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I'm guessing you're all familiar with the classic intuitive way of explaining time dilation: with a light clock traveling at velocity v directed at a parallel direction to the mirrors that make up the light clock.

Now, what happens if we have a light clock that goes upwards at v? That is to say, what if its velocity is in a direction perpendicular to the mirrors' orientations? As far as I can tell, this situation wouldn't present any time dilation. Thoughts?

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  • $\begingroup$ An instrument which makes a direct experimental comparison of the two cases is a Michelson interferometer, and if you build a big enough one and turn it about a vertical axis you are performing the Michelson-Morley experiment. $\endgroup$ – dmckee Oct 2 '17 at 17:20
  • $\begingroup$ Theorem: $4\times 2=8$. Proof: Take 4 piles of 2 stones each and count the stones; there are 8 of them. Objection: But what if instead of taking 4 piles, you'd taken 3 piles? It seems that in that situation $4\times 2$ would not equal 8. The point: Once you've proven something, it's proved. The fact that some other argument does not prove the same thing is irrelevant. Whatever you've proved with one set of light clocks stays proved, even if some other set of light clocks fails to provide that proof. $\endgroup$ – WillO Nov 9 '17 at 1:30
  • $\begingroup$ @WiillO that's not a very relevant comment. I was asking about the self-consistency of the theory, not trying to question the validity of the conclusion given by the other "proof" (a sketchy word to use to describe a thought experiment). Besides, the point of SR is that this should be valid for all inertial observers, so it should indeed be able to account for "some other set of light clocks"; it doesn't really make sense to accept that it's ok if it just works with a specific one. $\endgroup$ – Physics Llama Nov 13 '17 at 11:44
  • $\begingroup$ Something like this? m.youtube.com/watch?v=aNEryiOKkrc $\endgroup$ – Bill Alsept May 10 '18 at 6:00
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In that case, time dilation still occur, of course. In order to show this using t=d/v, you'd have to take into account the space contraction in the direction of motion. Mathematically, if d is the height of the clock, then the time taken from a photon at the bottom to reach the top of the clock isn't $\frac{d+vt}{c}$ but $ \frac{d/\gamma+vt}{ c}$. When you calculate the time taken for that photon to get back to the bottom of the clock and add it up to the time previously calculated, you get the exact same time dilation than the clock moving parallelly to the mirrors.

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  • $\begingroup$ Why do you divide the vt by gamma as well? I understand that the clock should be Lorentz-contracted, but why should the distance travelled by the elevator suffer this change as well (considering that in the "moving" frame this distance is zero, and in the "at rest" frame it is simply vt). $\endgroup$ – Physics Llama Nov 12 '15 at 0:49
  • $\begingroup$ @PhysicsLlama, whoops, you're right. I've edited/corrected it. $\endgroup$ – thermomagnetic condensed boson Nov 12 '15 at 0:59
  • $\begingroup$ Regardless, I think that with Lorentz contraction this does end up working. Thanks. $\endgroup$ – Physics Llama Nov 12 '15 at 1:04
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yes, time dilation still occurs. The reason is that the mirrors are there to provide an intuitive view of how/why time dilation occurs, not to to create it. This time dilations still occurs regardless of the existence of the mirrors or the direction of motion. But the drawing will no longer have explanatory power.

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The two parallel mirrors, A and B, are traveling together such that the axis of reflection is parallel to the direction of travel, and in such a direction that A trails behind B as both travel at the same velocity near the speed of light relative to our observatory. A laser attached to A fires a pulse at B which returns at some time $x/c$ later. Because B is moving rapidly away from the initial pulse location, it takes much longer for the pulse to reach B and reflect back in our frame than it does in the clock frame: time dilation still occurs.

What this arrangement introduces in addition to the textbook example (with the axis of reflection perpendicular to the axis of motion) is the relativity of simultaneity: in the textbook example, if a laser is mounted on each mirror and both fire at the same time, then both beams reach a sensor on the other mirror at the same time as well. In your example, the beam from B will reach A long before the beam from A reaches B, from our observational perspective; though in the frame of the ship carrying the light clock, they will still happen simultaneously.

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  • $\begingroup$ not quite so, from stationary view the pulse will take longer to reach B but much less time to return to A, the only way is to use length contraction $\endgroup$ – Adrian Howard Jun 7 at 7:28
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By A and B I am referring to the two mirrors. Mirror A is the rearmost mirror; Mirror B is the foremost. As you say, the distance is increased when the light travels from A to B, but is reduced by the same amount when travelling from B to A. Thus (ignoring, for a moment, the Fitzgerald contraction) the total path length for a complete cycle remains the same as for a stationary system.

