# Invariant under Lorentz transformation

1. Apr 28, 2004

### Spook

HI guys first post

I need to show that

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

now:

$$B^2-E^2/C^2=B^2_x+B^2_y+B^2_z-E^2_x/C^2-E^2_y/C^2-E^2_z/C^2)$$

and

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

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

$$B'^2-E'^2/C^2=B^2_x+B^2_y+B^2_z-E^2_x/C^2-E^2_y/C^2-E^2_z/C^2$$

Can anyone help me out? Basically because of the $$\gamma^2$$ term im tring to factorise out a $$1-\frac{v^2}{C^2}$$ ie $$(1/\gamma^2)$$ but im having no joy.

Last edited: Apr 28, 2004
2. Apr 28, 2004

### turin

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
3. Apr 28, 2004

### Spook

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 $$B^2-E^2/C^2=B'^2-E'^2/C^2$$

4. Apr 29, 2004

### robphy

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

Write $$v = \tanh\theta$$ and $$\gamma =\cosh\theta$$, where $$\theta$$ is the rapidity. Of course, $$\gamma v = \sinh\theta$$.

$$\exp\theta$$ has physical significance, as well.

5. Apr 29, 2004

### DW

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

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

Adjust the signs and try again.

6. Apr 30, 2004

### Spook

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.

7. Apr 30, 2004

### DW

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