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In Feynman's Lecture 27 on Vol. II it is written that

There is an important theorem in mechanics which is this: whenever there is a flow of energy in any circumstance at all (field energy or any other kind of energy), the energy flowing through a unit area per unit time, when multiplied by $1/c^2$, is equal to the momentum per unit volume in the space.

I have never heard before of this theorem. Does it have a special name? In which classical work is this derived, or which textbook covers it?

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2 Answers 2

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It is simply the statement that in General relativity the energy-momentum tensor $T^{\mu\nu}$ is symmetric: $T^{\mu\nu}=T^{\nu\mu}$. The quantity $T^{0\mu}$ is the density of the $\mu$-th component of momentum, while $T^{\mu0}$ is the $\mu$-th component of the energy flux.

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  • $\begingroup$ How does the c^2 enter then? $\endgroup$
    – lalala
    Commented May 14 at 9:15
  • $\begingroup$ I am using natural units in which $c=1$. Look at the Wiki article I linked to for how $c^2$ comes in SI units. $\endgroup$
    – mike stone
    Commented May 14 at 12:13
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It comes from the theory of relativity, not Newtonian mechanics, and it is not a theorem, but more of a generalization of what holds for energy and momentum of ordinary matter and EM field, to other forms of matter or field.

In relativity theory, energy of a moving body is $\gamma mc^2$, and its momentum $\gamma m \mathbf v$. Per unit volume in the moving body, where mass exists with density $\rho$, we have energy and momentum densities

$$ \epsilon= \gamma \rho c^2, $$ $$ g = \gamma \rho v. $$

Thus we can see momentum density is related to energy density and velocity: $$ g = \frac{1}{c^2}\epsilon v. $$

But Feynman was talking about energy flow. Let the body be a cuboid whose length in direction of motion is $L$, face area $A$, and let $\Delta t $ be the time it takes for the body to shift by this length.

Energy passing through a unit area perpendicular to direction of motion, per unit time ("energy flow" or more correctly, energy flux density), is then

$$ S = \frac{E}{A\Delta t} = \frac{E}{A L/v} = \epsilon v. $$

Thus we can see momentum density $g$ can be expressed using the energy flow:

$$ g = \frac{S}{c^2}. $$

Similar derivation can be made for the Poynting energy flux density and momentum density of a plane EM wave, so the last relation holds both for moving matter, and for EM radiation (assuming energy and momentum are given by the Poynting expressions).

Now we can turn this on its head and make the generalization: since this relation is such a nice property, we expect that for all matter or field, when they are ascribed energy and momentum, it should be possible to be done in such a way that this relation still holds (this is somewhat connected to the topic of why the energy-momentum tensor is symmetric and whether it has to be symmetric, but that discussion is more complicated).

Dark matter? We don't know much about it, except it modifies gravity, but we assume its energy and momentum are described by some $\mathbf S, \mathbf g$ obeying the above relation.

In other words, the usual doctrine is, we will try to define energy flux and momentum density of any new stuff discovered or mathematically defined (matter, field) in such a way so that this relation holds. As far as I know, we do not know of any matter/field which would have $\mathbf S/c^2 \neq \mathbf g$ and thus break this doctrine.

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  • $\begingroup$ Thanks. Then Feynman is arguing a bit in circles if he uses it to derive the momentum of the EM field $\endgroup$
    – lalala
    Commented May 14 at 9:16
  • $\begingroup$ @lalala: I think "derive" is not quite the correct word here. To us mathematicians, that would be the ideal. But in this situation a very physicsy thing is happening: a special law holding in a special situation is being generalized. $\endgroup$
    – Lee Mosher
    Commented May 14 at 12:23

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