- #1
LightPhoton
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- TL;DR
- 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 $$(ds)^2=(Kdt)^2+(Pdx)^2$$
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[2]:
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> @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 $$(ds)^2=(Kdt)^2+(Pdx)^2$$
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