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Applying time dilation and length contraction to measure parralel c

  1. Mar 9, 2012 #1
    The title isn't quite clear, because the question was a little too long. Here it is in full:

    How to apply length contraction and time dilation to a moving observer to measure the speed of light of a light beam moving parallel to their reference frame.

    Imagery and familiarity are my learning tools.

    A vessel, whose pilot is named Cleopatra, is travelling at 2/3c. At an arbitrary moment and location, Cleo fires a beam of light from the nose of the ship. This event defines t=0 and x=0 for a nearby stationary observer, Anthoney. This event also defines t(prime)=0 and x(prime)=0 for Cleopatra.

    At t=1 for Anthoney, he sees that Cleopatra has travelled 200000km and the light beam has travelled 300000km. He also sees that the time according to Cleo is t/(gamma). where gamma is approx. 1/0.74 approx. 1.34.

    My current understanding is that, removing time dilation and length contraction, at this moment, t=1, Cleo will see that light has travelled just 100000km beyond the nose of the ship. Further, that by some process of length contraction and time dilation, this distance of 100000km can be adjusted such that when divided by the dilated time (0.74s) we will get some ratio of c. Specifically, it will need to be in the region of 223000km. Finally, that the process here is no different than if a beam of light had been emitted in the direction of motion at x=0 and t=0, by some 3rd party. i.e. we are only interested in the portion of the beam that has past the nose of the ship after any given time.

    The first check is whether or not this understanding of mine is in fact correct.

    I created a minkowski diagram and a light clock diagram for v=2/3c. The light clock I find to be more intuitive, despite the use of a clock 300000km long. However, I'm no closer to revealling the means of resolving the original question/statement.
    Last edited: Mar 9, 2012
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  3. Mar 9, 2012 #2


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    hi salvestrom! :smile:

    i'm not entirely following this

    remember that cleopatra and anthony are not measuring the distance between the same pair of events

    the far event for anthony is at what he says is the same time as the near event

    the far event for cleopatra is at what she says is the same time as the near event :wink:
  4. Mar 9, 2012 #3
    I'm not really sure what you are referring to with your last two sentences. The distance between the nose of the ship and the end of the beam?

    The minkowski diagram I have, I did manage to add a few lines of constant time for Cleopatra. They clearly transect her timeline and the lightline at very different locations than for Anthoney. Infact, the diagram, which is to scale, shows that what Cleopatra is measuring is a line some 2.3 times longer than what anthoney sees (which is in the ballpark of the figure we should be expecting approx. 223000km).

    Would it help if I provide the minkowski diagram with an arrow that says "I want to know how long this line is?" :) I r scared of imageshack!

    EDiT: I would also like confirmation regarding the point of the main paragraph, partiularly that Cleopatra will only measure the length of the light she sees ahead of her ship when determing c.

    EDiT: While looking at my minkowski diagram again to take another look at the line of constant time for Cleo I got hit by a moment of inspiration. I had already used a formula involving sqr root of c^2 - v^2 multiplied by gamma and time etc and gotten the answer, but I put it aside because it seemed only to relate to the lightclock. However, I drew two spaures, corner to corner, one of length v and the other of length c. Over laying this the ends of the hypotenuse match the slope of the line of constant time, but not the length of the slope. So I resized the squares by gamma and discovered they fitted quited precisely so that the beginning of the hypotenuse was at t(prime)=1 and the end of same was at the precise point that the t(prime)=1 timeline crossed the lightline. Essentially, I seem to have finally resolved for myself how to derive c for the moving observer.

    It seems a lorentz transformation must be performed first? There's no direct way to get from the 100k that Anthoney sees to the 223k Cleo sees?

    Is it correct that they are essentially not really measuring the same line?
    Last edited: Mar 9, 2012
  5. Mar 10, 2012 #4
    Cleopatra will see the beam to travel at c, Anthony will also see the beam to travel at c. When Anthony looks at his clock he sees 1s has elapsed he looks at his measuring rod and sees that the beam has reached 300000 km. But when he looks at Cleopatra's clock he will see 1/γ s and when he looks at the measuring rod attached to Cleopatra's spaceship he sees 300000/γ km.
  6. Mar 13, 2012 #5
    I think the best way to think about this is to look more closely at your two events, particularly the second event.

    Event A: Light is emitted. tA’ = tA = 0; xA’ = xA = 0
    Event B: Light reaches a distance in Anthony’s frame 300,000,000 m away

    For Anthony, the co-ordinates of Event B:

    tB = 1
    xB = 300,000,000 m

    For Cleopatra, the co-ordinates of Event B:

    The corresponding Lorentz transformation for the time of this event determined in Cleopatra’s frame, t’, is:

    tB’ = γ(tB – vxB/c2)


    xB is the distance, in Anthony’s frame from his origin to the point where the light is i.e. = 300,000,000 m
    γ = 1.34
    tB = 1 s
    v = 200,000,000 m/s

    Therefore tB’ = 0.447 seconds.

