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Time dilation, length contraction, but velocity invariant

  1. Jun 15, 2015 #1
    If a frame is moving at constant velocity relative to an observer, this observer perceives a time dilation and a length contraction. But in this case how the velocity (length/time) can appear constant ? It is expected to be contracted.. Thank you in advance for the explanation
  2. jcsd
  3. Jun 15, 2015 #2
    Almost certainly you overlooked (or don't even know) about relativity of simultaneity.
    If so, please search this forum for relativity of simultaneity and read §1 and 2 of http://fourmilab.ch/etexts/einstein/specrel/www/
    Next, in particular this animation by Janus is very useful: https://www.physicsforums.com/threads/twin-paradox-and-asymmetry.816455/#post-5126008
  4. Jun 15, 2015 #3
    Thank you, but I don't see precisely the answer in these refs (except that Enstein confirms that V=D/t)

    What is wrong in my question: D contracted?, t dilated? or D/t constant?
  5. Jun 15, 2015 #4


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    Both length contraction and time dilation have the same factor. If we have an object moving with speed ##v## with respect to you, from point A to point B, then:

    You measure the distance from A to B as ##d## and the time as ##t##, so calculate the speed as ##v = d/t##.

    The moving object calculates the distance from A to B as ##d/\gamma## and the time as ##t/\gamma##, so it calculates the speed as ##v = d/t##

    So, you both agree on the speed, but not on the distance travelled or elapsed time.
  6. Jun 15, 2015 #5
    that would be nice but d/gamma looks a length dilation?
  7. Jun 15, 2015 #6


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    To the moving object, the distance from A to B is contracted by a factor of ##\gamma = \frac{1}{\sqrt{1- v^2/c^2}} > 1##.

    And to you (at rest with respect to A and B), the moving clock is dilated by a factor of ##\gamma##.

    So, it is nice!
  8. Jun 15, 2015 #7
    OK I did not remember the meaning of gamma ( I thought it was the inverse)

    In this case, d/gamma is indeed a length contraction, but t/gamma is now a time contraction..

    d and t should be both either dilated or contracted to maintain the ratio constant. Correct?
  9. Jun 15, 2015 #8


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    The moving clock runs slow from your reference frame. That means time in that reference frame is "stretched" or "dilated". The maths and the transformation are the important things. "Contraction" and "dilation" are just words and need a context to know what they mean.

    Time in your frame: 0...1...2...3
    Time in moving frame: 0......1......2......3
    Last edited: Jun 15, 2015
  10. Jun 15, 2015 #9
    These words seem to be used inversely for lengths and time. This is very misleading.

    Do you confirm that viewed from my reference frame (arbitrarily considered immobile) : (1) the lengths in the moving frame appear contracted and (2) the time marked by the moving clock appears contracted (runs faster)?
  11. Jun 15, 2015 #10


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    (1) the lengths in the moving frame appear contracted and (2) the time marked by the moving clock appears dilated (runs s-l-o-w-e-r).

    But, here it's your lengths as viewed in the moving frame that are important. The moving object measures your distances as shorter than you do.
  12. Jun 15, 2015 #11
    Thanks a lot for your patience but I confess I am lost.
    All these effects are supposed to be reciprocal because the inertial frames can of course be permuted

    With respect to your post #4, ##t/\gamma## gives an apparent time shorter than ## t ##. To me shorter means contraction
  13. Jun 15, 2015 #12


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    Words are not important. Less elapsed time can mean contraction or dilation, whatever, according to taste. This is why maths trumps words in the end. The maths is unambiguous.

    But, there is something deep here and, in fact, understanding this little problem fully will take you a long way to understanding these two effects of SR.

    Let's start again:

    You set up two points A and B, a distance ##d## apart. You observe an object travel from A to B at speed ##v## and measure time ##t = d/v##. That's just standard kinematics. You also notice that the object's clock has only measured ##t' = t/\gamma## while it travelled from A to B.

    Now, what does the moving object observe? It must measure ##t' = t/\gamma## for its journey from A to B (although from its perspective A and B are moving towards it).

    There are then two ways to look at this. If we agree that the relative velocity must be the same for both frames (I won't go into the argument here, but let's accept this). Then, it must conclude that the distance from A to B is only ##d' = d/\gamma##

    Alternatively, if it measures the distance from A to B and finds this to be ##d' = d/\gamma##, then it agrees that its speed (or the speed of A and B) is ##v##.

    That's the basic symmetry of the situation. It's worth batting your head against this for a bit. Just keep going until it's clear. It may take some time but it's important.
  14. Jun 15, 2015 #13
    your sentence :

    seems contradictory with :

    In fact, my main question can be summarized: what means t' < t ?
    according to me, if seconds' are shorter than seconds, time t' runs faster than time t, not slower
  15. Jun 15, 2015 #14


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    If you have two clocks and one is running slower than the other. Let's say clock C' is running slower than clock C. And, if you start them both a 0, then: ##t' < t##

    For example, if C' is running at half the speed of C, then ##t = 2t'##.
  16. Jun 15, 2015 #15
    OK, I understand the misunderstanding! I reasoned in "time units"
    many thanks
  17. Jun 16, 2015 #16
    There are different ways to measure velocity, and it wasn't very clear from your question what velocity you meant. As you used the word "invariant", I supposed that you were thinking of measuring from the point of view of each frame as "rest frame", the change of position in that frame of a moving point that has a fixed position in the other frame (e.g. the motions of the origins of each frame as measured in the other frame).

    If so, then you compare the time as read on one clock, with the time as read on another clock at a distance. When you write D/t you probably mean D/(t2-t1): t2 is read on one clock, and t1 on another clock; that depends on clock synchronization as you saw. And as Einstein explained there, the different systems synchronize their clocks differently.

