Electromagnetic field 4-momentum density and inertial frames

[USeptim]

Electromagnetic field has a density of energy

U = ε/2*E²+ μ/2* H²

And a density of momentum, given by the Poynting vector

S = E x H

For an element of volume dV you have a four vector of energy and momentum which is

[E,P] = dV * [U, S]

Being E the energy in the element of volume and P the momentum of inertia.

If you measure this four vector from another inertial frame Fr' which moves with velocity v respect the initial frame, Fr, you can get [E',P'] by making a boost over [E,P]. Because of the contraction of lengths, the dimension in which the boost is made shortens in a factor 1/γ, γ = 1/(1-(v/c)²)^1/2, so the density of energy and momentum must be multiplied by γ.

So: [U', S'] = γ * Boost( [U, S], v )

Where Boost is the Lorentz Boost function.

You can transform the fields E and H themselves to the new frame Fr' by using the transformations described here:

https://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity

I would expect that after transforming the fields I would get the same values for [U', S'] that the ones get by the previous way, however, nothing farther to reality than this:

If I have a E = E_o x and I apply a boost in x direction, according to the transformations of fields I get the same fields:

E' = E H' = H = 0

So I will have U' = U and S' = S whereas if I transform the 4-momentum vector, I will get:

U' = γ² * U

S' = - γ² * U * v

Which of these results is the good one? Can anybody help me in finding what is wrong with this reasoning?

 

[Dale]

I would expect that after transforming the fields I would get the same values for [U', S'] that the ones get by the previous way, however, nothing farther to reality than this:

This is correct. The energy density and momentum density do not transform as components of a four vector, they transform as components of a rank 2 tensor. Specifically, they are components of the stress energy tensor.

 

[USeptim]

Thank you DaleSpam for your answer, so the correct four vector would be the one obtained by transforming the fields.

If want to transform the energy and momentum density I have to do it as a energy stress tensor, which transforms on boosts as explained in this link:

http://physics.stackexchange.com/qu...-momentum-tensor-under-lorentz-transformationSergio

 

[Dale]

If want to transform the energy and momentum density I have to do it as a energy stress tensor,

Yes.

so the correct four vector would be the one obtained by transforming the fields.

I don't think that there is a correct four vector. It is just a tensor. Trying to find the correct four vector to represent energy momentum density is like trying to find the right scalar to represent momentum. It just doesn't fit.

 

[vanhees71]

This is a very important point in relativistic field theory. You have a energy-momentum-stress tensor $T^{\mu \nu}$, which is symmetric. For a closed system energy-momentum conservation holds (I stick to special relativity; in GR it's more subtle). Then you have the continuity equation
$$\partial_{\mu} T^{\mu \nu}=0.$$
Now, using the four-dimensional Gauss integral theorem, you can show that
$$p^{\nu} = \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} T^{0\nu}$$
is conserved and a four-vector under Lorentz transformations.

For an open system, where the continuity equation does not hold, the so defined $p^{\nu}$ is neither a four vector nor conserved. For details, see Jackson, Classical Electrodynamics.

 

[stedwards]

Now, using the four-dimensional Gauss integral theorem, you can show that
$$p^{\nu} = \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} T^{0\nu}$$
is conserved and a four-vector under Lorentz transformations.

As I understand it, $T^{0 i}$, in terms of matter, is "the flux of relativistic mass across the surface $x^i$". This seems to be shorthand for the infinitesimal surface $dx^j \wedge dx^k = {\epsilon^{jk}}_i dx^i$.

Shouldn't the integral be over an area; $p^i = \int_{R^2} T^{0 i} dx^j \wedge dx^k$?

 

[vanhees71]

The energy-momentum four-vector wrt. to an arbitrary inertial reference frame has the components
$$(p^{\mu})=\begin{pmatrix} E/c,\vec{p} \end{pmatrix},$$
where it is a matter of convention, where to put the conversion factor $c$ (speed of light in vacuum).

There is no such thing as a relativistic mass. This is an old-fashioned idea from a time, when the full mathematical structure of SRT has not yet been fully understood. This was corrected very soon after Einstein's ground-breaking paper of 1905 by Minkowski (1908). The invariant mass of an object is given by
$$m^2 c^2=p \cdot p=\eta_{\mu \nu} p^{\mu} p^{\nu}=\frac{E^2}{c^2}-\vec{p}^2.$$
Right now, I'm in the process of writing an FAQ/Insights article on SRT. So stay tuned ;-).

 

[stedwards]

There is no such thing as a relativistic mass.

OK, no such thing as $E/c^2$.

The reference to relativistic mass is straight out of the Wikipedia Stress-Energy Tensor article. You may feel inspired to change it.

 

[vanhees71]

No $E/c^2$ is just energy, written in units of mass but not mass (at least not according to the modern use in contemporary high-energy-particle and -nuclear physics). It is misleading to mix the concepts up, because invariant mass, which is exclusively used to define the mass of a particle or composite object, is a Lorentz scalar (which is why it's called "invariant"), and energy is the time component of the energy-momentum four-vector wrt. an arbitrarily chosen frame of reference.

What Wikipedia concerns, it's a great place to start reading about a subject, but it's by no means totally trustworthy. You don't even know who's the author(s) of an article!

 

[stedwards]

No $E/c^2$ is just energy, written in units of mass but not mass (at least not according to the modern use in contemporary high-energy-particle and -nuclear physics).

Of course, however if you tried, I think you could find a model where E/c^2 servers as inertial mass. How did I get talked into beating this dead horse?

 

[vanhees71]

Unfortunately the horse is not dead. I never understood why!

 

[Jano L.]

There is no such thing as a relativistic mass. This is an old-fashioned idea from a time, when the full mathematical structure of SRT has not yet been fully understood. This was corrected very soon after Einstein's ground-breaking paper of 1905 by Minkowski (1908).

That sounds like concept police. Of course the relativistic mass is a valid and unambiguous concept. It is given by the formula

$$
m_{rel} = \frac{m}{\sqrt{1-\frac{v^2}{c^2}}}
$$
where $m$ is rest mass and $v$ is speed of the body. It allows momentum to be written as

$$
\mathbf p = m_{rel} \mathbf v
$$

similarly to classical mechanics.

It may be an old-fashioned concept, but it is a concept. Don't impose newspeak on everybody. For people to understand what is written in old books on relativity, they need to understand this concept.

 

[vanhees71]

No, it gives rise to misunderstandings, and it's not used in the community anymore. If you wish to stick to old-fashioned ideas, it's your choice, but don't expect to be understood easily by other people working in relativistic physics.

Again: What you quote is the on-shell energy of a classical particle or asymptotic free quantum (divided by $c^2$) and thus a temporal component of the energy-momentum four-vector (divided by $c$). The usually used quantity when one speaks about particle masses is the invariant mass (or for particles which have a non-zero invariant mass the rest mass). Usually one refers to this invariant mass simply as mass.

 

[Jano L.]

No, it gives rise to misunderstandings, and it's not used in the community anymore. If you wish to stick to old-fashioned ideas, it's your choice, but don't expect to be understood easily by other people working in relativistic physics.

I do not want to stick to any ideas. I refuse to accept a mindset that present-day fashion is the limit of my vocabulary. That's auto-censorship, not science. Understanding does not come easily, especially when talking about non-fashionable things.

 

[vanhees71]

Understanding comes easier with clear concepts rather than delving into problems solved for more than 100 years!