Geodesics in a constant gravitational field

1. Nov 25, 2012

tom.stoer

I want to interpret geodesics in a constant gravitational field.

$$ds^2 = -dT^2 + dX^2 + dY^2 + dZ^2$$

with a geodesic (in terms of coordinate time T)

$$X^\mu(T) = (T, X=A, 0, vT)$$

where A is an arbitrary constant and v ≤ c.

Now I transform this geodesic to Rindler coordinates. These coordinates can be understood as the effect of a constant gravitational field due to a constant acceleration g. However there is the problem that they still describe vacuum spacetime, so they do not provide a gravitational field produced by any non-zero mass distribution (but this is a problem I want to address later).

$$t = \frac{1}{g}\text{artanh}\left(\frac{T}{X} \right)$$
$$x = \sqrt{X^2 - T^2}$$
$$y = Y$$
$$z = Z$$

So for the geodesic we find

$$x^\mu(T) = \left(\frac{1}{g}\text{artanh}\left(\frac{T}{A} \right), \sqrt{A^2-T^2}, 0, vT\right)$$

still expressed in terms of T.

Now from x(T) we see that the yz-plane crosses the geodesic with constant X(T)=A at T=A; this can be interpreted as a particle falling in a gravitational field from X=A to the moving yz-plane in coordinate time T=A.

But in Rindler coordinates we find

$$t(T\to A-) = \frac{1}{g}\,\lim_{T\to A-}\,\text{artanh}\left(\frac{T}{A} \right) \to \infty$$

So measured in Rindler time t the particle will never cross the yz-plane.

The reason why I work with coordinate time t and T instead of proper time is that I want to include light-like trajectories for v=1. The problem is that I do not see any reference frame S' for which I get a reasonable expression for the coordinate time interval Δt' where the particle falls from the initial position at x'(X=A) to the y'z'-plane with coordinate x'=0.

Remark: what I am really interested in is not the Rindler spacetime but a non-vacuum spacetime produced by a mass distribution with a gravitational field in x-direction which is approximately homogeneous for a region [A-h,A+h]. But first I have to understand the Rindler case.

Last edited: Nov 25, 2012
2. Nov 25, 2012

tom.stoer

I forgot to mention that I doubt my reasoning regarding the particle never crossing the yz-plane in Rindler coordinates; something must be wroing with my reasoning, that's why I am posting this calculation

3. Nov 25, 2012

martinbn

Why should there be anything wrong? Isn't this the same as the test object never (t->infinity) crosses the black hole horizon.

4. Nov 25, 2012

tom.stoer

OK, perhaps I stopped my calculation too early. I will express the geodesic in terms of the Rindler coordinate x, i.e.

$$x^\mu(x) = \left(\frac{1}{g}\text{artanh} \sqrt{1-\frac{x^2}{A^2}}, x, 0, vA\sqrt{1-\frac{x^2}{A^2}}\right)$$

So I can calculate the Rindler time t for any x < |A|. Using this expression I can calculate the Rindler time it takes for a particle to fall from X=A to X=A-h.

$$t(h) = \frac{1}{g}\text{artanh} \sqrt{1-\frac{(A-h)^2}{A^2}} \simeq \frac{1}{g}\text{artanh} \sqrt{\frac{2h}{A}} \simeq \sqrt{\frac{2h}{Ag^2}}$$

which is valid for small h/A.

What is interesting is that this holds independent from the motion in Z-direction, i.e. it is valid for all Z(T) = vT; this is not to be expedcted for general spacetimes - and this is what I would really like to investigate!

So all this was just a warm-up ;-)

5. Nov 25, 2012

pervect

Staff Emeritus
Note that the Rindler coordinate system has an event horizon at z = -c^2/g. It's rather similar to the event horizon of a black hole.

Because of this, one expects it to take an infinite amount of Rindler time to cross the RIndler horizon, while the inertial observer and the falling observer see a finite amount of time.

