Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Changing frames of reference

  1. Jul 11, 2017 #1
    Suppose I'm an observer out in space and 30 km from me there's another observer, who is static with respect to me. Let's say my name is A and the other is B. We're both far far away from all the cosmical objects that might have a gravitational influence on us.

    We both notice a spaceship flying with a constant speed - it flies past point A then reaches to B in 30 seconds. Since A and B are static with respect to each other, I suppose we perceive time identically.

    The question is, how much time passed for the guy in the spaceship, and what is the distance between A to B with respect to him?

    I appreciate all of your answers!
  2. jcsd
  3. Jul 11, 2017 #2


    User Avatar
    Science Advisor

    What do you think?
  4. Jul 12, 2017 #3
    I think there will be more than 30 km for the spaceship's reference, and it will experience time going faster, so less than 30 seconds. Is this right? It is kind of confusing when the concept of speed comes in.
  5. Jul 12, 2017 #4


    User Avatar
    Science Advisor

    Well, relativity states that inertial observers are equivalent and as such there is no way you can tell the difference between different inertial observers. This means that between inertial observers (say A and B), time dilation (and hence also length contraction) appear symmetrically. A measures the clock of B running slower, and vice versa! If this would not be the case, one could differentiate himself from the other.

    The mistake some people make, is that if A measures the clock of B running slower, then B must measure the clock of A running faster. This mistake even appeared in a popular Dutch science program some time ago (I did my scientific duty and wrote a letter, no worries). But this would contradict the principle of relativity.
  6. Jul 12, 2017 #5
    It is not only Dutch television thinks so. Richard Feynman, for example, concludes that if “the man on earth” measures distances with squashed ruler, they appear to be greater.

    „When we said the horizontal distance is ##vt##, the man on the earth would have found a different distance, since he measured with a “squashed” ruler„

    34-8 Aberration. http://www.feynmanlectures.caltech.edu/I_34.html

    It would be good to write him a letter too. Sadly, he passed away long ago.

    On the other hand it is obvious. In classical case maximum angle of aberration will be ##arctan(1)## = 45 grad. In relativistic case due to aberration moving obsever tilts telescope more and more „into front“, since according to him space stretches. In relativistic case at veloctiy near that of light his telescope will be directed almost in the direction of travel, since ##arsin (1)## = 90 grad.

    There is another wrongdoer - Wikipedia. It claims that an observer (at rest) sees moving clock dilates gamma times, and moving observer sees that clock at rest is ticking gamma times faster.

    „The transverse Doppler effect is the nominal redshift or blueshift predicted by special relativity that occurs when the emitter and receiver are at the point of closest approach. Light emitted at closest approach in the source frame will be redshifted at the receiver. Light received at closest approach in the receiver frame will be blueshifted relative to its source frequency.“

    It even provides very convincing calculation!

    Moreover, a number of other sources claim exactly the same.

  7. Jul 12, 2017 #6


    User Avatar
    Science Advisor

    No. Less than 30km, since A and B are in motion in this frame and hence the distance between them is length contracted. Therefore less than 30s since the speed is the same. Edit: although note that at 1km/s, relativistic effects are pretty much negligible.

    This has been tested - relativistic muons produced in the upper atmosphere make it to the surface in large numbers due to this effect.
    Last edited: Jul 12, 2017
  8. Jul 12, 2017 #7


    User Avatar
    Staff Emeritus
    Science Advisor

    We have observers A and B who are stationary, and a flying space-ship S, that is moving with some velocity v with respect to A, and the same velocity v with respect to B.

    In the frame of A and B, the elapsed time in the AB frame is ##\Delta t_{AB} = D_{AB}/v##, where ##D_{AB}## is the distance between A and B in the AB frame, and v is the velocity. A and B share a common frame of reference, that I'll call the AB frame.

    When you say "S flies past point A and then reaches B in 30 seconds", it's unclear which frame this time is being measured in. I will assume that you mean that the time is measured in the AB frame, in which case we have ## \Delta t_{AB}## = 30 seconds, and ##D_{AB} = \Delta t_{AB} \, v##. I will not in passing that we need to be able to synchronize clocks in the AB frame in order to measure ##t_{AB}##. The general procedure for measuring ##t_{AB}## is to have one clock at A, and one clock at B, and to synchronize these clocks in the AB frame. Then ##\Delta t_{AB}## is the difference on the reading of clock A when the spaceship S passes it, and the reading on clock B when the spaceship passes it, and we've assumed the clocks are synchronized. This last point is often glossed over, but it can become important.

    The time measured by the spaceship S, which we will call ##\Delta t_{S}## will be smaller than ##t_{AB}##. Given the time dilation factor factor ##\gamma = 1/\sqrt{1-v^2/c^2}##, we can say that ##\Delta t_{S} = \Delta t_{AB} / \gamma##.

    Operationally, we can note that we need only a single clock on S to measure this time, and hence we don't need to know how to synchronize clocks to measure ##\Delta t_S##.

    We can also say that the distance between A and B in the S frame ##D_{s}## is given by ##\Delta t_{S} \, v##, which we see will be ##D_{AB} / \gamma##. So the distance is also smaller.

    We usually say that the ##t_{S}## is shorter than ##t_{AB}## due to 'time dilation', and that ##D_{s}## is shorter than ##D_{AB}## due to 'Lorentz contraction' or 'Length contraction'.
  9. Jul 13, 2017 #8
    Got it! Thanks!

    Trying to explain this effect intuitively, I came to a paradox:

    Imagine the spaceship has wheels and there is a road connecting A and B. The story remains the same, the ship travels from A to B with 1 km/s, except that now the wheels are touching the road, so that we can measure the distance.

