I'm reading Hartle, Gravity and read bout the Twin Paradox. I understand why there isn't really a contradiction, one frame is inertial, the other isn't. However, I'd like to know how to calculate time dilation when neither frame is inertial.

One twin is on a massive planet. The other is synchronizes clocks with the first twin, and then gets on a rocket goes off into space, escapes gravity, then comes back. Which twin is older when they both meet up again?

My first guess is to say that the one who didn't go anywhere will be oldest, as in the standard twin paradox. However, I've watched Interstellar and certainly know that gravitational time dilation is a thing too. So my guess is that if the gravitational potential is less than some critical value compared to the speed of the interstellar twin, the first twin will be older.

How do you calculate the time each twin feels has passed, and the amount they think the other twin has experienced, during this scenario up to when they reunite?

For simplicity the space faring twin travels up and then back down, relative to the first twin. $y(t)=\lambda \cdot t \cdot (\beta-t)$ is the height as observed by the first twin at their relative time. The gravitational potential as a function of $y$ is $G_p=-\cfrac{G \cdot M \cdot m}{r}$. If I ignore gravity, I get,

$$\tau=\int_{0}^{\beta} \sqrt{1-\cfrac{\lambda^2 \cdot (\beta-2 \cdot t)^2}{c^2}} dt=\cfrac{\sin^{-1}(\cfrac{\beta \cdot \lambda}{c}) \cdot c^2+\beta \cdot \lambda \cdot \sqrt{c^2-\beta^2 \cdot \lambda^2}}{2 \cdot c \cdot \lambda}$$

How do I adjust for gravity? In addition, and what I'm most interested in, how do I calculate all of this from the space faring twin's perspective?

  • $\begingroup$ If you assume that the spacefaring twin is freely falling under the influence of gravity for the whole time of his/her trip, there's actually a very simple answer to which twin will be older: The spacefaring twin is traveling along a geodesic in spacetime, and the proper time between any two timelike-separated events is maximized along the geodesic that connects them. Thus, the spacefaring twin will be older. $\endgroup$ – Michael Seifert Mar 4 '16 at 20:51
  • $\begingroup$ @Michael Yes, but the space faring twin is in a rocket, that works...and the other twin is on the surface of a massive planet. $\endgroup$ – Zach466920 Mar 4 '16 at 21:08

First choose your coordinate system. For simplicity we'll choose the rest frame of the stationary twin, and we'll assume the stationary twin is floating freely in space. This is an important assumption because it means that the metric for the stationary twin is the Minkowski metric. You'll see why this matters in a moment.

In the coordinate system of the stationary twin the velocity of the accelerating twin is some function of time, $v(t)$, where $v$ is the velocity in the stationary twin's coordinates and $t$ is the time in the stationary twin's coordinates.

Now, if the accelerating twin moves a distance $dx$ in a time $dt$ then the proper time is given by:

$$ c^2d\tau^2 = c^2dt^2 - dx^2 $$

This is where the Minkowski metric comes in, because this equation is just the Minkowksi metric where we've assumed the accelerating twin is moving on the $x$ axis so $dy = dz = 0$. The coordinate velocity $v$ is just $dx/dt$, so $dx = vdt$. If we substitute for $dx$ and rearrange we get:

$$ d\tau = \sqrt{1 - \frac{v(t)}{c^2}}dt $$

And to get the proper time for the accelerating twin we just integrate this:

$$ \Delta\tau = \int_{t_i}^{t_f}\,\sqrt{1 - \frac{v(t)}{c^2}}\,dt $$

where $t_i$ is the coordinate time the twin leaves and $t_f$ is the coordinate time the twin returns. The velocity $v(t)$ can be any arbitrary function of time, but it should be obvious that $\Delta\tau \lt t_f - t_i$ i.e. the accelerating twin's clock registers less time that the stationary twin's clock.

So far so good, this is all very straightforward, but when the stationary twin is in a gravity well their metric isn't Minkowski and that makes the calculation a lot harder. I'd probably do the calculation using the Schwarzschild coordinate frame and calculate the proper times for the two twins in this frame. That assumes you know the velocity $v(t)$ for the accelerating twin in the Schwarzschild frame, which may not be simple to calculate.

Finally you ask how to do the calculation in the accelerating twin's frame, and that's even harder. If the accelerating twin uses a constant proper acceleration, i.e. leaves at velocity $v$ and returns at velocity $-v$ while keeping a constant acceleration in between, then the accelerating twin's metric is the Rindler metric:

$$ c^2d\tau^2 = \left(1 + \frac{a}{c^2}x \right)^2 c^2 dt^2 - dx^2 $$

So you need to write $v(t)$ in the Rindler coordinates then integrate to calculate $\tau$. I have to confess that I've never done this so I don't know what the answer is.

  • $\begingroup$ I didn't think to incoroprorate the accelerations into the metric. It makes a lot of sense though, thanks! $\endgroup$ – Zach466920 Mar 4 '16 at 19:39

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