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Index notation: Find F_(μν) given F^(μν)?

  1. May 17, 2016 #1
    1. The problem statement, all variables and given/known data
    Find Fμν, given Fμν=
    (0 Ex Ey Ez)
    (-Ex 0 Bz -By)
    (-Ey -Bz 0 -Bx)
    (-Ez By -Bx 0)
    2. Relevant equations
    gμν = Diag( -1, 1, 1, 1)

    3. The attempt at a solution
    I tried computing Fμν = gμνFμν
    but this gave
    (0 -Ex -Ey -Ez)
    (-Ex 0 Bz -By)
    (-Ey -Bz 0 Bx)
    (-Ez By -Bx 0)
    Which seems wrong to me, since the next question is to find FμνFνμ and this is meant to come out as the Lagrangian and from my notes I don't see this working.
    I think I'm messing up my index notation, I think I may need an extra metric in there because there are 2 indices downstairs but I don't have the best understanding.

    Any help and advice on dealing with pesky index notation would be greatly appreciated, Thanks!
     
    Last edited: May 17, 2016
  2. jcsd
  3. May 17, 2016 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    There's a typo of a wrong sign for one of the terms.
    The convention is to sum over repeated indices. So the expression gμνFμνon the right side would imply a sum over μ and a sum over ν. The result would not depend on a specific value of μ and ν; whereas, the left side Fμν represents a specific component for specific values μ and ν. So, something is wrong.

    The correct way to lower the two indices is Fμν = gμαgFαβ which involves a sum over α and a sum over β. The μ and ν appearing on the right side have the same values as the μ and ν appearing on the left side.

    Because the metric has components gμν = Diag( -1, 1, 1, 1), you should be able to show that whenever you lower a 0 index, you just change the sign. For example A0 = -A0 for any 4-vector. When you lower any of the other indices, there is no change in sign. For example A2 = A2. The same holds for tensors with more than one index. See if you can show, for example, that F03 = -F03 and F23 = F23. Once you get the hang of it, you will be able to lower (or raise) indices very quickly.
     
  4. May 17, 2016 #3
    Hi and thank you for your response, you've cleared up a lot! Just a quick question, is gμα effectively g, because I know that if for instance you switched the indices you get the -g for example? And how would you know when they are the same or not?
    Thanks
     
  5. May 17, 2016 #4

    TSny

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    Homework Helper
    Gold Member

    No, these symbols represent specific components of the the metric tensor and they won't be equal in general.
    Actually, when you switch the order of the indices on the metric tensor, it does not introduce a negative sign. Instead, g = gβv. The metric is said to be "symmetric". If you put the components of g into a matrix, the matrix will be a symmetric matrix. More specifically, it will be a diagonal matrix where all of the off diagonal terms are zero. So, g = gβv = 0 if v ≠ β.

    You should probably practice with some simple examples. Suppose you had a 4-vector A with components A0 = 5, A1 = 6, A2 = 7, and A3 = 8. Then, if I want to find A2 I would apply the general relation Aμ = gμν Aν. Letting μ = 2(on both sides),

    A2 = g Aν

    The repeated index ν means I have to sum over ν for ν = 0, 1, 2, and 3. So,

    A2 = g20A0 + g21A1 + g22A2 + g23A3
    = 0⋅A0 + 0⋅A1 +1⋅A2 + 0⋅A3
    = A2
    = 7

    You can try using Aμ = gμν Aν to find A0 and A3.
     
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