# Is Poynting vector $E \times B$ and energy density $(E^2+B^2)/2$, a four vector?

As the title suggests, does the 00th and 0i the components of the electromagnetic energy-momentum tensor viz. Poynting vector and energy density form a four-vector?

Answers with both theoretical and mathematical explanations both, preferred.

• How much do you know about the relationships between two-index tensors (e.g. the energy-momentum tensor), the one-index tensors that we usually call vectors, and tensors with other numbers of indices? Are you working through a particular textbook?
– rob
Commented Jul 23 at 1:57

No, it isn't. Because the Poynting density and Poynting energy flow density 3-vector together are 0-th row of the matrix of components $$T^{\mu\nu}$$ of the energy-momentum tensor:

$$S^\nu = T^{0\nu}$$

and this matrix $$T^{\mu\nu}$$ transforms differently (with two Lorentz transformations, and mixing in other components not on the row) than a four-vector components does (with one Lorentz transformation, and not using any other components, just those on the row).

The transformation for components of $$T$$ is $$T'^{\mu\nu} =\Lambda^\mu_{~~\alpha} \Lambda^\nu_{~~\beta }T^{\alpha\beta}$$ so zeroth row transforms as

$$T'^{0\nu} = \Lambda^0_{~~\alpha} \Lambda^\nu_{~~\beta} T^{\alpha\beta},$$ or $$T'^{0\nu} = C^\nu_{\alpha\beta} T^{\alpha\beta},$$ where $$C$$ depends only on mutual velocity of the frames, not on $$T$$.

We can see that zeroth row in primed frame depends on components of $$T$$ in unprimed frame that are not on the zeroth row in unprimed frame, so this is not a Lorentz transformation. We do not have a Lorentz transformation between components $$T^{0\beta}$$ and components $$T'^{0\nu}$$.

An example: a charged capacitor. Consider a parallel plate capacitor, with distance between the plates much smaller than the plate dimensions, being charged in empty space (not external fields are present). In the unprimed frame, it is moving in direction of electric field that exists in between its plates, and in the primed frame, it is standing still. If Poynting density $$w$$ and Poynting momentum density $$g_x$$ transformed as a four-vector, then densities in the unprimed frame would obey the relations

$$w = \gamma w' + \frac{\gamma v}{c^2}g_x'$$ $$g_x = \gamma v w' + \gamma g_x'.$$

However, this is inconsistent with how fields transforms, and thus is incorrect.

In both frames, magnetic field in between the plates vanishes. In the frame where the capacitor is at rest, this is obvious - no currents are present. In the frame where the capacitor moves, we expect some magnetic field, because we have moving charge. However, from well-known transformation of EM fields, in the region where the electric field is parallel to velocity, $$B=0$$, in other words, magnetic field vanishes. Magnetic field is present only close to the plate edges outside the capacitor, where electric field has a component perpendicular to velocity.

Since $$g_x$$ is linear in $$B$$, $$g_x$$ vanishes in both frames, and the above implies

$$w = \gamma w'$$ $$0 = \gamma v w'.$$ The second equation is clearly incorrect, as both $$v$$ and $$w'$$ are non-zero. The first equation is also incorrect. From definition of $$w$$, we have $$w = \frac{1}{2}\epsilon_0 E^2$$ $$w' = \frac{1}{2}\epsilon_0 E'^2$$ From well-known transformation of EM fields, we know that $$E'=E$$, thus we have $$w'=w$$.

The Lorentz transformation for the four-tuple $$(cw,g_x,0,0)$$ gives incorrect results, thus the four-tuple $$(cw,g_x,0,0)$$ does not transform as a four-vector.

Another related quantity that transforms as a four-vector. To find something that transforms as a four-vector, instead of the Poynting volume densities of energy and momentum, we should check the actual Poynting energy and momentum associated with some "field packet" in the whole space. Thus we can integrate the zeroth row over some large enough spatial region $$V$$ (or the whole space):

$$P^\nu(t) = \int_{V} S^\nu dV.$$ Further detailed check shows that this four-tuple transforms as a four-vector, if we integrate over the whole empty space where the Poynting densities do not vanish, and if there is no exchange of energy and momentum happening between EM field and matter. If there is such an exchange in the integration region, so the Poynting energy and momentum change in time, there isn't a unique way to pair $$P^\nu$$ and $$P'^\nu$$ between two different frames, as these things depend on times $$t,t'$$ we use to evaluate the Poynting densities. Knowing $$t$$ does not define uniquely time $$t'$$, due to finite dimensions of the integration region. Thus in general, the integral does not transform as a four-vector, because there isn't any one-to-one relation between $$t$$ and $$t'$$ we use. But for a "free field packet", $$t,t'$$ do not affect the values of tuples $$P^\nu,P'^\nu$$, and it can be shown such tuples are related one to another as four-vectors are.

