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

[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?

 

[PeterDonis]

Easy to check that $P^{\mu}P_{\mu}$ is positive definite

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 $P^\mu P_\mu = 0$.

I wonder what is the meaning of this last quantity?

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

 

[bcrowell]

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.

 

[PeterDonis]

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.

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.

 

[andresB]

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 $P^\mu P_\mu = 0$.

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.

 

[bcrowell]

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.

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

 

[bcrowell]

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.

 

[my2cts]

Check out https://en.wikipedia.org/wiki/Poincaré_group#Details .

 

[bcrowell]

Check out https://en.wikipedia.org/wiki/Poincaré_group#Details .

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

 

[PFfan01]

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?

$P^{\mu}$ 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.

 

[bcrowell]

$P^{\mu}$ 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.

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.

 

[PFfan01]

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.
...

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...

 

[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).

 

[bcrowell]

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).

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.

 

[bcrowell]

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".

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

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...

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.

 

[PFfan01]

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.

But this article (Can. J. Phys. 93: 1510–1522 (2015) dx.doi.org/10.1139/cjp-2015-0167) claims:

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.

 

[PFfan01]

... it "a most remarkable anomaly" that the density of energy-momentum in an electromagnetic field doesn't transform as a four-vector.

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.