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B Clarifying relative velocity

  1. May 16, 2016 #1
    I have been told that relative velocity can be negative on this forum, so i would like to ask a few questions based on this.

    scenario 1) Two observers O and O' are facing each other and are approaching each other.

    O sees O' approaching at v=0.5c and vice versa. The vector of the velocity is however pointing towards the back/negative x axis. So would the relative velocity be rather v=-0.5c in this case?

    Or does the sign of the relative velocity depend on if they are approaching or moving away from each other?
    Should this be the case, then is the relative velocity negative as they approach each other and then switches to being positive when they move away of each other?

    What about the case when they exactly meet. What would be the relative velocity in this case?


    scenario 2) Observer O is looking at observer's O' back while approaching. O' has his back turned towards O while he is approaching from behind.

    Is the sign of the relative velocity the same for both observers, or is the relative velocity different for each of them, as in having the sign switched?
     
  2. jcsd
  3. May 16, 2016 #2

    Ibix

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    It depends which way you wish to define it. The signs do not matter in any absolute sense - it's all just book keeping. If you flip your directions around the +v will become -v, but nothing changes about the actual situation you are describing. So pick a convention and stick with it. I'd recommend sticking with standard conventions where available because then you can use standard results and standard tools.

    The system you were proposing in your other thread was not a standard system and was substantially more complicated (because you have to keep track of all the parity flips), and didn't add anything over the standard set up. In fact, if I understood Peter correctly, it rules out the use of some more powerful mathematical tools. But, as long as you are happy to stick to simple cases and to accept a higher error rate from book keeping trouble, it'll work because it's just a different book keeping convention.
     
  4. May 16, 2016 #3

    PAllen

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    It is really simple. If they are comoving, their relative velocities are 0. If they are not, then in the linear case, their relative velocities have opposite sign. Which you pick as + and which as -, does not matter. Whether they are approaching or receding does not matter as to the fact their relative velocities have opposite sign (it does matter for other things, like Doppler). For Doppler, if the the position is positive and velocity is positive, or both are negative (by standard convention) the Doppler will be redshift; otherwise it will be blue shift.
     
  5. May 16, 2016 #4

    Nugatory

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    Let's start with the velocity of observer two relative to observer one. Using coordinates (we'll call them ##x## and ##t##) in which observer one is at rest, write the position of observer two as a function of time: ##x(t)##. The relative velocity of observer two relative to observer one is (by definition!) ##v=\frac{dx(t)}{dt}##. This will be positive or negative depending on whether ##x(t)## is increasing or decreasing with time, and it is the same no matter which direction the two observers are facing and no matter whether one is behind or in front of the other.

    Now let's try the velocity of observer one relative to observer two. Using coordinates in which observer two is at rest (these are different coordinates so I need to call them something different - we'll use ##x'## and ##t'##) write the position of observer one as a function of time: ##x'(t')##. The relative velocity of observer one relative to observer two is (by definition!) ##v'=\frac{dx'(t)}{dt'}##. This will be positive or negative depending on whether ##x'(t')## is increasing or decreasing with time, and it is the same no matter which direction the two observers are facing and no matter whether one is behind or in front of the other.

    It is easy to see that if the direction of increasing ##x## is the same as the direction of increasing ##x'##, then ##v'=-v## (or equivalently ##v=-v'##). The observers are free to choose which direction is the increasing direction, and as the previous thread has made clear, everything is a lot easier if they choose ##x## and ##x'## to both be increasing in the same direction no matter which way they happen to be looking.
     
  6. May 16, 2016 #5
    Alright then. Let's have a look at how it is defined in the case of minkowski diagrams.

    Wx3o2kC.png

    Three observers, O, O' and O'', using the event e1(5,3). O' measures e1'(4.333.., -1666..) , O'' measures e2''(7.5..,6.35)

    The relative velocity can be negative obviously in this case, but it is a fixed value. It does not change depending on if any observers are approaching or moving apart from each other.

    The relative velocity as measured by O is the velocity vector of any at rest to O' or O''. Any observer at rest to O' would be approaching O if he was behind him and be moving away of him once he passed towards the positive x axis. Behind meaning the negative x axis.
    The vector is therefore positive 0.8c.
    Accordingly, the vector is negative -0.5c in the latter case.

    The two diagrams at the bottom, show the scenario from the perspective of O' and O'' respectively.

    O' would be an observer who is facing the back of O while approaching from behind him, does the measurements just when he passes by O and then moves away having his back turned against O.
    Hence, O is moving towards the negative x axis from the perspective of O. The velocity vector is negative.

