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Q: A&B moving in opposite directions @ 0.6 c comparison with sound.

  1. Oct 19, 2011 #1
    I'm trying to understand relativity on a simple example: 2 objects A & B moving in exactly opposite directions at 0.6c each, starting from frame C at the center of things that remains stationary.

    So, we have:

    W <----0.6c A ....................... C ......................... B 0.6c----> E


    1. From frame C, A is moving at 0.6c "West" and B is moving at 0.6c "East". (west and east here only denote the opposite directions in a shorter form, the space they are moving in is flat.)

    2. From frame A, assumed to be stationary, C is moving at 0.6c "East" and B is moving "East" ... at what speed?

    I realize that A cannot see B moving faster than c. From the C frame, A and B are moving apart at 1.2c, which is implied, because C can see only A or B at a time.

    My questions:

    1. How does relativity deals with A&B moving at 1.2c relative to each other?

    2. At what speed does A see B moving? Also, does not this mean that A will see B with a time delay of 0.2 c?

    3. How different things would be if instead of light in space there would be two sirens moving in opposite directions at 0.6 of the speed of sound in air? (Let's make a siren more interesting, with Morse-like count of each second embedded in its sound: 1, 2, 3, .....N.... And let A count seconds on his end too. So, when would A hear B's count N and what will be the difference with his own count?)

    PS

    And please, lets make A and B in light example above count seconds in a similar way (say, with change in frequency each second).

    Thank you very much for your input :)
     
  2. jcsd
  3. Oct 19, 2011 #2

    ghwellsjr

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    A and B with both see and measure the other one traveling away from himself at .882c. Look up velocity addition in wikipedia to see the formula.
    Nothing is moving at 1.2c. That is simply the result of adding two numbers together. It is no more complicated than if you were to add the speed of light in one direction to the speed of light in the other direction and get 2c. It is sometimes called the closing speed but that makes more sense when the two objects are traveling towards each other. Just remember, nothing is actually traveling at greater than c with respect to anything else or in any frame.
    I already answered the first question. I don't know what you are asking in the second question because 0.2c is not a time delay--it's a speed. If you're asking how fast each one will see the other one moving, it's the same answer, 0.882c.
    For the siren example, you are depicting normal Doppler (look it up in wikipedia). They would each detect the other one's time codes as coming in at 25% of their own.

    This is different from light where Relativistic Doppler is in effect and they each would see the other one's time coded messages transmitted via light signals at 50% of their own.

    EDIT: I made a mistake here. I calculated the Relativistic Doppler factor between C and A or B. The correct value between A and B is .25, just like the normal Doppler factor in air. More about this later.
     
    Last edited: Oct 19, 2011
  4. Oct 19, 2011 #3
    Thank you ghwellsjr very much :smile:

    So, if I work the calculations backwards, say A gets info of a B moving at speed x but appearing to him moving at 0.882c with a frequency shift of 0.5 (is that right?) then would not he be able to calculate that B is actually moving at 1.2c away from him?

    EDIT: I changed the scenario a bit. Say, you're a galactic cop A moving in your spaceship and suddenly you get info on your radar that a thief B is moving away from you at apparent speed 0.882c. You want to alert the galactic cops in a proper sector to intersect him. So, in the process of your calculations you gotta know that B is moving at 1.2c away from you (otherwise you may miss B's actual location, no?).

    Also, I find this Doppler shift very interesting. Why can't it be used to determine one's own "absolute" speed?
     
    Last edited: Oct 19, 2011
  5. Oct 20, 2011 #4

    ghwellsjr

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    [Please note in my previous post, I mistakenly said the frequency shift would be 0.5 but the correct value is 0.25.]

    If A and B detect a frequency shift is 0.25, then they both know that at some time in the past and maybe up until the present, the relative speed between them was 0.882c. This is not just an appearance or an optical illusion that is different from the actual speed. The actual relative speed is not 1.2c or any value other than 0.882c. But keep in mind that this only true if both observers have been maintaining their constant speeds for some time. For example, if A has maintained a constant speed but the distance to B is one light year, then B could have changed his speed any time within the last year, and A would have no way of knowing it until the accelerating signal from B finally gets to him.
    Again, there's no speed of 1.2c. If you see thief B moving away with a Doppler shift of 0.25, then his correct speed away from you is 0.882c but you will still have the problem of alerting the galactic cops because your radio signal will take a long time to get to them compared to the thief who already has a head start.
    Normal Doppler can be used to determine an absolute speed with respect to the medium that is carrying the signal. Previously, we assumed that A and B were both moving with respect to the air at 0.6s (where s is the speed of sound in air) in opposite directions and calculated the Doppler shift to be 0.25 which means the pitch of the siren that each one hears is one quarter of their own siren. This is based on the Doppler factor being:
    (1-VA)/(1+VB)
    where the velocities (V) are fractions of the speed of sound in air.

