Transform to accelerating coordinates

[etotheipi]

It's a silly example, but hopefully it will help me to understand the maths. Two guys $A$ and $B$ are initially at the same spacetime event $O$, and then $B$ receives an impulse along the $x$-direction giving him an initial coordinate velocity $\dot{x}_B = v_0$ as measured by $A$ [unbarred coordinates denote coordinates in the $A$ rest frame]. Also, $B$ has a rocket or something that he can use in such a way that his coordinate acceleration is $\ddot{x}_B = -k$, where $k$ is constant.

Their worldlines will again coincide at $t = 2v_0/k$, which I call event $P$. The proper time along $A$'s trajectory between $O$ and $P$ is $\tau = 2v_0/k$. As exercise, I want to show that $B$ also calculates this same value for the proper time $A$ experiences.

But, I don't know how to use accelerated coordinates! We need a transformation $\bar{x} = f(x, t)$ and $\bar{t} = g(x,t)$ and from that we should be able to work out the trajectory in $(\bar{t}, \bar{x})$ space, as well as the new metric, since we could then just use$$\bar{g}_{\mu\nu} = \frac{\partial x^{\rho}}{\partial \bar{x}^{\mu}} \frac{\partial x^{\sigma}}{\partial \bar{x}^{\nu}}g_{\rho \sigma}$$and finally we need to calculate$$\tau = \frac{1}{c} \int_{\lambda_1}^{\lambda_2} \sqrt{\bar{g}_{\mu \nu} \frac{d\bar{x}^{\mu}}{d\lambda} \frac{d\bar{x}^{\nu}}{d\lambda}} d\lambda$$and we can choose a worldline parameter $\lambda$ once we set up the problem. But so far I'm falling at the first hurdle, how do you do a transformation of coordinates into a frame undergoing constant coordinate acceleration? Thank you ☺

 

[Ibix]

Edit: this is irrelevant - I need to read the question properly.

You look up "Rindler coordinates". Turns out that a family of hyperbolae with a common focus are invariant under Lorentz boost in the same way concentric circles are invariant under rotation in Euclidean space, and the path of an observer under constant acceleration is a hyperbola.

 

[etotheipi]

You look up "Rindler coordinates". Turns out that a family of hyperbolae with a common focus are invariant under Lorentz boost in the same way concentric circles are invariant under rotation in Euclidean space, and the path of an observer under constant acceleration is a hyperbola.

Ohh that rings a bell, I hope I haven't asked this question before ... also, am I making more work for myself by setting it up so that $B$ has constant coordinate acceleration? I'm thinking it might be easier if he has constant proper acceleration. Anyway I'll try it and see how it goes, thanks ☺

 

[Ibix]

Ah! Sorry, I mis-read. Rindler coordinates are for constant proper acceleration.

Constant coordinate acceleration is harder because there isn't a symmetry to take advantage of, so there's no unique "right" way to set up coordinates. Basically you need to invent a system. A method I like is radar coordinates, which essentially equips some reference observer (your B) with a radar set and works out the coordinates that they would assign to radar echoes.

It might be rather messy in this case.

 

[etotheipi]

