This post goes to the heart of the E = m*c*^{2}, equation. It’s kinda* *funny, because Feynman just compresses all of it in a sub-section of his *Lectures*. However, as far as I am concerned, I feel it’s a very crucial section. *Pivotal*, I’d say, which would fit with its place in all of the 115 *Lectures *that make up the three volumes, which is sort of mid-way, which is where we are here. So let’s get go for it. 🙂

Let’s first recall what we wrote about the Poynting vector **S**, which we calculate from the magnetic and electric field vectors **E** and **B** by taking their cross-product:

This vector represents the *energy flow, *per unit area and per unit time, in electrodynamical situations. If **E** and/or **B **are zero (which is the case in electrostatics, for example, because we don’t have magnetic fields in electrostatics), then **S** is zero too, so there is no energy flow then. That makes sense, because we have no moving charges, so where would the energy go to?

I also made it clear we should think of **S** as something physical, by comparing it to the heat flow vector ** h**, which we presented when discussing vector analysis and vector operators. The heat flow out of a surface element

*da*is the area times the component of

*h*perpendicular to da, so that’s (

*•*

**h***)·da =*

**n***h*·

_{n}*da*. Likewise, we can write (

*•*

**S***)·da =*

**n***S*·

_{n}*da*. The units of

**S**and

**are also the same:**

*h**joule per second and per square meter*or, using the definition of the

*watt*(1 W = 1 J/s), in

*watt per square meter*.

*In fact, if you*

*and*

**h****S**are referred to as a

*flux density*:

- The heat flow vector
**h***heat flux density*vector, from which we get the heat flux through an area through the (•**h**)·da =**n***h*·_{n}*da*product. - The energy flow
is the*S**energy flux density vector*, from which we get the energy flux through the (•**S**)·da =**n***S*·_{n}*da*product.

So that should be enough as an introduction to what I want to talk about here. Let’s first look at the energy conservation principle once again.

**Local energy conservation**

In a way, you can look at my previous post as being all about the equation below, which we referred to as the ‘local’ energy conservation law:

Of course, it is *not *the *complete *energy conservation law. The local energy is not only in the field. We’ve got *matter* as well, and so that’s what I want to discuss here: we want to look at the energy *in the field* as well as the energy that’s *in the matter*. Indeed, field energy is conserved, and then it isn’t: if the field is doing work on matter, or matter is doing work on the field, then… Well… Energy goes from one to the other, i.e. from the field to the matter or from the matter to the field. So we need to include matter in our analysis, which we didn’t do in our last post. Feynman gives the following simple example: we’re in a dark room, and suddenly someone turns on the light switch. So now the room is full of *field energy*—and, yes, I just mean it’s not dark anymore. :-). So that means some *matter *out there must have radiated its energy out and, in the process, it must have lost the equivalent mass of that energy. So, yes, we had matter losing energy and, hence, losing mass.

Now, we know that energy and momentum are related. Respecting and incorporating relativity theory, we’ve got two *equivalent *formulas for it:

- E
^{2 }− p^{2}*c*^{2}= m_{0}^{2}*c*^{4} - p
*c*= E·(v/*c*) ⇔ p = v·E/*c*^{2 }= m·v

The E = m*c*^{2} and m = ·m_{0}·(1−v^{2}/*c*^{2})^{−1/2} formulas connect both expressions. So we can* *look at it in either of two ways. We *could *use the energy conservation law, but Feynman prefers the conservation of momentum approach, so let’s see where he takes us. If the field has some *energy *(and, hence, some equivalent *mass*) *per unit volume*, and if there’s some *flow*, so if there’s some *velocity *(which there is: that’s what our previous post was all about), then it will have a certain *momentum *per unit volume. [Remember: momentum is *mass *times *velocity*.] That momentum will have a *direction*, so it’s a *vector*, just like **p** = m**v**. We’ll write it as **g**, so we *define* **g** as:

**g** is the momentum of the field per unit volume.

What units would we express it in? We’ve got a bit of choice here. For example, because we’re relating everything to energy here, we may want to convert our *kilogram* into eV/*c*^{2 }or J/*c*^{2 }units, using the mass-energy equivalence relation E = m*c*^{2}. Hmm… Let’s first keep the kg as a measure of inertia though. So we write: [**g**] = [m]·[**v**]/m^{3 }= (kg·m/s)/m^{3}. Hmm… That doesn’t show it’s energy, so let’s replace the kg with a unit that’s got newton and meter in it, cf. the **F** = m**a** law. So we write: [**g**] = (kg·m/s)/m^{3 }= (kg/s)/m^{2 }= [(N·s^{2}/m)/s]/m^{2 }= N·s/m^{3}. Well… OK. The newton·second is the unit of momentum indeed, and we can re-write it including the joule (1 J = 1 N·m), so then we get [**g**] = (J·s/m^{4}), so what’s that? Well… Nothing much. However, I do note it *happens* to be the dimension of **S**/*c*^{2}, so that’s [**S**/*c*^{2}] = [J/(s·m^{2})]·(s^{2}/m^{2}) = (J·s/m^{4}). 🙂 Let’s continue the discussion.