    So when both velocities are 0.6s in opposite directions, the calculation is:
    (1-0.6)/(1+0.6)
    0.4/1.6
    0.25

    But now consider that there is a wind blowing at 0.3s in the direction from B to A. Now VA will be 0.3s and VB will be 0.9s and we calculate the Doppler factor as:
    (1-0.3)/(1+0.9)
    0.7/1.9
    0.368

    This value is considerably different than before, even though the relative speed between A and B is the same as it was before.

    Or instead of a wind, we could have actually made A to travel at a slower speed (from 0.6s to 0.3s) and B to travel at a faster speed (from 0.6s to 0.9s) maintaining the same relative speed as the previous two cases and we would calculate the same Doppler factor, 0.368.

    Now by repeating this experiment with different relative speeds and doing it in different directions, it is possible to measure the absolute stationary state of the air.

    But with light, things are different. If you look at the Relativistic Doppler Factor formula:
    √[(1-β)/(1+β)]
    you will see that it involves just one speed, β, the relative speed between the two observers, unlike the normal Doppler formula which does not depend on the relative speed but the absolute speed between the observers with respect to the medium, air, in this case.

    So if we carry out the same exercise that we did with the normal Doppler, we start with both observers traveling away from a common point in opposite directions at 0.6c. Now I made a mistake in the previous post by calculating the Relativistic Doppler (RD) as being 0.5 but that was actually the RD for both A and B with respect to C. You will note that I could have arrived at the correct RD between A and B by multiplying each of the RD's with respect to C, 0.5 times 0.5 to get the correct RD of 0.25.

    So let's pick another pair of speeds relative to C such that A and B still have the same speed relative to each other. Let's say that A is now moving away from C at 0.3c. Using the velocity addition formula (in subtraction mode) we can calculate that B must be moving away from C in the opposite direction at 0.792c. If you apply 0.3 and 0.792 to the velocity addition formula, you will get the same relative speed between A and B of 0.882c just like before.

    So now let's calculate the RD between C and A:
    √[(1-β)/(1+β)]
    √[(1-0.3)/(1+0.3)]
    √[(0.7)/(1.3)]
    √[0.538]
    0.7338

    And the RD between C and B:
    √[(1-β)/(1+β)]
    √[(1-0.792)/(1+0.792)]
    √[(0.208)/(1.792)]
    √[0.116)]
    0.3407

    Now if we multiply these two RD we get the RD between A and B:
    0.7338 * 0.3407
    0.25

    This is the same value we got before for the relative speed between A and B of 0.882:
    √[(1-β)/(1+β)]
    √[(1-0.882)/(1+0.882)]
    √[(0.118)/(1.882)]
    √[(0.0625)]
    0.25

    Note that we are doing the same kind of exercise that we did for normal Doppler but there we got a different answer. So since with Relativistic Doppler, we always get the same answer, we cannot use Doppler to determine an absolute speed with respect to some alleged stationary medium.
     
    Last edited: Oct 20, 2011
  6. Oct 20, 2011 #5
    Dear ghwellsjr, thank you very much for your explanation! I need time to absorb it all. There are 2 things which are still unclear to me. One is that counting-seconds signal that B transmits, say, by a peak in frequency every second. That should arrive from B to A with ever growing delay, as the distance between them keeps increasing. (Since the set up is symmetrical, for simplicity, I only care about A "reading" or "trying to catch" B.)

    The other thing is the actual location of B "in the galaxy". Somehow I have doubts that the cop A will get the right location of thief B, if he will go only by the "apparent" speed of 0.882c instead of "actual" speed of 1.2c. Did not we start with a real speeds of 0.6c for both A & B in frame C? This maps to actual location for both A & B in that frame. I'm afraid the cop will miss the thief's location by a large margin.

    I will try to do my own calculations and post it for you to check (or compare with yours if you do it sooner, which you probably will, since it's so easy for you).

    Thank you very much again! :smile: All this is very interesting.
     
  7. Oct 21, 2011 #6

    ghwellsjr

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    You're right, there is an ever growing delay but the distance between the peaks in frequency every second maintain a constant interval in time and a constant distance in space. What is increasing is the difference in the clock readings between the two observers. But as long as the two observers continue moving apart at the same relative velocity, the pattern of the incoming signals will continue the same from start to finish.
    If you want to have the C frame be the one that the observers use to communicate positions and speeds, then it's an easy enough thing for them to do using the Lorentz Transform. But you should not consider the distances or times defined according to this frame to be "actual" and other ones "apparent". They're all subject to the same limitation of requiring a defined Frame of Reference to give them meaning.
     
  8. Oct 21, 2011 #7
    Say they traveled for one light year, then A assumes it is at rest. Then B would be .882 light years away from A. But, then how would C measure them being .6 light years away from it? It would seem they would be 1.2 light years and .882 light years away from each other depending on the frame of reference. C can measure each one seperatly to be traveling .6 light years away from it, but then A and B can't measure each other as being 1.2 light years apart after 1 year.

    It would be like saying that no two galaxies could ever become more than one light year apart from each other in a year. But, they do...

