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I Simple Velocity Relativity Problem

  1. Dec 20, 2017 #1
    A problem I struggle to conceptualise as I have began leanring about relativity and seek your guidance and knowledge on:

    Consider a stationary observer at point A. Relative to the observer there is a train travelling at -0.6c. A passenger on board the train has a particle gun, and fires it in the same direction of the trains travel. The particle is fired at a speed of 0.6c relative to the train. Ignoring all logical rules as to why this couldnt happen, why does the particle fired not have a velocity relative to the observer of 1.2c?
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  3. Dec 20, 2017 #2


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    Because that is not how velocities add in special relativity. You need to use the relativistic formula for velocity addition. The point is that the postulate that the speed of light is invariant (the same in all frames) already tells you that velocities cannot add like ##v' = v + u## in relativity. If you do the math properly, the relativistic formula for addition of velocities drops out.
  4. Dec 20, 2017 #3


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    Another way of looking at it is to ask how the gunman knows his bullet is doing 0.6c? Basically, he has to use a ruler to measure the distance it travels in a fixed time measured by his clock.

    But according to the guy outside the train, the gunman's rulers are length contracted and his clocks are time dilated. So he does not expect the answer to be 0.6c+0.6c because one of those 0.6cs was measured by the gunman using rulers and clocks that don't measure right.
  5. Dec 20, 2017 #4


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    Can you draw a spacetime diagram of the situation?
    How would you "define" (or measure) the velocity of a particle?
  6. Dec 20, 2017 #5

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    Because adding velocities is not the right way to combine them.
  7. Dec 23, 2017 #6
    remember, the relativistic formula for distance is

    x' = (x+(vt/c))/sqrt( 1- (v/c)^2)

    t' = (t +( vx)/c)/ sqrt( 1- (v/c)^2)

    That v implies both observers measure the other as moving at velocity v, which in your example is 0.6 c. Let the x' observer fire a particle gun at 0.6 c measured in HIS frame.
    Then x/t is velocity u in the x/t frame

    x'/t' ' =(x+vt//c))/sqrt( 1- (v/c)^2)
    ____________________ =( x+vt)/(t+vx) since the 1/ (1 - (v/c)^2) in numerator and denominator cancel out
    (t + vx/c)/ sqrt( 1- (v/c)^2)

    divide numerator and denominator by t and you get, remembering x/t= u
    ((x/t +v)/ (1 +(v* x/t) = (u+v)/(1 + uv/c^2) = ( 0.6+0.6)/(1.36)

    So the fellow on the train measures the stationary observer as moving BACK at the rate of 0.6c and the bullet moving FORWARD at 0.6 c, for a
    difference of 1.2 c.

    Observer A sees the train moving at 0.6c and the bullet moving at (1.2/1.36) c. From HIS point of view, the train obvserver's measured speed is off because both his watch is off, and the distance measured is wrong.

    Here's an example of how two observers can measure each other slow. Say Han Solo is leading a group of 3 ships at a velocity less than c, with their clocks synchronized in their own frame. Luke Skywalker is leading a group of 3 ships in the opposite direction.

    Princess Leia is in a ship halfway between the two.

    Here's what Princess Leis sees at time 0:00 for Solo and Skywalker

    Han Solo Group Ship C Ship B Ship A
    clocks 4:00 2:00 0:00

    Skywalker Group Ship A' Ship B' Ship C'
    clocks 0:00 2:00 4:00

    Ship A and A' pass each other and they see both their clocks read 0:00

    Here's what Princess Leis sees at time 2:00 for Solo and Skywalker
    Han Solo Group Ship C Ship B Ship A
    clocks 6:00 4:00 2:00

    Skywalker Group Ship A' Ship B' Ship C'
    clocks 2:00 4:00 6:00
    From Princess Leia's point of view, clocks on both groups of ships are running at the same (slow) rate, but the clocks on A,B,C and A',B',C' are NOT synchronized.
    From Han Solo's point of view, clocks on A,B,C ARE synchronized. At time zero he sees Skywalker's clock read 0:00
    At time 2:00 he sees ship B' clock read 4:00 so he thinks B' is NOT synchronized with A'.
    Solo later gets a message from Ship B, travel speed limited by the speed of light, that Ship B, synchronized with Ship A, saw ship A's clock read 2:00 at B's time, 4:00, so Skywalker's clock is running at half speed.

    Naturally, Skywalker sees the same in reverse. From HIS point of view, A'B',C' clocks are synchronized and Han Solo's ship clocks run at half speed, but give bogus results because their clocks are UNSYNCHRONIZED from Skywalker's point of view.

    I just thought of a puzzle. From that added velocity formula above, (u+v)/ (1+(uv/^c2 ))
    And the fact that Solo and Skywalker see each other's clocks as running at half speed,
    how fast is each fleet of ships moving relative to Princess Leia?
  8. Dec 23, 2017 #7


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    Note that "(0.6+0.6)/(1.36)"=(15/17).

    Here's a spacetime diagram on rotated graph paper to help visualize
    why 15/17 is the relative velocity.

    (This diagram is taken from my post on https://physics.stackexchange.com/questions/326272/applying-time-dilation-twice )

    Alice measures Bob's velocity as 3/5, using the ratio PQ/OP "displacement/elapsed time according to Alice" ,
    where OP is along Alice's worldline and PQ is simultaneous according to Alice (that is PQ is Minkowski-perpendicular to OP).
    P is on Alice's worldine and Q is on Bob's worldline. PQ is "how far away Bob is according to Alice after elapsed time OP according to Alice".

    Similarly, Alice measures Carol's velocity as -3/5.

    Carol measures Bob' velocity as 15/17,
    since after 17 "ticks" on Carol's wristwatch, Carol says Bob is 15 spaceticks ("sticks") away.
    (This choice of 17 is out of convenience because the corresponding displacement is 15...
    there are Pythagorean triples lurking when velocities like 3/5 are used.)
    (btw, You can also read off the time-dilation factors 17/8 [relating Carol and Bob], and 5/4 [relating Carol and Alice, and relating Bob and Alice]. )

    The quantity NQ/OP (which you can form)
    is NOT "the velocity of Bob according to Carol" since OP is not on Carol's worldline.
    In addition, NQ is not Minkowski-perpendicular to ON on Carol's worldline.

    In Galilean physics, this situation below doesn't occur because
    ON=OP=OQ according to absolute time (so the diamonds in the diagram below have to be replaced by ticks that have the same vertical step)
    all observers would regard NQ as simultaneous (i.e. as Galilean-perpendicular to OP, to ON, and to OQ).

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