Now, momentum is conserved, and each *component *of it is conserved. So let’s look at the x-direction. We should have something like:

If you look at this carefully, you’ll probably say: “OK. I understood the thing with the dark room and light switch. Mass got converted into field energy, but what’s that second term of the left?”

Good. Smart. Right remark. Perfect. […] Let me try to answer the question. While all of the quantities above are expressed *per unit volume*, we’re actually looking at the *same *infinitesimal volume element here, so the example of the light switch is actually an example of a ‘momentum outflow’, so it’s actually an example of that second term of the left-hand side of the equation above kicking in! 🙂

Indeed, the first term just sort of reiterates the mass-energy equivalence: the energy that’s in the matter can become field energy, so to speak, *in our infinitesimal volume element itself*, and vice versa. But if it doesn’t, then it should get out and, hence, become ‘momentum outflow’. Does that make sense? No?

Hmm… What to say? You’ll need to look at that equation a couple of times more, I guess. But *I* need to move on, unfortunately. [Don’t get put off when I say things like this: I am basically talking to myself, so it means I’ll need to re-visit this myself. :-/]

Let’s look at all of the three terms:

- The left-hand side (i.e. the time rate-of-change of the momentum of matter) is easy. It’s just the
*force*on it, which we know is equal to**F**= q(**E**+**v**×**B**). Do we know that? OK… I’ll admit it. Sometimes it’s easy to forget where we are in an analysis like this, but so we’re looking at the electromagnetic force here. 🙂 As we’re talking infinitesimals here and, therefore, charge*density*rather than discrete*charges*, we should re-write this as the force*per unit volume*which is ρ**E**+**j**×**B**. [This is an interesting formula which I didn’t use before, so you should double-check it. :-)] - The first term on the right-hand side should be equally obvious, or… Well… Perhaps somewhat less so. But with all my rambling on the Uncertainty Principle and/or the wave-particle duality, it should make sense. If we scrap the second term on the right-hand side, we basically have an equation that is equivalent to the E = m
*c*^{2}equation. No? Sorry. Just look at it, again and again. You’ll end up understanding it. 🙂 - So it’s that second term on the right-hand side. What the hell does
*that*say? Well… I could say: it’s the local energy or momentum conservation law. If the energy or momentum doesn’t stay in, it has to go out. 🙂 But that’s not very satisfactory as an answer, of course. However, please just go along with this ‘temporary’ answer for a while.

So what *is *that second term on the right-hand side? As we wrote it, it’s an x-component – or, let’s put it differently, it is or was part of the x-component of the momentum density – but, frankly, we should probably allow it to go out in any direction really, as the only constraint on the left-hand side is a *per second *rate of change of something. Hence, Feynman suggest to equate it to something like this:

What a, b and c? The components of some vector? Not sure. We’re stuck. This piece *really *requires *very *advanced math. In fact, as far as I know, this is the *only* time where Feynman says: “Sorry. This is too advanced. I’ll just give you the equation.* Sorry*.” So that’s what he does. He explains the *philosophy* of the argument, which is the following:

- On the left-hand side, we’ve got the time rate-of-change of momentum, so that obeys the
**F**= d**p**/dt = d(m**v**)/dt law, with the force**F**,*per unit volume*, being equal to**F**(unit volume) =**E**+**j**×**B**. - On the right-hand side, we’ve got something that can be written as:

So we’d need to find a way to ρ**E**+**j**×**B** in terms of **E **and **B ***only – *eliminating ρ and **j** by using Maxwell’s equations or whatever other trick * – *and then juggle terms and make substitutions to get it into a form that looks like the formula above, i.e. the right-hand side of that equation. But so Feynman doesn’t show us how it’s being done. He just mentions some theorem in physics, which says that the energy that’s flowing through a unit area per unit time divided by *c*^{2} – so that’s E/*c*^{2 }*per unit area and per unit time* – ** must **be equal to the momentum per unit volume in the space, so we write:

**g** = **S**/*c*^{2}

He *illustrates *the general theorem that’s used to get the equation above by giving two examples:

