Throwing and catching a clock in general relativity

Question

I'm new to this Q&A site, so sorry if my formulas don't look great!

I'm trying to create a function that shows the time on a clock B that's been thrown up from and caught in position A, with respect to the mass of the object it's on (the function would be TB(m)). According to a clock A in position A, clock B has been in the air for 10s, but due to GR, the time shown on clock B should be shorter (how much shorter depends on the mass). Seen as 10s isn't very long, I figured that even if position A is on a mass with a very high curvature, the velocity of the object wouldn't be enough for SR to make much of a difference, so I'm just calculating using GR for the moment (although if SR is important then please let me know!).

So far I've taken the Schwarzschild metric for clock A and B

$$TF=\frac{TA}{\sqrt {1-\frac{2Gm}{rc^2}}}$$

$$TF=\frac{TB} {\sqrt{1-\frac{2Gm}{(r+h)c^2}}}$$

equated them and solved for TB

$$TB(m)=TA\sqrt {\frac {r(c^2(r+h)-2Gm)}{(r+h)(rc^2-2Gm)}}$$

Where r is the radius of the object (which I'm keeping constant), h is the height of clock B, TF is the time elapsed on a clock 'far away', TA is the time elapsed on clock A and TB is the time elapsed on clock B.

However this is only for if the clock were always at one height h, so if I integrate it from 0 to hmax (the height at which clock B starts coming down) with respect to h, and times it by 2 (because the clock has to come down too), then that should equal TB(m).

$$TB(m)= 2\cdot \int_0^{hmax} TB(h)$$

This would work if h increased and decreased constantly, however it doesn't because gravity is acting on the clock.

This is how far I've got with my problem. Would I be right in saying that to account for gravity I need to multiply TB(h) by the inverse of the h-t function of clock B?

Aside from the calculations I've done- am I even going about this the right way? I'm about to start my final year of school, so I'm trying to find a way to work with general relativity that isn't way over my head, and the Schwarzschild formula seems like something much more manageable than most other calculations I'm seeing related to GR. If there's a better way of calculating this though, then I'm more than happy go down another route if it isn't too complex!

Thanks in advance!

Answer 1

I think it is a bit messier than you would like. You cannot ignore the "Special Relativity part" because the time dilation caused by your launch speed will be of a similar order of magnitude to the gravitational effects you consider in your question.

You need to adopt a more "holistic" approach and start with the proper time interval in the Schwarzschild metric for a radial trajectory. $$c^2 d\tau^2 = \left(1 - \frac{r_s}{r}\right) c^2\ dt^2 - \left(1 - \frac{r_s}{r}\right)^{-1}\ dr^2 $$

$\tau$ here is the proper time measured on a clock and $r_s = 2Gm/c^2$. For clock A then $dr=0$ and the expression reduces to the one you have used. Note that, by convention, expressions like $dr^2$ mean $(dr)^2$.

For clock B \begin{eqnarray} c\tau_B & = & \int \sqrt{ \left(1 - \frac{r_s}{r}\right) c^2\ dt^2 - \left(1 - \frac{r_s}{r}\right)^{-1}\ dr^2} \\ & = & \int \sqrt{ \left(1 - \frac{r_s}{r}\right) c^2\ \left(\frac{dt}{dr}\right)^2 - \left(1 - \frac{r_s}{r}\right)^{-1}}\ \ \, dr \end{eqnarray}

where the integral runs from $r_A$ to $r_A+h$ and then back again.

From there you need an expression for $(dr/dt)^2$ for your ballistic object, which is the same as that for an object released from rest in the Schwarzschild metric at $r_A+h$, where the negative root would correspond to inward motion. $$ \left( \frac{dr}{dt}\right)^2 = \left(1 - \frac{r_s}{r_A+h}\right)^{-1}\left(1 - \frac{r_s}{r}\right)^2 \left( \frac{r_s}{r} - \frac{r_s}{r_A+h}\right)\ . $$

Note that if this is supposed to be a problem for something on Earth then you have the added complexity that everything is rotating with the Earth, although that can be neglected to first order at rotation speeds typical of the Earth's surface.

