What Are the Steps to Solve Faraday Tensor and Index Notation Problems?

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The discussion focuses on solving problems related to the Faraday tensor and index notation in the context of electromagnetism. Participants explore the relationships between the electric field (E), magnetic field (B), and the four-vector potential, emphasizing the importance of correctly applying tensor notation and derivatives. There is a detailed examination of the components of the Faraday tensor, particularly F^{01}, and how they relate to E and B fields. The conversation highlights the significance of negative signs in tensor calculations and the need for clarity in mathematical expressions. Overall, the thread serves as a collaborative effort to clarify complex concepts in electromagnetism and tensor calculus.
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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.

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Homework Equations



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.
 
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unscientific said:
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}##?
 
TSny said:
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..
 
You're not calculating the correct derivatives. What does \partial^{a} stand for?
 
dextercioby said:
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}##
 
unscientific said:
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}##
 
TSny said:
##\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}##.
 
unscientific said:
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}+...##
 
Matterwave said:
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}
 
  • #10
unscientific said:
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.
 
  • #11
Matterwave said:
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}
 
  • #12
unscientific said:
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##?"
 
  • #13
Matterwave said:
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}
 
  • #14
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
TSny said:
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} ##.
 
  • #16
unscientific said:
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.
 
  • #17
Matterwave said:
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}
 
  • #18
unscientific said:
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}##.
 
  • #19
Matterwave said:
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)
 
  • #20
unscientific said:
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...
 
  • #21
Matterwave said:
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.
 
  • #22
unscientific said:
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.
 
  • #23
Matterwave said:
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!
 

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