Why is time orthogonal to space?

[Dale]

In a Galilean transformation : $x'=x-vt$. If $x'=0$, hence the time primed axis, the equation is $x=vt$, which describe a straight line not orthogonal to x ?

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.

 

[robphy]

In a Galilean transformation : $x'=x-vt$. If $x'=0$, hence the time primed axis, the equation is $x=vt$, which describe a straight line not orthogonal to x ?

Just adding to Dale's comment...
recall from my earlier post above that
"the tangent line to the circle [the metric]" is (following Minkowski) "orthogonal to the timelike radial direction".
That characterization is an extension of the Euclidean case and can be extended further to the Galilean case.
So, in a Galilean spacetime, the spatial direction [where t=constant] is Galilean-perpendicular (but not necessarily Euclidean-perpendicular) to the timelike-worldline. [This corresponds to the "temporal metric" with signature +000.]

 

[robphy]

Using a classical Galileo transform also implies that time inclines over space by an angle $\arctan(v)$.

No, with a Galilean transformation, the "direction" of time doesn't change at all, because the Galilean transform only transforms space, it does nothing to time.

In the Galilean case, it turns out that the "Galilean-rapidity" (the "Galilean-angle"= "arc-length [using the degenerate spatial metric mentioned by Dale] along the Galilean-circle) is equal to the ordinary velocity v. [There is a "Galilean-analogue" of the tangent function---called $tang(\theta)$ by the mathematician I.M Yaglom... and it's defined as $tang(\theta)\equiv \theta$... using "dimensionless units". Along these lines, $cosg(\theta)\equiv 1$.]

(The additivity of rapidity (angles = "spatial arc-lengths on unit circles) always holds...
but the corresponding slopes [physically, spatial-velocities] are not generally additive...
except in the Galilean case. Unfortunately, we hold on so tightly to the special-case Galilean additivity of spatial velocity that we have trouble with the more general non-additivity of slopes [velocities].)

Note that, in a boost, all timelike worldlines are transformed [in accordance with the principle of relativity... no preferred reference frame... no timelike eigenvectors].

• So, in this sense, the "direction of time" (direction of the observer's 4-velocity) does change [e.g. it can be transformed to being the "observer at rest" in the spacetime diagram]...
  however,
  in the Galilean case, the "time-coordinate" doesn't change: $t'=t$... that is "1 sec after event O" is still "1 sec after event O".
• Along these lines, in Galilean boost, the spatial hypersurfaces are mapped onto themselves... the set of events with "t=1 sec" are the same after a Galilean boost. However, the coordinate labeling of those events are different $x'=x-vt$.

 

[The Bill]

It is orthogonal because you can fix your position in space and measure time independently.

Similarly x is orthogonal to y and z because you can fix your position in y and z and measure x independently.

A coordinate system with linearly independent but not orthogonal x, y, and z axes would satisfy your condition while not being orthogonal.

 

[arydberg]

But we can go foward and backward in space. In time we can only go in one direction.

 

[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 ?