A GW Binary Merger: Riemann Tensor in Source & TT-Gauge

MrFlanders
Messages
2
Reaction score
0
TL;DR Summary
There seems to be a difference in the Riemann tensor when you compute the R_{0101} component of the Riemann tensor along the x-axis for the source and TT-gauge. After a lot of reading and thinking I am unable to find why this might be the case.
In the book general relativity by Hobson the gravitational wave of a binary merger is computed in the frame of the binary merger as well as the TT-gauge. I considered what components of the Riemann tensor along the x-axis in both gauges. The equation for the metric in the source and TT-gauge are given in 18.19 and 18.21 respectively. In the source gauge R_{0101} = -0.5*(w/c)^2 h_11 and in the TT-gauge R_{0101} = 0 since the Riemann tensor is invariant they should be the same though. I can seem to find out why ?
 
Physics news on Phys.org
MrFlanders said:
since the Riemann tensor is invariant they should be the same though.
No, this is not correct. The Riemann tensor being "invariant" ("covariant" would be a better term) does not mean each individual component of the tensor is the same in any coordinate chart. It only means that scalar invariants derived by contracting the Riemann tensor (the simplest being the Ricci scalar) are the same in any coordinate chart.
 
If you'd say "tensor components", everything would be clear. Of course, tensors are by definition invariant objects, but tensor components transform when changing the (tangent) basis and co-basis.
 
  • Like
Likes PhDeezNutz and malawi_glenn
PeterDonis said:
No, this is not correct. The Riemann tensor being "invariant" ("covariant" would be a better term) does not mean each individual component of the tensor is the same in any coordinate chart. It only means that scalar invariants derived by contracting the Riemann tensor (the simplest being the Ricci scalar) are the same in any coordinate chart.
In general this would not be the case but in the linearised theory of gravity where the Riemann tensor is approximated to first order in the metric perturbation h. It can be shown directly from the transformations laws that each component of the Riemann tensor is Invariant to first order in the metric perturbation h.
 
MrFlanders said:
In general this would not be the case but in the linearised theory of gravity where the Riemann tensor is approximated to first order in the metric perturbation h. It can be shown directly from the transformations laws that each component of the Riemann tensor is Invariant to first order in the metric perturbation h.
I don't get what your are claiming. Even in SR, the components of a 4-vector do not stay the same after a Lorentz transform. Only scalars constructed from them stay the same.
 
  • Like
Likes vanhees71 and malawi_glenn
I asked a question here, probably over 15 years ago on entanglement and I appreciated the thoughtful answers I received back then. The intervening years haven't made me any more knowledgeable in physics, so forgive my naïveté ! If a have a piece of paper in an area of high gravity, lets say near a black hole, and I draw a triangle on this paper and 'measure' the angles of the triangle, will they add to 180 degrees? How about if I'm looking at this paper outside of the (reasonable)...
From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by ##g_{\alpha\beta}## to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation ##\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}## should be a tensor...
Back
Top