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FLRW metric tensor

  • Thread starter trv
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  • #1
trv
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Homework Statement



Not really homework, but thought this might be the best place to get a quick answer.

Question
Calculate g^{ca}g{ab} for the FLRW metric.

I would have thought this would be

[itex]
g^{ca}g{ab}=\delta^c_b=4
[/itex]

I thought 4 because I assumed there should be "1" for each non-zero element in the metric tensor g_{ab}.

Apparently however it should be g^{ca}g_{ab}=1

Can anyone please explain why?


Homework Equations



The Attempt at a Solution

 

Answers and Replies

  • #2
Cyosis
Homework Helper
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This is just the Kronecker delta which is 1 if b=c and zero if not, there is no summation going on. If it had been [itex]\delta_b^b[/itex] Einstein summation would have been implied, which would yield 4.

Mathematically:

[tex]
A^\gamma=g^{\gamma \alpha}A_\alpha= g^{\gamma \alpha}g_{\alpha \nu}A^\nu=\delta_\nu^\gamma A^\nu=A^\gamma
[/tex]

If it were four, you would get [itex]A^\gamma=4A^\gamma[/itex] which would make little sense.
 
Last edited:
  • #3
trv
73
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Thanks again Cyosis that clears things up.
 
  • #4
trv
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Actually, another quick question.

So the tensor, g^a_b for the FLRW metric, simply has components 1 along the diagonal?

Or for that matter, to generalise it further. Given any metric tensor r_{ab} with all elements being non-zero, r^a_b has components 1 everywhere?

To add to that, given,

[itex]
L=-0.5(\frac{d\phi}{dt})^2+V(\phi)
[/itex]

does g^a_bL = L?
 
Last edited:
  • #5
Cyosis
Homework Helper
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trv said:
So the tensor, g^a_b for the FLRW metric, simply has components 1 along the diagonal?
Correct.

trv said:
Or for that matter, to generalise it further. Given any metric tensor r_{ab} with all elements being non-zero, r^a_b has components 1 everywhere?
No, proof:

[tex]
r^\alpha_{\;\;\beta}=r^{\alpha \gamma}r_{\gamma \beta}=\delta^\alpha_\beta
[/tex]

In words, any mixed metric is equal to the Kronecker delta. You basically have a matrix with 1's on the diagonal and zeros elsewhere.
 
  • #6
trv
73
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Oh yeh, true the kronecker delta would imply zeros everywhere apart from the diagonal. I think where I'm going wrong is assuming we multiply each element by its inverse. Giving 1 at all places where the metric tensor had non-zero components. Obviously that's not how it works.
 
  • #7
cristo
Staff Emeritus
Science Advisor
8,107
72
Oh yeh, true the kronecker delta would imply zeros everywhere apart from the diagonal. I think where I'm going wrong is assuming we multiply each element by its inverse. Giving 1 at all places where the metric tensor had non-zero components. Obviously that's not how it works.
Think of these tensors as matrices, where g^{ab} is the inverse of g_{ab}. Just as multiplying a matrix with its inverse yields the identity matrix (AA^{-1}=I), multiplying the metric with its inverse yields the kronecker delta.
 
  • #8
trv
73
0
Thanks the two of you.
 

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