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

Frames of reference for speed?

  1. Feb 8, 2013 #1
    So sense speed is relative, could I observe something traveling faster than light? Like in a car, when I see oncoming traffic it appears much faster, so could I not be traveling at, say 1/2c and observe somebody approaching at 1/2c therefore moving at 1c relative to me?
     
  2. jcsd
  3. Feb 8, 2013 #2

    ghwellsjr

    User Avatar
    Science Advisor
    Gold Member

    No, look up Velocity Addition in wikipedia.
     
  4. Feb 9, 2013 #3

    ghwellsjr

    User Avatar
    Science Advisor
    Gold Member

    I decided to draw some spacetime diagrams to illustrate this scenario. We start with diagrams for the Inertial Reference Frame (IRF) of someone stopped on the road like a radar cop who is shown as a thick black line with dots every second of the cop's time. I am using scales such that the speed of light (and radar signals) is 1 to make it easy to do calculations. This makes the radar signals travel along lines on a 45 degree diagonal.

    You are shown in going to the right at one half the speed of light. You should be able to confirm that for every two seconds of Coordinate Time, you travel one light-second of Coordinate Distance. But since you are traveling at such a high speed, time for you is dilated by the factor gamma which is 1.1547 at 0.5c and your one-second blue dots are spaced farther apart than the coordinate time. The dots always show the Proper Time for each observer and are only the same as the Coordinate Time if the observer is not moving in the IRF. In that case the observer's path will be shown as a vertical line.

    Coming towards you from the right is a green car also traveling at -0.5c and with one-second green dots spaced farther apart, just like yours.

    In this first diagram, the cop is using radar signals shown in red to measure your speed:

    attachment.php?attachmentid=55553&stc=1&d=1360435102.png

    The cop sends two red radar signals and receives two echoes. The first signal is sent at Coordinate Time of -6 and received at -2. (Remember, since the cop is not moving, the Proper Time on his clock matches the Coordinate Time.) From this data, the cop can conclude that halfway between these two points, at his Proper Time of -4, your distance was one-half of the difference between his two Proper Times which is (-2+6)/2 = 4/2 = 2 light-secs. You can see that at Coordinate Time -4, your distance from the cop was 2 light-secs.

    The second red radar signal is sent at the cop's Proper Time -3 and received at -1. So applying the same process we conclude that at Coordinate Time -2, your distance was (-1+3)/2 = 2/2 = 1 light-sec which you can also confirm.

    Now to determine your speed, the cop has to take the difference between your two distances that he measured and divide it by the difference in the two times that he placed those measurements at. This calculates as (2-1)/(-4+2) = 1/-2 = -0.5c. This means that the cop measures that you are coming towards him at 0.5c.

    Now we look at how you in the blue car can do the same thing with your radar gun aimed at the cop as depicted in this diagram:

    attachment.php?attachmentid=55554&stc=1&d=1360435102.png

    Since you are traveling with respect to the Coordinate Time, you need to use your own Proper Time as indicated by the blue dots. If you follow what the cop did with his Proper Time, you can see that your measurement is exactly the same as what the cop did and you will conclude that the cop is moving towards you at 0.5c.

    However, if you look at the diagram, the distances don't appear to be correct as depicted on the drawing. But you can't tell that, all you can tell is what the data from your measurements tell you.

    But if use the Lorentz Transformation process to create a new diagram for an IRF in which you are stationary, we can see how the distances would come out consistent with your measurements. Since you are traveling to the right at 0.5c we use an IRF that is traveling to the left of the first one at -0.5. This is what we get:

    attachment.php?attachmentid=55555&stc=1&d=1360435102.png

    Now it is easy to see that the distances you measured match the Coordinate Distances but remember, you can't tell the difference as all your observations are dependent on radar signals traveling at c.
     

    Attached Files:

  5. Feb 9, 2013 #4

    ghwellsjr

    User Avatar
    Science Advisor
    Gold Member

    Now that we know how to measure the speed of another object, we are ready for you to measure the speed of the green car that is traveling on the road at -0.5c toward you while you are traveling on the road at 0.5c. You do the same thing that you did to measure the speed of the cop relative to you but of course you will get different numbers. Here is the diagram for the original IRF with the measurements you make:

    attachment.php?attachmentid=55558&stc=1&d=1360438329.png

    Now the distance calculation for the first radar signal is (-2+18)/2 = 16/2 = 8 light-seconds at a time that is at the midpoint between when you sent and received the signals, at -10 seconds. For the second radar signal these calculations are (-1+9)/2 = 8/2 = 4 light-seconds and the midpoint is at -5 seconds. Taking the differences as we did before we get (8-4)/(-10+5) = 4/-5 = -0.8c which means you see the green car approaching you at 0.8c.

