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

Homework Statement

Find Fμν, given Fμν=
(0 Ex Ey Ez)
(-Ex 0 Bz -By)
(-Ey -Bz 0 -Bx)
(-Ez By -Bx 0)

Homework Equations

gμν = Diag( -1, 1, 1, 1)

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:

TSny
Homework Helper
Gold Member
(0 Ex Ey Ez)
(-Ex 0 Bz -By)
(-Ey -Bz 0 -Bx)
(-Ez By -Bx 0)
There's a typo of a wrong sign for one of the terms.
I tried computing Fμν = gμνFμν
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.

Poirot
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.
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

TSny
Homework Helper
Gold Member
is gμα effectively g
No, these symbols represent specific components of the the metric tensor and they won't be equal in general.
because I know that if for instance you switched the indices you get the -g for example?
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.