Help with the basics of Maxwell and Lorentz equations


by abotiz
Tags: basics, equations, lorentz, maxwell
abotiz
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#1
Feb23-10, 07:46 PM
P: 58
Hi there.

I have just started to read about the Relativity part of Maxwell's equations and would like to get some clarification from any one here were I can discuss and/or get answers in a more " every day life " way of writing.

1) What is the most simple way of transforming Maxwells equations into a Covariant form? I know that it is done by using four-vectors/tensors but how is it actually done?

2) I know that from Lorents Transformation, that magnetic and electric fields are simply different aspects of the same force. Also that when looking from a frame either in rest or moving, in reference to charged particles we get 2 different results, an electric field and an magnetic field. What I would like to know is if there is a quick way of showing that this is true ? Like a set of equations so that when put in a moving frame you get the Dimensions of a Magnetic field.

3) How do I prove in the most simple way how the Lorentz invariance ( invariant ) is true for Maxwell's Equation.
What I know is this:

Quantities which remain the same under Lorentz transformations are said to be Lorentz invariant.
According to the representation theory of the Lorentz group, Lorentz covariant quantities are built out of scalars, four-vectors, four-tensors, and spinors.

How do I use this knowledge ( if I even should ) to confirm that Maxwell's equations is L. invariance

Thanks!
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jtbell
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#2
Feb23-10, 09:04 PM
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See the following page and related pages on the same site:

http://farside.ph.utexas.edu/teachin...s/node121.html
bcrowell
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#3
Feb24-10, 01:07 AM
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This may be helpful: http://www.lightandmatter.com/html_b...ch11/ch11.html

JustinLevy
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#4
Feb24-10, 03:05 AM
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Help with the basics of Maxwell and Lorentz equations


Quote Quote by abotiz View Post
I know that from Lorents Transformation, that magnetic and electric fields are simply different aspects of the same force. Also that when looking from a frame either in rest or moving, in reference to charged particles we get 2 different results, an electric field and an magnetic field.
...
What I would like to know is if there is a quick way of showing that this is true ?
Okay, this may be oversimplifying what you are asking for, but for a really quick way to show that electric and magnetic fields can transform into each other when changing frames:

Consider a moving charge in a frame where there is only a magnetic field. The charge feels a force.
Now imagine the same situation from the charge's (instantaneous) rest frame. The only way that it could feel a force (since it is at rest) is if there was an electric field in this frame.

So even without getting into the nitty gritty details, it is clear that relativity demands electric and magnetic fields must be able to transform into each other when changing frames.

Does that help? If you want all the details, the links already provided look pretty good.
Mentz114
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#5
Feb24-10, 03:10 AM
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Quick answer -

The EM field has a relativistically invariant scalar [itex]F^{ab}F_{ab}=-2(E^2-B^2)[/itex].
The field tensor is [itex]F^{ab}=\partial^aA^b-\partial^bA^a[/itex].

The invariance is manifest, because if we boost the field-tensor the scalar becomes,

[tex]\lambda_p^a\lambda_q^bF_{ab}\lambda_a^p\lambda_b^qF^{ab}=(\lambda_p^a\l ambda_a^p)(\lambda_q^b\lambda_b^q)F^{ab}F_{ab}=F^{ab}F_{ab}[/tex]

The [itex]\lambda[/itex] are not tensors but Lorentz transformation matrices, and the pairs in parentheses are inverses.
Mentz114
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#6
Feb24-10, 03:50 AM
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How to make a magnetic field appear -

If we start with an electric field in the y direction,

[tex]
F=\left[ \begin{array}{cccc}
0 & 0 & -E_y & 0 \\\
0 & 0 & 0 & 0 \\\
E_y & 0 & 0 & 0 \\\
0 & 0 & 0 & 0 \end{array} \right]
[/tex]
and boost in the x direction using this matrix ( twice )
[tex]
\left[ \begin{array}{cccc}
\gamma & \beta\gamma & 0 & 0 \\\
\beta\gamma & \gamma & 0 & 0 \\\
0 & 0 & 1 & 0 \\\
0 & 0 & 0 & 1 \end{array} \right]
[/tex]
the field tensor becomes
[tex]
F'=\left[ \begin{array}{cccc}
0 & 0 & -\gamma E_y & 0 \\\
0 & 0 & -\beta\gamma E_y & 0 \\\
\gamma E_y & \beta\gamma E_y & 0 & 0 \\\
0 & 0 & 0 & 0 \end{array} \right]
[/tex]
which has a magnetic field in the z direction. Note that [itex]-E^2+B^2=-E_y^{2}\gamma^{2}+\beta^{2}E_y^{2}\gamma^{2}=-E_y^2[/itex].

I transcribed my last two posts from some Latex notes I already had. I hope it's correct in every particular but not guaranteed. Right in principle, though.
Phrak
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#7
Feb24-10, 05:41 AM
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Quote Quote by abotiz View Post
1) What is the most simple way of transforming Maxwells equations into a Covariant form? I know that it is done by using four-vectors/tensors but how is it actually done?
The field tensor is key, of course. But you want things in terms of the covariant electromagnetic field tensor Fmu nu = F[mu nu] rather than Fmu nu. I didn't see that in any of the above help. I don't mean to slight Mentz114. It's a good thing to know that the electromagnetic field tensor is invariant under Lorentz transforms: if the field tensor is invariant so is its covariant form.

