Lorentz invariant mass of electromagnetic field?

da_willem

An photon has mass zero by virtue of its momentum canceling its energy in

[tex] m^2c^4 = E^2-p^2c^2[/tex]

But in electromagnetism a field configution only has momentum when both a magnetic field and an electric field are present, e.g. in an electromagnetic wave. Now when there is only an electric or magnetic field present, doesn't the field have an invariant rest mass E/c^2 with E the total energy stored in the field? Does it make any sense to think of it like that?

(Problem is maybe that for e.g. a point charge this mass is infinite...so it can't be the correct picture gravitationally right?)

*pmb_phy*

Quote by da_willem

An photon has mass zero by virtue of its momentum canceling its energy in

[tex] m^2c^4 = E^2-p^2c^2[/tex]

But in electromagnetism a field configution only has momentum when both a magnetic field and an electric field are present, e.g. in an electromagnetic wave.

I disagree. On what do you base this on? I worked out an example which gives the opposite of your conclusion. See

http://www.geocities.com/physics_wor..._mag_field.htm

Now when there is only an electric or magnetic field present, doesn't the field have an invariant rest mass E/c^2 with E the total energy stored in the field?

No.

Does it make any sense to think of it like that?

No.

(Problem is maybe that for e.g. a point charge this mass is infinite...so it can't be the correct picture gravitationally right?)

The mass of a point charge is finite even in the case of a point charge which has an infinite mass density.

Best wishes

Pete

Xezlec

Quote by pmb_phy

Now when there is only an electric or magnetic field present, doesn't the field have an invariant rest mass E/c^2 with E the total energy stored in the field?

No.

Wait, are you just saying no because he used the term "mass"? I mean, relativistically, the energy stored in an electric or magnetic field certainly behaves like a mass E/c^2. It has inertia and it gravitates. Right?

Xezlec

Oh, wait, you're just saying he can't say for a photon that that energy is a rest mass. Of course if you are at rest relative to the photon it has no mass. That zero mass gets dilated to finite mass when the photons speed becomes c, because at c, the mass is dilated by a factor of infinity. Which really makes no rigorous sense to say at all. But it makes intuitive sense.

I wonder if that was any help?

da_willem

Quote by pmb_phy

I disagree. On what do you base this on? I worked out an example which gives the opposite of your conclusion. See

http://www.geocities.com/physics_wor..._mag_field.htm

No.

No.
The mass of a point charge is finite even in the case of a point charge which has an infinite mass density.

Best wishes

Pete

I will take a look at your website later, but for now I would like to say a few things. Of course I figured a field configuration with only an electric or magnetic field has zero momentum, because the Poynting vector vanishes!

Now with zero momentum and a nonzero field energy density this would seem to imply a mass by the energy momentum relation. I know this would make no sense physically, but the equations do appear to indicate such a (sometimes infinite) mass, what's the deal here?

Xezlec

Quote by da_willem

Now with zero momentum and a nonzero field energy density this would seem to imply a mass by the energy momentum relation. I know this would make no sense physically, but the equations do appear to indicate such a (sometimes infinite) mass, what's the deal here?

Oh, I see what you're asking now: when you integrate the energy density of the E-field of a point charge, you get infinite energy. This makes no sense because it certainly doesn't behave as though it has infinite mass.

Here's what a website I found says:

Unfortunately, if our point charges really are point charges then $a\rightarrow 0$, and the self-energy of each charge becomes infinite. Thus, the potential energies predicted by Eqs. (585) and (594) differ by an infinite amount. What does this all mean? We have to conclude that the idea of locating electrostatic potential energy in the electric field is inconsistent with the existence of point charges. One way out of this difficulty would be to say that all elementary charges, such as electrons, are not points, but instead small distributions of charge. Alternatively, we could say that our classical theory of electromagnetism breaks down on very small length-scales due to quantum effects. Unfortunately, the quantum mechanical version of electromagnetism (quantum electrodynamics, or QED, for short) suffers from the same infinities in the self-energies of particles as the classical version. There is a prescription, called renormalization, for steering round these infinities, and getting finite answers which agree with experiments to extraordinary accuracy. However, nobody really understands why this prescription works. The problem of the infinite self-energies of elementary charged particles is still unresolved.

da_willem

Thanks! But, is it wrong to associate a 'rest mass' to the energy of a field, e.g. in the light of its gravitational influence? If so, what's the reason, as the equations (naively) seem to indicate such a mass?

*pmb_phy*

Quote by da_willem

I will take a look at your website later, but for now I would like to say a few things. Of course I figured a field configuration with only an electric or magnetic field has zero momentum, because the Poynting vector vanishes!

I disagree. As my derivation demonstrates you can have only a magnetic field in a frame S and still have a non-zero Poynting vector in a frame S' which is moving relative to S. The reason being is that in S' there will be a non-vanishing E field which, which crossed with the B field in S' will give a non-vanishing Poynting vector.

Pete

da_willem

Quote by pmb_phy

I disagree. As my derivation demonstrates you can have only a magnetic field in a frame S and still have a non-zero Poynting vector in a frame S' which is moving relative to S. The reason being is that in S' there will be a non-vanishing E field which, which crossed with the B field in S' will give a non-vanishing Poynting vector.

Pete

Right, I know that having an E or a B field is observer dependent (you do have an E field in the S' frame!), which might not cause a problem for the Lorentz invariance of the quantity

[tex] 'm'= \frac{1}{c^2} \sqrt{\frac{1}{2}\epsilon \int E^2 dV+ \frac{1}{2\mu} \int B^2 dV - \frac{c^2}{\mu} \int |\vec{E} \times \vec{B}| dV} [/tex]

So in the S frame the change (increase mainly due to the magnetic field) in field energy is probably cancelled by the arising of field momentum. If I find the time I will try to do the calculation using your example.

Xezlec

Quote by da_willem

Thanks! But, is it wrong to associate a 'rest mass' to the energy of a field, e.g. in the light of its gravitational influence? If so, what's the reason, as the equations (naively) seem to indicate such a mass?

By my reading of that reference, I think if you apply the right normalization, you can find out if a particular field actually has any energy or not, and if so how much. The electric field of an electron itself doesn't have energy, as the article says, no one really has figured out any "nice" explanation why. It's just an exception to the usual rule about energy densities of fields. So when you have true point charges lying around, you have to do that normalization thing to get the energy calculations right.

The electric field due to a continuous distribution of charge does carry energy in the usual way. And in this case, as I understand it, it is correct to say it has a rest mass. Point charges are the only weird thing, where the rules stop applying nicely.

Of course there are no continous charge distributions really, but if you want to approximate... meh.

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da_willem	Thanks! But, is it wrong to associate a 'rest mass' to the energy of a field, e.g. in the light of its gravitational influence? If so, what's the reason, as the equations (naively) seem to indicate such a mass?