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

[Poirot]

Homework Statement

Find F_μν, given F^μν=
(0 E_x E_y E_z)
(-E_x 0 B_z -B_y)
(-E_y -B_z 0 -B_x)
(-E_z B_y -B_x 0)

Homework Equations

g_μν = Diag( -1, 1, 1, 1)

The Attempt at a Solution

I tried computing F_μν = g_μνF^μν
but this gave
(0 -E_x -E_y -E_z)
(-E_x 0 B_z -B_y)
(-E_y -B_z 0 B_x)
(-E_z B_y -B_x 0)
Which seems wrong to me, since the next question is to find F_μνF^νμ 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!

 

[TSny]

(0 E_x E_y E_z)
(-E_x 0 B_z -B_y)
(-E_y -B_z 0 -B_x)
(-E_z B_y -B_x 0)

There's a typo of a wrong sign for one of the terms.

I tried computing F_μν = g_μνF^μν

The convention is to sum over repeated indices. So the expression g_μνF^μνon 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 F_μν represents a specific component for specific values μ and ν. So, something is wrong.

The correct way to lower the two indices is F_μν = g_μαg_vβF^αβ 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 g_μν = 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 g_μνF^μνon 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 F_μν represents a specific component for specific values μ and ν. So, something is wrong.

The correct way to lower the two indices is F_μν = g_μαg_vβF^αβ 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 g_μν = 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 g_μα effectively g_vβ, because I know that if for instance you switched the indices you get the -g_vβ for example? And how would you know when they are the same or not?
Thanks

 

[TSny]

is g_μα effectively g_vβ

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_vβ for example?

Actually, when you switch the order of the indices on the metric tensor, it does not introduce a negative sign. Instead, g_vβ = g_βv. 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_vβ = g_βv = 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 A_μ = g_μν A^ν. Letting μ = 2(on both sides),

A₂ = g_2ν A^ν

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 A_μ = g_μν A^ν to find A₀ and A₃.