Why is time orthogonal to space?

In summary, the concept of orthogonality in geometry is a matter of convention and does not change the fundamental properties of a plane. However, using orthogonal coordinates can make calculations and formulas easier to apply and understand. It is possible to use non-orthogonal coordinates, but it is not recommended for beginners.
  • #36
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.
 
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  • #37
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.
 
  • #38
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 eggs
 
  • #39
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.
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.
 
  • #40
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.

I got ##e_x=(1,0)^t## and ##e_t=(-v,1)^t##

So for metric (+0) we get ##e_x.e_t=-v## and for (0+) gives inner product 0.

But the equation ##t=t'## could hide a simplification like ##t'=a/a*t## where the numerator is due to inclination of the axis and denominator to dilatation ?
 

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