Faraday Tensor Matrix Element

Tags:
1. Dec 3, 2014

unscientific

The Faraday Tensor is given by:

Consider the following outer product with the 4-potential:

The Faraday Tensor is related to the 4-potential:

$$F^{mn} = \Box^{m} A^n - \Box^n A^m$$

For example, $F^{01} = -\frac{1}{c} \frac{\partial A^x}{\partial t} - \frac{1}{c}\frac{\partial \phi}{\partial x}$

How do I show that $\frac{E_x}{c} = -\frac{1}{c} \frac{\partial A^x}{\partial t} - \frac{1}{c}\frac{\partial \phi}{\partial x}$?

2. Dec 3, 2014

Staff: Mentor

How are the scalar and vector potentials defined?

3. Dec 3, 2014

unscientific

$$E = -\nabla \phi - \frac{\partial A}{\partial t}$$

How do I prove this?

4. Dec 3, 2014

Staff: Mentor

You don't prove it. All definitions are true by definition.

5. Dec 3, 2014

unscientific

Also, in einstein summation, why does the $\mu$ go to the bottom on the first term:

$$X \dot X = X_{\mu} X^{\mu}$$

6. Dec 3, 2014

unscientific

I thought usually electric field is simply the grad of potential? $E = -\nabla \phi$

7. Dec 3, 2014

Matterwave

This is true in electro-statics, but not true in electro-dynamics. In electro-dynamics Faraday's law tells you that a changing magnetic field will produce an electric field. That is where the $\frac{\partial \vec{A}}{\partial t}$ term comes from.

8. Dec 3, 2014

unscientific

Is there a derivation for this?

9. Dec 3, 2014

Matterwave

It is basically a definition, not so much a derivation. By Gauss's law for magnetism we know $\nabla\cdot\vec{B}=0$ so that we can define $\vec{B}\equiv\nabla\times\vec{A}$ for some vector potential $\vec{A}$, Faraday's law tells us $\nabla\times\vec{E}=-\frac{1}{c}\frac{\partial \vec{B}}{\partial t}=-\frac{1}{c}\nabla\times\frac{\partial \vec{A}}{\partial t}$ so we know that if we define $\vec{E}\equiv -\nabla\phi-\frac{1}{c}\frac{\partial \vec{A}}{\partial t}$ everything will work out (the first term will go away when you take the curl of it). Notice that because of Faraday's law, we can not simply define $\vec{E}\equiv\nabla\phi$ since this will imply $\nabla\times\vec{E}=0$ which is not true.

Last edited: Dec 3, 2014