OK. Two good examples. However, it’s still frustrating to *not *see how we get the **g** = **S**/*c*^{2} in the *specific *context of the electromagnetic force, so let’s do a dimensional analysis at least. In my previous post, I showed that the dimension of **S** *must *be J/(m^{2}·s), so [**S**/*c*^{2}] = [J/(m^{2}·s)]/(m^{2}/s^{2}) = [N·m/(m^{2}·s)]·(s^{2}/m^{2}) = [N·s/m^{3}]. Now, we know that the unit of mass is 1 kg = N/(m/s^{2}). That’s just the force law: a force of 1 *newton *will give a mass of 1 kg an acceleration of 1 m/s per second, so 1 N = 1 kg·(m/s^{2}). So the [N·s/m^{3}] dimension is equal to [kg·(m/s^{2})·s/m^{3}] = [(kg·(m/s)/m^{3}] = [(kg·(m/s)]/m^{3}, which is the dimension of momentum (**p** = m**v**) *per unit volume*, indeed. So, yes, the dimensional analysis works out, and it’s also in line with the **p** = v·**E**/*c*^{2}** ^{ }**= m·

**v**equation, but… Oh… We did a dimensional analysis already, where we also showed that [

**g**]

**= [**

**S**/

*c*

^{2}] = (J·s/m

^{4}). Well… In any case… It’s a bit frustrating to

*not*see the detail here, but let us note the

**once again:**

*the Grand Result*The Poynting vector** S **gives us the **energy flow** as well as the **momentum density g **= **S**/*c*^{2}.

But what does it all mean, *really*? Let’s go through Einstein’s illustration of the principle. That will help us a lot. Before we do, however, I’d like to note something. I’ve always wondered a bit about that dichotomy between energy and momentum. Energy is* force* times *distance*: 1 *joule *is 1 *newton *× 1 *meter *indeed (1 J = 1 N·m). Momentum is *force* times *time*, as we can express it in N·s. Planck’s constant *h *combines all three in the dimension of *action*, which is *force *times *distance *times *time*: *h *≈ 6.6×10^{−34} N·m·s, indeed. I like that unity. In this regard, you should, perhaps, quickly review that post in which I explain that *h *is the energy per cycle, i.e. per wavelength or per period, of a *photon*, regardless of its wavelength. So it’s really something very fundamental.

We’ve got something similar here: energy and momentum coming together, and being shown as one aspect of the same thing: some *oscillation*. Indeed, just see what happens with the dimensions when we ‘distribute’ the 1/*c*^{2 }factor on the right-hand side over the two sides, so we write: *c*·**g **= **S**/*c *and work out the dimensions:

- [
*c*·**g**] = (m/s)·(N·s)/m^{3 }= N/m^{2 }= J/m^{3}. - [
**S**/*c*] = (s/m)·(N·m)/(s·m^{2}) = N/m^{2 }= J/m^{3}.

Isn’t that nice? Both sides of the equation now have a dimension like ‘the force per unit area’, or ‘the energy per unit volume’. To get that, we just *re-scaled ***g** and **S**, by *c* and 1/*c* respectively. As far as I am concerned, this shows an underlying unity we probably tend to mask with our ‘related but different’ energy and momentum concepts. It’s like **E** and **B**: I just love it we can write them together in our Poynting formula **S **= ε_{0}*c*^{2}**E**×**B**. In fact, let me show something else here, which you should think about. You know that *c*^{2 }= 1/(ε_{0}μ_{0}), so we can write **S **also as **S** = **E**×**B**/μ_{0}. That’s nice, but what’s nice too is the following:

**S**/*c*=*c*·**g**= ε_{0}*c***E**×**B**=**E**×**B/**μ_{0}*c***S**/**g**=*c*^{2 }= 1/(ε_{0}μ_{0})

So, once again, Feynman may feel the Poynting vector is sort of counter-intuitive when analyzing specific situations but, as far as I am concerned, I feel the Poyning vector makes things actually *easier *to understand. Instead of two **E** and **B** vectors, and two concepts to deal with ‘energy’ (i.e. energy and momentum), we’re sort of unifying things here. In that regard – i.e in regard of feeling we’re talking the same thing really – I’d really highlight the **S**/**g **= *c*^{2 }= 1/(ε_{0}μ_{0}) equation. Indeed, the universal constant *c *acts just like the fine-structure constant here: it links everything to everything. 🙂

And, yes, it’s also about time we introduce the so-called *principle of least action* to explain things, because *action*, as a concept, combines force, distance and time indeed, so it’s a bit more promising than just energy, of just momentum. Having said that, you’ll see in the next section that it’s sometimes quite useful to have the choice between one formula or the other. But… Well… Enough talk. Let’s look at Einstein’s car.

