Using the transform of the electromagnetic tensor F between frames, verify that

1. Dec 24, 2011

blueyellow

1. The problem statement, all variables and given/known data

Using the transform of the electromagnetic tensor F between frames,

F'=RFR$^{T}$

verify that:

i) the perpendicular component of the magnetic field in the frame, S', moving with velocity v with respect to the frame S, can be found from the transform of B$_{\bot}$ in S by

B'$_{\bot}$=$\gamma B_{\bot}$ - $\frac{1}{c^2}$$\gamma$v $\times$E

and that
ii) the parallel component of the E-field, E(parallel), remains unchanged.

Remember that the parallel and perpendicular components are with respect to the direction of the velocity v of the S' frame in S.

The attempt at the solution:

$\begin{equation*} \mbox{R}=\left[ \begin{array}{rrrr} \gamma&0&0&i\beta\gamma\\ 0&1&0&0\\ 0&0&1&0\\ -i\beta\gamma&0&0&\gamma\\ \end{array} \right] \end{equation*}$

$\begin{equation*} \mbox{F}=\left[ \begin{array}{rrrr} 0&B_{3}&-B_{2}&\frac{-i}{c}E_{1}\\ -B_{3}&0&B_{1}&\frac{-i}{c}E_{2}\\ -B_{2}&-B_{1}&0&\frac{-i}{c}E_{3}\\ \frac{i}{c}E_{1}&\frac{i}{c}E_{2}&\frac{i}{c}E_{3}&0\\ \end{array} \right] \end{equation*}$

F'=
$\begin{equation*} \left[ \begin{array}{rrrr} \gamma&0&0&i\beta\gamma\\ 0&1&0&0\\ 0&0&1&0\\ -i\beta\gamma&0&0&\gamma\\ \end{array} \right] \end{equation*}$
$\begin{equation*} \left[ \begin{array}{rrrr} 0&B_{3}&-B_{2}&\frac{-i}{c}E_{1}\\ -B_{3}&0&B_{1}&\frac{-i}{c}E_{2}\\ -B_{2}&-B_{1}&0&\frac{-i}{c}E_{3}\\ \frac{i}{c}E_{1}&\frac{i}{c}E_{2}&\frac{i}{c}E_{3}&0\\ \end{array} \right] \end{equation*}$
$\begin{equation*} \left[ \begin{array}{rrrr} \gamma&0&0&-i\beta\gamma\\ 0&1&0&0\\ 0&0&1&0\\ i\beta\gamma&0&0&\gamma\\ \end{array} \right] \end{equation*}$

=

$\begin{equation*} \left[ \begin{array}{rrrr} 0&\beta_{3}\gamma - \frac{\beta\gamma E_{2}}{c}&-i\beta_{2}\gamma - \frac{\beta\gamma E_{3}}{c}&\frac{\beta^{2}\gamma^{2}E_{1}}{c} - \frac{\beta\gamma^{2}E_{a}}{c}\\ -\beta_{3}\gamma + \frac{\beta\gamma E_{2}}{c}&0&\beta_{1}&-i\beta\gamma\beta_{3} - \frac{i\gamma E_{2}}{c}\\ \beta_{2}\gamma + \frac{\beta\gamma E_{3}}{c}&\beta_{1}&0&-i\beta\gamma\beta_{2} - \frac{i\gamma E_{3}}{c}&\\ \frac{i E_{1}\gamma^{2}}{c} - \frac{i\beta^{2}\gamma E_{1}}{c}&-i\beta\beta_{3}\gamma + \frac{i\gamma E_{2}}{c}&-\beta\beta_{2}\gamma - \frac{i\gamma E_{3}}{c}&0\\ \end{array} \right] \end{equation*}$

But I do not see any perpendicular signs in the matrix above. And the minus and cross signs are not there either. So how is one supposed to go from that matrix to the equation:

B'$_{\bot}$=$\gamma B_{\bot}$ - $\frac{1}{c^2}$$\gamma$v $\times$E

I have looked through textbooks and the internet, but nowhere, it seems, will tell me how to do this question or a similar question.

