Perturbation to the flat space metric

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  • Thread starter Apashanka
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  • #1
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Main Question or Discussion Point

From the geodesic equation
d2xμ/dΓ2μ00(dt/dΓ)2=0,for non-relativistic case ,where Γ is the proper time and vi<<c implying dxi/dΓ<<dt/dΓ.
Now if we assume that the metric tensor doesn't evolve with time (e,g gij≠f(t) ) then Γμ00=-1/2gμs∂g00/∂xs.
If we here assume that the metric components of the curved part is a perturbation on the flat part
Then gμϑμϑ(flat part)+hμϑ(perturbation)
After which I got stuck in calculating the inverse components of the metric tensor gϑμ which is needed in Γμ00 above.
Can anyone please help me in sort put this.
Thank you.
 

Answers and Replies

  • #2
897
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because all components of ##h## are assumed to satisfy ##|h_{\mu \nu}| <<1## and also off diagonal terms are zero. Then the inverse is just $$g^{\mu \nu} = \frac{1}{\eta_{\mu \nu} + h_{\mu \nu}} \approx \eta_{\mu \nu} - h_{\mu \nu}$$ This follows from the observation: if ##a^2 - b^2 = (a + b)(a-b) = \approx 1## then ##1 / (a+b) \approx a-b##.
 
  • #3
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because all components of ##h## are assumed to satisfy ##|h_{\mu \nu}| <<1## and also off diagonal terms are zero. Then the inverse is just $$g^{\mu \nu} = \frac{1}{\eta_{\mu \nu} + h_{\mu \nu}} \approx \eta_{\mu \nu} - h_{\mu \nu}$$ This follows from the observation: if ##a^2 - b^2 = (a + b)(a-b) = \approx 1## then ##1 / (a+b) \approx a-b##.
Yes it is but if the off diagonal terms are non-zero for the general case what will be it??
 
  • #4
897
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Yes it is but if the off diagonal terms are non-zero for the general case what will be it??
For the general case, consider that $$g_{\kappa \sigma}g^{\sigma \rho} = (\eta_{\kappa \sigma} + h_{\kappa \sigma})(\eta^{\sigma \rho} \pm h^{\sigma \rho}) = \delta_\kappa{}^\rho \pm \eta_{\kappa \sigma} h^{\sigma \rho} + \eta^{\sigma \rho} h_{\kappa \sigma} + \mathcal O (h^2) \approx \delta_\kappa{}^\rho \pm h_\kappa{}^\rho + h_\kappa{}^\rho$$ this will be equal to ##\delta_\kappa{}^\rho## only if we use the minus sign.

(indices are raised and lowered with ##\eta##)
 
  • #5
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From matrix formulation if matrices A,B and C are given with their inverses $$A^{-1},B{^-1 }$$and $$C^{-1}$$ and given A=B+C
If $$A^{-1}=B^{-1}+C^{-1}$$ ,then $$I=2I+BC^{-1}+CB^{-1}$$$$I_{ik}+b_{ij}c^{jk}+c_{ip}b^{pk}$$$$\delta_{i}^{k}+\eta_{ij}h^{jk}+h_{ip}\eta^{pk}$$$$\delta_{i}^{k}+2h^k_i=0$$
Similarly for $$A^{-1}=B^{-1}-C^{-1}$$ then $$CB^{-1}-BC^{-1}=I$$$$h_{ij}\eta^{jk}-\eta_{ip}h^{pk}=I_{ik}$$ $$h_i^k-h_i^k=\delta_i^k$$......(1)
Now if $$g_{ij}=\eta_{ij}+h_{ij}$$ and $$g^{ij}=\eta^{ij}-h^{ij}$$( where ##\eta^{ij}## is the inverse element of ##\eta_{ij}## and similarly for h also),to satisfy equation (1) that hij is not the inverse element of hij? Is it??
 
  • #6
897
53
The notation ##h_{\kappa \sigma} h^{\sigma \rho}## means ##h_{\kappa \sigma} \eta^{\sigma \lambda} \eta^{\rho \mu} h_{\lambda \mu} = h_\kappa{}^\lambda h_\lambda{}^\rho##.
This means you are multiplying ##h## with itself: the ##\kappa##-th colunm of ##h## is being multiplied with the ##\rho##-th row of ##h##.

However ##\eta_{\kappa \sigma} \eta^{\sigma \rho} = \delta_\kappa{}^\rho## because ##\eta^{\sigma \rho}## is really the inverse of ##\eta_{\sigma \rho}##.
 
  • #7
haushofer
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I'm not sure what you want to calculate, but for static solutions you can always diagonalize the metric such that the line element is invariant under a time reflection. Then the inverse of the metric is easily found.
 

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