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Question about light clock.

  1. Jul 5, 2009 #1
    Often to describe relativity is used example with two ships. Ship B is traveling at 0.5c speed from ship A. Ship A is standing still and shooting out beam of ligth in direction of ship B. Ship B has this light clock wich is placed verticaly. I understand the time delation with vertical clock thing can be calculated with formula. c^2=c(h)^2+c(v)^2 , where c-speed of light, c(h)- the speed of ship B=0.5c compared to shipA. c(v)- the speed of light in vertical direction compered to ship A. So c^2=0.25c^2+c(h)^2 -> c(h)=0.8666. So i undertand time delation factor is 0.86 or 1/0.86=1.16. The time for ship B i slowed down by 1.16 sek compered to A. But if th time delation facotr is 1.16 how can he see the passing light (of ship A) at c speed? time delation factor should be c/0.5c = 2. So hes time compered to ship should be slowed be 2 not by 1.16. So where i am wrong? Hope you undertand my question and what i have written. My head is boiling from these questions since im interested in these Einstein's theories.
     
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  3. Jul 5, 2009 #2

    Doc Al

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    OK. This light clock experiment shows that ship A measures the clocks in ship B to run slow. (Of course, the reverse is also true: Ship B would measure the clocks in ship A to run slow by the same factor.)
    I don't understand how you deduced this, given the above. Can you explain your reasoning?
     
  4. Jul 5, 2009 #3

    Janus

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    Time dilation is not the whole picture. For light signals traveling along the length of ship B you also have to take into account length contraction and the Relativity of Simultaneity.

    Also, it is not a good idea to think of things in terms of "B's time dilation causes him to measure the speed of light as c", but rather " A and B both measure the speed of light as c, and thus measure time dilation in each other".
     
    Last edited: Jul 5, 2009
  5. Jul 6, 2009 #4
    '' But if th time delation facotr is 1.16 how can he see the passing light (of ship A) at c speed? time delation factor should be c/0.5c = 2. So hes time compered to ship should be slowed be 2 not by 1.16. So where i am wrong?

    I don't understand how you deduced this, given the above. Can you explain your reasoning? ''

    Well for the ship B to see light clock beam going up and down at speed c time delation must be 1.16 (se formula). But what time delation should be for ship B to see horizonatal passing by beam from ship A? c-0.5c(speed of ship)=0.5c therefore to see 0.5c at c speed time must slow down at c/0.5c = 2.

    ''Time dilation is not the whole picture. For light signals traveling along the length of ship B you also have to take into account length contraction and the Relativity of Simultaneity.''

    Thank you must be the right answer. Only i cant imagine the hole picture with length contraction and Simultaneity. would be very nice if there would be some good visualiztions with horizontal light clock to understand this.
     
  6. Jul 6, 2009 #5

    Saw

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    Archis, instead of the pure light clock, think of a Michelson-Morley-like device that passes by you: when the device meets you, two light beams depart at right angles to each other, hit their respective targets and reflect back, just to arrive simultaneously at the origin. This is a combination of the typical light clock (light moving transversally to the direction of motion of the device, relative to you) and a longitudinal light clock (light moving in the direction of motion of the device relative to you, in the go trip, and against it, in the return trip).

    Each arm of the device is 1 light-second long (= approximately 300,000 km), as measured in its rest frame (the device frame).

    You judge that light travels relative to you at c, which is = 1, if you measure distances in light-seconds. But in your opinion, relative to the device:

    * As to the transversal light, it travels at sqrt(1-v^2) c = 0.866 c. When this light returns to the origin, for you, 1.154 s * 2 = 2.309 s have elapsed.

    * As to the longitudinal light, it travels at the average between c-v (go trip) and c +v (return trip), which is (1-v^2) c = 0.75 c. However, you also assume that the horizontal or longitudinal arm of the device has contracted by sqrt(1-v^2), so in your frame it is only 0.866 light-seconds long or 1.732 light-seconds long counting the go and return trip. Thus the time it needs to complete the round trip = distance / velocity = (if you call x the distance as measured in the device’s frame = 2 light seconds) =

    x * sqrt(1-v^2) / (1-v^2) = x / sqrt(1-v^2) = 2/0.866 = 2.309 s

    So you agree that the two beams come back to the origin at the same time, since they do it after 2.309 s as measured in your frame. This measurement of yours assumes that the device suffers time dilation (his time is slower) and length contraction (the device measures that the length of its longitudinal arm is 1 light-second, while you measure it is shorter). Instead, the observer at the device judges that 2 s have elapsed, since he doesn’t appreciate any time dilation or length contraction for his own instruments: the length of each arm is 1 light-second and light takes 1 s to traverse such length, 2 s for the go-and-return trip.

    If both you and the device had displayed clocks at the target mirror of the longitudinal arm, what should they read? Yours should read 1.732 s and his should read 1 s. Why? Well, you might have used the very longitudinal light beam to synchronize your clocks: when this light returns to the device’s origin, the latter marks 2 s (1*2) and when it returns to you, your clock marks 3.464 s (=1.732 *2).

    The same results can be obtained with the fomulas for transformation of coordinates (the Lorentz transformations) and for transformation of intervals (those that relate proper with coordinate time and rest length with coordinate length). And the system also works the other way round: for the device, you have TD and LC and your synchronization is “wrong”.

    Attached is a spacetime diagram. It's a little crowded, but if you have difficulties, please ask.
     

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  7. Jul 6, 2009 #6

    Janus

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    Here, try this:

    http://home.earthlink.net/~parvey/sitebuildercontent/sitebuilderpictures/length_con2.gif
    We have light clocks in which the mirrors are positioned both vertically and horizontally.Each dot represents a light pulse bouncing back and forth between the mirrors. The expanding circle shows how each light pulse travels at the same speed as seen by the "stationary" light clock. The clocks have a 0.866c relative velocity.

