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Special Relativity Question (Lorentz)

  1. Sep 15, 2016 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data
    Synchronized clocks A and B are at rest in our frame of reference a distance 2 light minutes apart. Clock C passes A at a speed of c*4/5 bound for B, when both A and C read t =0 in our frame.

    a) What time does C read when it reaches B?

    b) How far apart are A and B in C's frame?

    c) In C's frame, when A passes C, what time does B read?

    2. Relevant equations

    Lorentz equations

    [tex] x' = x\sqrt{1-\frac{v^2} {c^2}} [/tex]
    [tex] t' = t\sqrt{1-\frac{v^2} {c^2}} [/tex]
    3. The attempt at a solution


    a) I don't believe we have to use lorentz equations for this.
    [tex] t = \frac{distance} {speed} = \frac{2c*mins} {0.8*c} = \frac{5} {2} mins [/tex]

    b) This is where we use Lorentz contraction so

    [tex] x^c = 2c*mins\sqrt{1-\frac{16}{25}} = 1.2c*mins[/tex]

    c) This is the tougher one for me.

    So, in C's frame, B is coming towards us at 4*c/5. We are at t=0.

    Does time B read 1.2 minutes?

    Thanks for your time / looking over my work.
     
    Last edited: Sep 15, 2016
  2. jcsd
  3. Sep 15, 2016 #2

    andrewkirk

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    Yes Lorentz is needed for (a). The question is what does the moving clock of C read, not what do clocks A and B read. Doing a calculation that divides the distance by the speed - both of which are given in the units of the frame A and B - will tell us what clocks A and B read, not what C reads. Also the arithmetic is incorrect: ##2\div 0.8## is not equal to ##4/5##.

    First do a correct calculation of the elapsed time on clock A. Then use Lorentz to convert that to an elapsed time on clock C.
     
  4. Sep 15, 2016 #3

    RJLiberator

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    Whoops! I fixed the arithmetic error.

    Let me analyze more deeply what we have:

    In "our" frame of reference:
    A, B are at rest
    velocity of clock c = (4/5)c
    d = 2 light minutes

    We need to find t in C's frame of reference as it reaches B
    so t in c we denote t' = t*sqrt(1-16/25)

    The problem is in finding what t is equal to?
    If I say x' = 2 light minutes (1-16/25) = 1.2 c minutes as I determined in b, then

    x' = 1.2 c*minutes

    Now I could use v = d/time ==> time = d/v

    [tex] t' = \frac {d} {v} = \frac {1.2c*mins} {\frac{4c} {5}} = \frac{3} {2} mins [/tex]



    And so, the answer for a would be 1.5 minutes, b would be 1.2 c minutes.

    Is this so far correct?
     
  5. Sep 15, 2016 #4

    andrewkirk

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    Yes, that looks correct to me.

    For (c) I don't think the equations you have written are enough, as they are simplifications of the full Lorentz equations. The simplified equations only cover scaling and tell us nothing about simultaneity.

    Question (c) asks for the frame-AB time coordinate of the spacetime point P that is the intersection of the worldline of B and the line of simultaneity in frame C that passes through the spacetime point Q at which C passes A.

    The full equations, that enable calculation for (c) are here.
     
  6. Sep 16, 2016 #5

    RJLiberator

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    Nice!

    Hm, I see.

    [tex] t' = \gamma(t-\frac{vx} {c^2}) [/tex] Should be the one we are looking for.

    t' = B's reading when A passes C.

    [tex] \gamma = \frac{1} {sqrt{1-\frac{v^2} {c^2}}} [/tex]

    Well, we know x should be 1.2 c minutes apart part b
    I suppose t = 3/2 minutes now from part a
    v = c*4/5


    These are all in C's frame.

    So what we get is the following

    [tex] t' = \frac{1.5-\frac{4*1.2c^2} {5c^2}} {\sqrt{1-\frac{16c^2} {25c^2}}} [/tex]

    And this equals 0.9 minutes.

    Although, I'm not really sure why using these values for t, v, x all work out for the initial time that it reads.
    This seems fishy, but, the answer appears to be OK (or at least make sense).
     
  7. Sep 17, 2016 #6

    RJLiberator

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    Can anyone confirm my reasoning for the last post?

