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Question about the x term in time dilation

  1. Aug 3, 2013 #1
    When I first learnt time dilation I was given this equation proven using the light clock example.

    t = γt'

    However, when I looked into the topic in more detail, I was given this equation.

    t = γ(t' + xv/c2)

    If x represents the distance between 2 frames, then does x have to change consistently as 1 frame moves away from the other?
     
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  3. Aug 3, 2013 #2

    ghwellsjr

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    Where'd you get that second equation from? It looks like a messed up version of the equation for the time coordinate for the Lorentz Transformation which is used to convert the coordinates of an event in one Inertial Reference Frame (IRF) into the coordinates of a second IRF moving at speed v along the x-axes.
     
    Last edited: Aug 3, 2013
  4. Aug 3, 2013 #3

    Nugatory

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    The first equation describes time dilation, the way that one observer's time interval may not be the same as another's. You could save yourself some confusion by writing it as ##\Delta{t}=\gamma\Delta{t'}## instead, to emphasize that it is about the interval between t values, not the t values themselves.

    The second equation is one of the Lorentz transformations (written a bit oddly). These tell us how to convert one observer's statement "At time t and position x something happened" into t' and x' values that would make sense to the other observer looking at the same event.

    The Lorentz transformations are far more basic and important than the time dilation formula. In fact, you can easily derive the time dilation formula from the Lorentz transformations by working with statements of the form "At time ##t##, a clock was at position ##x## and its hands were pointing to ...".

    Yes. I'm driving a car, I'll say that the tip of my nose is the point x=0. If you're sitting by the side of the road watching me drive past, you would say that the position of the tip of my nose is x'=vt' where x' and t' are your time and distance measurements and v is our relative speed; intuitively, the longer you watch me drive away, the farther away from you the tip of my nose is.
     
    Last edited: Aug 3, 2013
  5. Aug 3, 2013 #4
    So x is the distance between frames, then this is my problem.

    Suppose a spaceship travels past at 0.5c, the spaceship passes an observer on a planet.
    The observer measures 1 second on his clock and a clock on the spaceship will spin for 0.75s using the time dilation formula.

    However, after 1 second, the spaceship is 0.5c meters away from the planet. If the observer on the planet measures another second, does the clock on the spaceship pass 0.75s or 0.25s? 0.25s by using t = γ(t' + xv/c2) and making x = 0.5c.
     
  6. Aug 4, 2013 #5

    ghwellsjr

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    I cannot figure out what you are doing. Could you show your intermediate steps, including what your value for γ is and how you calculated it?

    In the mean time, this is how I would do the calculation based on the correct Lorentz Transformation equation for time:

    t' = γ(t - xv/c2)
    where γ = 1/√(1-v2/c2)

    At v = 0.5c, γ = 1/√(1-(0.5c)2/c2) = 1/√(1-0.25) = 1/√(0.75) = 1/0.866 = 1.1547

    t' = γ(t - xv/c2) = 1.1547(1 - (0.5c)(0.5c)/c2) = 1.1547(1 - 0.25) = 1.1547(0.75) = 0.866

    This, of course, does not match your value of 0.75 from the time dilation formula. But maybe if you would show how you calculated γ and how you applied the formula, we might be able to figure out why you got a different answer than I got.

    Also, I'd like to repeat my question: where'd you get that second equation?
     
  7. Aug 4, 2013 #6
    is just the usual time dilation (how fast does the clock run in another system).

    is the lorentz transformation for the time variable.

    The second part of the second equation (##xv/c^2##) has to do with simultaneity,
    which basically says that if something happens to be in order AB for observer1 it does not have to
    be AB for observer2. It could also be BA (if they are causually not related).

    ##x## is the coordinate of an event in the ##S## reference frame.

    This should be correct if you used the correct values.

    It should still be 0.75s.
    This has nothing to do with ##t = γ(t' + xv/c^2)##.
    This equation describes events, not time itself.

    Maybe you should check
    -simultaneity (http://en.wikipedia.org/wiki/Relativity_of_simultaneity)
    -synchronizing clocks
    -time dilation
    -lorentz transformations

    It should be clear then.
     
