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I'm trying to derive expressions for the EM-field of a charge moving at relativistic velocity.

Let's make a rest frame for the charge in which it's velocity is equal to zero.

The charge only produces an electric field with components:

[itex]E'(x)=k\frac{q}{R'^3}x'[/itex]

[itex]E'(y)=k\frac{q}{R'^3}y'[/itex]

Where x' and y' are components of R'.

Now to transform to another reference frame we have:

[itex]E(x)=E'(x)[/itex]

[itex]E(y)=\gamma E'(y)[/itex]

where [itex]\gamma[/itex] = gamma factor in relativity

We will also have:

[itex] x'= \gamma R cos(\theta)[/itex]

[itex] y'=y=R sin(\theta)[/itex]

[itex] R'=\gamma R {(1-\frac{v^2}{c^2}sin^2(\theta))^{(1/2)}}[/itex]

Where [itex]\theta[/itex] is the angle between R and x.

Putting this in above equations will give:

[itex]E(x)= \frac{kq(1-\frac{v^2}{c^2})}{R^3{(1-\frac{v^2}{c^2}sin^2(\theta))^{(1,5)}}}R cos(\theta)[/itex]

[itex]E(y)= \frac{kq(1-\frac{v^2}{c^2})}{R^3{(1-\frac{v^2}{c^2}sin^2(\theta))^{(1,5)}}}R sin(\theta)[/itex]

And the resulting electric field:

[itex]E= \frac{kq(1-\frac{v^2}{c^2})}{R^3{(1-\frac{v^2}{c^2}sin^2(\theta))^{(1,5)}}}R[/itex]

And the magnetic field can be obtained by:

[itex] B=v\times\frac{1}{c^2}E[/itex]

I doubt this is right because when I set [itex]\theta[/itex]=0 the electric field in the direction of motion is reduced by [itex]\gamma^2[/itex] even though E'(x) should be equal to E(x)?

I would gladly add a sketch of the situation if it isnt clear enough.

Thank you very much!

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# Elecromagnetic field of relativistic charge

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