I think I'm screwing up in a big way, I think it's becoming clear to me that I have no idea what I'm doing ... so far I found these transformations$$t = \frac{1}{\alpha} \text{artanh} \left(\frac{T}{X+ \frac{1}{\alpha}} \right)$$ $$x = \sqrt{\left((X + \frac{1}{\alpha}\right)^2 - T^2} - \frac{1}{\alpha}$$Now I worked out the partial derivative, changing the problem statement to now have a proper constant acceleration$$\frac{\partial t}{\partial T} = \frac{1}{\alpha X + 1 - \frac{\alpha T^2}{X + \frac{1}{\alpha}}}$$ $$\frac{\partial t}{\partial X} = \frac{-T}{\alpha(X+ \frac{1}{\alpha})^2 - \alpha T^2}$$ $$\frac{\partial x}{\partial X} = \frac{X + \frac{1}{\alpha}}{\sqrt{(X+ \frac{1}{\alpha})^2 - T^2}}$$ $$\frac{\partial x}{\partial T} = \frac{-T}{\sqrt{(X+ \frac{1}{\alpha})^2 - T^2}}$$ Now the components of the metric are$$\bar{g}_{00} = - \left(\frac{\partial t}{\partial T} \right)^2 + \left(\frac{\partial x}{\partial T} \right)^2$$ $$\bar{g}_{11} = - \left(\frac{\partial t}{\partial X} \right)^2 + \left(\frac{\partial x}{\partial X} \right)^2$$ $$\bar{g}_{10} = -\frac{\partial t}{\partial X} \frac{\partial t}{\partial T} + \frac{\partial x}{\partial X} \frac{\partial x}{\partial T} $$ $$\bar{g}_{01} = -\frac{\partial t}{\partial T} \frac{\partial t}{\partial X} + \frac{\partial x}{\partial T} \frac{\partial x}{\partial X} $$Now I also need to find trajectory in the $(T, X)$ space, so from the trajectory in the unbarred coordinates

 

[etotheipi]

Nah I'm done with this today, going to go and watch some memes instead. I'll read the thing you linked later @Ibix (hope it doesn't contain any 'Steve Gull errors!')

 

[Ibix]

So $\alpha$ is the instantaneous proper acceleration of B? I'm not quite sure how you got to your new coordinates (not saying they're wrong, just don't know if it's right).

To define B's rest coordinate system you need its worldline, which I believe is $x=v_0t-\frac 12kt^2$. This should let you write $\gamma(t)$ and hence $t(\tau)$ and $x(\tau)$. That's your time axis. Then I'd use radar coordinates: for an arbitrary event $(x,t)$, work out B's proper times $\tau_-$ and $\tau_+$ at which a radar pulse from B would be emitted and received if it bounced off that event. Assign coordinates $T=(\tau_++\tau_-)/2$ and $X=(\tau_+-\tau_-)/2$ to that event. Now you have a valid coordinate transform and you can write down the Jacobean and transform the metric (which is what I think you are doing above).

Note that this is likely to be messy because B's worldline is not always timelike - constant coordinate acceleration would exceed $c$. You will find that some of the transforms I'm proposing are non-existant outside some region of spacetime.

 

[Ibix]

(hope it doesn't contain any 'Steve Gull errors!')

I don't believe so.

 

[etotheipi]

So α is the instantaneous proper acceleration of B? I'm not quite sure how you got to your new coordinates (not saying they're wrong, just don't know if it's right).

Yeah, I zoned out a bit and I didn't realize until after I wrote it that I'd attempted to derive the metric components for constant proper acceleration, and not constant coordinate acceleration. So we can ignore #5, it's a load of garbage

I started reading, and I've got a question already... I don't get their definition of the "observer's hypersurface of simultaneity at $\tau_0$",$$\Sigma_{\tau_0} = \{x: \tau(x) = \tau_0 \}$$i.e. the set of all $x = (x^{\mu})$ in the spacetime where $\tau(x)$ is that specified value. I thought, $\tau$ parameterised the trajectory $\gamma$ of the observer, so I don't understand how it's a function of the position in the spacetime (e.g. for points not even on the curve)?

 

[Ibix]

My phone's refusing to download the PDF so I can't check, but I think they're just using $\tau$ as the timelike coordinate. I agree that's a little unconventional.

 

[etotheipi]

My phone's refusing to download the PDF so I can't check, but I think they're just using $\tau$ as the timelike coordinate. I agree that's a little unconventional.