• As @Mateo says, "Yes" it's a 4-vector.
• As @knzhou comments in @Mateo's answer, "This is quite misleading, as under a Lorentz transformation, $$n_\mu$$ transforms as well."
• I agreed with @knzhou and further commented
"...it's better to say that $$T^{\mu\nu}n_{\nu}$$ is an "observer-dependent 4-vector" since it depends on $$n_{\nu}$$."
This is similar to saying that (up to sign and conventions) $$F^{\mu\nu}n_{\nu}$$ is an observer-dependent 4-vector, namely, the "electric field seen by $$n_{\nu}$$". (As we know, an "electric field" is observer-dependent... and its tensorial construction displays the dependence.)

Usually, when one refers to a 4-vector [without further qualifications], that 4-vector is observer-independent.

• For example, the 4-momentum $$p^\mu$$ of a massive particle of mass $$m$$ is tangent to the particle worldline and has magnitude $$m$$.
These facts are true, regardless of the observer.
• Along these lines, the stress tensor $$T^{\mu\nu}$$ is a 4-tensor which is observer-independent.

However, the quantity $$T^{\mu\nu}n_{\nu}$$, by construction, depends on $$n_{\nu}$$.

To support my comment, here are two references:

• Misner Thorne Wheeler, Gravitation, p. 131 (Box 5.1)

where the last line says "for observer with 4-velocity $$u^a$$". (Emphasis mine)

• Wald, General Relativity, p.62-63

Equation (4.2.11) has an important physical interpretation. Consider a family of inertial observers with parallel 4-velocities $$v^a$$, so that $$\partial_b v^a=0$$. According to the above interpretation of $$T_{ab}$$, the quantity $$J_a = -T_{ab}v^b \quad(4.2.16)$$ represents the mass-energy current density 4-vector of the fluid as measured by these observers. (Emphasis mine.)

Note: this means all inertial observers will agree that:
this $$J_a= -T_{ab}v^b$$ is the "mass-energy current density 4-vector as measured by $$v^a$$".

• Oh I think I understand in what I am wrong. Indeed $-T_{ab}v^b$ is equivalent to what I have written and shown to be a 4-vector. That is not the same as $T_{a0}$ however, which holds indeed only for some specific frame/observer. Commented Jul 23 at 15:35

The volume integral of a conserved tensor, $$\partial_\mu T^{\mu\nu}=0$$, is a four vector. Not the density but the total quantity can be a four vector.

Yes. (Actually no and my answer is wrong. $$T^{\mu\nu}n_{\nu}$$ is, but $$T^{\mu 0}$$ is not. Read the edit below)

The simple answer is that the stress-energy tensor is a rank 2 tensor. The Minkowski product against any 4-vector will give you a 4-vector, in particular, with $$n_{\nu}=(1,0,0,0)$$, you get $$$$T^{\mu\nu}n_{\nu} = T^{\mu 0}\,.$$$$ Covariant notation works so nicely that you immediatly recognize anything with one free index $$\nu$$ as a 4-vector.

A more mathematically detailed explanaition may be needed if you are not used to it, or if it seems somewhat obscure. Then, you should convince yourself as follows:

