The discussion centers on the Minkowski metric, specifically the form ##ds^2 = -c^2 dt^2 + dx^2 + dy^2 + dz^2##, and the implications of introducing off-diagonal terms. Participants explore how these terms affect the coordinate speed of light, concluding that if off-diagonal elements are present, inertial observers would not measure light traveling at speed ##c##. The conversation also highlights the importance of coordinate systems and assumptions regarding the metric's properties, such as flatness and constancy of components.
PREREQUISITESThe discussion is beneficial for physicists, mathematicians, and students of general relativity who are interested in the properties of spacetime metrics and their implications for the behavior of light and inertial observers.
kent davidge said:the additional assumption that should be made is that light travels with the same speed in all directions. This is enough to kill off the non diagonal terms in the metric.
but what if say, ##dx^1 / d\tau = 0##? Then ##\Gamma^\alpha{}_{1 \delta}## could be different from zero.PeterDonis said:for all possible combinations of indices ##\beta## and ##\gamma##
kent davidge said:what if say, ##dx^1 / d\tau = 0##? Then ##\Gamma^\alpha{}_{1 \delta}## could be different from zero
Ah, I got it. But since the ##\Gamma## are a sum of derivatives of the metric, wouldn't ##\Gamma = 0## only imply that the sum of these terms is zero? (And not necessarily the terms themselves, that is, the derivatives).PeterDonis said:Free particles can travel in all directions, and you must meet the conditions for all possible 4-velocities for a free particle. That means you can't pick out any particular component of ##dx / d\tau## and say it's zero. You must allow for the possibility of all of them being nonzero (because there are free particle 4-velocities for which all of them are nonzero).
kent davidge said:since the ##\Gamma## are a sum of derivatives of the metric, wouldn't ##\Gamma = 0## only imply that the sum of these terms is zero?