- #1
Peeter
- 305
- 3
For electrostatics one can work out the energy of a charge configuration in terms of its associated field:
[tex]
U = \epsilon_0 \vec{E}^2/2
[/tex]
I've seen that the general energy of the field is given by:
[tex]
U = \epsilon_0 (\vec{E}^2 + c^2\vec{B}^2)/2
[/tex]
but am unsure how to show this. I suspect that a relativistic argument would be one way.
Another way, which doesn't work (or I did it wrong), is a calculation of the Hamiltonian for the (non-proper time formulation) of the Lorentz force Lagrangian. I get an energy that only includes the q\phi part as in electrostatics. This actually makes some sense since doing a line integral against a perpendicular field won't contribute.
I'd be interested to know two things:
1) A high level non-math description of where the B^2 term comes from.
2) Some hints for the math side of the question. What is a way (or some ways) that the E^2 + B^2 form of the field energy can be derived?
[tex]
U = \epsilon_0 \vec{E}^2/2
[/tex]
I've seen that the general energy of the field is given by:
[tex]
U = \epsilon_0 (\vec{E}^2 + c^2\vec{B}^2)/2
[/tex]
but am unsure how to show this. I suspect that a relativistic argument would be one way.
Another way, which doesn't work (or I did it wrong), is a calculation of the Hamiltonian for the (non-proper time formulation) of the Lorentz force Lagrangian. I get an energy that only includes the q\phi part as in electrostatics. This actually makes some sense since doing a line integral against a perpendicular field won't contribute.
I'd be interested to know two things:
1) A high level non-math description of where the B^2 term comes from.
2) Some hints for the math side of the question. What is a way (or some ways) that the E^2 + B^2 form of the field energy can be derived?