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A few simple questions about clocks moving slow relative to observers

  1. Aug 20, 2010 #1
    Hello,

    I want to make these questions as simple as possible, so that no one answering it will have to use any mathematics or formulas to explain it. I don’t mean this in a disrespectful way, I’m simply interested in the specific issues.

    Assume that I am an observer on earth, and a spaceship with a traveler onboard has just passed my location at a near-light velocity heading to a distant star. Assume that all of the spaceship’s acceleration to achieve such velocity has already occurred prior to its getting to my location, so that it is now moving at constant velocity relative to me.

    Further assume that both I and the traveler have some type of unique vision, so that we can each perceive any change (even a trillionth of a trillionth… of a second) in a clock movement no matter how far away the clock is.

    Given the above, I have these four questions:

    First, based upon everything that I’ve read, will we BOTH measure the other’s clock moving slow, relative to our own clock?

    Second, won’t each of us measure our own clocks as running normally?

    Third, since we are moving away from each other with CONSTANT velocity, won’t we both also measure the other’s clock moving SLOWER AND SLOWER relative to our own clock as the distance between us increases? In other words, won’t we each measure the other’s clock slowing down more and more as our distance increases and as the light takes longer and longer to reach us?

    Finally, if the reason that we both measure the other’s clock running slow is because with each increase of distance, the light from the clock will take longer to reach us, why do we need special relativity to teach us this? Wouldn’t this fact be a logical effect of two clocks moving away from each other in Newtonian space as well?

    Thanks
     
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  3. Aug 20, 2010 #2

    bcrowell

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    But keep in mind that this vision can only receive information at the speed of light. Therefore, it might be easier just to imagine radio signals instead of special vision.

    Yes.

    Yes.

    No. Say they're sending you signals at what they perceive as one-second intervals. You will perceive them as arriving at some longer interval, say 7 seconds. This 7-second interval stays constant.

    Each of you explains your observations as a combination of two effects: (1) time for signals to propagate, and (2) time dilation.
     
  4. Aug 20, 2010 #3
    On my third question, let us assume that it is not the traveler on board who perceives that he is sending me signals at what he assumes to be one second intervals, but rather that I am actually seeing the light from the clock as it ticks away. In such case, won't I notice that the intervals of time between ticks increases with each change of distance.

    And on my last question, why is not simply the fact of the distance increasing enough to account for the measured relative slowing of the clocks, without having to rely on time dilation for the effect?
     
  5. Aug 20, 2010 #4

    JesseM

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    No, each successive signal has the same amount of "extra" distance to travel (as compared to the previous signal) as every other signal, so the fact that the source is traveling away from you just causes the rate you see signals to be slower than the rate signals are "really" being emitted in your frame by a constant amount (this is the Doppler effect).

    For example, suppose the source is sending out signals once every 20 seconds according to its own clock, but it's traveling away from you at 0.6c, which means in your frame it only sends out signals once every 25 seconds due to the time dilation factor of 1/sqrt(1 - 0.6^2) = 1.25. Since it's moving at 0.6c, every 25 seconds its distance from you increases by 0.6c*25 = 15 light-seconds. So, each signal has an "extra" distance of 15 light-seconds to reach you, as compared to the previous signal, and thus the signals only reach your eyes once every 25 + 15 = 40 seconds. If you had three successive signals with the first emitted at T=0 in your frame at a distance of X=100 light-seconds from you, then the second would be emitted at T=25 in your frame at a distance of 115 light-seconds, and the third would be emitted at T=50 in your frame at a distance of 130 light-seconds. So, the first would take 100 seconds to reach you and so you'd receive it at T = 0 + 100 = 100, the second would take 115 seconds to reach you and so you'd receive it at T = 25 + 115 = 140, and the third would take 130 seconds to reach you and so you'd receive it at T = 50 + 130 = 180. You can see from this that the gap between your receiving successive signals is always a constant 40 seconds, double the rate they are being sent in the frame of the source (once every 20 seconds), which is what would be predicted by the relativistic Doppler shift equation with v=-0.6c.
     
