- #36
Dale
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What are you objecting to? You start with a “but”, but no quote to apply it to.arydberg said:But we can go foward and backward in space. In time we can only go in one direction.
What are you objecting to? You start with a “but”, but no quote to apply it to.arydberg said:But we can go foward and backward in space. In time we can only go in one direction.
remember this is strictly about time, geometry. not the breaking / "unbreaking" of eggsarydberg said:But we can go foward and backward in space. In time we can only go in one direction.
Right, it is a coordinate change not a metric, but you wrote it as a metric. If you are writing a linear coordinate transform then you can write it in the form of a matrix. Typically you would label it something like ##\Lambda_{\mu}^{\mu’}## to indicate that it changes between the primed and unprimed coordinates.jk22 said:I understood I made a mistake : I computed ##x^\mu=(t',x')=(t,x-vt)##
Then ##\vec{e}_i=\frac{\partial x^\mu}{\partial i}## and ##g_{\mu\nu}=\left(\begin{array}{cc} 1+v^2&-v\\-v &1\end{array}\right)##
But this makes no sense since it's a change of coordinates and not a manifold and that we can always use a background metric of some signature.
Dale said:It is still orthogonal to x. Orthogonality is based on some inner product. In relativity the metric that defines the inner product has a signature (-+++), but in Newtonian theory it is two degenerate metrics with signatures (+000) and (0+++) respectively. Thus the inner product of ##\hat x’## and ##\hat t’## is zero for both of the degenerate metrics.