Why is time orthogonal to space?

[Dale]

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.

 

[jk22]

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.

 

[nitsuj]

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

 

[Dale]

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.

 

[jk22]

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 ?