# What's the meaning of the norm of Poynting 4-vector?

1. Dec 16, 2015

### andresB

Assuming a four velocity
$u_{\nu}=(1,0,0,0)$

we can use the Maxwell Energy-Momentum Tensor to build a 4-vector in the following way

$P^{\mu}=u_{\nu}T^{\mu\nu}=\left(\frac{E^{2}+B^{2}}{2},\mathbf{E}\times\mathbf{B}\right)$

So, we have a vector whose time component is the energy density of the field and the spatial components the 3d Poynting vector.

Easy to check that $P^{\mu}P_{\mu}$ is positive definite, and I wonder what is the meaning of this last quantity?

2. Dec 16, 2015

### Staff: Mentor

Is it? Try it for a pure radiation field, where $\mathbf{E}$ and $\mathbf{B}$ are equal in magnitude and perpendicular to each other. You should get that $P^\mu P_\mu = 0$.

First you might want to think about the question, what is the physical meaning of $P^\mu$ itself?

3. Dec 16, 2015

### bcrowell

Staff Emeritus
A better way to think about it is that the manifestly relativistic version of the Poynting vector is simply the stress-energy tensor itself. In general, anything we express as a vector cross product in three dimensions is really a rank-2 tensor in relativity.

4. Dec 16, 2015

### Staff: Mentor

I like this suggestion, but there are two flies in the ointment:

(1) The 1-1 correspondence between cross products of vectors and rank-2 tensors only holds in 3 dimensions, not 4;

(2) The rank-2 tensor corresponding to the cross product (in 3 dimensions) is antisymmetric, whereas the stress-energy tensor is symmetric.

5. Dec 16, 2015

### andresB

Duh, you are correct, to think that the it was positive definite was my mistake, for some reasons I thought that the mass term will come up non-zero for radiation. So, nevermind.

6. Dec 16, 2015

### bcrowell

Staff Emeritus
Right, I mean "is" in a loose sense.

7. Dec 18, 2015

### bcrowell

Staff Emeritus
Hawking and Ellis has a nice discussion of this sort of thing on p. 62. In their discussion, the reason for singling out the vector that the OP notates as u is that it's a Killing vector. Minkowski space has 10 Killing vectors, so we have 10 different flux densities such as P. The reason we care about a flux is that we can integrate it and apply Gauss's theorem to it. We don't normally care about the square of a flux, so I don't see any strong physical reasons to believe that the square of P would have any useful physical interpretation.

8. Dec 19, 2015

### my2cts

9. Dec 19, 2015

### bcrowell

Staff Emeritus
I don't see the connection to the OP's question, which was about electromagnetism.

10. Dec 25, 2015

### PFfan01

$P^{\mu}$ is not a energy-power flow 4-vector even in free-space, because it is not consistent with the principle of relativity; in other words, $P^{\mu}$ is not invariant in form in all inertial frames of reference. Thus $P^{\mu}$ itself does not have physical meaning, just a pure math 4-vector.

11. Dec 25, 2015

### bcrowell

Staff Emeritus
This logic doesn't quite work. We can't conclude, just because something isn't Lorentz invariant, that it has no physical meaning. For example, velocity is frame-dependent, but it does have physical meaning.

In fact we do have a measure of the flow of energy-momentum, which is the stress-energy tensor. The stress-energy tensor isn't Lorentz invariant, but it does have physical meaning.

12. Dec 25, 2015

### PFfan01

I think you misunderstood what I mean. What I mean is that $P^{\mu}$ as a 4-vector in OP does not have physics meaning; namely $P^{\mu}$ can not taken as a "physical law". The words "velocity is frame-dependent" is ambiguous. For a massive particle, the velocity is given by the space component of a 4-velocity that is Lorentz invariant in form; however, a photon does not have a 4-velocity, which is not violating the principle of relativity, because the principle of relativity does not require every physical quantity to be a Lorentz invariant, 4-vector, or tensor...

13. Dec 25, 2015

### Jonathan Scott

The Poynting vector combined with the Maxwell energy density only transforms between frames of reference like a four-vector energy-momentum density in the absence of sources, although it works as a local description for relative flow of energy and momentum. For a discussion of this, see the paper "A Proposed Electromagnetic Momentum-Energy 4-Vector for Charged Bodies" by J.W.Butler, or section 17.5 "Covariant Definitions of Electromagnetic Energy and Momentum" in Jackson "Classical Electrodynamics", second edition (which references the Butler paper in a footnote on page 795).

14. Dec 25, 2015

### bcrowell

Staff Emeritus
Reading the abstract of the Butler paper, I don't understand why he considers it "a most remarkable anomaly" that the density of energy-momentum in an electromagnetic field doesn't transform as a four-vector. The density of energy-momentum shows up in relativity as the stress-energy tensor, which transforms as a rank-2 tensor. You can pick out the top row of the matrix and make it look like an object with four components, but of course it's not going to transform as a vector.

15. Dec 25, 2015

### bcrowell

Staff Emeritus
I don't know what you mean by this. A vector is a physical object. A physical law is a prediction.

Your logic doesn't make much sense to me here. Nobody claimed that the principle of relativity required every physical quantity to transform as a tensor. I don't see the relevance of your example about the photon.

16. Dec 25, 2015

### PFfan01

Traditionally, “Lorentz invariance” refers to all mathematical equations expressing the laws of nature must be invariant in form only under the Lorentz transformation, and they must be Lorentz scalars, four-vectors, or four-tensors... (see p. 540 of ref. 25, for example). In other words, the Lorentz invariance is a single requirement that combines the two postulates together, and it is equivalent to the two postulates [45].

25. J.D. Jackson. Classical electrodynamics. 3rd ed. John Wiley & Sons, Hoboken, NJ. 1999.
45. W. Pauli. Theory of relativity. Pergamon Press, London. 1958. p. 21.

Last edited: Dec 25, 2015
17. Dec 25, 2015

### PFfan01

It is exactly what I mean.
This article also claims the same thing: "... the momentum and energy densities themselves cannot directly constitute a four-vector, ..."
Can. J. Phys. 93: 1510–1522 (2015) dx.doi.org/10.1139/cjp-2015-0167.