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Invariant under Lorentz transformation

  1. Apr 28, 2004 #1
    HI guys first post :cool:

    I need to show that

    [tex]B^2-E^2/C^2[/tex] is invariant under Lorentz transformation (E and B are electromagnetic fields)

    now:

    [tex]B^2-E^2/C^2=B^2_x+B^2_y+B^2_z-E^2_x/C^2-E^2_y/C^2-E^2_z/C^2)[/tex]

    and

    [tex]E'_x=E_x[/tex]
    [tex]E'_y=\gamma(E_y-\frac{v}{c}B_z)[/tex]
    [tex]E'_z=\gamma(E_z-\frac{v}{c}B_y)[/tex]
    [tex]B'_x=B_x[/tex]
    [tex]B'_y=\gamma(B_y+\frac{v}{c}E_z)[/tex]
    [tex]B'_z=\gamma(B_z+\frac{v}{c}E_y)[/tex]

    but i cant manupilate it to give me the correct answer ie

    [tex]B'^2-E'^2/C^2=B^2_x+B^2_y+B^2_z-E^2_x/C^2-E^2_y/C^2-E^2_z/C^2[/tex]

    Can anyone help me out? Basically because of the [tex]\gamma^2[/tex] term im tring to factorise out a [tex]1-\frac{v^2}{C^2}[/tex] ie [tex](1/\gamma^2)[/tex] but im having no joy.
     
    Last edited: Apr 28, 2004
  2. jcsd
  3. Apr 28, 2004 #2

    turin

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    First of all, you can make things less complicated by rotating the coordinate system so that E = exEx (or, just as well, eyEy or ezEz). (EDIT: I TAKE IT BACK; NO YOU SHOULDN'T DO THIS.) I'll give it a more thorough look to see if I can find your problem.

    The first suspicion I have is that the problem is stated in a different unit system (SI) than you are using in your solution (Gaussian?). In the SI units, E has the same units as cB (or E/c as B), but your transformed components add E to (v/c)B, so the units don't seem like they agree.

    I get a (+) sign in front of the B term in my transformed Ez'. I think this makes sense, because the orientation of z to y is the opposite of the orientation of y to z (wrt rotation about the x axis).
     
    Last edited: Apr 28, 2004
  4. Apr 28, 2004 #3
    Dont make it anymore complicated than it needs to be :p. Its a direct substitution problem ie change E to E' and B to B' and using the transformed values show that [tex]B^2-E^2/C^2=B'^2-E'^2/C^2[/tex]
     
  5. Apr 29, 2004 #4

    robphy

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    Gold Member

    In my experience, identities involving [tex]\gamma[/tex] and [tex]v[/tex] are more recognizable if you use the hyperbolic functions.

    Write [tex]v = \tanh\theta[/tex] and [tex]\gamma =\cosh\theta[/tex], where [tex]\theta[/tex] is the rapidity. Of course, [tex]\gamma v = \sinh\theta[/tex].

    [tex]\exp\theta[/tex] has physical significance, as well.
     
  6. Apr 29, 2004 #5

    DW

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    A couple of your equations have wrong signs(and wrong factors of c depending on your choice of units). They should be

    [tex]E'_x=E_x[/tex]
    [tex]E'_y=\gamma(E_{y}-vB_{z})[/tex]
    [tex]E'_z=\gamma(E_{z}+vB_{y})[/tex]
    [tex]B'_x=B_x[/tex]
    [tex]B'_y=\gamma(B_y+\frac{v}{c^2}E_z)[/tex]
    [tex]B'_z=\gamma(B_z-\frac{v}{c^2}E_y)[/tex]

    Adjust the signs and try again.
     
  7. Apr 30, 2004 #6
    Yes thanks I forgot to report back that I had done it successfully. The annoying thing was I actually looked up the equations on a website so assumed they were correct.
     
  8. Apr 30, 2004 #7

    DW

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    LOL. Out of curiosity what was the website? About assuming website physics as accurate, check out the link
    http://www.crank.net/physics.html
     
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