# Question Regarding Electromagnetic Fields in Special Relativity

1. May 5, 2009

### Savant13

According to the http://en.wikipedia.org/wiki/Biot-Savart_Law" [Broken], the equation for the magnetic field around a charged particle moving with constant velocity is

$$\mathbf{B} = \frac{1}{c^2} \mathbf{v} \times \mathbf{E}$$

But then, http://en.wikipedia.org/wiki/Mathematical_descriptions_of_the_electromagnetic_field" [Broken], the relativistic description for the magnetic field, where B and E are the nonrelativistic magnetic and electric fields is

$$\mathbf{B} ' = \gamma ( \mathbf{B} - \frac{1}{c^2} \mathbf{v} \times \mathbf{E})- \frac{\gamma - 1}{v^2} ( \mathbf{B} \cdot \mathbf{v} ) \mathbf{v}$$

But this would mean that B' is always 0. Am I misunderstanding something? Is one of these equations the wrong one to use?

Would the B in the relativistic equation be zero to begin with?

Last edited by a moderator: May 4, 2017
2. May 5, 2009

### quZz

In the first equation v is velocity of the particle, while in the second one the same letter is used to mean relative velocity of reference frames. If these two velocities are equal then B' = 0 because the particle is at rest in the '-reference frame.

3. May 5, 2009

### clem

Read carefully, the Wiki article on BS says
"Note that the law is only approximate," for the equation B= qvXr/r^3.
To put it more clearly, that equation is wrong.
The correct equations for E and B are the last two equations in that section.
They do include B=vXE, but E is the complicated field, not the NR Coulomb field.
The second equation you give is for a Lorentz transformation of the fields, and refers to different things than your first equation.