    The inverse Lorentz formula was not needed at all. What changed is the relative velocity. The relative velocities are different depending on the perspectives. Both bottom minkowski diagrams use the non-inverse formula to arrive at the correct values.

    Using the same logic, i am going to attempt going from O'' to O' now.

    Since all three observers are moving in parallel, the scenario seen by O'' should be as follows:

    O'' has his back turned against O'. O' is facing the back of O'' and is approaching him from behind. When they pass by each other, they do their measurement and O' moves away towards the positive x axis. Hence, the velocity vector is positive from the perspective of O''.

    Unfortunately the site i drew those on, does not allow values higher than 0.8c so no pretty diagrams for this case.

    Their relative velocity is(from the perspective of O'') 0.92857....c with γ= 2,6943...

    Doing the Lorentz transformations i get for d=7.5056... and t=6.35

    d' = γ(d-vt) = 4.333 as expected
    t' = -1.666...


    Now the reason the relative velocities are different for observer O and O' or any other combination between any two of the observers O,O',O'' is because the cases are not symmetric.
    To be symmetric, they would have to be either both facing each other while approaching and having their backs turned against each other while moving away(or the other way around for the 2nd symmetric case).
    In the first case, the relative velocity would be negative for both while in the latter case it would be positive for both.

    In both cases, the transformations (not calling them LS transformations) should be involutions.

    Not sure it would be possible to draw minkowski diagrams like the one on top for the symmetric cases though. Doesn't seem impossible however. I would have to give this more thought.


    p.s

    I created those diagrams using this site http://www.trell.org/div/minkowski.html,

    However, it allows only two concurrent observers/frames within 1 diagram. I had to use other software to overlap two diagrams in order to create the one on top.
    Does anyone know of a site which allows you to draw similar diagrams but with arbitrary many concurrent observers/frames?
     
    Last edited: May 16, 2016
  7. May 16, 2016 #6
    When i use the term "facing" i mean he is facing towards the positive x-axis. It determines how the observer defines his positive and negative x-axis.(when he does a measurement)
     
    Last edited: May 16, 2016
  8. May 17, 2016 #7
    If we call transforming from O to O' as the (direct) Lorentz transform then changing the sign of the relative velocity to transform from O' to O is the inverse Lorentz transform.

    But you didn't just change the sign of the velocity, you also re-labeled your frames. In the bottom diagrams which you call the O' and O" perspective you're actually flipping the primed and unprimed frames.

    Edit: More specifically, in the bottom left diagram your black axes should be labeled t' and d' and the blue axes t and d. In the bottom right the black axes should be t" and d" and the blue axes t and d.
     
  9. May 17, 2016 #8
    Well then, one could see it as me being surprised that one would present a whole new set of formulas when you could work with just one set by just entering the proper relative velocity as a variable into the non-inverse LST, or to be more precise, the relative "relative velocity" as it seems, in opposition to absolute "relative velocity". The relative "relative velocity" depends on if we are measuring it from the perspective of O or O' when defining "relative velocity" the way it is used for Minkowski diagrams for example.


    You are right but i thought i would spare myself all the editing in some paint program and use the output i got from http://www.trell.org/div/minkowski.html directly. I thought it would be clear enough what is meant still.
     
  10. May 17, 2016 #9

    PeterDonis

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    It depends on how you define the coordinates of each observer, as was explained to you multiple times in the previous thread. If you define the two sets of coordinates so that they are related by a standard Lorentz transformation, then the relative velocities have opposite signs (B relative to A, vs. A relative to B). If you define the two sets of coordinates so that they are related by a Lorentz transformation plus a parity reversal, then the relative velocities have the same sign--but in that case zero relative velocity does not correspond to the identity transformation, it corresponds to a pure parity reversal.
     
  11. May 17, 2016 #10

    Nugatory

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    No, it does not. You will continue to confuse yourself until you learn how relative velocity is defined and then use the correct definition: Pick a coordinate system in which one of the objects we're considering is at rest; the relative velocity is the time derivative of the position of the other object. The "perspective" of various observers is irrelevant, as is the way that we draw Minkowski diagrams.

    Your best bet may be to go back through your previous thread and this one from the beginning, reading carefully until you find something that you either disagree with or don't understand. At that point you are being misled by a mistaken assumption - stop right there and try to identify that assumption.
     
  12. May 17, 2016 #11

    PeterDonis

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    The issue raised in the OP has been addressed. Thread closed.
     
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