6. Nov 25, 2012

DrGreg

Why's that so surprising? The time it takes to fall a small vertical distance h is independent of horizontal velocity v. Isn't that exactly what you'd expect?

(With your axes, x is "vertical" and y and z are "horizontal".)

Actually with the axes and origin used here it's at x=0.

7. Nov 25, 2012

tom.stoer

In Newtonian gravity you are right, but there are counterexamples in GR.

In the Rindler case a photon with vanishing x-velocity falls identically fast in x-direction as a massive test particle with vanishing initial x-velocity - OK.

But in Schwarzschild geometry there is a photon orbit at 3/2 rs where photons do not fall at all, i.e. dr/dt = 0, which cannot be true for any massive particle with vanishing initial r-velocity.

And there are spacetimes generated by infinite sheets or branes where the x-component of the geodesic equation looks like

$$\frac{d^2x}{dt^2} = a(x) + b(x,v_\perp)$$

(y and z are trivial)

So in these spacetimes photons and massive bodies cannot fall equally fast b/c for the photon the perpendicular velocity must never vanish (for dx/dt = 0 as initial condition) and therefore the correction term b will in general be non-zero.

http://arxiv.org/abs/0708.2906v1

That means that Einstein's Gedankenexperiment of a photon and a massive body in a box in a constant gravitational field = in a box with constant acceleration is a very special case (the Rindler solution) and that the main idea - that all bodies fall equally fast in a (nearly) homogeneous gravitational field - does not hold in general.

Last edited: Nov 25, 2012
8. Nov 25, 2012

bcrowell

Staff Emeritus
GR doesn't actually have anything that corresponds 100% correctly to what we would like for a uniform gravitational field:

http://www.lightandmatter.com/html_books/genrel/ch07/ch07.html#Section7.4 [Broken]

Minkowski space in Rindler coordinates is one possibility, but it has some unsatisfactory properties. Another candidate is the Petrov metric.

Last edited by a moderator: May 6, 2017
9. Nov 25, 2012

tom.stoer

I agree.

Intriguing

Interesting; I have never seen this metric before.

My observation was not so much related to a uniform field but to the fact that it seems to be possible to introduce yz-velocity which affects x-acceleration - even in rather simple situations.

Last edited by a moderator: May 6, 2017
10. Nov 26, 2012

tom.stoer

A simple starting point would be the geodesic equation:

$$\dot{u}^\mu + \Gamma^\mu_{\rho\sigma} = 0$$

Now I am interested in a "gravitational force in z-direction", w/o any force in x- or y-direction.

Would it be reasonable to set several components of the connetcion to zero ensuring that perpendicular velocities in x-y do not influence the "force" in z? And would it be possible to reconstruct the metric based on the connection?

11. Nov 26, 2012

grav-universe

I worked on something similar a while back and this is what I found.

By the equivalence principle, a constant "gravitational" field would be the same as with constant acceleration, so we can just use SR for an accelerating observer. With Rindler or for a ship of observers accelerating with Born rigidity, however, there is still some ambiguity about distances over the length of the ship during acceleration. Distances and times generally require two observers at different locations to measure, but we can still determine some things about what a single observer with constant acceleration 'a' might infer using only their own clock while also assuming symmetry between ascent and descent of a particle.

Rather than just letting the particle "fall" away from an accelerating observer, the accelerating observer will propel it forward at a locally measured speed u', measured over an infinitely small length and time, which we assume can be performed in the same way that light is measured at c locally over infinitesimals in GR, for instance. From there, we just calculate when the accelerating observer will catch up to the particle and assuming symmetry between ascent and descent of the particle, we can determine some aspects about the distance and time it takes the particle to fall in a constant gravitational field. One can imagine this as simply having an observer on Earth, for example, toss a ball into the air and having it fall back to them. By assigning a coordinate system that assumes this symmetry, we can cut the time in half to find the time for the particle to gain this speed in a uniform gravitational field while also assuming that a particle simply falling away from the accelerating observer would gain that same speed in this time as well.

