Light Clock Problem on the Train: Exploring Time Dilation in Special Relativity

[DAC]

Hello PF.
Re. the light clock on the train thought experiment. If the mirrors are one metre apart in both frames. And the speed of light is the same in both frames. Why isn't the time it takes light to travel one metre, the same in both frames?
Thanks.

 

[Orodruin]

Why isn't the time it takes light to travel one metre, the same in both frames?

It is.

 

[Ibix]

They aren't one meter apart in both frames, assuming we're talking about one clock here. In one frame they are length contracted and moving, so the distance from where one mirror was to where the other will be is not one meter.

 

[Orodruin]

They aren't one meter apart in both frames, assuming we're talking about one clock here. In one frame they are length contracted and moving, so the distance from where one mirror was to where the other will be is not one meter.

In the usual light clock setup the light clock is perpendicular to the direction of motion. The mirrors are definitely one meter apart in both frames with this setup.

The reason the light needs to travel more than one meter in the frame where the clock is moving is that the mirrors move.

 

[DAC]

Hello PF.
Re. the light clock on the train thought experiment. If the mirrors are one metre apart in both frames. And the speed of light is the same in both frames. Why isn't the time it takes light to travel one metre, the same in both frames?
Thanks.

doesn't that mean time is the same in both frames?

 

[Orodruin]

doesn't that mean time is the same in both frames?

No. Even if the clock is perpendicular to the direction of motion, the light needs to travel longer in the frame where the mirrors are moving (it needs to travel additional distance to catch up with the mirrors). Therefore, the clock is dilated in the frame where it is moving.

 

[Ibix]

In the usual light clock setup the light clock is perpendicular to the direction of motion. The mirrors are definitely one meter apart in both frames with this setup.

The reason the light needs to travel more than one meter in the frame where the clock is moving is that the mirrors move.

True. Scratch the "length contracted" from what I wrote. The distance from one mirror to the other at any given instant is 1m. The distance from where one mirror was when the light hit it to where the other will be when it gets hit is more than 1m.

 

[FactChecker]

Suppose you are an outside, stationary observer watching the light beam. You see the light travel along the hypotenuse of a right triangle. The triangle has one side of 1 meter. The other side will be the distance traveled while the light beam went from one mirror to the other. So the time you measure will be proportional to the length of the hypotenuse. It's a simple Pythagorean theorem calculation.

 

[Dale]

If the mirrors are one metre apart in both frames. And the speed of light is the same in both frames. Why isn't the time it takes light to travel one metre, the same in both frames?

Because the distance the light travels is not equal to the distance between the mirrors, in general.

 

[Janus]

doesn't that mean time is the same in both frames?

You have two light clocks, each with their mirrors 1 meter apart along a line perpendicular to the relative motion between the two clocks. You have an observer with each clock. At the moment the clocks pass each other, a light pulse leaves the bottom mirror of each clock. Each observer note how long it takes for his pulse to reach his upper clock. They both get the same answer. However when they consider the pulse for the other clock they note that it takes longer to reach its upper mirror than it did for their own pulse to reach their upper mirror.
This is because both observers measure the speed of both pulses as being the same relative to themselves. Observer 1 measures his pulse moving at c and notes that it takes after it has traveled m meter vertically it has reached his mirror. The pulse of the other clock also moves 1 meter in the same time. However it is moving at an angle and after traveling for 1 meter has not yet reached the top mirror of the second clock. Thus it takes longer for pulse of clock 2 to reach the top mirror as measured by observer 1. Conversely, as far as observer 2 is concerned, it is clock 1's pulse that takes longer to reach its top mirror.

So if you were to ask each observer how long it took for his pulse to reach his top mirror, they both will give the same answer, and if asked how long it took the othr clock's pulse to reach its top mirror, they will say longer than it took for their own pulse to reach their top mirror.

 

[peety]

Is

Suppose you are an outside, stationary observer watching the light beam. You see the light travel along the hypotenuse of a right triangle. The triangle has one side of 1 meter. The other side will be the distance traveled while the light beam went from one mirror to the other. So the time you measure will be proportional to the length of the hypotenuse. It's a simple Pythagorean theorem calculation.

Is there an easy way to show how you get from here to the v squared over c squared of Lorentz?

 

[Ibix]

Is there an easy way to show how you get from here to the v squared over c squared of Lorentz?

Say it takes the light clock a time $t$ to tick when it isn't moving, and the mirrors are a distance $L$ apart. Can you relate $L$ and $t$ using the speed of light?

Now say the light clock is moving. It takes a time $t'$ to tick. How far do the mirrors move in this time? So how far did the light pulse have to travel? This distance must be equal to $ct'$, by definition. Can you write down a relationship between $L$, $t'$ and $v$ involving the speed of light?

Using the last results of the last two paragraphs, eliminate $L$. That should be it.

 

[stevendaryl]

Is there an easy way to show how you get from here to the v squared over c squared of Lorentz?

Pick a rest frame $F$, and let's suppose that you have a light clock moving in the x-direction at speed $v$, according to frame $F$, and it is oriented perpendicularly to its direction of motion, in the y-direction. Let a pulse of light go from one side of the clock, at $y=0$, to the other side, at $y=L$. The position of the first mirror as a function of $t$ is given by $x=vt, y=0$. The position of the second mirror is $x=vt, y=L$. So if it takes $T$ seconds to travel from one mirror to the other, then the event of leaving the first mirror has coordinates

$x=0, y=0, t=0$

and the event of arriving at the second mirror has coordinates

$x=vT, y=L, t=T$

So the total distance traveled by the light is: $D = \sqrt{\delta x^2 + \delta y^2} = \sqrt{v^2 T^2 + L^2}$. Since light travels at speed $c$, it must be that $D = cT$. So we have:

$cT = \sqrt{v^2 T^2 + L^2}$, or $c^2 T^2 = v^2 T^2 + L^2$. Solving for $T$ in terms of $L$ and $v$ gives:

$T = L/\sqrt{1 - v^2/c^2}$

 

[Ibix]

$T = L/\sqrt{1 - v^2/c^2}$

Nitpick: this is out by a factor of c.

