# Dropping a ball in a train moving close to the speed of light?

Suppose a train is moving very close to the speed of light, say 0.999c relative to a stationary observer on Earth.

Now a stationary observer on Earth will observe clocks on the train to tick slower than usual.

Now suppose a boy within the train drops a ball.

The stationary observer measures how long it takes for the ball to hit the ground. Will the result simply be $\sqrt{\frac{2s}{g}}$ or will the result be a greater number due to time dilation?

• I guess this is meant to be a simplified version of your other question? This is just my opinion, but I think you cut out too much. The way you've written this question (v1), the answer is pretty obvious (of course time dilation applies). If you included some argument to think otherwise - i.e. some of the seemingly paradoxical material from the other post, or at least some reason you have to think time dilation might not apply - I think it could be a much more interesting question. Commented Apr 30, 2014 at 3:30
• @DavidZ The reason he's re-asked in this formulation is me and my big mouth. And now, I'm forced to think carefully again. Stay tuned. Commented Apr 30, 2014 at 4:11
• Ah, yes, asking is complicated sometimes ;-) Well I do think there's a good question in all this somewhere, and that the earlier version may have been closer to it. Commented Apr 30, 2014 at 4:21
• Here's equation that describes time dilation $\frac{t_0}{t_v} = \gamma$, this equation means that time in stationary frame(in this case earth) divided by time in moving frame of reference (in this case time passed in train) equals $\gamma$(Lorentz Factor), So know $t_v$ and you know that $\gamma$ is always bigger than 1 so it will take more time for someone on earth.
– user21420
Commented Apr 30, 2014 at 4:43
• @DavidZ, if a moving ball takes longer to fall than a stationary one, as Gigi says, then this implies that a moving elevator is not equivalent to a gravitation field, since both the moving ball and the stationary one will hit together in a moving elevator, but not in a gravitational field. Commented Apr 30, 2014 at 4:57

Rely on the invariance of the interval $\tau$ between the ball being dropped and the ball hitting the floor. Also using $c=1$ units to keep the typing down.

The moving frame is un-primed and the stationary frame primed.

$$t^2 - h^2 = \tau^2 = {t'}^2 - [{h'}^2 + (vt')^2] \,.$$

Because the velocity $v$ is horizontal both parties measure the same $h$, allowing us to write

\begin{align*} {t'}^2 - [h^2 + (vt')^2] - \tau^2 &= 0 \\ {t'}^2(1 - v^2) - h^2 - \tau^2 &= 0 \\ {t'}^2(1 - v^2) - h^2 - (t^2 - h^2) &= 0 \\ {t'}^2(1 - v^2) - t^2 &= 0 \\ t' = \frac{t}{\sqrt{1 - v^2}} \,, \end{align*}

so I owe Mew an apology for my comments in the earlier thread, but it does not violate the equivalence principle because we didn't use any information about the value of $h$ in the above work, so it works fine when $h=0$ as in the elevator case.

• $t$ is time and $h$ is height? It looks like you are subtracting meters (squared) from seconds (squared) in your first equation, or I missed something? Commented Apr 30, 2014 at 7:05
• @brightmagus: the time is being multiplied by $c$ to convert it to a distance, but dmckee is using units in which $c = 1$. This is commonly done in relativity to avoid factors of $c$ complicating the equations. Commented Apr 30, 2014 at 8:28
• @dmckee, I'm happy to acknowledge time dilation occurs here, but I disagree that this is applicable to the elevator case. I have re-posed the question in a better way, with pictures to come soon here: physics.stackexchange.com/questions/110606/… Commented Apr 30, 2014 at 8:30
• John Rennie: "This is commonly done in relativity to avoid factors of c complicating the equations." Is this why dmckee's answer has no $c$, which is kind of strange? Commented Apr 30, 2014 at 8:49
• and which means that for a simple case (with a ball in the train moving faster than $1 m/s$) we get in the denominator a root of a negative number expressed in what? $m/s$ or ...? Which, ultimately, gives us time expressed in $s^2/m$ or ...? I'm asking, because I don't know how to interpret this. Commented Apr 30, 2014 at 14:11

The stationary observer measures how long it takes for the ball to hit the ground. Will the result simply be $\sqrt{\frac{2s}{g}}$ or will the result be a greater number due to time dilation?

Of course, the the result will be time dilated and greater than $\sqrt{\frac{2s}{g}}$. It must be dilated if SR is a valid theory. You can simply consider the movement of the ball a kind of clock.

EDIT: The situation is equivalent to a similar problem. The boy is holding the ball in his hand. He lets it drop, bounce off the ground and come back to his hand (presuming it would return to the same height). Let's assume we cannot see the very ball (so we do not worry about the vertical movement), but we can see the boy open his hand (letting the ball drop), and then close his hand (to catch the ball when it is back).

Will the movements of his hand be time-dilated? Of course they will - according to SR - like everything else he does.

• I turned my above comment into a new question here: physics.stackexchange.com/questions/110606/…, which is hopefully more clear. Commented Apr 30, 2014 at 7:45
• OK, I'll take a look. See my EDIT in the meantime. Commented Apr 30, 2014 at 7:54