$$\partial_{\mu}$$ and $$A_{\mu}$$ are both 4-vectors, transforming as (note I will freely raise and lower indexes in what follows) $$$$A^{\mu'}=\Lambda^{\mu'}_{\;\,\mu}\;A^{\mu}\,.$$$$ So $$F^{\mu\nu}=\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}$$ transforms as $$$$F^{\mu'\nu'} = \Lambda^{\mu'}_{\;\,\mu}\Lambda^{\nu'}_{\;\,\nu}\;F^{\mu\nu}$$$$ and, similarly $$\eta^{\mu'\nu'} = \Lambda^{\mu'}_{\;\,\mu}\Lambda^{\nu'}_{\;\,\nu}\;\eta^{\mu\nu}$$ (which is actually invariant by definition of the Lorentz group). Both $$F^{\mu\nu}$$ and $$\eta^{\mu\nu}$$ are rank 2 tensors, by definition, because they transform in such way. From here it follows that $$T^{\mu\nu}=F^{\mu\alpha}F^{\nu}_{\;\,\alpha}-\frac{1}{4}\eta^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}$$ is a rank 2 tensor as well $$$$T^{\mu'\nu'} = \Lambda^{\mu'}_{\;\,\mu}\Lambda^{\nu'}_{\;\,\nu}\;T^{\mu\nu}\,.$$$$ Note that in order to verify this relation you need to use the fact that upper and lower indices transform with inverse matrices, i.e. $$\Lambda_{\mu'}^{\;\,\mu}=\left(\Lambda^{-1}\right)^{\mu}_{\;\,\mu'}$$. Finally, if $$n_{\nu}$$ is any 4-vector (you may choose $$n_{\nu}$$ as I mentioned before), you can verify that $$T^{\mu\nu}n_{\nu}$$ is a 4-vector. Let me show it a little bit more explicitly \begin{align} T^{\mu'\nu'}n_{\nu'} = \delta_{\nu'}^{\;\,\alpha'} T^{\mu'\nu'}n_{\alpha'} & = \delta_{\nu'}^{\;\,\alpha'} \Lambda^{\mu'}_{\;\,\mu}\Lambda^{\nu'}_{\;\,\nu}\Lambda_{\alpha'}^{\;\,\alpha} \; T^{\mu\nu}n_{\alpha} \\ & = \Lambda^{\mu'}_{\;\,\mu}\Lambda^{\nu'}_{\;\,\nu}\Lambda_{\nu'}^{\;\,\alpha} \; T^{\mu\nu}n_{\alpha} \\ & = \Lambda^{\mu'}_{\;\,\mu}\left(\Lambda^{-1}\right)^{\;\,\nu'}_{\nu}\Lambda_{\nu'}^{\;\,\alpha} \; T^{\mu\nu}n_{\alpha} \\ & = \Lambda^{\mu'}_{\;\,\mu}\delta_{\nu}^{\;\,\alpha} \; T^{\mu\nu}n_{\alpha} \\ & = \Lambda^{\mu'}_{\;\,\mu}T^{\mu\nu}n_{\nu} \end{align} That is, $$$$T^{\mu'\nu'}n_{\nu'} = \Lambda^{\mu'}_{\;\,\mu}T^{\mu\nu}n_{\nu}\,,$$$$ which means $$T^{\mu\nu}n_{\nu}$$ is -by definition- a 4-vector.

Physically, the Poynting vector is the linear momentum of the electromagnetic field and hence $$T^{\mu 0}$$ should be its 4-momentum. If you know that the 4-momentum $$(E,\textbf{p})$$ of a particle is a 4-vector, you should expect the 4-momentum of the electromagnetic field to be a 4-vector too, and indeed the electromagnetic theory is built in such a way that it is.

Edit: Ján's answer is the correct one. I have shown $$T^{\mu\nu}n_{\nu}$$ is a 4-vector, which is correct, and Ján has shown that $$T^{0\nu}$$ is not a 4-vector, which is also correct. What I have stated incorrectly is that these two objects are the same, since in general $$T^{\mu\nu}n_{\nu} \neq T^{\mu 0}$$. It may hold true for some specific frame of reference but, as stated in the comments, that is an observer/frame-dependent statement. Also as Ján said the 4-momentum is a four vector, this is what you expect physically, but not necesarilly for the density $$T^{0\nu}$$ before integration.

• This is quite misleading, as under a Lorentz transformation, $n_\mu$ transforms as well. Commented Jul 23 at 2:48
• @knzhou Yes, indeed. You need to use the fact that upper and lower indeces transform with inverse $\Lambda$ matrices, such that they cancel out. I should probably clarify that Commented Jul 23 at 2:52
• Elaborating on @knzhou's comment, it's better to say that $T^{\mu\nu}n_\nu$ is an "observer-dependent 4-vector" since it depends on $n_\nu$. Similarly (up to sign and conventions) $F^{\mu\nu}n_\nu$ is an observer-dependent 4-vector, namely, the "electric field seen by $n_\nu$". Commented Jul 23 at 4:10
• @robphy Of course any 4-vector is observer-dependent, since it transforms non-trivially from one frame of reference to another Commented Jul 23 at 4:52
• @Mateo The 4-momentum $p^\mu$ of a massive particle is observer-independent. It is tangent to the worldline of the particle and has a magnitude equal to its mass. All observers will agree on these statements. (Of course, components of 4-vectors are observer-dependent. But I'm talking about the 4-vector itself.) Commented Jul 23 at 5:08