    You will note that if you followed the Velocity Addition formula linked to in post #2, you would have calculated:

    (0.5+0.5)/(1+0.52) = 1/(1+0.25) = 1/1.25 = 0.8

    which is the same answer.

    Now one last diagram to show the IRF in which you are at rest:

    attachment.php?attachmentid=55559&stc=1&d=1360438329.png

    You should be able to confirm that both of the IRF's in this post present exactly the same information.
     

    Attached Files:

    Last edited: Feb 9, 2013
  6. Feb 9, 2013 #5
    no. velocities do not add EXACTLY as s = v + u....that is a low speed approximation.

    Under SPECIAL RELATIVITY in the above link from ghwells, use the first formula for all speeds, even those close to c.

    s = [v +u]/[1+ vu/c2]

    and substitute for v,c = 1/2c and u = 1/2c

    and figure out what answer you get...it will be less than c.
     
  7. Feb 9, 2013 #6

    How, then, do you know if you are using special relativity or normal relativity? Also, v and u just mean the speed of the observer and the speed of the object observed, correct?
     
  8. Feb 9, 2013 #7

    Nugatory

    User Avatar

    Staff: Mentor

    There's no such thing as "normal" relativity. There's "special relativity", called that because it is a special case of a more general theory, and there's that more general theory, which is naturally called "general relativity". They don't conflict. Any problem that can be solved using SR can also be solved using GR, and you'll get the same answer; but there are some problems (any in which the effect of gravity is significant) that can only be solved using GR.

    We're using special relativity in this thread because there are no gravitational effects involved.
     
  9. Feb 9, 2013 #8
    yes....my post was supposed to read 'substitute 1/2c for u, and for v.
     
  10. Feb 9, 2013 #9

    No, because observer moving at velocity 30 m/s, observing an object moving at velocity 30 m/s, will obviously observe the velocity of the object to be 0, relative to himself.

    But the velocity addition formula gives the result aproximately 60 m/s, if we plug those numbers in.
     
  11. Feb 10, 2013 #10
    My post #8 is relative to the originally posed question regarding my interpretation of the '1/2c' velocities used in the example.
     
  12. Feb 10, 2013 #11
    I guess there is a formula to calculate what velocity a moving observer observes when observing a moving object.

    I guess the formula is:
    s = [-1*v +u]/[1+ -1*vu/c2]


    This relativistic velocity addition formula here:

    s = [v +u]/[1+ vu/c2]

    is good for adding speeds, when the observer and the observed object have opposite velocities.

    And it's also good for adding the velocity of a ship, observed by a person standing on the shore, and the velocity of a fly, observed by a person sitting inside the ship.
     
  13. Feb 10, 2013 #12
    Unless the object is moving towards the observer, and the observer towards the object. Then the velocity of the object would be 60m/s relative to the observer(assuming they are both moving 30m/s towards each other) right?
     
  14. Feb 10, 2013 #13

    Nugatory

    User Avatar

    Staff: Mentor

    No. As you've written it, this is an impossible situation because an observer is never moving relative to himself, so the observer cannot be moving at 30 m/s towards the object. As far as the observer is concerned, he's at rest and the object (and perhaps everything else nearby, even the ground under his feet) is moving towards and past him.

    However, I think that you may really be asking about a different but similar situation: There's a second observer, he's at rest relative to himself (of course!), he sees the object approaching from the left at 30 m/s, and he sees the vehicle carrying the first observer approaching from the right at 30 m/s. This is basically the same situation as if the second observer were standing at the side of a road, watching two cars approach from opposite directions at 30 m/s.

    This second observer will see the object and the first observer both approaching him at a speed of 30 m/s each, and therefore moving towards each other at 60 m/s.

    The first observer will consider himself to be at rest while the second observer is moving towards him at 30 m/s. However, he will not see the object approaching him at 60 m/s; it will be something less and you'll have to use the relativistic velocity addition to calculate what it is. Notice especially that the two observers do not agree about the speed of the object relative to each other.
     
  15. Feb 11, 2013 #14

    ghwellsjr

    User Avatar
    Science Advisor
    Gold Member

    That's exactly how I interpreted the situation except I made the second observer be a radar cop in a patrol car parked at the side of the road and, of course the speeds are 0.5c, not 30 m/s. Where'd that come from?

    Note that I showed the calculation of the resultant speed between the two diagrams in post #4 as 0.8c.
     