    If the "moving" system with contracted rulers and slow clocks used the same synchronization as the "rest" system, they would measure as speed a longer distance / less time. Observers of the "stationary" system interpret the fact that observers of the "moving" system measure the same speed as themselves as due to the wrong clock synchronization of the "moving" system.
  18. Jun 16, 2015 #17


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    Good question. I think some spacetime diagrams may help to answer your question.

    Let's take a scenario where an observer sends out a light pulse to a reflector 6 feet away and measures with his clock how long it takes for the reflection to get back to him. Since light travels at 1 foot per nanosecond, it will take 12 nsecs for him to see the reflection and he will validate that the light is traveling at 1 foot per nsec for the 12 feet of the round trip that the light takes. If he is following the precepts of Special Relativity, he will define the time at which the reflection took place at 6 nsecs according to his clock and I have made that dot black:


    In this diagram, the observer is shown as the thick blue line with dots marking off one-nanosecond increments of time according to his clock. Since he is at rest in this Inertial Reference Frame (IRF), the horizontal grid lines align with his clock and the vertical grid lines align with his measurement of distance.

    The reflector is shown as the thick red line with similar dots. The thin blue line represents the pulse of light that the observer emits and the thin red line represents the reflection. Hopefully, this all makes perfect sense to you.

    Now let's transform the coordinates of all the dots in the above diagram to a new IRF moving at 0.6c with respect to the original IRF:


    At 0.6c, gamma is 1.25 and you will note that the dots are placed 1.25 nsecs apart as defined by the gird lines. Also, the distance to the reflector is contracted to 4.8 feet (6/1.25) as defined by the grid lines. You will also note that the light pulse and its reflection continue to travel at 1 foot per nanosecond but it takes less time in this IRF (3 nsecs) for the light to reach the reflector and more time (12 nsecs) for the reflection to get back to the observer. The total Coordinate Time is 15 nsecs and the total Coordinate Distance the light traveled is 15 feet so the velocity is maintained at 1 foot per nanosecond.

    Just for the fun of it, let's see what would happen if we applied the method of your question to establish the speed of the light. Here is a diagram that depicts the parameters that would apply if we just took the Length Contraction and Time Dilation factors into account:


    Note that we would be using the length at the start of the scenario and the time as defined by the observer as when the reflection occurred to establish where the light reached at the time of the reflection which is way off base and would result in a contracted velocity as shown by the extra thin blue line. Instead, we need to use the Coordinate Distance of where the reflector has moved to when the light pulse reaches it and the Coordinate Time at which this occurred, not the time that the observer defines the reflection to occur according to his clock.

    Does this all make perfect sense to you? Any questions?
  19. Jun 17, 2015 #18

    Good! Very clear!
    You should use "Courier New"
  20. Jun 17, 2015 #19
    First,...##\gamma = \frac{1}{\sqrt{1-0.36}} = 1.25##
    Let's say BR Frame is the moving blue/red frame at 0.6c
    B is Blue
    R is Red
    W is the rest observer/watcher

    1) Why vertical line = 15?
    Because ##\sqrt{15^2-(-9)^2} = 12##, the time the event takes place. 12 seconds wrt BR. Hyptenuse in 4D, Is this right? 15 seconds wrt Rest, is this right?

    2) Why -9? from (-9)2
    When the light bounces from RED, it has to catch BLUE moving 0.6c 4.8ls away.
    So it takes ##ct = 4.8+vt; t = 4.8 + 0.6t; t = 12## seconds
    So ##3-12 = -9##, so it will catches BLUE at -9 wrt rest, is this right?

    3a) What is 3? from ##3-12 = -9##
    Because this is where and when wrt W, the light from (0,0) reaches Red WL and bounce back. Is this right?
    3b) Why 3?
    Because ##3 = \frac{4.8}{1+0.6}##, is this right?

    4)Why 4.8 not 6?
    It comes from the Lorentz contraction, length transformation, which is 200% proven, irefutable, undeniable. ##6*\sqrt{1-0.6^2}## I think this is right.

    5)Why the origin of the red line (6,0) wrt W, before now seems in (7.5,4.8)? Okay, lets say I don't know if it's in 7.5, 4.8, just say at (7.?,4.?). I think we should calculate when and where RED started.
    I can't draw, can I just type?
    Supposed, all this events from BR frames takes 12 seconds.
    There are 3 events.
    There are 3 object
    W, R, and B
    E0: Clocks are synchronized.
    E1: Blue emit light to Red, and BR moves.
    E2: The light touches Red and bounces back and R also sends a digital signal to W the clock wrt R this happen and the location wrt O this happen.
    R can send the location wrt W because R is entering W light cone. Can R do that?
    E3: The light touches Blue again.

    E1, starts
    E2: Will W know when E2 wrt R happens?. R, record its clock and W's clock from W light cone sends to W back.
    R clock wrt R -> 6
    E2 location wrt W -> 3 from the data that R collect from W light cones.
    Because R is entering W light cone, R also sends its speed to W

    Is this how W translate the data?
    R receive light signal when its clock is 6 wrt R.
    R is in 3 lys away from where W start move wrt W
    R speed is 0.6c based on R calculation, and doppler and Lorentz transformation. Can R calculate its speed?
    If it moves at 0.6c so gamma is 1.25.
    But I'm lost here. How can R clock read 6? Does R clock ticks twice? Does R clock start earlier?
  21. Jun 17, 2015 #20
    thank you all for your support.
    In fact, I better understand equations that diagrams supposedly didactic...
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