Last edited: Dec 24, 2011
2. Dec 26, 2011

Thaakisfox

Re: Using the transform of the electromagnetic tensor F between frames, verify that..

Write out the components of the cross product ;)

3. Dec 26, 2011

blueyellow

Re: Using the transform of the electromagnetic tensor F between frames, verify that..

Sorry for sounding stupid, but so what if v and E have perpendicular and parallel components? You still can't see 'parallel' and 'perpendicular' anywhere in the matrix. Maybe I would see it if I actually wrote out the components of the cross product, but I have to admit that I don't know how to do this. Cartesian coordinates are being used?

x y z

v$_{x}$ v$_{y}$ v$_{z}$

E$_{x}$ E$_{y}$ E$_{z}$

=(v$_{y}$*E$_{z}$-v$_{z}$*$E_{y}$)+...

Please can you or someone else explaining how this is supposed to help. I can't even see those components anywhere in the matrix. So still, back to square one: how are those matrixes used to derive that equation?

4. Dec 26, 2011

blueyellow

Re: Using the transform of the electromagnetic tensor F between frames, verify that..

The v's are not anywhere in the matrices anyway. Because stuff cancel out when you do the corss product? If so, how? But if things cancel out and v cross E is not there in the matrix anyway, then how can you justify putting it into the equation?

Are there any textbooks or websites that tell you how to do that sort of question? Or is this one of those things that one has to work out oneself?

5. Dec 26, 2011

blueyellow

Re: Using the transform of the electromagnetic tensor F between frames, verify that..

Do I have to use the matrices S and S', whatever they are? Then why are F and F' mentioned in the question? How are S and S' related to F and F'?

6. Dec 27, 2011

blueyellow

Re: Using the transform of the electromagnetic tensor F between frames, verify that..

So I found out somewhere that:

'The inverse transformation is obtained by interchanging the barred and the unbarred quantities and replacing β by -β. These are found to be:

E$_{x}$=E$_{x}$-bar
E$_{y}$=$\gamma$(E$_{y}$-bar + $\beta$B$_{z}$)
E$_{z}$=$\gamma$(E$_{z}$-bar - $\beta$B$_{y}$)

Note the components of E parallel to the relative velocity remain unchanged.'

How do the above equations show that the parallel components remain unchanged?
What are barred and unbarred quantities? Complex conjugates? Really? - why would they be used? Why is $\beta$ being replaced by -$\beta$?

7. Dec 27, 2011

blueyellow

Re: Using the transform of the electromagnetic tensor F between frames, verify that..

I made some mistakes previously while multiplying matrices. i got B's and betas mixed up, and some other mistake. I corrected them and I equated F and F' and got:

B1=B1

B2=B2γ+E3βγ/c

B3=B3γ-βγE2/c

B1=B1

B2=B2γ+E3γ/(c^2)

B3=B3γ-γE2/(c^2)

I noticed that this looks something like the equation that one is supposed to verify in i), except that I don't know how to take into account how it is perpendicular, and I don't understand by in B(subscript 2) there is a plus sign between the terms and in B(subscript 3) there is a minus sign between the terms. I also don't understand how the cross product sign came about or where the 'v' in the equation comes from when there are no 'v' terms in the matrices. Please help.

8. Dec 27, 2011

Thaakisfox

Re: Using the transform of the electromagnetic tensor F between frames, verify that..

the v is embedded in beta and gamma. Do u know wt beta and gamma determine?
The 3 part of magnetic field can be separated into perpendicular and parallel components with respect to the velocity.
Suppose the direction of the the velocity is along the z-axis. This means that the parallel components are the z components. And the perpendicular ones are the x and y components. These can be represented with numbers.
For example B3 means the z component of vector B and B2 means the y component, B1 means the x component.
Hence B perpendicular is the combination of B1 and B2.