  16. Feb 11, 2013 #15
    According to the observer the velocity of the object is very close to 60 m/s.


    But how do we calculate that result?

    Let's try relativistic velocity addition formula: s = [v +u]/[1+ vu/c2]
    v=30 m/s, u=-30 m/s
    result of calculation is 0 m/s

    Here's a warning for all newbies: If you ask this question about approaching cars in any forum, very likely you will be told to use the relativistic velocity addition formula, which will give wrong results, if you plug in the velocities of the cars.
     
    Last edited: Feb 11, 2013
  17. Feb 11, 2013 #16

    ghwellsjr

    User Avatar
    Science Advisor
    Gold Member

    You do raise an important point that you have to be careful about signs. However, you still didn't get it correct.

    The example in the wikipedia article quoted Galileo's example of v being the speed of a ship moving away from the shore and u being the speed of a fly moving away from the ship. So if you want to define the speed of an object relative to an observer as being positive when the object is moving away from the observer, then you would say that when traveling on a road, a radar cop parked on the road is moving at a negative speed relative to the driver in the car and an approaching car is also moving at a negative speed relative to the cop and the resultant speed is also negative.

    So you should have said:

    v=-30 m/s, u=-30 m/s
    result of calculation is very close to -60 m/s

    And if you look at my derivations in posts #3 and #4, you will see that I was consistent in this regard and I got -0.8c as the speed of the approaching car relative to the OP's car. I could have been more consistent when I plugged numbers into the formula by making both of the being negative but I was just concerned about the magnitude of the answer.

    The OP's question had to do with two very large speeds, 0.5c, which when added are not close to their algebraic sum. They add to 0.8c, not 1.0c or even close to 1.0c. By using very slow speeds, you obfuscate the issue that the OP was asking about because with your example the correct answer is so close to the algebraic answer as to be indistinguishable.

    Whenever I use the Velocity Addition formula, I don't worry about the signs of the velocities. It is always obvious whether the final result should be larger or smaller than either of its two inputs.
     
  18. Feb 11, 2013 #17

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    It's helpful to use descriptive subscripts.
    (All quantities are signed... think "x-components of velocities" (not speeds).)

    Relative-velocity formula
    [tex]
    v_{C\ wrt\ B}=\frac{v_{C\ wrt\ A}-v_{B\ wrt\ A}}{1-v_{C\ wrt\ A}v_{B\ wrt\ A}}
    [/tex]
    (You can drop the "wrt"="with respect to" when the notation is clear.)

    Velocity-addition formula
    Since [itex] v_{A\ wrt\ B} = -v_{B\ wrt\ A}[/itex], you can write
    [tex]
    v_{C\ wrt\ B}=\frac{v_{C\ wrt\ A}+v_{A\ wrt\ B}}{1+v_{C\ wrt\ A}v_{A\ wrt\ B}}
    [/tex]

    The Galilean versions are:
    Relative-velocity formula
    [tex]
    v_{C\ wrt\ B}=v_{C\ wrt\ A}-v_{B\ wrt\ A}=(v_{C\ wrt\ Z}-v_{A\ wrt\ Z})-(v_{B\ wrt\ Z}-v_{A\ wrt\ Z})=(v_{C\ wrt\ Z}-v_{B\ wrt\ Z})
    [/tex]

    Velocity-addition formula
    [tex]
    v_{C\ wrt\ B}=v_{C\ wrt\ A}+v_{A\ wrt\ B}=(v_{C\ wrt\ Z}-v_{A\ wrt\ Z})+(v_{A\ wrt\ Z}-v_{B\ wrt\ Z})=(v_{C\ wrt\ Z}-v_{B\ wrt\ Z})
    [/tex]

    To do the analogue of this for special relativity, use rapidities [itex]\theta_{C\ wrt\ B}[/itex] (which are additive like the Galilean velocities are), where velocity=tanh(rapidity). You'll make use of the identity for the tanh of a difference.
     
  19. Feb 11, 2013 #18
  20. Feb 11, 2013 #19
    That was not my point, and I don't think I made errors worth mentioning. Here's what I thought, quite simple isn't it?

    Car1 ---> <---Car2
    v=30 m/s v=-30 m/s



    If the signs of the inputs of a formula, or the result, must be adjusted using intuition, then it's not so good formula. That's just my opinion. But here's a problem: what do you answer when a student asks "what exactly do I plug into this formula"?
     
  21. Feb 11, 2013 #20

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I would tend to agree.
    I tell often tell my students... How you would write a computer program [lacking physical intuition] to get the correct answer?
    One could use a universal formula using signed-components.. or else use special cases depending on whether the ships are approaching or receding.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook