- #1
eoghan
- 210
- 7
Hi there
I'm studying GR and I am confused about coordinate transformations.
In my understanding, if I want to study a rotating reference system this is what I do.
In my inertial system the object trajectory is described by
$$
x = r\cos(\theta - \omega t)\\
y = r\sin(\theta - \omega t)
$$
This means that if I make the change of coordinates
$$
t = \tau\\
x = r\cos(\theta - \omega \tau)\\
y = r\sin(\theta - \omega \tau)
$$
then I switch to a frame where the object is at rest, i.e. I switch to the rotating reference frame. The problem is that this transformation is not a Lorentz transformation, so the metric in the rotating frame is not going to be Minkowskian, and indeed it is
$$
ds^2 = (1-\omega^2r^2)d\tau^2-2\omega r^2 d\tau d\theta -dr^2 - r^2 d\theta^2
$$
But why can't I do the same to describe a uniformly moving reference frame? I.e. if an object is in uniform motion along the x-axis, then it's trajectory is
$$
x = vt
$$
and so with the coordinate transformation
$$
t = \tau\\
x = v\tau - x'
$$
I switch to a reference frame where this body is at rest. The metric is obviously not Minkowskian
$$
ds^2 = (1-v^2)d\tau'2 - dx'^2 + 2v dx' d\tau
$$
Clearly this is not the metric of an inertial reference system, so what is this metric describing? How do I relate this metric to the metric of an inertial observer?
One answer I can come up with is that I'm not considering length contraction and time dilation, but I'm not considering them in switching to the Rindler coordinates either!
Another possibility is that in SR we are obliged to make only Lorentz transformations. But in GR we can do whatever coordinate transformations we want, so the transformation above should be allowed.
I'm studying GR and I am confused about coordinate transformations.
In my understanding, if I want to study a rotating reference system this is what I do.
In my inertial system the object trajectory is described by
$$
x = r\cos(\theta - \omega t)\\
y = r\sin(\theta - \omega t)
$$
This means that if I make the change of coordinates
$$
t = \tau\\
x = r\cos(\theta - \omega \tau)\\
y = r\sin(\theta - \omega \tau)
$$
then I switch to a frame where the object is at rest, i.e. I switch to the rotating reference frame. The problem is that this transformation is not a Lorentz transformation, so the metric in the rotating frame is not going to be Minkowskian, and indeed it is
$$
ds^2 = (1-\omega^2r^2)d\tau^2-2\omega r^2 d\tau d\theta -dr^2 - r^2 d\theta^2
$$
But why can't I do the same to describe a uniformly moving reference frame? I.e. if an object is in uniform motion along the x-axis, then it's trajectory is
$$
x = vt
$$
and so with the coordinate transformation
$$
t = \tau\\
x = v\tau - x'
$$
I switch to a reference frame where this body is at rest. The metric is obviously not Minkowskian
$$
ds^2 = (1-v^2)d\tau'2 - dx'^2 + 2v dx' d\tau
$$
Clearly this is not the metric of an inertial reference system, so what is this metric describing? How do I relate this metric to the metric of an inertial observer?
One answer I can come up with is that I'm not considering length contraction and time dilation, but I'm not considering them in switching to the Rindler coordinates either!
Another possibility is that in SR we are obliged to make only Lorentz transformations. But in GR we can do whatever coordinate transformations we want, so the transformation above should be allowed.