Oh yeah, they defined it literally two lines above, they said $\tau(x) = \frac{1}{2}(\tau^{+}(x) + \tau^{-}(x))$, which is like you said the radar time coordinate of a particular point in this coordanite system. I think the rest of it makes a little more sense now

 

[etotheipi]

Here's a status update,$$x = v_0 t - \frac{1}{2}kt^2 \implies \frac{dx}{dt} = v_0 - kt \implies \gamma(t) = \frac{1}{\sqrt{1-(v_0-kt)^2}}$$We can get the proper time as a function of A's coordinate time,$$\tau(t) = \int_0^t \frac{d\xi}{\gamma(\xi)} = \int_0^t d\xi \sqrt{1-(v_0 - k \xi)^2}$$We use a substitution $v_0 - k\xi = \sin{\theta}$, which transforms the integral to $$\tau(t) = -\frac{1}{k}\int_{\arcsin{(v_0)}}^{\arcsin{(v_0 - kt)}} d\theta \left[ \frac{1}{2} + \frac{1}{2} \cos{(2\theta)} \right]$$When we insert the limits,$$\tau(t) = -\frac{1}{2k} \left( \left[ \arcsin{(v_0 - kt)} -\arcsin{v_0} \right] + (v_0 - kt) \sqrt{1-(v_0 - kt)^2} - v_0 \sqrt{1-v_0^2}\right)$$I am not certain right now how to invert this to obtain $t(\tau)$. But once I figure out how to do that (and subsequently also get $x(\tau)$), we can use the approach in the article to define intermediate null coordinates $u = x+t$, $v = x-t$ such that B's trajectory is $(u_B(\tau),v_B(\tau)) = (x_B(\tau)+t_B(\tau), x_B(\tau)-t_B(\tau))$. The coordinate lines of the null coordinates are parallel to the worldline of a light ray, so for some arbitrary spacetime point $x = (u, v) = (h,k)$ represented in the $u,v$ coordinates, by definition $u_B(\tau^+(x)) = h$ and $v_B(\tau^-(x)) = k$. Then, finally we can get the $X, T$ radar coordinates, derive the metric components, and finally compute the original integral in $B$'s coordinates.

But first step, is to invert $\tau(t)$! Maybe I will see this in the morning

 

[PeterDonis]

I think I'm screwing up in a big way, I think it's becoming clear to me that I have no idea what I'm doing ... so far I found these transformations

Those are for Rindler coordinates, i.e., constant proper acceleration, not constant coordinate acceleration.

 

[pervect]

I think I'm screwing up in a big way, I think it's becoming clear to me that I have no idea what I'm doing ... so far I found these transformations$$t = \frac{1}{\alpha} \text{artanh} \left(\frac{T}{X+ \frac{1}{\alpha}} \right)$$ $$x = \sqrt{\left((X + \frac{1}{\alpha}\right)^2 - T^2} - \frac{1}{\alpha}$$

You might check out Misner, Thorne, Wheeler, "Gravitation", chapter 6, "Accelerated observers".

MTW's treatment starts with finding T as a function of (t,x) and X as a function of (t,x), if I'm understanding your notation correctly.

You write
$$\bar{x} = f(x,t) \quad \bar{t} = g(x,t)$$
defining the new coordinates in terms of the old. But it's easier if you look at the inverse, the old coordinates defined in terms of the new

$$x = p(\bar{x}, \bar{t}) \quad t=q(\bar{x}, \bar{t})$$Let
$$p_x \frac{\partial p}{\partial x} \quad p_t = \frac{\partial{p}}{\partial t} $$
and similarly for $q_x$ and $q_t$.

Then you can just use algebra to find the line element in the new metric

$$dx = p_x d\bar{x} + p_t d\bar{t}\quad dt = q_x d\bar{x} + q_t \bar{t}$$

Then, modulo the sign convention

$$ds^2 = -dt^2 + dx^2 = \left( q_x d\bar{x} + q_t d\bar{t} \right)^2 - \left( p_x d\bar{x} + p_t d\bar{t} \right)^2$$

I'm rushing this a bit, hope I haven't made too many typos.

 

[etotheipi]

Thanks! I wondered, if this should be

$$dx = p_x d\bar{x} + p_t d\bar{t}\quad dt = q_x d\bar{x} + q_t \bar{t}$$

instead$$dx = p_{\bar{x}} d\bar{x} + p_{\bar{t}} d\bar{t}, \quad dt = q_{\bar{x}} d\bar{x} + q_{\bar{t}} d\bar{t}$$with the partial derivatives w.r.t. the barred coordinates. But in any case, that's a nicer method than my way of working out the metric components individually!

