# Homework Help: Faraday Tensor and Index Notation

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1. Jan 10, 2015

### unscientific

1. The problem statement, all variables and given/known data

(a) Find faraday tensor in terms of $\vec E$ and $\vec B$.
(b) Obtain two of maxwell equations using the field relation. Obtain the other two maxwell equations using 4-potentials.
(c) Find top row of stress-energy tensor. Show how the b=0 component relates to j.

2. Relevant equations

3. The attempt at a solution

Part (a)

The relations between the potentials and fields are:

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

The four-vector potential is given by $A = \left( \frac{\phi}{c}, \vec A \right)$.

From the relation given: $F^{ab} = \partial^{a} A ^b - \partial^b A^a$, it looks something like $\nabla \times A$. How do I show this? I've read the basics of tensor notation and seems alright, but I can't seem to apply the knowledge.

2. Jan 10, 2015

### TSny

Just start writing it out explicitly. For example, what do you get for $F^{01}$?

3. Jan 10, 2015

### unscientific

$$F^{01} = \frac{\partial A_x}{\partial \phi} - \frac{1}{c} \frac{\partial \phi}{\partial A_x}$$

I don't see the point of writing out everything, as it only gives terms like $A_x, A_y, A_z$ and partial derivatives..

4. Jan 10, 2015

### dextercioby

You're not calculating the correct derivatives. What does $\partial^{a}$ stand for?

5. Jan 13, 2015

### unscientific

Am I missing out a factor of $c$? I think $\partial^a = c \frac{\partial}{\partial \phi}$

6. Jan 13, 2015

### TSny

$\partial^a$ is a compact notation for a derivative with respect to a space or time coordinate: $\frac{\partial}{\partial x^a}$

7. Jan 13, 2015

### unscientific

Yes, in this case $a=0$, so we're taking the first coordinate in the 4-vector $(\frac{\phi}{c}, \vec A)$ which is $\frac{\phi}{c}$.

8. Jan 13, 2015

### Matterwave

The notation $\partial^a$ means it's a derivative with respect to space-time coordinates. It has nothing to do with the 4-vector potential. It has to do with the space-time coordinates $(t,\vec{x})$. Specifically $\partial^a=\eta^{ab}\partial_b\equiv\eta^{ab}\frac{\partial}{\partial x^b}=\eta^{at}\frac{\partial}{\partial t}+\eta^{ax}\frac{\partial}{\partial x}+...$

9. Jan 13, 2015

### unscientific

Alright, so

$$F^{01} = \frac{1}{c}\frac{\partial A_x}{\partial t} - c\frac{\partial t}{\partial x}$$

10. Jan 13, 2015

### Matterwave

Why are you taking the derivative of $t$ in the second term? Also, your factors of $c$ look off to me. Maybe write it out in index notation first if you are still getting confused. You're almost there though.

11. Jan 13, 2015

### unscientific

$$F^{01} = \partial^0 A^1 - \partial^1 A^0$$

The space-time four-vector is $(ct, \vec r)$. The four-vector potential is given by $(\frac{\phi}{c}, \vec A)$.

$$F^{01} = \frac{\partial A_x}{\partial ct} - \frac{\partial \frac{\phi}{c}}{\partial x}$$
$$F^{01} = \frac{1}{c}\frac{\partial A_x}{\partial t} - \frac{1}{c}\frac{\partial \phi}{\partial x}$$
$$F^{02} = \frac{1}{c}\frac{\partial A_y}{\partial t} - \frac{1}{c}\frac{\partial \phi}{\partial y}$$
$$F^{03} = \frac{1}{c}\frac{\partial A_z}{\partial t} - \frac{1}{c}\frac{\partial \phi}{\partial z}$$

12. Jan 13, 2015

### Matterwave

Ok. I think all that's left is you are missing a negative sign that should have came in when you raised the indices in $\partial^a=\eta^{ab}\partial_b$. See post #8 above. Depending on what metric signature you are using, one or the other term should have an additional negative in it.

After that, ask yourself "what is $E^x$?"

13. Jan 13, 2015

### unscientific

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

$$E_x = -\frac{\partial \phi}{\partial x} - \frac{\partial A_x}{\partial t}$$
$$F^{01} = \frac{1}{c}\frac{\partial A_x}{\partial t} - \frac{1}{c}\frac{\partial \phi}{\partial x}$$

14. Jan 13, 2015

### TSny

As matterwave has pointed out, I was wrong when I said in post #8 that $\partial^a = \frac{\partial}{\partial x^a}$. I should have said $\partial_a = \frac{\partial}{\partial x^a}$ with the index in the lower position on the left side of the equation. Matterwave has shown how to relate $\partial_a$ and $\partial^a$. Sorry for the confusion.

15. Jan 13, 2015

### unscientific

$$E_x = -\frac{\partial \phi}{\partial x} - \frac{\partial A_x}{\partial t}$$
$$F^{01} = - \frac{1}{c}\frac{\partial \phi}{\partial x} -\frac{1}{c}\frac{\partial A_x}{\partial t}$$

This implies that $F^{01} = \frac{E_x}{c}$.

16. Jan 13, 2015

### Matterwave

Ok, now do this for $F^{02},~F^{03},~F^{12},...$ and you will have your answer. Notice that $F^{ab}$ should have 6 independent components because it is anti-symmetric, so only the top right (or bottom left) triangle is independent. Be careful with the negative signs though.

17. Jan 13, 2015

### unscientific

$$F^{01} = -\frac{1}{c}\frac{\partial A_x}{\partial t} - \frac{1}{c}\frac{\partial \phi}{\partial x}$$

$$F^{02} = -\frac{1}{c}\frac{\partial A_y}{\partial t} - \frac{1}{c}\frac{\partial \phi}{\partial y}$$

$$F^{03} = -\frac{1}{c}\frac{\partial A_z}{\partial t} - \frac{1}{c}\frac{\partial \phi}{\partial z}$$

$$F^{12} = \frac{\partial A_y}{\partial x} - \frac{A_x}{\partial y}$$

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

18. Jan 13, 2015

### Matterwave

Like I said, be careful with the negative signs. Why do you have a negative sign in the first term here?

There's 2 other components of $F^{ab}$ that matters. Now match these expressions to $\vec{E},\vec{B}$.

19. Jan 13, 2015

### unscientific

By inspection, the time-like part is electric in nature, while the space-like part is magnetic in nature?

$$F = ( \frac{1}{c}\vec E, \vec B)$$

20. Jan 13, 2015

### Matterwave

The statement is correct, but this mathematical notation doesn't really mean anything...

21. Jan 13, 2015

### unscientific

How do I express it mathematically? I'm having some trouble translating this concept into some notation.

22. Jan 13, 2015

### Matterwave

You can either simply enumerate $F^{01}=E^x, F^{02}=E^y,...$ or, usually for simplicity, you can write out a 4x4 matrix and then put in the corresponding $\vec{E},\vec{B}$ components.

23. Jan 13, 2015

### unscientific

I think writing out the 4x4 matrix is simpler and better, since this is the presentation adopted in the text as well.

I'll have a go at parts (b), (c) and (d) meanwhile!