    Last edited: Aug 20, 2010
  6. Aug 20, 2010 #5

    bcrowell

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    The classic demonstration of this is the twin "paradox," in which the rocket ship comes back to earth, and less time has passed for the travelers than for the people back on earth. This has been verified in the Hafele-Keating experiment: http://www.lightandmatter.com/html_books/6mr/ch01/ch01.html [Broken]
     
    Last edited by a moderator: May 4, 2017
  7. Aug 20, 2010 #6

    jtbell

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    The classical (non-relativistic) Doppler effect is exactly this. The relativistic Doppler effect starts with this same principle, and in addition includes the effect of time dilation ("slowing down") of the moving clocks.

    http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/reldop2.html

    The difference between classical and relativistic Doppler effects is measurable at relativistic speeds, and experiments confirm the relativistic version.

    In addition, if the source moves at right angles to your line of sight, there is no classical Doppler effect because the distance between you and the source is constant. There is still a relativistic Doppler effect (called the transverse Doppler effect) in this case, because of time dilation, and it has been observed.
     
  8. Aug 20, 2010 #7
    The correct answer hinges on how you define "measure". If you define measure to mean the actual values you get from a measuring instrument, in this case a device measuring the rate of incoming light pulses, then here is an example:

    Suppose we have two identical spaceships that are sending one light pulse a second. They are approaching (and later retreating) a point in the center between them with a speed of 1/3c.

    Then each spaceship will:
    • measure the speed of the other ship as 0.6c.
    • measure the rate of the pulse of the other ship to be 2 pulses a second when approaching.
    • measure the rate of the pulse of the other ship to be 0.5 pulses a second when retreating.
    • conclude that the clock rate in the other ship is 0.8.
    • conclude that both ships have a clock rate of about 0.94 ([itex]\sqrt{8}/3[/itex] to be exact) with respect to a clock in the center.
    Now the conclusion that the other clock is going slower cannot be directly measured but is inferred by neutralizing the standard (not relativistic) Doppler effect and taking into account the finite speed of light.

    A noteworthy speed in this scenario is if the relative speed between the point in the center is [itex]\sqrt{2}-1[/itex].

    Then each spaceship will:
    • measure the speed of the other ship as [itex]1/\sqrt{2}[/itex].
    • measure the rate of the pulse of the other ship to be [itex]\sqrt{2}+1[/itex] when approaching.
    • measure the rate of the pulse of the other ship to be [itex]\sqrt{2}-1[/itex] when retreating.
    • conclude that the clock rate in the other ship is [itex]1/\sqrt{2}[/itex].

    Talking about [itex]\sqrt{8}/3[/itex], if the relative speed between the spaceships is [itex]\sqrt{8}/3[/itex] then the relative speed between the point in the center and any spaceship is [itex]1/\sqrt{2}[/itex].

    Now who wants to guess what the root and fraction of the relative speed between the spaceships is when the relative speed between a point in the center is [itex]\sqrt{8}/3[/itex] :tongue:
     
    Last edited: Aug 21, 2010
  9. Aug 22, 2010 #8
    After [itex]\sqrt{8}/3[/itex] the next term is [itex]\sqrt{288}/17[/itex] and then [itex]\sqrt{332928}/577[/itex].
    Of course much more fun would be to stay inline with [itex]\sqrt{2}[/itex] by having:
    [itex]2\sqrt{2}/3[/itex], [itex]12\sqrt{2}/17[/itex] and [itex]408\sqrt{2}/577[/itex].
    Then watch the numbers if you take the square of the Doppler shift which gives the same results.

    [itex]2\sqrt{2}+3[/itex], [itex]12\sqrt{2}+17[/itex] and [itex]408\sqrt{2}+577[/itex]

    The same numbers used in a different way.
     