You might begin by reading a short diversion into covariant fields within Sean Carroll's Text, Notes on General Relativity chapter 2. He derives one covariant form and but not the hardest. Each of two pairs of Maxwell's equations combine to one covariant form. It can take a few pages to get from Maxwell's three dimensional equations to the fully covariant 4 dimentional form for the two charge and current equations.

The first step is to recognize that the elements Bi are equal to Bi and Ei are equal to Ei in orthonormal coordinates. The second point to understand is that [rho, J] is a 4-vector in Minkowski space, and that [-rho, J] is it's covariant form. The entire derivation is not too obtuse, just tedious.

(The really cool thing that drives all this, is that what is true in one coordinate system for tensors, is true in any other well behave coordinate system--without singularities 'n stuff. We can derive things in orthonormal coordinates and they are equally true in other coordinate systems. This means that the covariant forms of Maxwell's equations are just as good in most of general relativity as they are in nice flat Minkowski space. This is an awfully nice thing to know. It means that the simple impression of a vector bundle upon the spacetime manifold has simple global rules. And this simple impression can be coerced into a metric (but I degress). Even better for mathematical simplicity, the connection coefficients of Riemannian geometry vanish for p-forms (antisymmetric covariant forms, such as the electromagnetic field tensor).)

Ultimately, you obtain two compact equations, dF=0 and d*F=J, or just one physical equation d*F=J, where dF=0 is a mathematical identity that rules out magnetic charge (All exact forms are closed on well behaved manifolds. This is applicable when F=dA, and A is the so-called vector potential. For the dizzying extremes where exact forms are not closed, see de Rham cohomologies).

I find this all very exciting, or I wouldn't be blathering like this. Anyway.

I think the undoctored derivation is in my notes somewhere that should be alot less obtusive than the above. But you have to be willing to take the risky step of asking for an MS word file that I would send to you.
clem
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#8
Feb24-10, 06:46 AM
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abotiz:
I think you would do better looking at a textbook than trying to wade through lots of posts.
abotiz
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#9
Feb24-10, 12:20 PM
P: 58
Hi and thank you for your replies

Quote Quote by JustinLevy View Post
Consider a moving charge...
This seems to be a very short and easy to understand formulation. That is exactly what Iam looking for. The example in it self is something I have come acrossed many times, but the way you wrote the last part is what I was looking for.
Quote Quote by JustinLevy View Post
it is clear that relativity demands electric and magnetic fields must be able to transform into each other when changing frames.
I have not mentioned it but Iam a student in a Swedish university and although i know some english, its very hard for me to understand something in words that are not used in " every day life "

This is my answer to the other replies wich I did not understand.

Quote Quote by clem View Post
abotiz:
I think you would do better looking at a textbook than trying to wade through lots of posts.
This is exactly what I have done, thats why I wanted to ask the things I did not udnerstand in this forum, were there are people who perhaps can explain things in simple English, not in English wich is written by Professors in Educational books.
Of course, these questions Iam asking cant always be explained in simple English, you have to sometime use words that are not common, Iam well aware of that. But surely there has to be a way.

Regarding 1), From wikipedia
" The covariant formulation of classical electromagnetism refers to ways of writing the laws of classical electromagnetism in particular, Maxwell's equations in a form which is "manifestly covariant" (i.e. in terms of covariant four-vectors and tensors) "

Iam looking for a very simple explanation, something like this:
By using the four-vectors wich has a component of "ct" and can be transformed to other inertial system without changing the values ( covariant ) and by writing the magnetic/electric contribution of a system into a Tensor, you can write Maxwell's equation in this form : Click image for larger version

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Regarding 3) Iam looking for an explanation in very simple equation form, this is generally done by using words together with equation, so that the math does not get overwhelming. For example sonething like this :
by using this equation and knowing this, you can see the values does not change wether the reference frame is moving or in rest, cause when you put in the maxwell's equation you see that the value is the same.

Thank you Very Much
abotiz
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#10
Feb26-10, 04:34 AM
P: 58
Anyone?
Fredrik
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#11
Feb26-10, 08:15 PM
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It sounds like you want an explanation both without math and without words that you may have to look up at Wikipedia or dictionary.com. I think that's expecting too much. You should probably try to follow a derivation from a book or a web site, and ask about the specific details that you find difficult.

I was going to copy and paste some stuff from my personal notes, but it looks like I messed up the signs somewhere and I'm not sure where yet. Can someone tell me if there's a standard convention for how to use the Levi-Civita symbol with a +--- metric? Is [itex]\varepsilon^{123}=1[/itex] or [itex]\varepsilon_{123}=1[/itex]?
Phrak
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#12
Feb27-10, 03:10 AM
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Quote Quote by Fredrik View Post
Can someone tell me if there's a standard convention for how to use the Levi-Civita symbol with a +--- metric? Is [itex]\varepsilon^{123}=1[/itex] or [itex]\varepsilon_{123}=1[/itex]?
Interesting. I been working in the -+++ metric where this doesn't come up. It keeps conversions to, from and within the 3Dspace subspace easier. You can simply low the index on Bi for instance and the equation is still true.

I can't tell you if there's a dominant standard in +---, but I can make a guess. The Levi-Civita tensor is usually introduced as a tensor with all lower indices, where the elements of even permutations are equal to one.


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