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Faraday Tensor and Index Notation

  1. Jan 10, 2015 #1
    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.

    w0ow7o.png

    2. Relevant equations

    3. The attempt at a solution

    Part (a)

    The relations between the potentials and fields are:

    [tex] \vec B = \nabla \times \vec A [/tex]
    [tex]\vec E = -\nabla \phi - \frac{\partial \vec A}{\partial t} [/tex]

    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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  3. Jan 10, 2015 #2

    TSny

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    Just start writing it out explicitly. For example, what do you get for ## F^{01}##?
     
  4. Jan 10, 2015 #3
    [tex]F^{01} = \frac{\partial A_x}{\partial \phi} - \frac{1}{c} \frac{\partial \phi}{\partial A_x} [/tex]

    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..
     
  5. Jan 10, 2015 #4

    dextercioby

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    You're not calculating the correct derivatives. What does [itex] \partial^{a}[/itex] stand for?
     
  6. Jan 13, 2015 #5
    Am I missing out a factor of ##c##? I think ##\partial^a = c \frac{\partial}{\partial \phi}##
     
  7. Jan 13, 2015 #6

    TSny

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    ##\partial^a## is a compact notation for a derivative with respect to a space or time coordinate: ##\frac{\partial}{\partial x^a}##
     
  8. Jan 13, 2015 #7
    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}##.
     
  9. Jan 13, 2015 #8

    Matterwave

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    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}+...##
     
  10. Jan 13, 2015 #9
    Alright, so

    [tex]F^{01} = \frac{1}{c}\frac{\partial A_x}{\partial t} - c\frac{\partial t}{\partial x} [/tex]
     
  11. Jan 13, 2015 #10

    Matterwave

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    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.
     
  12. Jan 13, 2015 #11
    [tex]F^{01} = \partial^0 A^1 - \partial^1 A^0 [/tex]

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

    [tex]F^{01} = \frac{\partial A_x}{\partial ct} - \frac{\partial \frac{\phi}{c}}{\partial x} [/tex]
    [tex] F^{01} = \frac{1}{c}\frac{\partial A_x}{\partial t} - \frac{1}{c}\frac{\partial \phi}{\partial x} [/tex]
    [tex] F^{02} = \frac{1}{c}\frac{\partial A_y}{\partial t} - \frac{1}{c}\frac{\partial \phi}{\partial y} [/tex]
    [tex] F^{03} = \frac{1}{c}\frac{\partial A_z}{\partial t} - \frac{1}{c}\frac{\partial \phi}{\partial z} [/tex]
     
  13. Jan 13, 2015 #12

    Matterwave

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    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##?"
     
  14. Jan 13, 2015 #13
    [tex] \vec E = -\nabla \phi - \frac{\partial \vec A}{\partial t} [/tex]

    [tex]E_x = -\frac{\partial \phi}{\partial x} - \frac{\partial A_x}{\partial t} [/tex]
    [tex]F^{01} = \frac{1}{c}\frac{\partial A_x}{\partial t} - \frac{1}{c}\frac{\partial \phi}{\partial x} [/tex]
     
  15. Jan 13, 2015 #14

    TSny

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    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.
     
  16. Jan 13, 2015 #15
    [tex]E_x = -\frac{\partial \phi}{\partial x} - \frac{\partial A_x}{\partial t}[/tex]
    [tex]F^{01} = - \frac{1}{c}\frac{\partial \phi}{\partial x} -\frac{1}{c}\frac{\partial A_x}{\partial t} [/tex]

    This implies that ##F^{01} = \frac{E_x}{c} ##.
     
  17. Jan 13, 2015 #16

    Matterwave

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    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.
     
  18. Jan 13, 2015 #17
    [tex]F^{01} = -\frac{1}{c}\frac{\partial A_x}{\partial t} - \frac{1}{c}\frac{\partial \phi}{\partial x}[/tex]

    [tex]F^{02} = -\frac{1}{c}\frac{\partial A_y}{\partial t} - \frac{1}{c}\frac{\partial \phi}{\partial y}[/tex]

    [tex]F^{03} = -\frac{1}{c}\frac{\partial A_z}{\partial t} - \frac{1}{c}\frac{\partial \phi}{\partial z}[/tex]

    [tex]F^{12} = \frac{\partial A_y}{\partial x} - \frac{A_x}{\partial y} [/tex]

    [tex]\vec B = \nabla \times \vec A[/tex]
    [tex]\vec E = -\nabla \phi - \frac{\partial \vec A}{\partial t}[/tex]
     
  19. Jan 13, 2015 #18

    Matterwave

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    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}##.
     
  20. Jan 13, 2015 #19
    By inspection, the time-like part is electric in nature, while the space-like part is magnetic in nature?

    [tex]F = ( \frac{1}{c}\vec E, \vec B) [/tex]
     
  21. Jan 13, 2015 #20

    Matterwave

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