    Last edited: Aug 22, 2010
  10. Aug 23, 2010 #9
    Hi Lbrandt,
    Special Relativity Exits for good reasons and time dilation is necessary.
    The slowing of the clock can be well accounted by calculating light travel times in the classical way. (however note that even in the classical method, the slowing will be constant, say you will always see the clock tick at 1 tick per 5 (your) seconds. So, you need to rethink your thought about slowing down more and more with increased distance)
    Back to the point. But your calculated slowing using only classical methods (This calculation is what has been called the Doppler Effect (classical) Calculation), will be slightly off for small velocities and highly off for relativistic Velocities.

    When you make calculation using classical approach you assume that 'actually' both clocks tick at the same rate, and account the dilation only to the light travel time.

    In the relativistic Dopler effect calculation; You chose one of the observer, then you proceed as previous calculation for light travel times, but this time you take the other clock to 'actually' tick quite slowly (as predicted by time dilation formula).

    And one final thing that must be told is
    For this scenario, Your classical approach (light travel time calc) OR Time Dilation formula "ALONE" won't give the answer.
    YOU MUST USE BOTH. AND THIS IS THE RELATIVISTIC DOPPLER EFFECT. AND THAT WILL GIVE THE RIGHT ANSWER.
     
  11. Aug 23, 2010 #10
    Light travel time is simply the square of the relativistic Doppler formula. To calculate (as opposed to philosophize) you do not need to 'think' about the rate of the other clock.

    If you know the velocity you can calculate the frequency shift.
    If you know the frequency shift you can calculate the velocity.
    If you know the change in radar roundtrip time you can also calculate the velocity.

    There is no need to 'think' about the clock speed at the other end, or 'think' that the relativistic Doppler formula is some combination of things.

    Simple formulas, simple answers!
    Relativity is very simple, I think it is the 'explanation' of unmeasurable things that confuses most.
     
    Last edited: Aug 23, 2010
  12. Aug 23, 2010 #11

    JesseM

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    That doesn't seem to make sense, light travel time depends on the distance at which a signal was emitted, but the relativistic Doppler formula doesn't depend on the distance of the source, only on the velocity of the source. Take my earlier example:
    Do you agree that for the three signals in the example, the first signal had a travel time of 100 seconds in the receiver's frame, the second signal had a travel time of 115 seconds, and the third had a travel time of 130 seconds? If not, how are you defining the phrase "light travel time"? Are you defining it in some other way than (time coordinate signal was received) - (time coordinate signal was sent)?
     
  13. Aug 23, 2010 #12
    JesseM if you want to measure the radar roundtrip time there is no need to consider the clock rate at the reflection point. One can easily determine the velocity of a moving mirror by looking at the increase or decrease of light travel time, what I am saying is that this is directly related to the relativistic Doppler shift.
     
  14. Aug 23, 2010 #13

    JesseM

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    Gotcha, I thought you were talking about one-way travel times, it is true that if you send a series of time-coded signals to a mirror which bounces them back at you, you can figure out the roundtrip time for each signal just by (time signal returns) - (time it was sent). Note however that in this case the Doppler shift between the frequency at which you send signals and the frequency at which you receive them back from the mirror you will not be given by the relativistic Doppler shift formula, instead it will be given by the classical Doppler formula [tex]f_{received} = f_{sent} \frac{c}{c-v} = f_{sent} \frac{1}{1 - v/c}[/tex] (where v is the velocity of the mirror, positive v indicating it's moving towards you and negative v indicating it's moving away)--only if we consider the rate at which the mirror is bouncing back signals in its own frame (according to its own clock, which is time-dilated in our frame) would the relativistic Doppler formula apply. Do you disagree?

    edit: Actually I made a mistake here, the above formula would be correct if we were comparing the frequency I was receiving signals with the frequency they were being bounced back by the mirror (in my frame), i.e. it'd be true that [tex]f_{received} = f_{bounced} \frac{1}{1 - v/c}[/tex], but the frequency at which I am sending signals is related to the frequency at which the signals bounce back from the mirror by [tex]f_{bounced} = f_{sent} * (1 + v/c)[/tex] (again with v positive if the mirror is coming towards me), so the correct formula (still different than the normal relativistic Doppler shift formula) would be:

    [tex]f_{received} = f_{sent} \frac{1 + v/c}{1 - v/c}[/tex]

