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Einstein's light clock with mirrors experiment involves mirrors on top and bottom of a moving train and light reflecting back and forth. To a stationary observer, it would seem that the light is traveling more distance and since the speed of light is the same for both stationary and moving observors, we conclude that moving clock runs slower relatively. My doubt is what if we are placing the mirrors outside the train horizontally to the train? Assume that the train is made of glass. Now to the stationary observer, the light beam would be traveling in a straight line but to the observer in the train, the light beam would move in a curved path and hence covers more distance. Hence to the stationary observer, time moves at the same speed inside the train but for the observer inside the train, time moves slower. Could you please explain the difference in the point of view of the stationary observer in the 2 experiments and doesn't it conflict with each other?

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    $\begingroup$ Why would the light move in a curved path in the train’s frame? Putting the mirrors inside or outside doesn’t change anything. $\endgroup$
    – Dale
    Commented Nov 2, 2019 at 3:27
  • $\begingroup$ By "outside the train" do you mean that the light clock is stationary? $\endgroup$ Commented Nov 2, 2019 at 4:01
  • $\begingroup$ @Dale since the observer inside the train would see the light move diagonally due to the motion of the train $\endgroup$
    – Shamy
    Commented Nov 2, 2019 at 4:58
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    $\begingroup$ @probably_someone yes the light clock in stationary $\endgroup$
    – Shamy
    Commented Nov 2, 2019 at 4:58
  • $\begingroup$ Hey @Shamy, I don't understand your reply to Dale's question. If a person at rest inside the train turned on a torch briefly to let out a photon, then that person would observe the photon to go up and down only (Train frame is an inertial frame : therefore, the emitted photon would behave the same way it behaves when a person on the ground turns on the torch briefly). Mirror's motion is not going to determine the photon's path. $\endgroup$
    – Ajay Mohan
    Commented Nov 2, 2019 at 6:33

3 Answers 3

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To the observer standing on the ground, time moves slower on the train (since the light bouncing in a light clock on a train, which is moving in this observer's frame, has to cover a longer distance between bounces). To an observer standing in the train, time moves slower on the ground (since the light bouncing in a light clock on the ground, which is moving in this observer's frame, has to cover a longer distance between bounces).

There is no contradiction because you can't be in two reference frames at once. You cannot be both moving and not moving in any one reference frame.

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  • $\begingroup$ but shouldn't time always move slower then you are in motion? $\endgroup$
    – Shamy
    Commented Nov 2, 2019 at 17:14
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    $\begingroup$ @Shamy How do you know you're in motion? What experimental test can you do to verify this? $\endgroup$ Commented Nov 2, 2019 at 17:27
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Your comment 'but shouldn't time always move slower when you are in motion' is the clue to why you might be struggling to understand what is happening, so let's see if we can help...

Firstly, let's imagine we have a vertical light clock on the train and an identical vertical clock on the platform.

In the frame of the platform, the light in the train's clock takes a diagonal path from the bottom of the clock to the top, so the time difference between the start and end of the light's path must be greater on the platform than on the train.

However, in the frame of the train, the light in the platform's clock takes a diagonal path from the bottom to the top, so the time difference between the start and end of the light's path must be greater on the train than on the platform.

You should be able to see from what I have just said, that the effect is totally symmetrical. It doesn't matter whether you consider the train to be moving and the platform to be stationary or vice versa. So how can time on the train be running more slowly than time on the platform when the time on the platform is running more slowly than the time on the train? It seems contradictory. The resolution to the puzzle is that what is really happening is that time is not running more slowly but out of synch.

When you have two inertial frame moving relative to each other, they have sloping planes of constant time. What that means is that when it is a constant time everywhere along the train, it is a different time everywhere along the platform- if you had clocks set at regular intervals along the platform they would be increasingly out of synch in either direction, each clock being a little ahead of the previous one in the direction of travel of the train and each one being a little behind the previous one in the other direction.

It is the fact that the clocks are increasingly out of synch that causes the appearance of time dilation. Imagine you had two clocks a certain distance apart on the platform and one is running a second ahead of the other. You pass one clock when it reads 0s, and 5s later, say, by your watch you reach the next clock. Because that next clock had been running a second ahead, it will read 6s According to your watch 5s have passed, but according to the clock you have just reached 6s have passed. That is not because your watch and the clock are running at different rates, but because the clock is running a second out of synch with the first clock you passed. That is the key to understanding time dilation- it is an effect related to the relativity of simultaneity.

You can derive the time dilation formula by considering horizontal light clocks, but it is a lot more tricky to do it that way. Interestingly, however, considering horizontal light clocks forces you to consider the relativity of simultaneity more explicitly than is the case when vertical light clocks are used, so you might find it a helpful exercise to try.

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A train moving with relative speed $v$ to ground. A person on ground, groundman sees train moving with speed $v$ and a person sitting in the train, trainman, seen by groundman as moving with same speed $v$ and thus no relative speed between train and trainman. Now if a ball or a light is push in direction of motion within from the train with speed $c$ by trainman. Now groundman sees that light has more speed than the train because it goes from back end to front end, so the speed must $c+v$.

A reader should not assume that if speed of anything, projectile is either increase or decrease depending upon direction of projectile in relative frame. The groundman sees speed of projectile is $c+v$ but also sees that train moves distance in time $t$, with relative speed,$$x+vt=(c+v)t\implies x=ct$$So there is no difference in speed of projectile measured by trainman and groundman. So there is no need of introducing relative time and distance. Time is universal variable and this is evident from relativity that there is proper time, which is same in relative frames. It means that relativity has no different time for frame in motion measured from within frame, relativity is symmetric.

So if same projectile is on ground, then trainman calculate same speed of projectile from the train as groundman, but find speed of projectile as $c-v$ but distance traveled by projectile is still same, $x=ct$.

In case of projectile in perpendicular direction of relative motion, we can use pythogorean theorem or trignometric relation, and reach on same conclusion that even speed of prohectile is different in relative motion but length reamin same. So length and time is conserved in relative motion and they are fundamental quantities while speed is derived quantity. The term relativity comes from relative speed, so why bother to conserve speed on the cost of length, mass and time.

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