Answer 2

I'm trying to create a function that shows the time on a clock B that's been thrown up from and caught in position A, with respect to the mass of the object it's on (the function would be TB(m)). According to a clock A in position A, clock B has been in the air for 10s, but due to GR, the time shown on clock B should be shorter (how much shorter depends on the mass) ...

The answer by @ProfRob uses the Schwarzschild metric for the spacetime outside a spherical mass, which is the "right" way to go. But as he showed it did not lead to a closed-form expression for the quantity you want. However, since you are only talking about 10 s we can probably neglect tidal gravity and instead use the Rindler metric, which is not the "right" way to go, but is an approximation to the Schwarzschild metric that should work here. It does result in a closed-form expression.

First, we start with the metric in Rindler coordinates: $$ ds^2 = -c^2 d\tau^2 = -\frac{g^2}{c^2} x^2 \ dt^2 + dx^2 + dy^2 + dz^2 $$ where $g$ is the gravitational acceleration at the starting point $x_0=c^2/g$ and the acceleration is along the $x$ direction. For convenience we can set $y$ and $z$ to some constants and suppress them from now on.

This gives us the Lagrangian $$ \mathcal{L} = -\frac{g^2}{c^2}x^2 \dot t^2 + \dot x^2 $$ which does not contain $t$ so we get a conserved quantity $$ C_t=\frac{2 g^2}{c^2}x^2 \dot t$$ We can determine this conserved quantity by requiring that the proper time for clocks at rest at $x_0$ is equal to the coordinate time, meaning $\dot t=1$ at $x=x_0$. Then we get $C_t=2c^2$. Substituting this in to the expression above and solving for $\dot t$ gives us $$\dot t = \frac{c^4}{g^2 x^2}$$

Now, we calculate the Euler Lagrange equation for $x$ and substitute in the above expression for $\dot t$ to get $$-\frac{2c^6}{g^2 x^3}-2 \ddot x = 0$$

This differential equation can be solved, with initial conditions of $x(0)=x_0$ and $\dot x(0)=v_0$ to get $$x(\tau)=-\frac{\sqrt{(c^2-cg\tau+gv_0\tau)(c^2+g \ (c+v_0)\tau))}}{g}$$ This is the $x$ location of a clock thrown upwards from $x_0$ at a speed of $v_0$ as a function of the proper time $\tau$ displayed on that clock. So, we can set $x(\tau)=x_0$ and solve for $\tau$ to find the final time on the thrown clock when it is caught back at $x_0$ when we do that we get $$\tau_f = \frac{2 c^2 v_0}{g\ (c^2-v_0^2)}$$

Now, we need to find the time on the clock that does not get thrown. Since that clock remains always at rest at $x_0$ the proper time for that clock is equal to the coordinate time (recall that is how we set $C_t$). So we simply need to find the coordinate time for when the thrown clock returns in order to find the proper time for the clock that stayed. We can do this simply by integrating the expression above that we already found for $\dot t$. This gives us $$t(\tau)=\frac{c}{g} \tanh^{-1}\left( \frac{v_0}{c} \right) + \frac{c}{g} \tanh^{-1}\left( \frac{-c^2 v_0+c^2 g \tau - g v_0^2 \tau}{c^3} \right) $$ $$t(\tau_f)=\frac{2c}{g}\tanh^{-1}\left( \frac{v_0}{c} \right) $$

Finally, our difference between the two clocks is simply the difference $$\Delta \tau = \tau_f - t(\tau_f)$$ If we assume, $v_0=50 \mathrm{\ m/s}$, $g=10 \mathrm{\ m/s^2}$ and $c=299792458 \mathrm{\ m/s}$ then we get that $\Delta \tau = 1.85 \ 10^{-13}\mathrm{\ s}$