  8. Aug 4, 2013 #7

    ghwellsjr

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    I'm going to show you how you can use the Lorentz Transformation to analyze your scenario from post #4:

    Let's start in the rest IRF for the spacecraft:

    attachment.php?attachmentid=60682&stc=1&d=1375623483.png

    The black dots represent one-second intervals of time aligned with the Coordinate Times starting at -7 seconds to +7 seconds. Each dot represents an event with both a time and a location coordinate (along the x axis).

    Now we want to transform these events into the rest IRF for the observer on the planet. To do this, we will note that as far as the spacecraft is concerned, the planet is traveling to the left at 0.5c so we want to use v=-0.5c. I like to use units where c=1, in this case seconds and light-seconds. This allows us to use simplified equations based on β=v/c and β=-0.5. The equation for gamma is:

    γ = 1/√(1-β2)

    I already evaluated this in post #5 so I won't do it again:

    γ = 1.1547

    Now we have to use both Lorentz Transformation equations:

    t' = γ(t-xβ)
    x' = γ(x-tβ)

    Note that on the right side of the equations there are no primed terms, they are both on the left sides. The way you use these equations is you pick an event in the original IRF and plug its x (location) and t (time) coordinates into the equations to calculate the x' and t' coordinates for the same event in a second IRF moving at β with respect to the original IRF.

    So let's use the uppermost event with coordinates of x=0 and t=7. Plugging these values into the two equations we get:

    t' = γ(t-xβ) = 1.1547(7-0) = 8.0829
    x' = γ(x-tβ) = 1.1547(0-7*(-0.5)) = 1.1547(3.5) = 4.04145

    If we calculate the events for the bottom event we will see that it has the same values but they are the negatives.

    Now you could repeat the calculations for all the other events or you could just realize that the dots will be equally spaced in the new IRF and just mark them in if you are drawing a diagram by hand. I use a computer program so we will get:

    attachment.php?attachmentid=60683&stc=1&d=1375623649.png

    As you can see, the dots representing 1-second intervals are spaced farther apart--they are dilated. We call the time for the spacecaft, the Proper Time of the spacecraft. For convenience, we will call the Proper Time when the Coordinate Time is zero to also be zero and count the dots upwards or downwards from that point accordingly to determine the Proper Time of any other dot.

    Now since this is the rest IRF of the observer on the planet, we can just draw him in using blue for his color:

    attachment.php?attachmentid=60684&stc=1&d=1375625708.png

    Now I think you can see that at the observer's Coordinate Time of 1 second, the Proper Time of the spacecraft is a little less than 1 second. This is where your calculation comes in.

    The event (not shown by a dot) that you want to consider has coordinates of t=1 and x=0.5 and β is now 0.5 (because the spacecraft is moving to the right in the planet's rest IRF). You just plug those into the same two equations we used before and get coordinates of:

    t' = 0.866
    x' = 0

    I previously did the calculation for t' in post #5 so I won't repeat it here and you can see that the x' value evaluates to zero (as it must since we are transforming to the rest IRF of the spacecraft).
     

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  9. Aug 4, 2013 #8
    I think I am starting to understand this now, the second equation doesn't describe an interval of time, but a moment in time.

    This equation t' = γ(t-xβ) (should this be t' = γ(t-tβ) instead?) calculates the time an event happens in the frame of s' given the time and position of the event happening in s. In the equation that I posted I forgot the ', it should say t = γ(t' + x'v/c2), hopefully this clears up my mistake. :frown:

    I am not seeing where this part appears in the diagram, would the proper time of the spacecraft be the 0.866s calculated? The distance between 2 events in the spaceship appear to be the value of the Lorentz transform 1.1547.
     
  10. Aug 4, 2013 #9

    ghwellsjr

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    No, it is correct as I gave it. It's important to realize that events in Special Relativity require coordinates in both space and time and changing either part in one IRF will change both of the parts in the transformed IRF. That's why we refer to "spacetime".

    That form of the Lorentz Transformation is used for converting the coordinates of an IRF back to its original IRF without changing the sign of the velocity. You should use the form as I presented it with the negative sign between the terms when going from s to s'. Then you can use the form with the plus sign to go back from s' to s (without having to change the sign of v or β).