    Say for example, you look through a telescope to each edge of the visable universe. You look in two opposite directions and find galaxies moveing away at .9c. Then you assume that you are looking into space at Earth from one of these galaxies. You would find that Earth is on the edge of the visable universe traveling at .9c. But what about the other galaxy? With the velocity addition you would find that it was only traveling at .9999999c and should be very close to Earth that was seen to travel at .9c since it can't travel FTL. But, here on Earth we know that it is not true, but is really about the same distance from Earth as the other galaxy.
     
  9. Oct 21, 2011 #8

    ghwellsjr

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    You have to be very precise when you set up situations like this. You have to say which frame you are assuming the statements apply to. Questions arise:

    What do you mean, "they traveled for one light year"? From the rest of your comments, I think you meant to say, "they traveled for one year", but even that is ambiguous unless you state which frame it applies to and from the rest of your comments, it appears that you are applying it to all three frames simultaneously, which of course will lead to paradoxes.

    What do you mean, "then A assumes it is at rest"? Do you mean that A actually comes to rest in Frame C? If not, why do you suggest that A makes this assumption only after one year? It has always been at rest in Frame A.

    This is true in Frame A after one year.
    In Frame C, A and B are .6 light years away after one year. But this is not something that C can measure. After one year, he will see A and B as being only .3 light years away from him. He has no way of knowing whether they are actually at their locations .6 light years away from him at that time. For all he knows, they may decide to turn around and come back to him during the course of the next year or maybe they blew up and end their existences. He has to actually wait another year (for a total of two years) to see and measure the positions of where they were after one year.
    Yes. I can't tell whether you see this as a solution or a problem. It's a solution, not a problem.
    They can if they each stop after one year in Frame C and wait until the light signals catch up to them which will take more than another year for A and B but only one more year for C (as I said before).
    When we're talking about galaxies in our real world, we have to bring in General Relativity which doesn't apply to this discussion about Special Relativity.

    Also, where'd you get .9999999c as the velocity addition of .9c and .9c? I get .9945c. Are you thinking that it has to be limited to something just under c in order to avoid FTL?
     
  10. Oct 21, 2011 #9
    one light year in who's frame in C's frame of reference or A's? Then A assumes it's at rest in C's frame of reference?

    Yes the difference will be different depending on the frame of reference. You are also only looking at part of the equation. Also look at length contraction and time dilation. There is no preferred frame of reference where length, time, and velocity are calculated for. Instead you take a frame of reference, you started with C and saying A is moving at .6c with in C's frame of reference.

    The length of things(distance between things), velocity that things are moving at and speed that a clocks is ticking will be different in A's frame of reference. They all can be calculated since you know what's going on in C's frame of reference. But nether is these are right or wrong, it just depends on which frame of reference you chose.
     
  11. Oct 22, 2011 #10

    ghwellsjr

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    I have to apologize and correct a mistake I made here. I said that C will see A and B as being only .3 light years away after one year but the correct distance is .375 light years. Then I said he has to wait another year to see them arrive at .6 light years away but the correct time is .6 years for a total of 1.6 years.

    Sorry for any confusion this may have caused.
     
  12. Oct 23, 2011 #11
    I find it surprising that being in two different locations at once compared to two different frame of references would be an actual solution. I did mean to say they just started out traveling for one year. It makes it sound to me that velocity addition supports the many worlds thoery, since two frame of reference's could become completely different from each other. Or, would it be just how the velocities where added depending on what frame of reference was considered? Then if those velocities become different depending on the frame of reference, then they would expereince different amounts of spacetime dialation that depends on their velocity, so then their measurements of the speed of light would also become different.

    Sorry, the .999999c was an approxiamation, since when dealing with relativistic velocity addition at speeds close to the speed of light you can never get a value that is equal to or greater than the speed of light. I don't think there is anything seen in nature that this theory can be applied to, since we don't see it as a description of galaxies in the universe.
     
  13. Oct 23, 2011 #12

    ghwellsjr

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    Special Relativity does not say that anything is in two different locations at the same time. It merely provides a consistent means for defining the coordinates, not just for the location, but also for the time, of events. So the same event according to one Frame of Reference can have different location and time coordinates according to a second Frame of Reference.
    Velocity addition has nothing to do with the many worlds theory and it isn't restricted to the Theory of Special Relativity. It's what happens in the real world, independent of any theory. But keep in mind that it only applies to multiple observers who are traveling along a straight line at different constant relative speeds. As each observer measures the relative speed of each of the other observers, by any means that they desire, the velocity addition formula is what agrees with all of their measurements, no matter how many observers there are or how fast they are going or which direction they are going in. Simple addition won't work.
    Yes, very true. But just remember, no matter what frame of reference we use, this in no way affects the measurements that each observer makes of the relative speed of each of the other observers, they will still make the same measurements, but an arbitrary Frame of Reference will provide another set of relative speeds with respect to each observer and these can be different for each Frame of Reference.
    Please don't think that since Special Relativity isn't enough to explain what happens with galaxies, that it has no application in nature.
     
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