Is Poynting vector the electromagnetic density of momentum?

[SergioPL]

I learned that the Poynting vector was the electromagnetic density of momentum but recently, while reading the Electromagnetic_stress–energy_tensor article at Wikipedia, I thought about the implications of the momentum conservation equation and arrived to an inconsistency, this equation is:

∂_tp_EM - ∇·σ + ρE + J × B = 0

Where p_EM is the electromagnetic density of momentum and ∇·σ is the divergence of the Maxwell stress tensor,

This equation is similar to the equation for the flow of EM energy: ∂_tu_EM + ∇·S + J·E = 0. Where u_EM is the EM energy and S is the Poynting vector.

For a static point charge it is easy to see that ∇·σ=0 everywhere (excluding the charge itself), however, let's imagine that we want to measure ∂_tp_EM in a point which a point charge far away at -R x and moving with speed +v x, in this case, ignoring constants and taking c=1, we would have:

∇·σ = ∂_xσ_xx + ∂_xσ_xy + ∂_xσ_xz

With:
∂_xσ_xy = + 1 / r^5
∂_xσ_xz = + 1 / r^5

∂_xσ_xx = (1 + v/(1-v)) (1/2) ∂_r(1/r^4) = -2 (1 + v/(1-v)) * 1 / r^5

Therefore we have:
∂_xp_EM = ∇·σ = - v/(1-v) * 2 / r^5

The term v/(1-v) is caused because a modification in the distance to the source causes the retarded time to change, causing an additional contribution to the variation of position from the charge to the point (that contribution is the difference in the retarded time multiplied by the particle's speed).

Because each change dt in time, the source position has moved dt v/(1-v), the change in the momentum from infinite (p=0) to a position X can be calculated as:

p(X) = ∫ ∂_xp_EM * ∂_r/∂_t dt = (1/2)E^2

Where E^2 is just the squared electric field evaluated at X (let it be X the vector from the charge to that point.

This way we have obtained a momentum that is independent on the velocity, on the other hand, it is easy to check that the magnetic field of this particle would be zero in the x axis and therefore, the Poynting vector will also be 0.

This makes me to conclude that the Poynting vector and the EM momentum are not the same.

Can anybody tell me what may I have done incorrectly?
Sergio

 

[SergioPL]

Sorry, I did a mistake in my calculations, p(X) is really 0, there is nothing weird on it.

 

[Meir Achuz]

The Poynting vector is NOT the electromagnetic density of momentum.
They differ in magnitude and units, and have different physical meanings.

 

[SergioPL]

The Poynting vector is NOT the electromagnetic density of momentum.
They differ in magnitude and units, and have different physical meanings.

The electromagnetic density of momentum and the Poynting vector differ in a 1/c^2 factor but everywhere I look I see p = S/c^2.

The Poynting vector contains the intensity of EM energy flowing through a surface, it makes sense that the magnetic momentum is the intensity of "relativistic electromagnetic mass" flowing through a surface and hence it would be 1/c^2 times the Poynting vector since m = E/c^2.

Is there any other know difference between Poynting vector and EM density of momentum?

My concern arises because when I do a boost to the energy momentum vector [u, S], I get inconsistent results except if the EM field is a radiated wave (i.e. E, B orthogonal and with equal magnitude in CGS units). That makes me to think that the Poynting vector and the EM density of momentum may differ in something else than a constant.

 

[Meir Achuz]

The electromagnetic density of momentum and the Poynting vector differ in a 1/c^2 factor but everywhere I look I see p = S/c^2.

If they "differ", they are different.
The only possible bilinear combination of E and B that is a vector is EXB. This means that all possible combinations different only by a constant factor. Are they all the same?