Okay, so if we are to assume symmetry, we should first verify that an accelerating observer will receive the particle at the same speed at which it was emitted. From the perspective of an inertial observer, let's say that the accelerating observer is measured to be travelling at an instantaneous speed v1 upon emitting the particle and with a speed v2 upon catching up to the particle, and the particle will be measured to be travelling inertially with a speed u. The particle is emitted at t1 and received at t2 according to the inertial observer, and the distance both the accelerating observer and the particle travels during this difference in time must be the same, since the accelerating observer and particle coincide at both of these two times, so

u (t2 - t1) = (c^2/a) [sqrt(1 + (a t2 / c)^2) - sqrt(1 + (a t1 / c^2)]

where (v/c) = (a t / c) / sqrt(1 + (a t / c)^2) for v = 0 at t = 0, so by re-arranging we get a t / c = (v/c) / sqrt(1 - (v/c)^2), giving

(u/c) [(v2/c) / sqrt(1 - (v2/c)^2) - (v1/c) / sqrt(1 - (v1/c)^2)] = 1/sqrt(1-(v2/c)^2) - 1 / sqrt(1 - (v1/c)^2)

v1/c = (u/c - u'/c) / (1 - (u /c) (u'/c))

sqrt(1 - (v1/c)^2) = sqrt[1 - (u/c - u'/c)^2 / (1 - (u/c) (u'/c))^2]

= sqrt[(1 - u u' / c^2) - (u/c - u'/c)^2] / (1 - (u/c) (u'/c))

= sqrt[1 + u^2 u'^2 / c^4 - (u/c)^2 - (u'/c)^2] / (1 - (u/c) (u'/c))

= sqrt(1 - (u/c)^2) sqrt(1 - (u'/c)^2) / (1 - (u/c) (u'/c))

From which we can find

(v1/c) / sqrt(1 - (v1/c)^2) = (u/c - u'/c) / (sqrt(1 - (u/c)^2) sqrt(1 - (u'/c)^2))

And substituting from before,

(u/c) [(v2/c) / sqrt(1 - (v2/c)^2) - (u/c - u'/c) / (sqrt(1 - (u/c)^2) sqrt(1 - (u'/c)^2))] = 1/sqrt(1 - (v2/c)^2) - (1 - u u' / c^2) / (sqrt(1 - (u/c)^2) sqrt(1 - (u'/c)^2))

(1 - (u/c) (v2/c)) / sqrt(1 - (v2/c)^2) = [(1 - u u' / c^2) - (u/c - u'/c) (u/c)] / (sqrt(1 - (u/c)^2) sqrt(1 - (u'/c)^2))

(1 - (u/c) (v2/c)) / sqrt(1 - (v2/c)^2) = sqrt(1 - (u/c)^2) / sqrt(1 - (u'/c)^2)

And squaring both sides and multiplying by the denominators,

[1 - 2 (u/c) (v2/c) + (u/c)^2 (v2/c)^2] (1 - (u'/c)^2) = (1 - (u/c)^2) (1 - (v2/c)^2)

(u'/c)^2 (1 - (u/c) (v2/c))^2 = (v2/c - u/c)^2

u'/c = (v2/c - u/c) / (1 - (u/c) (v2/c))

So one can see so far that the particle will be measured locally to be received at the same speed that it was emitted.