    In Cleopatra’s frame, the time the light has been travelling for is:
    t’B – t’A = 0.447 s so Cleopatra’s time appears to Anthony to be running more slowly than his own.

    Distance between front of Cleopatra’s ship and the point the light has reached

    In Anthony’s frame, the distance travelled by the light is 300,000,000 m (So Anthony does indeed determine that the light is 100,000,000 m ahead of the ship).

    But in Cleopatra’s frame, the location of this event is given by the Lorentz transformation:

    xB’ = γ(xB – vtB)

    γ = 1.34
    xB = 300,000,000 m
    tB = 1 s
    v = 200,000,000 m/s

    Therefore xB’ = 134,164,000 m. Therefore, Cleopatra reckons the light has travelled a distance:

    x’B – x’A

    i.e. to her the light has travelled 134,164,000 m since it left her ship.
    To her, it is this distance away from her origin – and her origin is the spaceship itself. In her frame of reference, she isn’t moving at all.

    So Cleopatra’s value for the velocity of light would be:
    134,164,000 / 0.447 = 300,000,000 m/s
  7. Mar 13, 2012 #6
    Just an afterthought.

    Cleopatra's dtermination of how far event B is in front of her ship was 134,164,000 m.

    But, due to Lorentz contraction, this reduces to 100,000,000 m as viewed from Anthony's frame.
  8. Mar 13, 2012 #7
    Hmm. I've been using 0.745 as the time in t(prime) (when t=1). Essentially t(prime) = t/gamma.

    The t(prime) of 0.447 is equaly to what the unprimed observer sees if we take account of the travel time of light. I.e. any signal emitted by Cleopatra at 0.447 seconds arrives at Anthoney at 1.0s. Anthoney therefore sees Cleo's clocks running almost 2.25 times slower than his own. I was under the impression that it was actually 1.34 times slower.

    I think I understand, though. They are each constantly looking into the others past. At t=1 for Anthoney the pecise same instant is t=0.74 for Cleo. However, neither can instantly see that, and are bound only to witness what light shows them. In this case, that at t=1 for Anthoney, he can only see information from t=0.447s for Cleo.

    So, the time dilation formula will tell you what the time difference between the clocks would be if they were brought back together, while the transformations project from present to past or vice versa. This would explain why my original measurements appeared to require cleo to measure a point on the lightline in both their futures.

    It's all rather cunning.
    Last edited: Mar 13, 2012
  9. Mar 13, 2012 #8
    First, gamma is defined as [itex] 1/\sqrt{1-v^2/c^2}[/itex] which in this case is approximately 1.341 and so the time according to Cleo is t/(gamma) = 1/1.341 = 0.745 seconds approx (when the time according to Anthony is 1 sec).

    Stay with that.

    Don't use light travel times. Lorentz transformations do not use them and if you use them you will become totally confused. Anthony does see Cleo's clock running 1.34 times slower.

    In these type of problems when we talk about "an observer" we actually mean an array of observers that are all at rest with respect to each other who all have synchronised clocks and note the time of any events that happen at their immediate location. This collective set of observers then compare notes at a later time to analyse what has happened. For example we might have Anthony0 at the origin, Anthony1 at Cleo's location at t=1 and Anthony2 at the location of the photon at t=1. When we say the time of an event according to Anthony, we mean the time according to the version of Anthony standing right next to the event so light travel times are not an issue. Similarly we we have Cleo0, cleo1 and Cleo2 etc. all at rest with respect to each other and all with clocks synchronised with each other.

    To make things simpler and clearer the photon will now be emitted from the back of Cleo's ship where Cleo0 is located and Cleo1 is somewhere midway along the ship and Cleo2 is at the nose of the ship and the ship is going to the right according to Anthony.

    Now consider the events in detail.

    Event 0

    This is the emission of the photon at t=t'=x=x'=0. Straight forward enough.

    Event 1

    This is where Cleo0 is at t=1 according to Anthony1 who happens to be standing 200,000 km to the right of the origin and he notes that the time showing on Cleo0's clock is 0.745 seconds as she passes close to him.

    Event 2

    Anthony2 is standing 300,000 km right of the origin and he sees the photon wizz by him at t=1 and calls this event 2 with coordinates, x=300000, t=1. At exactly the same time he sees Cleo1 in the mid section of the rocket pass right next to him. (It is a very long rocket).

    What time is showing on Cleo1's clock as she passes Anthony1? You will need the Lorentz transform to calculate this.

    According to (the) Anthony(s) events 1 and 2 happened simultaneously at t=1 and the events are spatially separated by a distance of 100,000 km.

