In layman's terms why is the time component of the stress-energy tensor associated with energy density?

Question

Why is the time component of the stress energy tensor $T^{00}$ associated as energy density?

Answer 1

Consider a stationary box of volume $V$ filled with total electric charge $Q$. The charge density of the box is $\rho_0 = \frac{Q}{V}$, and the box is stationary so the current density is zero. If you boost to a frame moving with relative velocity $-\mathbf v$, then the volume of the box will be contracted, and so the charge density will increase; the new charge density will be $\rho = \gamma \rho_0$, where $\gamma \equiv (1-v^2/c^2)^{-1/2}$. At the same time, the observed current density would become $\rho \mathbf v = \gamma \rho_0 \mathbf v$.

The charge density and current density transform like the time and space components of a 4-vector, so we introduce $$\mathbf J = (c\rho,\rho \mathbf v)$$

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Now consider a stationary box of volume $V$ filled with mass $m$. The energy density of the box is $u_0=\frac{mc^2}{V}$, and the box is stationary, so the momentum density of the box is zero. If you boost to a frame moving with relative velocity $-\mathbf v$, then the volume of the box will be contracted, and the energy of the box will increase, so the energy density will increase twice; the new energy density will be $$u = \frac{\gamma mc^2}{V/\gamma}=\gamma^2 u_0$$ and the new momentum density will be $\frac{1}{c^2}\gamma^2 u_0 \mathbf v$.

Energy density and momentum density (unlike charge density and current density) do not transform like the components of a 4-vector. They do, however, transform as though we defined a (2,0)-tensor with components

$$T^{\mu \nu} = \pmatrix{u_0 & 0 & 0 & 0\\0&0&0&0\\0&0&0&0\\0&0&0&0}$$ and performed a Lorentz transformation on it. Let $\mathbf v = v\hat x$, so

$$\Lambda^\mu_{\ \ \nu} = \pmatrix{\gamma & \beta\gamma & 0 & 0 \\ \beta\gamma&\gamma&0&0\\0&0&1&0\\0&0&0&1}$$

We then have

$$\Lambda^\mu_{\ \ \rho}\Lambda^\nu_{\ \ \sigma}T^{\rho\sigma}=\Lambda^\mu_{\ \ 0}\Lambda^\nu_{\ \ 0}u_0 = \pmatrix{\gamma^2 u_0& \frac{1}{c}\gamma^2 u_0v&0&0\\ \frac{1}{c}\gamma^2u_0 v& \frac{1}{c^2}\gamma^2 u_0v^2& 0 & 0\\0&0&0&0\\0&0&0&0}=T'^{\mu \nu}$$

$T^{00}$ is identified with energy density, while $T^{0i}=T^{i0}$ are identified with momentum density (times $c$). The physical interpretation of the $T^{ij}$ term is not so obvious; its interpretation is borne out when one performs the same analysis on a relativistic fluid, in which case the $T^{ij}$ terms are identified with the fluid's stress tensor.

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What why is energy density in the time component of the tensor? Or to put it another way what does energy density have to do with time?

Energy has a very deep connection with time. In this case, it is the $00$-component of the stress-energy tensor ultimately because it is the $0$-component of the 4-momentum.

The deeper explanation for this is that energy is the conserved Noether charge associated with time-translation symmetry. If a theory with Lagrangian $L$ is invariant under spacetime translations $x^\mu \rightarrow x^\mu + dx^\mu$, then a quantity $p^\mu$ (the 4-momentum) is conserved; $x^0$ is time and $p^0$ is energy. The continued association between energy and time extends to the stress-energy tensor.