Now let's find the time that the accelerating observer will measure for the particle to ascend and then descend back to be received at a locally measured speed u'. One can see that regardless of whatever inertial frame observes these two events of emitting and receiving the particle, the times upon the accelerating observer's clock at each event must be an invariant for a particular locally measured speed u' (locally referring to the accelerating observer), so this time for simplicity, we will just have the accelerating observer accelerate from rest in the inertial observer's frame, making v1 = 0 at t1 = 0, and so v2 and t2 according to the inertial observer are just v and t, and u' = u as well in this case. The time that passes upon the accelerating observer's clock, then, is

t'_u = (c/a) log[sqrt(1 + a t / c) + a t / c]

= (c/a) log[1 / sqrt(1 - (v/c)^2) + (v/c) / sqrt(1 - (v/c)^2)]

And addition of speeds again from the earlier result upon receiving the particle,

u'/c = (v/c - u/c) / (1 - (v/c) (u/c)) , where u' = u upon emission, so also upon reception, giving

v/c= (2 u/c) / (1 + (u/c)^2)

1/sqrt(1 - (v/c)^2) = (1 + (u/c)^2) / sqrt[(1 + (u/c)^2)^2 - (2 u/c)^2]

= (1 + (u/c)^2) / (1 - (u/c)^2)

a t / c = (v/c) / sqrt(1 - (v/c)^2) = (2 u/c) / (1 - (u/c)^2)

And substituting,

t'_u = (c/a) log[(1 + (u/c)^2) / (1 - (u/c)^2) + (2 u/c) / (1 - (u/c)^2)]

= (c/a) log[(1 + u/c)^2 / (1 - (u/c)^2)]

= (c/a) log[(1 + u/c) / (1 - u/c)]

So now we have the time that the accelerating observer says it will take for the particle to coordinately accelerate to a speed u (u' = u), which is half of the ascend and return time, so

t' = (c/a) log[(1 + u/c) / (1 - u/c)] / 2

Vice versely, the speed the particle will achieve after a time t' according to the accelerating observer, then, is

e^(2 a t' / c) = (1 + u/c) / (1 - u/c)

e^(2 a t' / c) - (u/c) e^(2 a t' / c) = 1 + u/c

u/c = [e^(2 a t' / c) - 1] / [e^(2 a t' / c) + 1]

From this coordinate system, the only conditions for which is assuming symmetry between ascending and descending of the particle and that these measurements can be applied anywhere within a constant and continuous field, we can also gain the distance the accelerating observer infers the particle to have travelled in a time t' starting from rest at zero time with

dr' = u dt' using the instantaneous speed u at time t',

r' = int u dt'

= int c [e^(2 a t' / c) - 1] dt' / [e^(2 a t' / c) + 1]

= (c^2/a) log[-2 (e^(2 a t' / c) + 1)] - c t' - (c^2/a) log[-4]

= (c^2/a) log[(e^(2 a t' / c) + 1) / 2] - c t'

Last edited: Nov 26, 2012
12. Nov 26, 2012

grav-universe

Oh, looking back at earlier posts, for shorter trigonometric solutions (and a shorter post :) ), we would have

t' = (c/a) arctanh(u'/c)

u'/c = tanh(a t' / c)

r' = (c^2 / a) log[(e^(2 a t' / c) + 1) / 2] - c t'

t' = (c/a) log[e^(a r' / c^2) + sqrt(e^(2 a r' / c^2) - 1)]

Last edited: Nov 26, 2012
13. Nov 26, 2012

grav-universe

By re-arranging the above, we can also find the speed of the particle after falling some distance with

a r' / c^2 = log[1 / sqrt(1 - (u'/c)^2)]

u'/c = sqrt[1 - 1 / e^(2 a r / c^2)]

Also, the time that passes upon a freefalling clock in relation to the time upon the accelerating observer's clock is

u'/c = tanh(a t' / c)

sqrt(1 - (u'/c)^2) = sqrt[(e^(2 a t' / c) + 1)^2 - (e^(2 a t' / c) - 1)^2] / (e^(2 a t' / c) + 1)

= 2 e^(a t' / c) / (e^(2 a t' / c) + 1)

d_tau = dt' sqrt(1 - (u'/c)^2) = dt' 2 e^(a t' / c) / (e^(2 a t' / c) + 1)

using the instantaneous speed u' at time t', giving

tau = (c/a) [2 atan(e^(a t' / c)) - pi / 2]

Okay, that's all for now :)

Last edited: Nov 26, 2012