    However, if you just use the same form all the time and consider s to be the "from" frame and s' to be the "to" frame, then you can use the same equations for going both ways as long as you are careful to adhere to the correct sign of v or β. One reason I prefer to do it this way is if you ever want to transform between three frames, then you have to invent new forms of the equations--why bother? Just understand what you're doing and you'll never have a problem.
    Yes.

    Yes, the Coordinate Distance in seconds of time between 2 events that are separated by 1 second of Proper Time is the Lorentz Factor, gamma, which is 1.1547. More precisely, the difference in Coordinate Time divided by the difference in Proper Time is gamma.
     
  11. Aug 4, 2013 #10
    The equation t = γ(t' + x'v/c2) creates t = γ(t' + t'β) with β=v/c and t' = x'/c, or t' = γ(t - xv/c2) into t' = γ(t - tβ).
    I am not sure where t' = γ(t - xβ) comes from.

    Are you talking about t = γ(t' + x'v/c2) or t' = γ(t - xβ)?
     
  12. Aug 4, 2013 #11

    ghwellsjr

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    If you start with the standard Lorentz Transformation equations as given by just about everyone, including wikipedia (just below the subheading, "Boost in the x-direction"), you have:

    t' = γ(t - vx/c2)

    But if we substitute β for v/c and use units where c=1 (seconds for time and light-seconds for location) then it becomes:

    t' = γ(t - βx) or γ(t - xβ)

    I'm talking about t = γ(t' + x'v/c2). This is also mentioned in the wikipedia article (just above the subheading, "Boost in the y or z directions"), where they talk about the inverse transformations.

    I still haven't seen your original calculations. How did you get 0.25s in the last sentence of post #4?
     
    Last edited: Aug 4, 2013
  13. Aug 4, 2013 #12
    So t' = γ(t - vx/c2) and t' = γ(t - xβ) are the same except x in the second equation is in the unit of light seconds. I assume this also applies to x' = γ(x-tβ), where x' = γ(x-vt), v = β/c and c = 1.

    As for the 0.75s and 0.25s, I realized I forgot the square root when I solved for γ, so those values are wrong. :frown:

    Going back to the diagram with the spaceship example, the proper time in the spaceship is 0.866s, but I don't see 0.866 in the diagram. Is 0.866 the distance between 2 blue dots if the distance between 2 black dots is 1?
     
  14. Aug 5, 2013 #13

    ghwellsjr

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    Yes, except we could use a bunch of other pairs of units such as years and light-years or even one of my favorites, nanoseconds and feet.

    But even if you hadn't forgotten the square root, I still don't understand how you got the second number significantly less than the first number. Can you please show me your calculation?

    I think you may have the right idea, but you really should think in terms of the Coordinate Time, not the blue Proper Time for several reasons:

    Firstly, the blue Proper Time applies just to the blue observer on the planet which remains localized at x=0 in this IRF, whereas the spaceship is only at x=0 for an instant in time (t=0). On the other hand, the Coordinate Time applies everywhere.

    Secondly, the blue observer doesn't have to remain stationary, he could take off in his own spaceship at some high speed, in which case his own Proper Time world be dilated and no longer related to the Proper Time of the black spaceship.

    Thirdly, we don't even need the blue observer as is the case in the second diagram in post #7.

    Fourthly, we can transform to a different IRF in which both the blue observer and the black spaceship are traveling at new speeds with new gamma values and new time dilations with new spacings of the Proper Time dots which will always have the correct ratios with respect to the Coordinate Time.

    So the Proper Time interval from t=0s to t=0.866s, is 86.6% of the vertical distance from the position on the graph of the black dot (which is obscured by the blue dot at the same position) to the next black dot upwards and occurs exactly at the Coordinate Time of 1s.

    I have drawn a zoomed in diagram and added a red dot at the coordinates of t=1 and x=0.5:

    attachment.php?attachmentid=60689&stc=1&d=1375678468.png

    Now I let my program redraw the diagram with β=0.5. This transforms the rest IRF of the blue observer on the planet to the rest IRF of the black spaceship:

    attachment.php?attachmentid=60690&stc=1&d=1375678468.png

    Now the red dot has the coordinates that were calculated previously and I think you can see that it is 86.6% of the way from the Proper Time of 0s to the Proper Time of 1s for the black spaceship.