 

[DAC]

It is.

Is one metre the same in both frames?

 

[Orodruin]

Is one metre the same in both frames?

It is the distance light travels in 1/299792458 s regardless of the frame.

That is not saying that an object which is one meter will be one meter in all frames though.

 

[DAC]

Apart from length contraction are there any instances.

 

[Orodruin]

Apart from length contraction are there any instances.

Of what? Please separate the concepts of the length of an object being different in different frames from the fact that a meter (the unit!) is the same in all frames.

 

[DAC]

Of what? Please separate the concepts of the length of an object being different in different frames from the fact that a meter (the unit!) is the same in all frames.

Apart from length contraction does the length of an object differ in different frames?

 

[Orodruin]

Apart from length contraction does the length of an object differ in different frames?

So what you are asking is essentially "apart from the fact that objects have different length in different frames, do objects have different lengths in different frames?" It is impossible to give a reasonable answer to this type of question.

 

[DAC]

In which case I apologise,
Under what circumstances do objects have different lengths in different frames

 

[Ibix]

Apart from length contraction does the length of an object differ in different frames?

As Orodruin says, this is nonsensical. However, imagine that you are standing beside the back of a car that is 5m long. How far do you have to walk to get to the front? 5m. How far do you have to walk to get to the front if it is moving? More than 5m, because you've got to play catch-up.

The light clock we have been discussing is a transverse light clock. In this case the question is how far do you have to walk to get from one back corner of the car to the other. It's 2m wide. If the car is not moving, then the answer is 2m. If the car is moving you'll have to walk slightly diagonally and the distance is more than 2m, although the width of the car doesn't change.

 

[Ibix]

In which case I apologise,
Under what circumstances do objects have different lengths in different frames

Objects have different lengths in different frames due to length contraction unless the length being measured is perpendicular to the direction of motion.

However, there is a distinct phenomenon in moving objects which occurs in both Newtonian and Einsteinian relativity. The distance traveled by something traversing a moving object is not the same as its length, because in the time it takes to traverse the object the end has moved.

I suspect that the latter concept is what is eluding you as, coupled with the invariance of the speed of light, it is the "cause" of time dilation in this derivation.

 

[DAC]

Thanks.
If the object being measured is less than perpendicular to the direction of motion, is the contraction also less.

 

[Orodruin]

Thanks.
If the object being measured is less than perpendicular to the direction of motion, is the contraction also less.

The contraction is only in the direction of motion and it is always the same. You can work from there using Pythagoras' theorem.

 

[DAC]

Many Thanks.

 

[Dale]

Apart from length contraction does the length of an object differ in different frames?

In addition to what others have said, I would like to point out that the length of the light clock is not the important factor. The important part is the distance that the light travels. This is, in general, different from the length of the clock. Do you understand that?

Have you ever played football or some other sport where you have to chase somebody and tackle them? They may never be more than 5 m away from you and yet you may have to run more than 50 m to catch them.

 

[DAC]

In addition to what others have said, I would like to point out that the length of the light clock is not the important factor. The important part is the distance that the light travels. This is, in general, different from the length of the clock. Do you understand that?

Have you ever played football or some other sport where you have to chase somebody and tackle them? They may never be more than 5 m away from you and yet you may have to run more than 50 m to catch them.

In addition to what others have said, I would like to point out that the length of the light clock is not the important factor. The important part is the distance that the light travels. This is, in general, different from the length of the clock. Do you understand that?
If by the length of the clock, you mean the distance the mirrors are apart, then yes I realize the lights path as seen by the platform observer is longer than the distance the mirrors are apart,when the train is moving.

 

[vanhees71]

It's also easily resolved using Lorentz transformations or, even simpler, drawing the Minkowski diagram.

Let the mirrors making up the light clock be in the $xz$ plane of a Cartesian coordinate system. In their rest frame the wave goes forth and back along the $y$ axis. The trajectory of the wave front is given by (using natural units with $c=1$)
$$x(t)=z(t)=0, \quad y(t)= \begin{cases} t & \text{for} \quad 0<t<L,\\ L-T & \text{for} \quad L<t<2L.\end{cases}$$
Now to get the trajectory in the frame, where the mirror moves with velocity $v$, $|v|<1$, in $x$ direction you need the corresponding Lorentz transform
$$t'=\gamma(t-v x) ,\quad x'=\gamma(x-v t), \quad y'=y, \quad z'=z.$$
The trajectory there is thus given by
$$t'=\gamma t, \quad x'=-v \gamma t, \quad y'(t)=y(t), \quad z'=0.$$
At $$t=2L$$ you have $$t'=2L \gamma=\frac{2L}{\sqrt{1-v^2}}.$$

This can also be seen entirely in the frame $\Sigma'$. The distance to be traveled in the first half of the trajectory is given by
$$D=\sqrt{L^2+v^2 t'^2},$$
but
$$t'=D\; \Rightarrow \; D^2=L^2+v^2 D^2 \; \Rightarrow \; D=\gamma L.$$
The way back is just symmetric, i.e., the total path traveled by the wave head is
$$2D=2 \gamma L.$$