Acceleration does not come into it. The light clock thought experiment is a demonstration of Special Relativity, in which the clock and the observer are travelling at constant relative speed.

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Time dilation would occur in this scenario.

This diagram depicts two light clocks. The left light clock is stationary. The right light clock is traveling to the right at 60% of the speed of light.

enter image description here

The green and red lines represent the mirrors of each light clock.

The blue lines represent photons bouncing between the mirrors of each light clock. All 45 degree lines represent the speed of light. The numbers indicate ticks of the light clocks, which are counted when the photon strikes a clock's left mirror.

The right light clock ticks slower because half of the time, its right mirror is moving away from the photon, which takes longer to catch up with it.

After 5 ticks of the left light clock, the right light clock has recorded only 4 ticks, indicating that time dilation has taken place.

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  • $\begingroup$ This doesn't deal with the case described in the question. I.e. a motion perpendicular to the mirrors. You considered a motion parallel to the mirrors. $\endgroup$ – thermomagnetic condensed boson May 3 at 12:51
  • $\begingroup$ @thermomagneticcondensedboson I believe that is incorrect, and that I did in fact illustrate motion perpendicular to the mirrors. The vertical axis in my diagram represents time, and not an additional dimension of space. $\endgroup$ – Drew May 4 at 15:04
  • $\begingroup$ I am sorry then Drew. Would it be possible to illustrate the case of when the motion is parallel to the mirrors in your diagram? I have a hard time to mentally picture it. And that would give me an opportunity to transform my downvote to a positive vote (which I am not allowed to change unless an edit is made). $\endgroup$ – thermomagnetic condensed boson May 4 at 18:24
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    $\begingroup$ @thermomagneticcondensedboson Motion parallel to the mirrors doesn't really work in that diagram with space and time as the two axes, because the paths of the mirrors and the photon would appear coincident. This page has a light clock with motion parallel to the mirrors: pitt.edu/~jdnorton/teaching/HPS_0410/chapters/… Don't worry about changing your downvote. It doesn't really matter. : - ) $\endgroup$ – Drew May 5 at 20:26
  • $\begingroup$ this is not dealing with a light clock horizontal to line of motion, merely one at a slight angle $\endgroup$ – Adrian Howard Jun 7 at 7:33
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If we imagine that clock is rotated 90 degrees from the textbook example, and now the clock light travels back and forth on the same axis that object is moving. For a stationary clock, time needed for a photon to go from one mirror and back would be:

τ(stat)=2*d0/c

For a moving clock and a stationary observer, we first know that photon must go over distance d, increased by the distance that mirror 2 has traveled:

c*t1 = d + v*t1

Notice that we're assuming that distances d are not equal for both cases. Then it must go back, but this time the distance it has to go will be reduced by the amount that mirror 1 travels in it's direction:

c*t2 = d - v*t2

Total amount of time needed for photon to go from mirror 1 to mirror 2 and back would be:

τ(mov)= t1 + t2
      = d/(c-v) + d/(c+v)
      =2cd/(c^2 - v^2)

Now, the distance d here is not the same as distance d0 that we have in the first equation, because we have the effect of length contraction. Motion distance will be:

 d=d0/γ
 d=τ(stat)*c/(2*γ)

Where we know:

 γ^2= c^2/(c^2 - v^2)

If we replace that in our equation, we get:

τ(mov)= [2*c*d]/(c^2 - v^2)
      = [2*c*τ(stat)*c/(2*γ)]/(c^2 - v^2)
      =  τ(stat)*c^2/[γ*(c^2 - v^2)]
      =  τ(stat)*γ^2/γ
      =  τ(stat)*γ

This should be a time dilation formula for longitudinal light clock.

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In the textbook version it is easy to see how the path length increases in accordance with Pythagoras' theorem. With the light travelling parallel to the direction of motion, the pulse will travel further in order to get from $A$ to $B$ than from $B$ to $A$, but the total path length over a complete cycle remains the same regardless of the relative speed. So the clock runs at the same rate, whatever the relative speed. But hang on: the Fitzgerald-Lorenz contraction will shorten the distance between the mirrors by a factor of $\gamma$. So the clock will run faster with increasing relative speed.

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    $\begingroup$ You've neglected the fact that the distance traveled is increased as the light travels one direction and decreased as the light goes the other way. The clock will still run slower with relative speed (and by the same factor). Otherwise you would observe a fringe shift in a Michelson interferometer as you accelerate relative the device, but that violates the principle of relativity because a stationary observer would see no such effect. $\endgroup$ – dmckee Oct 2 '17 at 17:23

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