Tensor Covariant Derivative Expressions Algebra (Fermi- Walk

binbagsss
Messages
1,291
Reaction score
12

Homework Statement



Hi
I am looking at part a).

fermiwalker.png


Homework Equations



below

The Attempt at a Solution



I can follow the solution once I agree that ## A^u U_u =0 ##. However I don't understand this.

So in terms of the notation ( ) brackets denote the symmetrized summation and the [ ] the antisymmetrized, both come with a factor of 1/2, for a 2 indices tensors.
So I agree with the second equality if I were to ignore the parentheses, and I agree that the covariant derivative acting on 1 is of course zero for the last equality.

But I thought ##A^u## is such that the covariant derivative acts on the ##U^v## in that expression and that is it, once you multiply it by another vector eg ##A^u V^v ## the covariant derivative does not act on ##V^v ##, it does not act on everything to the right side? So I don't understand how we've changeed the parentheses as in the second equality (solution here: )

fermiwalkersol.png


Thanks
 

Attachments

  • fermiwalker.png
    fermiwalker.png
    17.9 KB · Views: 882
  • fermiwalkersol.png
    fermiwalkersol.png
    9.5 KB · Views: 367
Physics news on Phys.org
What is
$$
\frac{d(f(x)^2)}{dx}?
$$
 
Orodruin said:
What is
$$
\frac{d(f(x)^2)}{dx}?
$$

## 2 f(x) d(f(x)) / dx ##
 
So ... what is ##\nabla_\nu (U_\mu U^\mu)##?
 
  • Like
Likes Matter_Matters
Orodruin said:
So ... what is ##\nabla_\nu (U_\mu U^\mu)##?

Sorrry are the brackets here indicating the symmetrized sum or simply what the covariant deriviatve is acting upon?

Well I conclude that we have ##2\nabla_v U^u = \nabla_v U^u U_u ## (if the parantheses are indicating what the covariant derivative is acting on only), however this looks like nonsense since the indices are not consistent each side...
 
binbagsss said:
Sorrry are the brackets here indicating the symmetrized sum or simply what the covariant deriviatve is acting upon?
Full brackets never indicate symmetrisation. Also, the expression has no indices of the same typy (covariant/contravariant) that can be symmetrised.

binbagsss said:
Well I conclude that we have 2∇vUu=∇vUuUu2∇vUu=∇vUuUu2\nabla_v U^u = \nabla_v U^u U_u (if the parantheses are indicating what the covariant derivative is acting on only), however this looks like nonsense since the indices are not consistent each side...
You are missing one of the Us and your indices do not match because of it.
 
Orodruin said:
Full brackets never indicate symmetrisation. Also, the expression has no indices of the same typy (covariant/contravariant) that can be symmetrised.You are missing one of the Us and your indices do not match because of it.

I'm fully aware that is why my indices do not match but was trying to follow the f(x)^2 logic example, and here ofc ##U^u U_u ## denotes the ' ^ 2 ' so I'm a bit confused. ta.
 
binbagsss said:
I'm fully aware that is why my indices do not match but was trying to follow the f(x)^2 logic example, and here ofc ##U^u U_u ## denotes the ' ^ 2 ' so I'm a bit confused. ta.
Look at your expression in #3 again and compare it with what you did.
 
Back
Top