Why is My Relativity Light Clock Calculation Incorrect?

[PASta95]

Homework Statement

A clock is built using two mirrors (A and B) mounted horizontally (x-direction), between which a pulse of light is repeatedly reflected. The proper length between the two mirrors is 5.00 m. The spacecraft’s velocity (0.60c relative to our galaxy; γ = 1.25) is in the positive x direction.

1. According to an observer in the frame of our galaxy, what is the distance between the mirrors?

2. One period of this clock is the time taken for light to travel from A to B and back to A. According to an observer in the frame of our galaxy, what is the period of the clock?

Relevant Equations

t=t0*γ (time dilation)
L0=L*γ (length contraction)

1. I'm fine with this one, I simply calculated L = L0/γ = 5/1.2 5= 4m.

2. This is the one I'm having problems with.

My approach was to say that both observers would measure the light beam to have velocity of ‘c’. Therefore, if 4m is the distance between the mirrors as observed in the frame of the galaxy, then the light beam traverses a distance of L=8m in one period, and the period T as observed in the frame of the galaxy should be T=L/c=8/(3E8)=26ns.

However the correct answer is 42 ns. This was determined by calculating the proper time as observed by someone in the spacecraft as t0 = 2L/c = 10/(3E8), then using the time dilation formula to calculate the dilated time, i.e. t=t0*gamma = (3.33E-8)*1.25 = 42ns.

So i) I don’t understand why my answer is incorrect, and ii) I also don’t understand how 42ns could be correct when, for L=8m, the observer would measure v = 8/(42E-9) = 2E8 m/s (which doesn’t equal c).

Appreciate any assistance you could provide, thank you.

 

[timetraveller123]

i think your method might also be right but your computation is wrong

how did you calculate time required in galaxy frame it is not as simple 2L/c
can you think how you can modify this

 

[Janus]

So i) I don’t understand why my answer is incorrect, and ii) I also don’t understand how 42ns could be correct when, for L=8m, the observer would measure v = 8/(42E-9) = 2E8 m/s (which doesn’t equal c).

Appreciate any assistance you could provide, thank you.

You forgot to properly account for the fact that, in the galaxy frame, the mirrors are moving in the x direction.
So, for example, Light leaving the trailing mirror, after 13ns will have traveled a distance of 4 m in the x direction ( the distance between the mirrors as measured in the galaxy frame.) However, in that 13ns, the leading mirror will have also moved in the x direction by 2.4 m, and thus after 13ns, the light will still be 2.4m short of reaching the leading mirror. It will require additional time for it to "catch up" with the leading mirror.
Going from the opposite direction, the "return" trip will be shorter.
Try working out the time for each leg of the "round trip" and summing them up.

 

[PASta95]

Thank you both for your replies.timetraveller123: I agree that the time required in the galaxy frame is not simply 2L/c, and in fact this is recognised in the solution to the problem where it calculates the proper time as observed by someone in the spacecraft as t0 = 2L/c = 10/(3E8), then uses the time dilation formula to calculate the dilated time, i.e. t=t0*gamma = (3.33E-8)*1.25 = 42ns. So effectively, the equation is t = (2L*gamma)/c.

Janus: I did forget this, so I made an attempt to consider the velocity of the mirrors. I calculate that in the 13ns the light needs to make up a distance of (0.6c)*(13E-9)=2.34m. Therefore, the total distance of the first leg is 4+2.34 = 6.34 m which would take 21.1ns. Then for the return leg, the lights falls short of the 4m by 1.5m (see note below), therefore the light travels 2.5m before hitting the returning mirror. This takes 8.3ns. The total journey is then 21.1+8.3=29.4ns. Unfortunately, still not the answer I expected!

I figure that I’m either overcomplicating this or missing something that’s staring me right in the face.

Note: I calculated that if L is the distance light travels before it hits the A (return journey) mirror, and x is the distance mirror A travels before the lights hit it on the return journey, then L=4-x and I get t=x/0.6c=(4-x)/c. Solving for x gives me the 1.5m.