    When you have done the transformations you will find that Cleo0 and Cleo1 do not agree that event 1 and event 2 happened simultaneously. The location of the photon in the Cleo rest frame when Cleo0 passes Anthony1 at t=1 is not next to Anthony2 but somewhere further along. Let say that conveniently the photon passes the nose of the rocket (where Cleo2 is located) at t'=0.745 seconds and call this event 3. See if you can work out the time and location of event 3 in Anthony's reference frame and the rest length of the rocket. I am sure I and others here will assist you if you have a good go at the problem and get stuck.

    It will help to visualise things if you plot all events on a spacetime diagram from Anthony's point of view and then plot all the same events on a spacetime diagram from Cleo's point of view. Remember that in Anthony's rest frame, simultaneous events are on a horizontal line (assuming the t axis is the vertical one) and events that are simultaneous according to the Cleos will not be a horizontal line on Anthony's spacetime diagram. The reverse is true when you plot events on Cleo's spacetime diagram.
  10. Mar 13, 2012 #9
    As far as I can tell, you're wrong about this. The Lorentz Transformations give you numbers that correspond quite precisely to what Anthoney or Cleopatra see when light travel time is taken into consideration.

    Prior to making the original post I already had a spacetime diagram and have since added a large number of light lines. These match the numbers. At 1 second for Anthoney he will recieve light/information that was emitted or reflected by Cleopatra at 0.447s. Although, when the clocks are brought back together the differece will be determined purely by dividing t by gamma, at no point on their journey can they see this relationship. t=1 for Anthoney and t=0.74 for Cleo are seperated by 200000km, and so Anthoney cannot "see" what Cleo's clock says at this exact moment. As far as I can tell, based purely on the information they recieve, both will consider that the others clocks are running 2.23 times slower. The actual difference is lost beneath the every increasing distance between them.

    EDiT: Infact the entire transformation for t can be substituted with t/2.236 and you will always get t(prime). 2.236 is obviously only valid for this particular velocity.

    EDiT(2): For reasons that escape me at the moment 2.236 is equal to 3/gamma. It seems slightly odd that the entirre transformation can be substituted as [itex]\gamma[/itex]t/3.
    Last edited: Mar 14, 2012
  11. Mar 14, 2012 #10
    Yes, things can get very tricky if you try to work out what people actually ‘see’, as opposed to ‘observe’, due to the finite travel time of light. For this reason, special relativity best concerns itself with determining the co-ordinates of a particular event (or series of events) as 'observed' in each frame of reference. Observations, in the sense of special relativity, are to do with the readings on clock and measuring stick immediately adjacent to an event. You can work with how things would actually appear to an individual, but I find I always get into a mess. The distinction between ‘see’, in the everyday sense, and ‘observe’ in the relativity sense, is crucial, as they are two completely different things.

    Some texts imagine vast numbers of observers in a frame of reference, strewn out across space, reporting back to a guy at the origin the clock and ruler readings of an event that has occurred right next to them.

    Another thing to remember is that in Anthony’s frame, time in Cleopatra’s frame is asynchronised (and, this being relativity, vice versa).

    When Cleopatra flies past Anthony and emits the light, both their personal clocks read zero. Other clocks that are strung out in front of and behind Cleopatra, but moving with her (i.e. stationary with respect to her in her frame), will have different readings as observed by Anthony. Clocks further ahead of her in space would be running progressively further and further behind Cleopatra’s clock. Whereas clocks behind Cleopatra would be running ahead of hers and already showing times greater than zero at the instant she passes Anthony.

    When the light reaches the particular position after 1 second has passed in Anthony’s frame, Anthony would determine the time showing on Cleopatra’s personal clock to be:

    t’ = γ(t – vx/c2)


    t = 1s
    v = 200,000,000 m/s
    x = distance from Anthony to Cleopatra = 200,000,000 m

    i.e. t’ on Cleopatra’s personal clock would read 0.745s.
    So while 1 second has passed for Anthony (t), only 0.745s (t’) have passed for Cleopatra, confirming your earlier statement about time dilation, t’ = t/γ

    As mentioned, at the very same instant in Anthony’s frame, the light has already reached a distance of 300,000,000 m from his origin, and, as pointed out earlier, a clock moving along in Cleopatra’s frame that is immediately next to this event would be observed by Anthony to read 0.447 s.

    If Anthony had noted the reading on this particular clock at the instant Cleopatra passed him (t = t’ = 0), he would have observed that it was displaying 0.447 – 0.745 = - 0.298 s.
  12. Mar 14, 2012 #11
    I always forget something.

    So, Cleopatra's clock has advanced by 0.745 seconds while Anyhony's has advance by 1 s.

    And, the last clock I was talking about (the one that ends up right next to where the light has reached) has, also advanced by 0.745 s during the second that passes in Anthony's frame (going from -0.298s to 0.447s). It has demonstrated, if you like, exactly the same degree of time dilation as Cleopatra's personal clock.
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