    To see that this formula works, consider a mirror moving away at 0.6c, and suppose I am sending a signal once every 20 seconds. Then when one signal reaches the mirror and bounces back the next signal will be (20 seconds)*c = 20 light-seconds behind it, but since the next signal is moving at 1c and the mirror is moving at 0.6c the "closing speed" between them is only 1c - 0.6c=0.4c, so the next signal will take a time of (20 s)*c/(1c -0.6c)=(20 s)/(1 - 0.6) = 50 seconds to reach the mirror and bounce back towards me after the previous one hit the mirror (so if the period of my sending signals is 20 seconds and the period of them being bounced back is 20/(1 - 0.6), the frequency of sending signals is 1/20 while the frequency of signals being bounced back is (1 - 0.6)/20, which agrees with the formula [tex]f_{bounced} = f_{sent} * (1 + v/c)[/tex] keeping in mind that a velocity of 0.6c away from me means v=-0.6c). And at the time the second signal bounces off the mirror, the first signal has moved a distance of 50 seconds*1c = 50 light-seconds closer to me than the position it hit the mirror, and meanwhile the mirror has moved 50*0.6c=30 light-seconds in the opposite direction, so the distance between the two signals at the moment the second one bounces back is 50*(1c + 0.6c)=80 light-seconds. And 50 seconds was obtained by taking the time between signals being sent (i.e. 1/fsent), 20 seconds, multiplying by c to get 20 light-seconds, and then dividing by (1c - 0.6c)...since v=-0.6c here, the more general version of this would be [(c/fsent)/(c + v)]*[c - v], or (c/fsent)*[(c-v)/(c+v)]. And this is the distance between signals coming back towards me after being bounced back from the mirror, so the period between receiving signals is this distance divided by c (80 seconds in the example), or (1/fsent)*[(c-v)/(c+v)]. And of course frequency = 1/period, so the frequency at which I receive signals back is fsent * [(c+v)/(c-v)].
     
    Last edited: Aug 23, 2010
  15. Aug 24, 2010 #14
    You brought me some doubts :)

    So I started calculating. For a retreating mirror with velocity v (x0, t0 = initial mirror and emitter position = 0, and instant acceleration to v) I get:

    [tex]\frac{1}{ (v-1)^2} t[/tex]

    So for instance a signal sent from x0 at:
    • t=10 for v=0.5 will return at x0 at t = 40 seconds.
    • t=10 for v=0.6 will return at x0 at t = 62.5 seconds.
    • t=10 for v=0.8 will return at x0 at t = 250 seconds.
    Can you confirm?

    Edited to correct for v=0.5: instead of t=50 I got t=40
     
    Last edited: Aug 24, 2010
  16. Aug 24, 2010 #15

    JesseM

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    If the mirror is moving at v=0.5, and started from x=0 at t=0, then the mirror has x(t) = 0.5*t. Meanwhile if the light signal is sent from x=0 at t=10 and is moving at velocity 1 (in units where c=1), then the light signal has x(t)=1*t - 10. So, the light will reach the mirror when 0.5*t = 1*t - 10, or 0.5*t = 10, or t=20. At that moment, both the mirror and the light will be at position x=10. So, it will take another 10 seconds for the light to return to x=0, meaning the signal will return at t=30.

