Centre of Energy in Special Relativity

Question

In Griffith's Introduction to Electrodynamics (4th edition), Chapter 12 on Special Relativity, the author says on page 546 that analogous to the center of mass in Newtonian mechanics, one can define the center of energy in relativity as given. In the footnote, he also says that the proof of this expression is not trivial.

I tried deriving this expression by considering total momentum as the sum of individual momenta and assuming that energy varied with time. I got an expression similar to the one in the book, but with two extra terms involving time derivatives of energy, which are not canceling out. So, should we assume that total energy is a constant throughout the derivation? Also, I checked the references, but couldn't find any sort of derivation there. Are there other methods to derive this expression? Edit: I think I will also share my attempt at the solution. In the last step, after the desired term, two extra terms are coming in, which are not canceling out. Can someone please check if the derivation is correct? Consider an n-particle system. For the particle i, one may write the momentum as $$\vec p_i = \frac{m_i\vec u_i}{\sqrt{1-\frac{u_i^2}{c^2}}}$$ For the collection of particles, net momentum is then, $$\vec p = \sum_{i=1}^n \frac{m_i\vec u_i}{\sqrt{1-\frac{u_i^2}{c^2}}}$$ if one multiplies and divides this expression on RHS by $c^2$ the equation can be rewritten as $$\vec p = \frac{1}{c^2}\sum_{i=1}^n E_i\vec u_i$$ where $E_i$ is energy of particle i. Also, $\vec u_i = \frac{d\vec r_i}{dt}$ Thus, $\vec p = \frac{E}{c^2}(\frac{1}{E}\sum_{i=1}^n E_i \frac{d\vec r_i}{dt})$ Now comes the trouble. I was trying to relate the center of mass's derivation in classical mechanics to this derivation: if individual energies do not vary with time, one gets the expression in the book. However, if this happens, or if an external force is being applied, E is no longer constant in time, and we cannot simply multiply and divide by E to get the result from the book. ($E=\sum_{i=1}^n E_i$, the total energy of the system). Applying the product rule of differentiation to this gives: $$\vec p = \frac{E}{c^2}\frac{d\vec R}{dt} + \frac{1}{Ec^2}(\frac{dE}{dt}\sum_{i=1}^n E_i\vec r_i - E\sum_{i=1}^n\vec r_i\frac{dE_i}{dt})$$ where $\vec R = \frac{\sum_{i=1}^nE_i\vec r_i}{E}$ represents the centre of energy. The first term on RHS is all we need, but I don't see how the other terms cancel: so, is there any fault with this derivation, or any other elegant way to prove this relation?

Answer 1

If we refer to the Coleman, Van Vleck article cited in your excerpt from Griffith, we can use the stress-energy tensor $T^{\mu\nu}$ to derive the relation. First, we know that this tensor is symmetric ($T^{\mu\nu}=T^{\nu\mu}$) and that it is conserved ($\partial_{\mu}T^{\mu\nu} = 0$).

This tensor describes the density and flux of energy and momentum in a region of spacetime. The energy of a system (considering a closed volume) is: \begin{equation} E = \int T^{00}d^{3}\vec{x} \end{equation}

Now according to the conservation equation, $\partial_{0}T^{0\nu}=0$, meaning then that $\partial_{0}T^{00}=0$. Thus, the total energy of the system is conserved: \begin{equation} \frac{dE}{dt} =0 \end{equation}

Indeed, in your calculations you get the correct result if you assume E independent of time (which is the correct assumption).

If you want to know more about the stress-energy tensor.