B Why is time orthogonal to space?

PeterDonis

Mentor
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.

if the equivalence principle were used that would also give a curved classical spacetime for gravitation and hence a local description of classical gravity ?
No. You can describe Newtonian gravity in geometric terms, as Cartan showed [1], but not the way you're describing here.

[1] https://en.wikipedia.org/wiki/Newton–Cartan_theory

jk22

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 ?

Ibix

The Galilean transforms are$$\begin{eqnarray*}t'&=&t\\x'&=&x-vt\end{eqnarray*}$$So the $t'$ axis is not $x'=0$.

In relativity, because space and time are part of spacetime, changing your notion of "rest" changes your coordinates on spacetime, which includes changing your notion of time.

But Newtonian physics models time as something completely distinct from space. Changing your notion of rest doesn't change your notion of time - there is an absolute time parameter that is the same for all observers. So the time axis doesn't change when you change frames.

jk22

Yes took the line of the origin of the x-axis as the time axis, a misunderstanding of my self.

Or is it from differential geometry that $\vec{e}_{t'}=\frac{\partial\vec{r'}}{\partial t}$ ?

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FactChecker

Gold Member
2018 Award
We naturally think of time and position as being independent and unrelated because, in our experience, positions do not generally move uniformly with time. Consider what would happen if you lived all your life floating in a moving stream of water. Then you would associate the passage of time with a uniform motion of all objects in the stream. Predicting the position of an object would always change as time passed. Your concept of time and space would not be orthogonal.
But that is not our experience. When one goes back home after a day of work, he goes to the same positional coordinates that he left home in the morning. Our concepts of time and position are orthogonal. Our mathematical definition of spacetime is made to match our natural concepts. That is practical, but not forced by mathematics.

[EDIT] I think we can see that in a Minkowski diagram, where the axes of a moving reference frame are not orthogonal.

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Dale

Mentor
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

Homework Helper
Gold Member
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 ?
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

Homework Helper
Gold Member
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 $x'=x-vt$.

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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

Mentor
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

Mentor
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 ?

"Why is time orthogonal to space?"

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