    From the AB reference frame the spaceship seems smaller, so the wheels seem smaller too. Let's say that from A to B the wheels make N rotations.

    Now if we're in the shaceship's reference frame, the distance from A to B is smaller, so the wheels make a number of rotations that is smaller than N.

    This contradicts any common sense, since the number of rotations must be the same from any reference frame. There's something wrong, but what?
  10. Jul 13, 2017 #9


    User Avatar
    Science Advisor

    The wheels are circular in the rocket's rest frame. What shape are they in the road's rest frame?
  11. Jul 13, 2017 #10
    Thank you for the explanation and the mathematical description! It's much clearer now.

    However, how do we know that ΔtS=ΔtAB/γ and not ΔtAB=ΔtS/γ ? If we were initially given the distance Ds and if I had to calculate DAB, I would have probably made a mistake.
  12. Jul 13, 2017 #11
    I never thought about this. Are they linear?
  13. Jul 13, 2017 #12


    User Avatar
    Science Advisor

    Length contraction and time dilation are how something is described in a frame where it is moving. Something that is 1m long in a frame where it is at rest is <1m in a frame where it is moving. A clock that has s between ticks in its rest frame has >1s between ticks in a frame where it is moving. Given that ##\gamma\geq 1##, you can figure out where it must go.

    A and B are 30km apart in the frame where they're not moving. Their clocks tick off 30s but the spaceship clock is moving. Apply the rules above.

    If you aren't sure, you can always apply the Lorentz transforms. In fact, I strongly recommend doing it that way rather than learning a collection of rules of thumb. Then learn the rules of thumb.
  14. Jul 13, 2017 #13


    User Avatar
    Science Advisor

    No. What does motion do to lengths measured parallel to it? What about perpendicular?
  15. Jul 13, 2017 #14


    User Avatar
    Science Advisor
    Gold Member

  16. Jul 13, 2017 #15
    In the reference frame of the road the wheel rotates slower due to time dilation though angular velocity remains finite. Also, the rim of the wheel Lorentz – contracts.

    The rest length of the rim of the wheel must remain constant. This means that the rim Lorentz contracts, and that the radial extension of the wheel contracts accordingly. The result is that the wheel become infinitely small in the limit that the wheel moves with the velocity of light.

    If ##v## is velocity of the rim in the rest frame ##K## of the wheel, we have ##\Omega=v/R##, where ##R=R_0/\gamma## is the contracted radius of the rotating wheel, and ##R_0## is their radius when they are at rest. The angular velocity of the rotating wheel is then ##\Omega = \gamma v /R_0##
    hence, in this case the angular velocity ##\Omega## must approach an infinitely great value in ##K## when the speed of the spaceship approaches that of light.

    As observed in the road’s frame ##K'##, the distance between the marks on the road each time a point on a rim of the wheel leaves it is

    ##l'=\gamma 2 \pi R = 2\pi R_0##

    and this distance is independent of the speed of the wheel, even if the radius of the wheel decreases with increasing velocity, because the distance between the marks depends upon the rest length of the rim of the wheel and not their Lorentz contracted length. Also in this frame the angular velocity of the wheel remains finite even if the wheel have a vanishing radius when the velocity of the wheel approaches that of light,

    ##\Omega'=\gamma^{-1} \Omega = v/R_0##

    and hence ##\lim\limits_{v \to c} \Omega' =c/R_0##, which is finite.

    It looks like "Relativistic Trolley Paradox"

    Related: page 39, fig. 8

    K. Voyenli - Alternative derivation of the circumference of a relativistic, rotating disk,
    Last edited: Jul 13, 2017
  17. Jul 14, 2017 #16
    The lenght is contracted if it's measured parallel to it and is constant if it's measured perpendicularly. However, the circumference is still smaller.


    "Without going into technical details we therefore equip the bicycle with wheels that are assembled in rotation. This is done in such a way that in stationary rotation the wheels have the geometric shape of ordinary wheels at rest."

    Interesting! But could this be possible?
  18. Jul 14, 2017 #17
    Thank you!
  19. Jul 14, 2017 #18
    But then it is kind of confusing, the clock in the spaceship show less than 30s because it is moving with respect to the AB reference frame. But if we choose the spaceship as reference, then the rest of space is moving with respect to it, so it seems like the clocks outside must show less time then the one in the spaceship, which is contradictory...
  20. Jul 14, 2017 #19


    User Avatar
    Science Advisor

    That depends on the distance between the observers. At very large distanced these velocities can give considerable time dilation.
  21. Jul 14, 2017 #20


    User Avatar
    Science Advisor
    Gold Member

    First note note that Feynman's phrase does not match your paraphrase. Where he said "different" you say "greater". Second, while it could be argued this phrase possibly promotes misunderstanding, the text around it and the following equations are all correct and match yours below. I find your criticisms unfair.
    Again your paraphrase is wholly inaccurate statement of what wikipedia is saying.

    Here, you are wrong and (in this case) wikipedia is right, and these facts are quite well known. The key point is that as two bodies approach and move past each other, the following pairs of events are different pairs:

    1) emission at closest point in emitter's frame, reception at an event that is NOT closest for receiver
    2) emissions at an event not closest approach in emitters frame that is received at closest approach in the receiver's frame.

    These are not symmetric. In particular, the relative velocity (direction included) between the respective emission and reception events of these pairs of events are different, so different Doppler results.
    And this source is also correct.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Changing frames of reference
  1. Frames of Reference (Replies: 37)

  2. Frames of reference (Replies: 3)

  3. Frames of reference? (Replies: 10)

  4. Reference frames (Replies: 5)