# Maxwell equations vs Lorentz transformation of fields

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I don't know how to best ask this boring question...

A moving observer observes a charged particle. We could use field-transformation equations to compute the field that the moving observer will see [various sources say that the electric field will look somewhat oblate and there will be a magnetic field around the particle].

But what if the moving observer simply uses Maxwell equations to compute the field of the particle (the particle is moving in his frame of reference) - will he again obtain equally oblate electric field and the magnetic field?

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2 hours ago, Danijel Gorupec said:

A moving observer observes a charged particle. We could use field-transformation equations to compute the field that the moving observer will see [various sources say that the electric field will look somewhat oblate and there will be a magnetic field around the particle].

But what if the moving observer simply uses Maxwell equations to compute the field of the particle (the particle is moving in his frame of reference) - will he again obtain equally oblate electric field and the magnetic field?

What do you mean by "the particle is moving in his frame of reference" ?

Can you rephrase this properly identifying the frames you wish to employ?

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By this remark I only wanted to point out that when computing fields using Maxwell equations, the observer will be dealing with the moving particle (as he sees it). I suppose that Maxwell equations will provide different results if you compute fields of a moving particle than if you compute field of a stationary particle.

In the original post, this approach (computing field of a moving particle directly by Maxwell equations) is contrasted to the approach quoted in the first paragraph (taking a field of a stationary particle and then using transformation equations to obtain the field as seen by a moving observer).

I am not sure if I made it clear enough. [ In my question there is only one particle and one observer that are in relative motion. ]

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13 hours ago, Danijel Gorupec said:

I don't know how to best ask this boring question...

A moving observer observes a charged particle. We could use field-transformation equations to compute the field that the moving observer will see [various sources say that the electric field will look somewhat oblate and there will be a magnetic field around the particle].

But what if the moving observer simply uses Maxwell equations to compute the field of the particle (the particle is moving in his frame of reference) - will he again obtain equally oblate electric field and the magnetic field?

Yes, they should. They have to come up with the same answer.

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3 hours ago, Danijel Gorupec said:

In my question there is only one particle and one observer that are in relative motion.

3 hours ago, Danijel Gorupec said:

I am not sure if I made it clear enough.

Yes I understood that, which is why I asked for more detail.

Whenever you have relative motion you must, of necessity, have (at least) two frames.
One for each of the objects in relative motion.
The rest frame of one of these is chosen as the 'rest frame' and the rest frame of the other, the moving frame.

Furthermore the observer's frame also needs some charge to be able to interact with (and thus observe) any fields generated by the moving charge.

You have introduced 'relativity'.

Have you considered whether the observers in the observer frame will measure the same charge as observers in the frame of the moving charge?
This is indeed the case - charge is invariant under the Lorenz transformation so both will agree the value of the charge.

However it is variation of charge density that produces the fields.
Charge density is not Lorenz invariant so each will measure a different charge density.

This is often handled by using Maxwell's equations to introduce what are known as vector potentials.
Now Maxwell's equations are (simultaneous) partial differential equations and when we solve them for the fields or potentials we do not obtain unique solutions.
Solutions differ by arbitrary (vector) functions.

The required solutions are picked out by applying further conditions (boundary conditions) and inparticular what is known as

The Lorenz Condition.

Note this is not the Lorenz transformation.

Classically this leads to what are known as 'retarded potentials'.

When we introduce relativity, it is often easier to use four-vectors than work directly.

Here is a simple non four-vector explanation due to Grant and Phillips that may help.

Notice their footnote as to how positive and negative charge do not necessarily transform the same under Lorenz.

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Thanks - I think these answers pointed me in the right direction. I think I understand now where I was wrong... Following equations bothered me (field transformation from Wikipedia):

My error was because I failed to understand that E' and B' refer to the field at coordinates that are also 'contracted'. That is, if the E and B are, say, one meter in front of particle, then E' and B' are not one meter in front of particle but 1/gamma meters in front of particle.

This also resolves my previous question (unanswered) about possible mismatch regarding Lorentz transformation of fields.

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59 minutes ago, Danijel Gorupec said:

Thanks - I think t﻿hese answers pointed me in the right dir﻿ectio﻿n﻿﻿﻿.

Yes I think it is oft forgotten that relativity acts (directly) only on space and time so affects things only insofar as they link to have a connection to these variables.

So for electrodynamics the charges remain the same but the space and tinme interval between them alters, thus altering density (scalar) and rate of change properties (vector).

I can post another short extract (from Grant and Phillips) detailing the oblateness you referred to if you like.

Edited by studiot

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