A Spacetime interval in Galilean relativity

LightPhoton
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If we define metric in Galilean spacetime we can get a notion of spacetime interval. But is this correct?
In this [1] video, EigenChris mentions that there is no spacetime distance in Galilean relativity. And in this [2] video he replies to a comment as follows:

> @hariszachariades8299: If the "spacetime separation vector" has components ##(\Delta x,\Delta t)##, there is no combination ##a\Delta x^2
+ b\Delta x\Delta t + c\Delta t^2## that remains invariant under the Galilean transformation ##\Delta x' = \Delta x - v\Delta t, \Delta t' =
\Delta t##. Maybe this is the reason that we do not usually see discussions of Galilean spacetime.

> @eigenchris: Yeah. The only options are to set either "##a##" to zero or "##c##" to zero. So you end up with a metric for only time or only space.



But isn't this claim false?

We can always define spacetime distance as

$$(ds)^2=g_{\mu\nu}dx^\mu dx^\nu$$

Where ##g_{\mu\nu}## are the components of metric defined as

$$g_{\mu\nu}=\vec e_\mu\cdot\vec e_\nu\tag1$$

Now again from this [3] video of EigenChris, we have

$$\vec{e_t}'=\vec{e_t}+v\vec{e_x}$$
$$\vec{e_x}'=\vec{e_x}$$

Thus in nonprime coordinates, the metric becomes identity*, whereas in prime coordinates it becomes

$$g'=\begin{pmatrix}
1+v^2 & v \\
v & 1
\end{pmatrix}$$


Now, since from equation ##(1)##, in nonprime coordinates we get
$$(ds)^2=(dt)^2+(dx)^2\tag3$$

And using prime coordinates along with Galellian transformations, we get

$$(ds)^2=(1+v^2)(dt')^2+2v(dx')(dt')+(dx')^2$$
$$(ds)^2=(1+v^2)(dt)^2+2v(dx-vdt)(dt)+(dx-vdt)^2$$
$$(ds)^2= (dt)^2+(dx)^2\tag 3$$

Thus equations ##(2)## and ##(3)## refer to the same spacetime interval.
So did EigenChris said something wrong, or am I missing something?

*If we don't assume ##\vec{e_x}## and ##\vec{e_t}## to have a magnitude of one, but instead choose:

$$\vec{e_t}\cdot\vec{e_t}=K^2$$
$$\vec{e_x}\cdot\vec{e_x}=P^2$$

For some functions, ##K## and ##P##, even then the distance would be the same in both coordinates, just changed to $$(ds)^2=(Kdt)^2+(Pdx)^2$$

[1]:
[2]:
[3]:
 
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haushofer said:
Just a quick reply: in

https://research.rug.nl/en/publications/newton-cartan-gravity-revisited

section 2.8 I treat the Galilean "spacetime interval". Maybe that helps.
Thanks for the quick reply!

It is well above me for now (and a similar answer I got on StackExchange is as well, https://physics.stackexchange.com/questions/842622/spacetime-interval-in-galilean-relativity#842629) so I am not able to fully appreciate the content you shared.

Can you kindly explain to me, why my approach ultimately fails, or even to begin to understand the answer to such a question I need group theory?
 
LightPhoton said:
Thanks for the quick reply!

It is well above me for now (and a similar answer I got on StackExchange is as well, https://physics.stackexchange.com/questions/842622/spacetime-interval-in-galilean-relativity#842629) so I am not able to fully appreciate the content you shared.

Can you kindly explain to me, why my approach ultimately fails, or even to begin to understand the answer to such a question I need group theory?
You don't need group theory; you just need a bit of tensor analysis. What I did in that section is just assume there is such an invariant Galilean interval, and apply the Galilei-transformations to both the supposed metric (components) and its inverse (components) and solve. If you've followed a course on GR, you should be able to follow that argument; it's nothing fancy, just applying the tensor transformation law with Galilean coordinate transformations.
 
LightPhoton said:
We can always define spacetime distance as

$$(ds)^2=g_{\mu\nu}dx^\mu dx^\nu$$

Where ##g_{\mu\nu}## are the components of metric defined as

$$g_{\mu\nu}=\vec e_\mu\cdot\vec e_\nu\tag1$$

Now again from this [3] video of EigenChris, we have

$$\vec{e_t}'=\vec{e_t}+v\vec{e_x}$$
$$\vec{e_x}'=\vec{e_x}$$

Thus in nonprime coordinates, the metric becomes identity*, whereas in prime coordinates it becomes

$$g'=\begin{pmatrix}
1+v^2 & v \\
v & 1
\end{pmatrix}$$
The issue is that while what you have defined here is a perfectly valid tensor on Newtonian spacetime, it does not take the same form in all inertial frames (related by Galilean transformations) as the Minkowski metric does under Lorentz transformations. As such, your lime element singles out a particular frame as a rest frame.
 
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Orodruin said:
The issue is that while what you have defined here is a perfectly valid tensor on Newtonian spacetime, it does not take the same form in all inertial frames (related by Galilean transformations) as the Minkowski metric does under Lorentz transformations. As such, your lime element singles out a particular frame as a rest frame.
Ah ok, you mean that it is a well defined tensor field as geometric object at each point on Newtonian spacetime (i.e. a coordinate-independent bilinear map). However, looking at its form in coordinates, one recognizes that it doesn't take the same form in all inertial frames/coordinates, so basically it picks/singles out a particular one of them as a/the rest frame.
 
Yes.
 
haushofer said:
You don't need group theory; you just need a bit of tensor analysis. What I did in that section is just assume there is such an invariant Galilean interval, and apply the Galilei-transformations to both the supposed metric (components) and its inverse (components) and solve. If you've followed a course on GR, you should be able to follow that argument; it's nothing fancy, just applying the tensor transformation law with Galilean coordinate transformations.
Now that I have had a chance to read it carefully, I understand your point along with Orodruin's. However, how do we reconcile this with the idea that ##g_{\mu\nu} = \vec e_{\mu} \cdot \vec e_{\nu}##? As you showed, only the time-time component of ##g_{\mu\nu}## survives, which implies that ##\vec e_{i} \cdot \vec e_{j} = 0##
 
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