# Energy-momentum vs charge-current

Is there a simple intuitive way to explain why energy-momentum density requires a tensor, while charge-current density is a vector?

$$\partial_{\mu} J^{\mu} = 0$$ is a statement, in effect, that the integral of $$\rho$$ over all space is time-independent: that is, charge is conserved. In classical physics, energy is conserved. Energy density together with energy flow density would seem to be a 4-vector very much like the flow of charge. Feynman gives a very similar argument when he introduces the concept of electromagnetic field energy and momentum density. I'm looking for an intuitive explanation of why energy current density and charge current density can't be treated in similar ways.

• What is mass-momentum?
– SRS
Commented Mar 19, 2019 at 18:42
• @SRS. He's referring to the four-momentum and the corresponding density. Commented Mar 19, 2019 at 18:51
• Energy-momentum tensor is the density of the 4-momentum vector. While the current density is the density of charge(scalar) and current (psuedo vector). In the first case we are taking the density of a vector hence it is two degree, while in the second we are taking the density of scalars hence it is a vector. There is a much better Mathematical derivation of Energy momentum tensor from Noether's theorem. In that we start by assuming a vector density but in order for the quantity to be independent of transformation we get a 2° tensor. Commented Mar 20, 2019 at 7:13
• It makes intuitive sense that density of a conserved vector quantity (momentum) should be a tensor; and that density of a scalar quantity (charge) should be a vector. What needs an intuitive handle is energy, taken alone: it is a scalar, and it is conserved. If a student takes that fact is along with the preceding sentence, he/she will conclude that there should be a 4-vector density associated with energy, consisting of (energy density and energy current density). Something more is needed to get over that intuituve obstacle. Commented Mar 20, 2019 at 13:49

Is there a simple intuitive way to explain why mass-momentum density requires a tensor, while charge-current density is a vector?

Yes. The four-current is the density associated with a conserved scalar, the charge. The stress energy tensor is the density associated with a conserved four-vector, the four-momentum. In each case the density is one rank higher than the corresponding conserved quantity.

I'm looking for an intuitive explanation of why energy current density and charge current density can't be treated in similar ways.

Although both are conserved, charge and energy are very different kinds of quantities. Charge is a scalar quantity (a tensor of rank 0) meaning that it is the same in all reference frames. Energy is not, different frames disagree on the value of energy. Instead, energy is a component of a vector (a tensor of tank 1).

There are very few circumstances where a scalar can be treated in a similar way as a component of a tensor.

• Why is the density one rank higher than the conserved quantity? Commented Mar 19, 2019 at 19:05
• In systems in which energy is conserved, the energy is a scalar. It is postulated (and confirmed by experiment) that charge is frame-independent, and I realize that energy density (or equivalently, mass density) is not frame independent. On the other hand, if all mass were, say, neutrons, then the neutron number and therefore the neutron rest mass could be conserved; and then one would expect mass density and mass flow density to form a 4-vector. That 4-vector would, I imagine, be a component of the total energy-momentum tensor describing the "neutron mass 4-current field". Am I wrong? Commented Mar 19, 2019 at 19:31
• @Dale, I edited the question to try to make it clearer. Can you edit your answer to more directly address the question? Commented Mar 20, 2019 at 0:23
• @S. McGrew done. Hopefully that helps.
– Dale
Commented Mar 20, 2019 at 1:38

Let $$\Omega^{\nu_1 \cdots \nu_r}$$ be some quantity of rank $$r$$. The density $$\omega^{\nu_1 \cdots \nu_r \mu}$$ is of rank $$r+1$$ because it will be contracted against a 3-dimensional element $$dV_\mu,$$ which in Minkowski space can be considered being a rank 1 tensor, e.g., with time being the first dimension, $$dx \, dy \, dz \sim (dx \, dy \, dz, 0, 0, 0)$$: $$\Omega^{\nu_1 \cdots \nu_r} = \int_{\mathbb{R}^3} \omega^{\nu_1 \cdots \nu_r \mu} \, dV_\mu.$$