    Note that this is also what is predicted by my formula [tex]f_{received} = f_{sent} \frac{1 + v/c}{1 - v/c}[/tex]--since frequency = 1/period, it can be flipped to give (period between receiving successive signals) = (period between sending successive signals)*(1 - v/c)/(1 + v/c). Now, just imagine that I am sending signals out with a period of 10 seconds, and I send the first out at t=0 when the mirror is right next to me so it quasi-instantaneously bounces off the mirror and returns to me at t=0. How long will it take for me to receive the next signal, according to the formula? With (period between sending successive signals) = 10, the formula says (period between receiving successive signals) = 10*(1 - v/c)/(1 + v/c), and with v/c=-0.5 (remembering that positive v means the mirror is moving towards me), this is equal to 10*(1 + 0.5)/(1 - 0.5) = 10*(1.5/0.5) = 10*3 = 30, so I should receive the second signal (the one sent at t=10) at t=30.
    Again using the same formula, the signal should return at 10*(1 + 0.6)/(1 - 0.6) = 10*(1.6/0.4) = 10*4 = 40 seconds. This is also what you conclude if you say the mirror has x(t) = 0.6*t, the light has x(t)=1*t - 10, so the light reaches the mirror when 0.6*t = 1*t - 10 which means 0.4*t = 10 so t=10/0.4=25. At t=25 the mirror and the light are 15 light-seconds away, so the light takes an additional 15 seconds to return, meaning it'll return at t=25+15=40.
    Using the formula, the signal returns at 10*(1 + 0.8)/(1 - 0.8) = 10*(1.8/0.2) = 10*9 = 90 seconds. And doing the full analysis again, the mirror has x(t) = 0.8*t, the light has x(t)=1*t - 10, so the light reaches the mirror when 0.8*t = 1*t - 10, meaning 0.2*t = 10, so t=10/0.2=50. And at t=50 the mirror and light are 40 light-seconds away, so the light takes an additional 40 seconds to return, meaning it returns at t=50+40=90.
     
  17. Aug 24, 2010 #16
    I noticed a mistake in transcribing the results for v=0.5 instead of 50 I got 40.

    But at any rate they are different from yours:
    • t=10 for v=0.5 will return at x0: I have 40 while you have 30.
    • t=10 for v=0.6 will return at x0: I have 62.5 while you have 40.
    • t=10 for v=0.8 will return at x0: I have 250 while you have 90.
    It looks like our calculations agree on the trip towards the mirror but not from the mirror back to the source.

    As soon as the light ray bounces the mirror and wants to go back you say it takes 10 seconds because the mirror is 10 light seconds from the source right? However from the perspective of the mirror the source is moving away at 0.5 right? So I included that in my calculations, you are saying that is not necessary? Are you saying we do not need to consider they are not at rest with respect to each other as soon as the photon bounces?

    If you are right, and it looks like you are, I should not have doubted my memory as it was indeed what I claimed before: the square of the relativistic Doppler factor! :biggrin:
     
    Last edited: Aug 24, 2010
  18. Aug 24, 2010 #17
    Could you explain what you mean by the square of the Doppler factor??
    I think it seems clear the relative v of the source is not relevant after the bounce as the radar distance is in the frame where the emitter is at rest , no???
    Thanks
     
  19. Aug 24, 2010 #18
    Well the relativistic Doppler factor is:

    [tex]\sqrt{ \frac{1 + v}{1 - v}}[/tex]

    So the square is simply:

    [tex]\frac{1 + v}{1 - v}[/tex]

    Yes, I agree. I was wrong!

    So the formula becomes:

    [tex]\frac{1 + v}{1 - v}t[/tex]
     
    Last edited: Aug 24, 2010
  20. Aug 24, 2010 #19
    Yep, JesseM is right.

    From the point of view of the mirror the source is moving away at 0.5c but the initial distance is L' = 10*sqrt(1-0.5^2) = 8.66 ls due to length contraction. The catch up time in the mirror frame is t' = L'/(c-v) = 8.66/(1-0.5) = 17.32 seconds and the final distance (reflection event to catch up with the source event) is x' = 17.32 ls. To calculate the time measured in the source frame we use the reverse Lorentz transformation t = (t' + vx')/sqrt(1-v^2) = (17.32+0.5*17.32)/sqrt(1-0.5^2) = 10 seconds.

    It is of course simpler to observe, as Jesse did, that the distance in the source frame is 10 ls and it takes the light 10 seconds to travel that distance.

    In the source frame, the speed of the light when it is reflected is unaffected by the velocity of the mirror, so the velocity of the mirror can be ignored. In the mirror frame, the velocity of the retreating source does make a difference in the catch up time and has to be factored in.

    {EDIT} Oops, I see you had already decided you were wrong while I was typing this response and our posts overlapped. Sorry!
     
    Last edited: Aug 24, 2010
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