# Invariance of E and B

1. Jun 4, 2009

### jsc01

Hello all, this is my first post so I hope the question I have is interesting, at least to a few, and that I can learn something from the discussion.

I had problem for a class not long ago in which I had to prove that:
$$\vec{E} \cdot \vec{B}$$ and $$E^2 - c^2B^2$$ were Lorentz invariant. I was able to show this without too much problem.

I was later told that these are the only bilinear combination of $$\vec{E}$$ and $$\vec{B}$$ that is Lorentz invariant. Is this true? What exactly does this mean?

JSC

2. Jun 4, 2009

### clem

It is true. Any other bilinear combination of E and B will change in a Lorentz transformation.

3. Jun 4, 2009

### jsc01

I believe you but how can this be proved?

JSC

4. Jun 4, 2009

### George Jones

Staff Emeritus
I don't know.

By the definition of "bilinear" with which I'm familiar, $E^2 - c^2 B^2$ is not a bilinear function of $\vec{E}$ and $\vec{B}$.

Hint: for mathematics that stands on it's own line, use the tex and /tex tags; for mathematics in line with prose, use the itex and /itex tags.

5. Jun 4, 2009

### George Jones

Staff Emeritus
A function $f$ is a bilinear function of $\vec{E}$ and $\vec{B}$ if for all real numbers $\alpha$ and all fields $\vec{E}_1$, $\vec{E}_2$, $\vec{B}_1$, $\vec{B}_2$,

$$f \left( \alpha \vec{E}_1 + \vec{E}_2 , \vec{B}_1 \right) = \alpha f \left( \vec{E}_1 , \vec{B}_1 \right) + f \left( \vec{E}_2 , \vec{B}_1 \right)$$

and

$$f \left( \vec{E}_1 , \alpha \vec{B}_1 + \vec{B}_2 \right) = \alpha f \left( \vec{E}_1 , \vec{B}_1 \right) + f \left(\vec{E}_1 , \vec{B}_2 \right).$$

Take

$$f \left( \vec{E} , \vec{B} \right) = E^2 - c^2 B^2,$$

and compare $f \left( \alpha \vec{E} , \vec{B} \right)$ and $\alpha f \left( \vec{E} , \vec{B} \right)$. Are these quantities, in general, equal?

6. Jun 4, 2009

### George Jones

Staff Emeritus
Are you sure this is the statement that you were told?

As I suspected, a true statement is

"Linear combinations of $E^2 - c^2 B^2$ and $\vec{E} \cdot \vec{B}$ are the only (quadratic) invariants of the fields."

These two statements are different.

7. Jun 4, 2009

### jsc01

Hello George,

I was told that $E^2-c^2B^2$ and $\vec{E} \cdot \vec{B}$ are the only Lorentz invariants composed of E and B bilinearly.

Could it be that if we let $\vec{E} \cdot \vec{B} = 0$, which is true for an electromagnetic wave, then the binlinear form is EB which is invariant?

Regards,

John

8. Jun 4, 2009

### clem

I guess GJ's 3 posts means he didn't like your use of bilinear for quadratic.
It might be that he is a mather and you physicer.
I took it to mean what you meant. Be more careful in the future.
The transformation properties E and B are given by their placement in the tensor F.
The only quadratic invariants come from full contraction of FF, which gives E^2-B^2 and the full contraction of F with its dual, giving E.B. There is no other way to get a quadratic invariant.

9. Jun 4, 2009

### jsc01

I am somewhat new to tensor algebra so I am not too familiar with the contraction operation. Here is what I found from Wolfram.

When $T^j^i$ is interpreted as a matrix, the contraction is the same as the trace.

Sometimes, two tensors are contracted using an upper index of one tensor and a lower of the other tensor. In this context, contraction occurs after tensor multiplication. This seems to be the case here.

Using Mathematica to do the matrix algebra I found $Tr(F^\alpha^\beta F_{\alpha\beta}) = 2E^2 - 2B^2$ and that $Tr(F^\alpha^\beta G_{\alpha\beta}) = -4\vec{E} \cdot \vec{B}$ where F is the field strength tensor and G is its dual.

Per Morse & Feshbach, a contracted tensor does not change its value when transformed. It is called a scalar or an invariant.

Is this what you mean? Contracting FF and FG ring out the invariants?

Regards,

John

10. Jun 5, 2009

### Bob_for_short

Exactly that. FF and FFG to be precise. And instead of TR it is better to use sum over all index values.

These invariants were discovered and made public by H. Poincaré.

Bob_for_short.

Last edited: Jun 5, 2009