the probability that our elementary particle is actually right there! Now that’s something real! Hence, let me say a few more things about that. The absolute square gets rid of the time factor. Just write it out to see what happens:

|reiθ|2 = |r|2|eiθ|2 = r2[√(cos2θ + sin2θ)]2 = r2(√1)2 = r2

Now, the gives us the maximum amplitude (sorry for the mix of terminology here: I am just talking the wave amplitude here – i.e. the classical concept of an amplitude – not the quantum-mechanical concept of a probability amplitude). Now, we know that the energy of a wave – anywave, really – is proportional to the amplitude of a wave. It would also be logical to expect that the probability of finding our particle at some point x is proportional to the energy densitythere, isn’t it? [I know what you’ll say now: you’re squaring the amplitude, so if the dimension of its square is energy, then its own dimension must be the square root, right? No. Wrong. That’s why this confusion between amplitude and probability amplitude is so bad. Look at the formula: we’re squaring the sine and cosine, to then take the square root again, so the dimension doesn’t change: it’s √J2 = J.]

The third reason why I think the probability amplitude represents some energy is that its real and imaginary part also interfere with each other, as is evident when you take the ordinary square (i.e. not the absolute square). Then the i2   = –1 rule comes into play and, therefore, the square of the imaginary part starts messing with the square of the real part. Just write it out:

(reiθ)2 = r2(cosθ + isinθ)2 = r2(cos2θ – sin2θ + 2icosθsinθ)2 = r2(1 – 2sin2θ + 2icosθsinθ)2 

As mentioned above, if there’s interference, then something is happening, and so then we’re talking something real. Hence, the real and imaginary part of the wavefunction must have some dimension, and not just any dimension: it must be energy, as that’s the currency of the Universe, so to speak.

Let me add a philosophical note here—or an ontological note, I should say. When you think we should only have one physical space, you’re right. This new physical space, in which we relate energy to time, is not our physical space. It’s not reality—as we know, as we experience it. So, in that sense, you’re right. It’s not physical space. But then… Well… It’s a definitional matter. Any space whose dimensions are physical, is a physical space for me. But then I should probably be more careful. What we have here is some kind of projection of our physical space to a space that  lacks… Well… It lacks the spatial dimension. It’s just time – but a special kind of time: relativistic proper time – and energy—albeit energy in two dimensions, so to speak. So… What can I say? Just what I said a couple of times already: it’s some kind of mixture between a physical and mathematical space. But then… Well… Our own physical space – including the spatial dimension – is something like a mixture as well, isn’t it? We can try to disentangle them – which is what I am trying to do – but we’ll never fully succeed.

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Some content on this page was disabled on June 16, 2020 as a result of a DMCA takedown notice from The California Institute of Technology. You can learn more about the DMCA here:



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