    Note the "moving" light clock is length contracted since we are viewing this from the the "stationary" clock's frame.

    Because of the length contraction, the "round trip" of the horizontal pulse for the moving clock is the same as the round trip for the vertical pulse, and the time dilation rates match.

    You will also note that the pulse while heading right takes longer to go from the left mirror to the right mirror than the reflected pulse takes to go from the right mirror to the left mirror.

    This is where the Relativity of Simultaneity comes in.

    Imagine there is a second digital time read out by the the right mirror in the moving frame. To someone in that frame, it always reads the same as the read out by the left mirror. To measure how long the light takes to travel from one mirror to the other, they would take the reading at the mirror it leaves, upon departure, and compare it to the reading at the second mirror upon arrival, taking into account the distance between the mirrors.

    But from the stationary frame, according to the Relativity of Simultaneity, the two readouts are not the same, when the right reads 0, the left readout already reads 0.48. Ergo, the pulse leaves the left mirror when it reads 0, and arrives at the right mirror when it reads 0.5 and when the left mirror reads 0.98. It then bounce back to the left mirror, arriving when it read 1.

    In the moving frame, this means that the pulse must also leave the left mirror when it reads 0, arrive at the right mirror when it reads 0.5 and return to the left when it reads 1.

    Thus for the observer in the moving frame it take a time of 0.5 for the light to go from left to right and another 0.5 for it to return, and taking into account the distance between the mirrors (which is not length contracted in this frame), he will get an answer of c for the speed at which the light travels between the mirrors.
     
  8. Jul 7, 2009 #7
    Thank you for your spend time and very good answers Saw and Janus. I just read last post by Janus and seems you have answerd the question wich i wanted to ask Saw. How can moving frame observer see light c when the light travels c+v and c-v the distance back and fort.

    ''But from the stationary frame, according to the Relativity of Simultaneity, the two readouts are not the same, when the right reads 0, the left readout already reads 0.48. Ergo, the pulse leaves the left mirror when it reads 0, and arrives at the right mirror when it reads 0.5 and when the left mirror reads 0.98. It then bounce back to the left mirror, arriving when it read 1.''

    Must be the key part. But cant figure out yet why for the ''stationary'' frame the two readouts are different.
     
  9. Jul 8, 2009 #8

    Janus

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    A part of your problem comes from the fact that you are trying to come at this from the wrong direction. You are trying to use the aspects of Relativity to explain why light is measured at c in all inertial frames, when this is in fact a postulate of Relativity. It is considered a fundamental fact, and all the aspects of Relativity are a consequence of it, not the other way around.

    Thus, as far as the difference in readings go:

    Imagine you have two clocks separated by a distance, you want them to start running at the same time, so you rig them to start when a light pulse reaches them and this pulse is emitted from a point halfway between them, like thus:

    http://home.earthlink.net/~parvey/sitebuildercontent/sitebuilderpictures/synch1.gif

    Now it doesn't matter whether or not these clocks are moving according to some other observer or not, this is what happens from the inertial frame of the clocks. The pulses reach both clocks at the same instant and they both start at the same instant, keeping time with each other from then on.

    From the inertial frame according to which the clocks are moving, things are different. By the light speed postulate, the light expands at c from a point that it halfway between the clocks at emission, but since the clocks are moving relative to this point, they are not equal distances from this point when the light arrives at each clock, like thus:

    http://home.earthlink.net/~parvey/sitebuildercontent/sitebuilderpictures/synch2.gif

    The light reaches one clock first and then the other, and one clock lags behind the other from then on.

    Remember, these animations show the exact same situation, just from different frames of reference.

    Thus in the frame of reference of the clocks, they always read the same time while running, and from the frame in which the clocks are moving, they always read different times once running.

    This is the Relativity of Simultaneity.
     
  10. Aug 12, 2009 #9
    A general point on light-clocks. I am bothered by what seems to be a paradox. Observers O1, O2 are in separate inertial frames having relative motion (no acceleration) toward each other. Each one has a light pulse bouncing back and forth between two reflective surfaces in a direction perpendicular to the relative motion. Each observer assumes he is stationery and sees his single light pulse (generated at an earlier time), hitting the same spots on the two reflective surfaces every time. If we were dealing witih billiard balls this would be expected, because any relative velocity component would be a component of the billiard ball velocities. Light pulses by contrast do not retain any velocity component of the device which generated them, in which case how can the contact points on the surfaces of the mirrors in both systems (moving toward each other always be the same?
     
  11. Aug 12, 2009 #10

    Doc Al

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    As seen by O1, O2's light pulse will have a velocity component in the direction of relative motion equal to that of O2.
     
  12. Aug 12, 2009 #11
    Thanks for the quick reply. I assumed that because carrying a flashlight does not add that forward component to the velocity of the beam, which is always 'c', that the same holds true when a laser moving horizontally emits a pulse. I thought that once a pulse has been emitted, that it continues in a straight line (barring gravitational fields), starting at the coordinates of the laser at the time of emission. Does the horizontal component in my example come from momentum due to the reflection from the moving mirrors?
     
  13. Aug 12, 2009 #12

    Doc Al

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    If in one frame (O2) you fire a laser, you can find the velocity of the beam as seen by a different frame (O1) by adding the velocity of the frame. (You must add the velocities relativistically, of course. The speed of light will always be the same, but the direction of travel will vary with frame.)

    Adding a velocity component in the direction of the light beam won't change anything.
    That's true.
     
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