    And, as I said earlier, we can also transform to any other IRF such as this one where both the blue observer on the planet and the black spaceship are traveling apart from each other at 0.268c:

    attachment.php?attachmentid=60691&stc=1&d=1375678468.png

    Now you can see that both the observer and the spaceship are time dilated by the same amount (1.038) and yet the red dot is still 86.6% of the way from the dot at t=0 to the next black dot up. However, the red dot has no significance in this diagram--it doesn't represent anything except to show that it is at the same Proper Time for the spaceship. Note that the red dot is just below the Coordinate Time of 1 second while the black dot above it is above the Coordinate Time of 1 second. The whole point of this exercise is to make clear that Time Dilation is different in different IRF's and is unobservable by the observers on the planet or spaceship. No observer has any awareness of their own Time Dilation or the Time Dilation of any other observer. How could they? It changes when we choose a different IRF to represent the scenario.
     

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    Last edited: Aug 5, 2013
  15. Aug 5, 2013 #14

    ghwellsjr

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    As I mentioned at the end of the previous post, no observer can see or have any awareness of the Time Dilation of anyone else and now I want to show you what they can observe. We do this by drawing in light signals along 45-degree paths from an event at one observer to the other observer and we will see that it doesn't matter what IRF we use, the Proper Time at which an observer gets the light signals from another observer remains the same.

    We start with the rest IRF of the blue observer on the planet as he is watching the spaceship approach, pass, and depart from him:

    attachment.php?attachmentid=60696&stc=1&d=1375708907.png

    The first signal I have drawn in starts at the black spaceship's Proper Time of -7s (you have to count the dots down from the dot representing 0s where the spaceship and the planet coincide) and it goes up and to the right to the blue observer's Proper Time of -4s. So the blue observer actually sees the clock on the spaceship displaying -7s when his own clock displays -4s.

    Then when the spaceship passes him, he sees both their clocks displaying 0s. So he would say that he is seeing the spaceships clock ticking 7/4 or 1.75 times the rate of his own. This is called Relativistic Doppler and there is a formula to calculate it which you can look up in wikipedia if you want. That formula gives a more exact value of 1.732051 but this is close enough for simply eye-balling values off a diagram.

    The second signal I drew in starts at the black spaceship's Proper Time of +4s and goes up and to the left to the blue observer's Proper Time of +7s. Now he would say that it appears that the spaceship's clock is ticking at 4/7 or 0.571 times the rate of his own. The formula gives about 0.57735.

    So neither one of these ratios is the correct Time Dilation of 1.1547 for this IRF. However, if we average the two Relativistic Doppler factors that we measured or calculated, we would get the correct Time Dilation factor:

    (7/4 + 4/7)/2 = (49/28 + 16/28)/2 = (65/28)/2 = 2.3214/2 = 1.1607

    Or more precisely using the Relativistic Doppler formula values:

    (1.732051+0.57735)/2 = 2.309401/2 = 1.1547

    Now let's look at the rest IRF for the spaceship:

    attachment.php?attachmentid=60697&stc=1&d=1375709032.png

    As you can see, the blue observer still sees the same comparative rates of the spaceship's clock and he makes all the same calculations but in this IRF, it is his own clock that is Time Dilated, not the spaceship's and so he gets the "wrong" answer. Or we could reinterpret what he is calculating and say that he is determining what the spaceship's Time Dilation would be in his own rest IRF even if we are analyzing the scenario from a different IRF.

    Finally, let's look at the IRF in which both the observer and the spaceship are traveling at the same speed in opposite directions (0.268c):

    attachment.php?attachmentid=60703&stc=1&d=1375737250.png

    Once again, all observations and calculations remain the same even though in this IRF the Time Dilations of the observer and the spaceship are identical.
     

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    Last edited: Aug 5, 2013
  16. Aug 5, 2013 #15

    Imager

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    @ghwellsjr,

    Thank you for the great explanation! I've been trying to get this to work right in an Excel spreadsheet for years and thanks to you it finally works!
     
