Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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)





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


    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


    User Avatar
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper
    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


    User Avatar

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


    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


    User Avatar

    LOL. Out of curiosity what was the website? About assuming website physics as accurate, check out the link
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook