# Time dilation in a moving frame of reference

According to the theory of relativity, time slows down in a moving frame of reference. But I am confused in the example given below.

Suppose there are two men $$A$$ and $$B$$. $$A$$ is standing in moving train ( say with speed 60% the speed of light) with a ball in his hand and $$B$$ is on the station. When $$A$$ drops the ball downward, it comes back to him in suppose 2 seconds. Now from $$B$$'s frame of reference, the ball would travel along the line as shown in the figure.

The ball will fall down as well as move forward. Since the height of the ball from the floor is not changed, the time it would take to hit the ground will be 1 second and at the same time it will move forward to point $$P$$ (as shown in fig), and in the return journey, it will again take 1 second and hence the total time of motion is 2 seconds which is the same as observed by $$A$$.

But we know that in a moving frame time slows down but in the above example, both the time is the same.

Fig. 1:

Fig. 2:

. How can we prove time dilation in this question? Please clarify my doubt. Use Lorentz transform after clarifying its importance in this question. Answer according to the question only ( don't use your own data)

• How did you take time taken to reach P as one second? – Krishna May 5 '20 at 9:13
• It is just like a dropped body and its speed in y direction is the same as observed by A . Only horizontal speed is the new one which doesn't affect its time of fall in the vertical direction. – user262060 May 5 '20 at 9:37
• Try imagining that the train is moving at near the speed of light. The ball's motion must be slowed down according to B, since the vertical and horizontal motion(near light speed) combined cannot be higher than the speed of light. When you assume the train is moving slowly, the time taken would indeed be close to the same – Philip.P May 5 '20 at 9:52
• You cannot do calculation with a speed of the ball. Speed of thr ball is not an ivarient quantity. But the speed of the light is. You also need to perform some numerical calculations to see the effect of time dilation. The situation that you are talking about is explanied in many SR or modern physics books. I would suggest you to look at them – Layla May 5 '20 at 9:58
• I think the title of this post should be edited to be more specific. Right now it may as well just be another tag on the post. – akozi May 5 '20 at 22:17

There is no law of physics that tells us that balls have to fall at the same rate among all inertial frames. It would obviously depend on the relationship between the inertial frame and the source of gravity. In the case of the inertial frame attached to the ground, the source is at rest with respect to the source of gravity. In the case of the inertial frame attached to the train, the source of gravity is moving. So, there is an asymmetry there--which tells us that we simply can't assume that balls have to fall at the same rate for both observers.

Once you remove this assumption, there is no paradox. The Lorentz transformations of acceleration tell us that acceleration in the direction perpendicular to the relative motion between two frames transform as $$a'=\frac{a}{\gamma^2}$$ where I chose the unprimed frame to be the frame of the train and the primed frame to be the ground frame. Since there is no length contraction in the direction perpendicular to the direction of relative motion, both the ground frame and the train frame will agree that the ball travels a vertical distance of, say, $$h$$. Since the motion will be constant acceleration motion in both frames (albeit with different accelerations), we can write $$\Delta t=\sqrt{\frac{2h}{a}}$$ $$\Delta t'=\sqrt{\frac{2h}{a'}}=\gamma\sqrt{\frac{2h}{a}}$$

Thus, we see that $$\Delta t' =\gamma \Delta t$$which tells us that as observed by the ground, the ball will take more time to hit the ground than it will take to do so as observed by the train. So, indeed you'd notice time dilation. The paradox arose because you simply assumed you wouldn't notice such an effect!

Clarification 1

Kindly notice that my answer is outside the context of general relativity where inertial frames have a nuanced meaning. In general relativity, my initial statement that balls don't have to fall at the same rate in inertial frames would not remain valid because inertial frames are the frames where gravitational acceleration is locally zero. So it'd be same in all inertial frames, namely, zero--but that would need a rewriting of the whole framework in which your question is posed. For example, we wouldn't be able to talk about Lorentz transformation between the train and the ground frame because neither would be an inertial frame given the fact that both experience gravitational acceleration locally.

Clarification 2

It should be strictly noticed that this analysis is only valid for small times after the release of the ball. Once the ball picks up a vertical velocity, the relationship between the accelerations in the two frames, $$a, a'$$ would change. Also, the acceleration due to gravity itself wouldn't remain $$g$$ in the train frame itself. See, Rindler coordinates for a discussion of how to describe constant acceleration motion over extended periods of time in special relativity.

• Thanks for your answer as well as comment. – user262060 May 10 '20 at 4:41
• @Aakankshasharma Glad it was helpful :) Kindly notice that my answer is outside the context of general relatively where inertial frames have a nuanced meaning. In general relativity, my initial statement that balls don't have to fall at same rate in inertial frames would not remain valid because inertial frames are the frames where gravitational acceleration is locally zero. So it'd be same in all inertial frames, namely, zero--but that would need a rewriting of the whole framework in which your question is posed. – Dvij D.C. May 10 '20 at 4:48

As others have pointed out, the impact of velocity on apparent time is really small on this scale. We can show the reason, however, if instead of a bouncing ball taking 1 second, we make the problem as follows:

Person A shoots a photon at light speed at a mirror on the floor 1.5 meters down. In his reference frame, the photon travels a total of 3 meters down and back.

The time is distance/velocity = 3 m/3e8 m/s = 1e-8 seconds.

Person B watches this experiment, but the train is going very fast (75% of light speed). So during the photon "bounce" the train moves sideways 1.125 meters while the photon falls 1.5 meters. By pythagorean theorem, Person B sees the total path length as: \begin{alignat}{7} p &~=~ 2 *\sqrt{1.125^2+1.5^2} = 3.75 \end{alignat} meters. Since Person B still sees the velocity as "c", he thinks the whole thing took a time of distance/velocity = 3.75 m/3e8 m/s = 1.25e-8 seconds.

Thus the time looks different to the two observers, because the path length looks different but the velocity looks the same.

• The photon also travelled the same vertical distance for both observers . So the time for vertical distance from both frame of reference should be same and for B the photon also travelled in horizontal direction in the same time . So again time should be the same. – user262060 May 6 '20 at 4:16
• @Aakanksha The total light path length in B's frame is longer than it is in A's frame. But they both agree that the light travels at c. So they have to disagree on the time it takes light to traverse the path. – PM 2Ring May 6 '20 at 5:22
• Suppose their are two bodies which are kept at the same height . Now if one body is dropped and the other body is thrown horizontally with some speed( comparable to speed of light) then which one will fall earlier ?? – user262060 May 6 '20 at 7:09
• You are thinking that horizontal and vertical motion are unrelated (as they are in Newtonian dynamics). But with light, the speed limit applies to the total velocity (to use the Pythogorean theorem again, the square root of the sum of the horizontal and vertical velocities squared must be 3e8 m/s). So while the vertical distance is the same, person B sees the vertical velocity as less. This also applies to masses near the speed of light. The acceleration due to gravity is reduced when there is a horizontal velocity, since the mass's combined velocity is limited to less than 3e8. – Ralph Berger May 6 '20 at 16:53
• [The acceleration due to gravity is reduced when there is a horizontal velocity, ] do you mean that gravity is reduced when we travel at light speed?? – user262060 May 8 '20 at 6:43

To explain this example you have to consider Lorentz transformations (so you understand how time changes between two observers). Relativity tells you that in this example both the observer see the other clock going slowly because it's equivalent to consider the train fixed and the ground moving or vice versa. If for the observer on the train (call him $$A$$) the ball takes one second to hit the ground, then, for the observer on the ground (call him $$B$$) it takes more time (but the difference is so small that he can't notice it). Indeed the observer $$B$$ see that while the ball fall the clock of $$A$$ measure 1 second, but since for $$A$$ the clock of $$B$$ is slower it means that for him it has passed takes more than 1 second.

One last thing, if in your example you change the ball with a ray of light and you assume that the speed of light is constant you can easily derive that both observes see the clock of the other going slowly. This is because the observer on the ground see the light travels more distance than the observer on the train but they both see the same speed of light. And you can repeat the reasoning considering the train fixed and the ground moving to see that everything is symmetric.

• Why should i consider Lorentz transformation? – user262060 May 5 '20 at 12:08
• I just want to know how time slows down using this example. – user262060 May 5 '20 at 12:24
• you should consider Lorentz transform because they tell you how $A$ see the time of $B$ and vice versa. I suggest you this video, there aren't formulas and there is the example with the ray of light youtube.com/… – SimoBartz May 5 '20 at 13:49
• In the example you did the observer on the ground see the clock of the observer in the train going slowly. If you want to understand why time is different for the two observer I'm afraid you can't with your example. Indeed this result came from the fact that light speed is the same for all the observer. Namely you need to involve light speed in your example to understand the time relativity – SimoBartz May 5 '20 at 13:53

The paradox lies in the fact that you are assuming that acceleration is equal in both systems, which is not.

We can use dimensional analysis to understand this: acceleration is $$[space]\over [time]^2$$, so if $$\Delta t' = \gamma \Delta t \qquad$$ because of time dilation, ($$\Delta t'$$ is the time in the moving frame and $$\Delta t$$ the time in the station frame), then the acceleration on the train drops by a factor $$\gamma^2$$ respect to the station: $$a' = {a\over \gamma^2}$$ (in our example $$a$$ is the gravitational acceleration $$g$$ so $$a'={g\over \gamma^2}$$)

Now if you try to compute the time $$\Delta t'$$ it takes for an object to fall with this (less) accelerated motion you will find that there is no more contradiction: $$\Delta t' = \sqrt{2\Delta s\over {g\over \gamma^2}} = \gamma \sqrt{2h\over g} = \gamma \Delta t$$

• If you take the primed coordinate to be the moving frame, the relation would be $a'=\frac{a}{\gamma^2\Big(1-\frac{u_xv}{c^2}\Big)^2}$ where $u_x$ is the velocity of the ball in the unprimed (ground frame). Since $u_x=v$, this would reduce to $a'=\frac{a}{\frac{\gamma^2}{\gamma^4}}=a\gamma^2$. This would change your final result to $\Delta t = \gamma \Delta t'$. – Dvij D.C. May 10 '20 at 3:48
• The symmetry between the two frames is broken by the fact that the ground that generates the gravitational field, at rest respect to the station observer, is moving respect to the train – Luca M May 10 '20 at 11:26
• I agree. I don't see how it's related to my comment tho. Unless it wasn't addressed to me, of course. – Dvij D.C. May 10 '20 at 11:33

$$t' = \gamma (t - vx)$$, taking $$c = 1$$

If $$t = 2$$, and $$x = 0$$, t' > 2s.

No matter what kind of event triggers the chronometer of the person inside the train.

$$x = 0$$ means: there are several clocks along the rail line, all synchronized by the station frame. The clock to be checked is that one in front of the passenger at each event.

The time interval between the events, measured at this clocks by the passenger, is greater than 2s.