    My answer of 0.9 minutes seems reasonable. Q: In C's frame, when A passes C, what time does B read?
    When A passes C, the time on B reads 0.9 minutes.

    My only concern is that I used the information for C over an elapsed time period, not the initial settings (as there I'm lost on how I'd go about that...).

    Why does this work ? (permitting this is the correct answer 0.9 minutes).
     
  8. Sep 17, 2016 #7

    PeroK

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    It's not 0.9 minutes. That's something else that is relevant here but not the answer to c).

    (It's really confusing calling those the "Lorentz" equations. Those are the time dilation and length contraction equations, not to be confused with the Lorentz Transformation. Where does that terminology come from?)

    There are several ways to do part c). You could ask the following questions:

    1) In the A-B frame, what do both clocks read when C passes B?

    2) In C's frame, what how much does B's clock advance between C passing A and B?

    3) Can you use these two calculations to get the answer to c)?
     
  9. Sep 17, 2016 #8

    RJLiberator

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    Well, we know that 1.5 minutes is the time elapsed when C reaches B in C's frame.

    [tex] 1.5mins = t\sqrt{1-\frac {16} {25}} => t = 2.5 minutes [/tex]

    That makes sense as the velocity = 4c/5 so you would expect it to be an elapsed time greater than the 2 minutes it takes speed to travel the distance.


    B's clock advances from 0 to 1.5 minutes in C's frame. B's clock advances 1.5 minutes.

    So, in c's frame, when A passes C that must been that B reads 1.0 minutes.

    A = 0 minutes when C see's A passing it (in c's frame).
    B = 1 minute when C see's A passing it (in c's frame)

    That way, when 1.5 minutes elapses (time elapsed) we get 2.5 total minutes.


    noted!
     
  10. Sep 17, 2016 #9

    PeroK

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    That's not right. The best approach is to analyse the problem entirely from the A-B frame first. In the A-B frame:

    C passes A at ##t = 0##

    C passes B at ##t = 2c.mins/0.8c = 2.5 mins##

    Therefore, when C passes B, both A and B's clocks must read 2.5 minutes in the A-B frame. And C's clock reads something else. I think you've worked this out, it's 1.5 minutes.

    Now, an important point. B's clock reads 2.5 minutes when C passes B full stop! And C's clock reads 1.5mins full stop. All other observers will agree on this: when that event took place (C passed B): the time in B's frame (at point B) was 2.5 minutes and the time in C's frame (at point C)was 1.5 minutes full stop.

    Now, can you analyse the problem from C's frame, starting when C is at A (##t = t' = 0##) until B reaches C?

    Hint: at ##t' = 0## you have an unknown time on B's clock: call it ##t_0## say. And work from there.
     
  11. Sep 17, 2016 #10

    RJLiberator

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    Okay, so with the hint from earlier to use the Lorentz transformations for part c, and with your help, here is what I have thus far.

    [tex] t = \gamma(t'+\frac {vx'} {c^2}) [/tex]

    Filling in with t' = 0, x' = 1.2 light minutes, and v = 4c/5

    we get the answer of 1.6 minutes.


    I see how I used t' = 0. That makes sense, and we are trying to find t_0 in C's frame. I'm using the correct distance and the correct velocity.

    I think the final answer is 1.6 minutes is what B reads.

    But, not sure how this makes sense, if at t=0 in A, t=1.6 minutes in B, in c's frame. I would think the answer would be closer to 0.
     
  12. Sep 17, 2016 #11

    PeroK

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    To answer your last question, you should go through the scenario in C's frame. You have used the Lorentz Transformation above to solve the problem directly. But, if you start with unknown ##t_0## on B's clock and calculate the time that elapses on B's clock during its motion towards C, then you should be able to confirm the answer you got and perhaps shed some light on why ##t_0 \ne 0##.
     
    Last edited: Sep 17, 2016
  13. Sep 17, 2016 #12

    vela

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    A spacetime diagram is really helpful to see what's going on as well. C's "now" when t'=0 corresponds to the x'-axis. You can see in the diagram, this axis intersects B's world line when t=1.6 min.
     

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  14. Sep 17, 2016 #13

    RJLiberator

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    The diagram does help out a lot. I imagine we are on the verge of learning how to develop those graphs.
     
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