# Faraday Tensor and Index Notation

## Homework Statement

(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. ## The Attempt at a Solution

Part (a)
[/B]
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.

## Answers and Replies

TSny
Homework Helper
Gold Member
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.

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

Just start writing it out explicitly. For example, what do you get for ## F^{01}##?
$$F^{01} = \frac{\partial A_x}{\partial \phi} - \frac{1}{c} \frac{\partial \phi}{\partial A_x}$$

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..

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

You're not calculating the correct derivatives. What does $\partial^{a}$ stand for?
Am I missing out a factor of ##c##? I think ##\partial^a = c \frac{\partial}{\partial \phi}##

TSny
Homework Helper
Gold Member
Am I missing out a factor of ##c##? I think ##\partial^a = c \frac{\partial}{\partial \phi}##

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

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

Matterwave
Science Advisor
Gold Member
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}##.

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}+...##

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}+...##

Alright, so

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

Matterwave
Science Advisor
Gold Member
Alright, so

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

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.

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

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

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

Matterwave
Science Advisor
Gold Member
$$F^{01} = \partial^0 A^1 - \partial^1 A^0$$

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

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

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##?"

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

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

TSny
Homework Helper
Gold Member
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.

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

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

Matterwave
Science Advisor
Gold Member
$$E_x = -\frac{\partial \phi}{\partial x} - \frac{\partial A_x}{\partial t}$$
$$F^{01} = - \frac{1}{c}\frac{\partial \phi}{\partial x} -\frac{1}{c}\frac{\partial A_x}{\partial t}$$

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

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.

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.

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

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

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

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

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

Matterwave
Science Advisor
Gold Member
$$F^{12} = -\frac{\partial A_y}{\partial x} - \frac{A_x}{\partial y}$$

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}##.

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

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

Matterwave
Science Advisor
Gold Member
By inspection, the time-like part is electric in nature, while the space-like part is magnetic in nature?

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

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

The statement is correct, but this mathematical notation doesn't really mean anything...
How do I express it mathematically? I'm having some trouble translating this concept into some notation.

Matterwave
Science Advisor
Gold Member
How do I express it mathematically? I'm having some trouble translating this concept into some notation.

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.

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.
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!