**Einstein’s car**

Einstein’s car is a wonderful device: it rolls without any friction and it moves with a little flashlight. That’s all it needs. It’s pictured below. 🙂 So the situation is the following: the flashlight shoots some light out from one side, which is then stopped at the opposite end of the car. When the light is emitted, there must be some recoil. In fact, we know it’s going to be equal to 1/*c* times the energy because all we need to do is apply the p*c* = E·(v/*c*) formula for v = *c*, so we know that p = E/*c*. Of course, this momentum now needs to move Einstein’s car. It’s frictionless, so it should work, but still… The car has some mass M, and so that will determine its recoil velocity: v = p/M. We just apply the general p = mv formula here, and v is *not *equal to *c *here, of course! Of course, then the light hits the opposite end of the car and delivers the same momentum, so that stops the car again. However, it *did *move over some distance x = vt. So we could flash our light again and get to wherever we want to get. [Never mind the infinite accelerations involved!] So… Well… Great! Yes, but Einstein didn’t like this car when he first saw it. In fact, he still doesn’t like it, because he knows it won’t take you very far. 🙂

The problem is that we seem to be moving the center of gravity of this car by fooling around on the inside only. Einstein doesn’t like that. He thinks it’s impossible. And he’s right of course. The thing is: the center of gravity did *not *change. What happened here is that we’ve got some *blob of energy*, and so that blob has some equivalent mass (which we’ll denote by U/*c*^{2}), and so that equivalent mass moved all the way from one side to the other, i.e. over the length of the car, which we denote by L. In fact, it’s stuff like this that inspired the whole theory of the field energy and field momentum, and how it interacts with *matter*.

What happens here is like switching the light on in the dark room: we’ve got matter doing work on the field, and so matter loses mass, and the field gains it, through its momentum and/or energy. To calculate how much, we could integrate **S**/*c* or *c*·**g **over the volume of our blob, and we’d get something in *joule *indeed, but there’s a simpler way here. The momentum conservation says that the momentum of our car and the momentum of our blob must be equal, so if T is the time that was needed for our blob to go to the other side – and so that’s, of course, also the time during which our car was rolling – then M·v = M·x/T must be equal to (U/*c*^{2})·*c *= (U/*c*^{2})·L/T. The 1/T factor on both sides cancel, so we write: M·x = (U/*c*^{2})·L. Now, what is x? Yes. In case you were wondering, that’s what we’re looking for here. 🙂 Here it is:

x = vT = vL/*c* = (p/M)·(L/*c*) = [U/*c*)/M]·(L/*c*) = (U/*c*^{2})·(L/M)

So what’s next? Well… Now we need to show that the center-of-mass actually did *not *move with this ‘transfer’ of the blob. I’ll leave the math to you here: it should all work out. And you can also think through the obvious questions:

- Where is the energy and, hence, the mass of our blob
*after*it stops the car? Hint: think about excited atoms and imagine they might radiate some light back. 🙂 - As the car did move a little bit, we should be able to move it further and further away from its center of gravity, until the center of gravity is no longer in the car. Hint: think about batteries and energy levels going down while shooting light out. It just won’t happen. 🙂

Now, what about a blob of light going from the top to the bottom of the car? Well… That involves the conservation of *angular *momentum: we’ll have more mass on the bottom, but on a shorter lever-arm, so *angular *momentum is being conserved. It’s a *very *good question though, and it led Einstein to *combine* the center-of-gravity theorem with the angular momentum conservation theorem to explain stuff like this.

It’s all fascinating, and one can think of a great many paradoxes that, at first, seem to contradict the Grand Principles we used here, which means that they would contradict all that we have learned so far. However, a careful analysis of those paradox reveals that they are *paradoxes* indeed: propositions which *sound *true but are, in the end, self-contradictory. In fact, when explaining electromagnetism over his various *Lectures*, Feynman tasks his readers with a rather formidable paradox when discussing the laws of induction, he solves it here, ten chapters later, after describing what we described above. You can busy yourself with it but… Well… I guess you’ve got something better to do. If so, just take away the key lesson: there’s momentum in the field, and it’s also possible to build up *angular *momentum in a *magnetic *field and, if you switch it off, the angular momentum will be given back, somehow, as it’s stored energy.

That’s also why the seemingly irrelevant circulation of **S** we discussed in my previous post, where we had a charge next to an ordinary magnet, and where we found that there was energy circulating around, is not so queer. The energy is there, in the circulating field, and it’s *real*. As real as can be. 🙂

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In the last diagram, what exactly would the energy flow look like in 3d? A sphere? Because if it has a circulation, then as a sphere it would need to conserve the flow in a really weird way. So is it possible that it looks like a spherical spiral? I’m just learning this stuff now, but I am comfortable with the equations so far.

Hi Ivan – I would need to look at that, but I have little time now. I have a busy day job ! I’ll try to look into it ! 🙂