![]() ![]() An inspection of the at spacetime metric suggests that the Christoffel symbols involving t as an index are zero because no metric coefcients are time dependent. All the afne coefcients are not expected to be zero. The restriction is that we have to have zero curvature, which depends on the derivatives of the Christoffel symbols. This qualitative argument is supported by calculation of the elements of these tensors. For example, you can do SR in spherical coordinates. Re the Christoffel symbols, there is no requirement that coordinate transformations be linear in SR, and you can have Christoffel symbols that don't vanish. And note that there is no such thing as a Lorentz transformation $\Lambda$ in general relativity. So that $ds^2$ stays invariant? So in that case, since $g$ may depend on time and position, the transformations $\Lambda$ are not linear anymore which gives rise to why the second term in the Christoffel symbol transformation is non-zero: Evaluating these formulas in terms of the quadrupole moment of a radiating source involves a lengthy calculation which we will not reproduce here. Since the flat-space DAlembertian has the form. We begin by considering the linearized equations in vacuum (6.23). In special relativity we have the conditionÄoes that mean that in general relativity, we get We begin with the Christoffel symbols, which are given by (6.4). This might be very fundamental, but I had a hard time finding this on google. I am trying to understand coordinate transformations in general relativity. ![]()
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