- #1
SergioPL
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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:
∂tpEM - ∇·σ + ρE + J × B = 0
Where pEM 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: ∂tuEM + ∇·S + J·E = 0. Where uEM 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 ∂tpEM 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 = ∂xσxx + ∂xσxy + ∂xσxz
With:
∂xσxy = + 1 / r^5
∂xσxz = + 1 / r^5
∂xσxx = (1 + v/(1-v)) (1/2) ∂r(1/r^4) = -2 (1 + v/(1-v)) * 1 / r^5
Therefore we have:
∂xpEM = ∇·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) = ∫ ∂xpEM * ∂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
∂tpEM - ∇·σ + ρE + J × B = 0
Where pEM 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: ∂tuEM + ∇·S + J·E = 0. Where uEM 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 ∂tpEM 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 = ∂xσxx + ∂xσxy + ∂xσxz
With:
∂xσxy = + 1 / r^5
∂xσxz = + 1 / r^5
∂xσxx = (1 + v/(1-v)) (1/2) ∂r(1/r^4) = -2 (1 + v/(1-v)) * 1 / r^5
Therefore we have:
∂xpEM = ∇·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) = ∫ ∂xpEM * ∂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