[N.B. I gave up with the constant coordinate acceleration version of the problem, since the algebra isn't nice. But I think I'll be able to make better progress with the problem statement reworded to include constant proper acceleration, instead.]

 

[PeterDonis]

You might check out Misner, Thorne, Wheeler, "Gravitation", chapter 6, "Accelerated observers".

By "accelerated" they mean proper acceleration, not coordinate acceleration.

I'm not aware of any textbook that treats the case where the observer's motion is defined by coordinate acceleration as opposed to proper acceleration.

 

[pervect]

Thanks! I wondered, if this should be
instead$$dx = p_{\bar{x}} d\bar{x} + p_{\bar{t}} d\bar{t}, \quad dt = q_{\bar{x}} d\bar{x} + q_{\bar{t}} d\bar{t}$$with the partial derivatives w.r.t. the barred coordinates. But in any case, that's a nicer method than my way of working out the metric components individually!

[N.B. I gave up with the constant coordinate acceleration version of the problem, since the algebra isn't nice. But I think I'll be able to make better progress with the problem statement reworded to include constant proper acceleration, instead.]

Yeah, that's what I meant :).

The Rindler transformation, which I suspect is the same as what you worked out, gives in MTW's notation

$$x^0 = \left(a^{-1} + \xi^1 \right) \sinh (a x^0 )\quad x^1 = \left(a^{-1} + \xi^1 \right) \cosh (a \xi^0)$$

The associated line elements are the cartesian coordinates $(x^0, x^1)$ with a line element of
$$ds^2 = -(dx^0)^2 + (dx^1)^2$$

and the Rindler coordinates $(\xi^0, \xi^1)$ with the line element
$$ds^2= \left(-1 + a \xi^1 \right) ^2 (d \xi^0)^2 + (d \xi^1)^2$$

You'll often see them written with a change of origin as

$$ds^2 = -\alpha^2 (\xi^1)^2 (d \xi^0)^2 + (d \xi^1)^2)$$

Wiki, for instance, writes https://en.wikipedia.org/w/index.php?title=Rindler_coordinates&oldid=987507644
writes
$$(-\alpha x)^2 dt^2 + dx^2$$

along with a couple of other versions, including a radar coordinate version.The above form sets the origin at (t=0, x=1). Wiki has three different commonly used line elements, one of which is the radar coordinates

 

[pervect]

By "accelerated" they mean proper acceleration, not coordinate acceleration.

I'm not aware of any textbook that treats the case where the observer's motion is defined by coordinate acceleration as opposed to proper acceleration.

Yes, the OP has already mentioned that they are aware of this:

Yeah, I zoned out a bit and I didn't realize until after I wrote it that I'd attempted to derive the metric components for constant proper acceleration, and not constant coordinate acceleration.

 

[etotheipi]

@pervect I get nearly the same as your line element... taking the differentials of the coordinate transforms you wrote in #17,$$dx^0 = (1+a\xi^1) \cosh{(a\xi^0)} d\xi^0 + \sinh{(a\xi^0)}d\xi^1$$ $$dx^1 = (1+a\xi^1) \sinh{(a\xi^0)} d\xi^0 + \cosh{(a\xi^0)}d\xi^1$$It follows that $$\begin{align*}

ds^2 = -(dx^0)^2 + (dx^1)^2 &= -(1+a\xi^1)^2 \left[\cosh^2{(a\xi^0)} - \sinh^2{(a\xi^0)}\right] (d\xi^0)^2 + \left[

\cosh^2{(a\xi^0)} - \sinh^2{(a\xi^0)}
\right](d\xi^1)^2 \\

&= -(1+a\xi^1)^2(d\xi^0)^2 + (d\xi^1)^2

\end{align*}$$But I think we can now go ahead with integrating the proper time over $A$'s trajectory in the $(\xi^0, \xi^1)$ coordinates, using that $(d\tau)^2 = -(ds)^2$ with our choice of signature.

Thanks @Ibix, @PeterDonis and @pervect!