 

[Janus]

Thank you both for your replies.timetraveller123: I agree that the time required in the galaxy frame is not simply 2L/c, and in fact this is recognised in the solution to the problem where it calculates the proper time as observed by someone in the spacecraft as t0 = 2L/c = 10/(3E8), then uses the time dilation formula to calculate the dilated time, i.e. t=t0*gamma = (3.33E-8)*1.25 = 42ns. So effectively, the equation is t = (2L*gamma)/c.

Janus: I did forget this, so I made an attempt to consider the velocity of the mirrors. I calculate that in the 13ns the light needs to make up a distance of (0.6c)*(13E-9)=2.34m. Therefore, the total distance of the first leg is 4+2.34 = 6.34 m which would take 21.1ns.

You didn't get this right. After 21.1 ns, the light will have indeed traveled 6.34 m. However, in 21.1 ns, the leading mirror will have moved 3.8 m, and will now be 7.8 m from where the light left the trailing mirror. In other words, it will still be 1.46 m ahead of the light.

This is like the race between Achilles and the tortoise.

In order to be fair, Achilles gives the tortoise a head start. By the time he reaches the point where the tortoise started, the tortoise will have moved on. If he now continues on to where the tortoise was at that time, when he get there, the tortoise again is further along...

We know that Achilles will eventually catch and pass the tortoise, the question is when. One way would be to add up all the segments where Achilles runs from one point where the tortoise was to the next point, and just keep doing this until the distance between Achilles and the tortoise becomes small enough to ignore.

Or, we could do something like this:
If L is the distance Achilles needs to cover to catch up to the tortoise and t is the time it takes to do this, then if Achilles runs at a velocity of V, then
L=Vt
If v is the velocity the tortoise travels, then the total distance the tortoise moves in the time t is vt. If d is the head start distance then this, plus the distance the tortoise moved in time t will also equal L
L= vt+d
Since both expressions equal L, we can equate them:

Vt = vt+d
Now it is just a matter of solving for t.
The same applies to your problem of how long it takes for the light to catch up with the leading mirror.

 

[timetraveller123]

timetraveller123: I agree that the time required in the galaxy frame is not simply 2L/c, and in fact this is recognised in the solution to the problem where it calculates the proper time as observed by someone in the spacecraft as t0 = 2L/c = 10/(3E8), then uses the time dilation formula to calculate the dilated time

No I didn't mean to say time dilation is the only way to go what I meant to say was your method was correct just that your computation was wrong ie not 2L/c but something else like what @Janus said
One another way to look at it would be what is the relative velocity between the mirrors and the photon when it traveling right and what is relative velocity when it is traveling left?

 

[jbriggs444]

One another way to look at it would be what is the relative velocity between the mirrors and the photon when it traveling right and what is relative velocity when it is traveling left?

Minor nitpick -- that's not a "relative velocity". That's a "closing velocity".

A "relative velocity" is the velocity of one thing in the rest frame of another. A "closing velocity" is the rate at which the distance between two objects changes from the perspective of a third party.

In ordinary Newtonian mechanics, there is no distinction to be made between the two. In relativity, there is.

 

[timetraveller123]

Minor nitpick -- that's not a "relative velocity". That's a "closing velocity".

A "relative velocity" is the velocity of one thing in the rest frame of another. A "closing velocity" is the rate at which the distance between two objects changes from the perspective of a third party.

In ordinary Newtonian mechanics, there is no distinction to be made between the two. In relativity, there is.

thanks for the clarification i always called the two relative velocity now i know the name

 

[PASta95]

Once again, thank you all. I got it.

For the first part of the light's journey, I calculate that it takes 33 ns to reach the leading mirror. This reconciles with the closing velocity being 0.4c over the 4m. For the second part of the journey, the light takes 8.3 ns to reach the trailing mirror, consistent with the closing velocity being 1.6c over the 4m. Then, the total journey time is 42 ns, which reconciles with the answer I was expecting.

Further, the total distance the light travels is 9.94 + 2.52=12.45 m. As it does this in 42ns, the resultant velocity is c also as expected.

It all makes sense, thanks also for your patience as my brain slowly made its way through this problem.