  17. Aug 5, 2013 #16
    So the 0.866s represents a fraction of distance between the points for the black line and not the total distance, but the black dots are a distance 1.1547 away. Does this mean that when the planet's clock tick 1 second, the observer sees an event, but when the clock ticks 1.1547 seconds, the spaceship sees an event?

    As for the 0.25s, I don't quite remember what my calculations were. However, I did the calculation by thinking that t = γ(t' + x'v/c2) was an expanded version of the time dilation formula since I wasn't quite sure what the equation meant. I thought the period of time dilation changed as one frame moved away from another frame.
     
  18. Aug 6, 2013 #17

    ghwellsjr

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    Take a look at this diagram which is basically the same as the first one in post #13, except I have added labels to show the Proper Times displayed on the blue clock traveling with the observer on the planet and the Proper Times displayed on the black clock traveling with the spaceship:

    attachment.php?attachmentid=60728&stc=1&d=1375795152.png

    So 0.866s (where the red dot is) represents the Proper Time on the black clock when it gets to the Coordinate Time of 1 second. The black dot at the Proper Time of 1 second is at the Coordinate Time of 1.1547s.

    When the planet's clock ticks 1 second, the observer sees the event of his own clock displaying 1 second, and he also sees the spaceship's clock at 0.577s which is approximately 4/7, the same ratio discussed in post #14 after the first diagram.

    When the observer's clock ticks 1.1547 seconds, the Coordinate Time is also 1.1547 seconds which is when the spaceship sees the event of its own clock displaying the Proper Time of 1 second. (But this is true only in this IRF, in other IRF's it may not be true.)

    Time Dilation is not an effect that involves multiple frames. It is an effect that applies to every clock in a single frame and is a function of the speeds of the clocks in that frame. When you transform all the events from one frame to another frame moving with respect to the first frame, then all the clocks may have different speeds and therefore different Time Dilations.
     

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    Last edited: Aug 6, 2013
  19. Aug 9, 2013 #18
    Thank you for all your help! =)

    So to summarize:

    When the observer's clock ticks 1 second, the spaceship's clock ticks 0.866s from time dilation.

    At 1 second, the observer sees an event and sees the spaceship's clock at 0.577s due the relativistic doppler effect (I couldn't find the equation for this when I searched it up, I only found equations dealing with frequency).

    When the observer's clock ticks 1.1547s, the spaceship's clock ticks 1 second and the spaceship sees an event.

    t' = γ(t-xβ) and x' = γ(x-tβ) are not time dilation and length contraction equations, they transform the time and position of an event from 1 frame to another.
     
  20. Aug 9, 2013 #19
    Yes, if you are in the frame of the observer.

    This has nothing to do with the doppler effect.
    The doppler effect describes the change of frequency (frequency shift of light).
    Just transform between the reference frames using the lorentz transformations and everything should be fine.

    Yes
     
  21. Aug 9, 2013 #20

    ghwellsjr

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    Whenever you make a statement like this, you need to state which frame it applies to as ProfDawgstein pointed out. In the rest IRF of the observer it is true but in the rest IRF of the spaceship, the opposite is true. And in other IRF's, such as the last one in post #13, both clocks tick at the same rate. Time dilation is frame dependent and you should really think of it as a coordinate effect as I tried to point out in post #13. In other words, in our example, it's the ratio of the coordinate time to the time on a clock, not fundamentally the times between two clocks. That's why it's so important to state which frame it applies to.

    But we are dealing with frequency, namely the frequency of the one-second ticks of the clocks, which tick at a frequency of 1 Hz. Knowing that, can you go back and see how the equation determines the rate that each observer sees the other ones clock ticking at compared to their own, both while the spaceship is approaching the observer and while the spaceship is receding from the observer?

    Note that the relativistic Doppler effect is not frame dependent--it doesn't matter what frame you are using to describe or analyze the effect.

    Again, this is true only in the rest IRF of the observer.

    Yes, but you didn't say what time on the observer's clock the spaceship sees when its clock ticks 1 second. Do you want to try to answer that question? (HINT: it's real easy.)

    Yes, again as ProfDawgstein pointed out.
     
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