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

Inertial Frames distinguished by proper times

  1. Mar 26, 2006 #1
    A first spaceship S1 departs from earth and quickly accelerates to a velocity V = c/2. S1 travels the shortest path (dead heading) toward a distant planet Alpha so that it arrives in 20 years as measured by a clock on S1. One year after S1 is launched from earth as measured by a clock on the earth, a second spaceship S2 is launched along the same trajectory (we assume neither earth nor alpha have moved during this experiment). S2 quickly accelerates to a greater velocity than c/2 and at some point S2 overtakes S1. When S2 arrives at Alpha, the clock aboard S2 will read a lapse time of 10 years since it departed from earth.

    Both S1 and S2 move at uniform velocity, so each is a valid inertial system, but the proper clock rate of the S2 clock is 1/2 the proper clock rate of the S1 clock. Based upon the difference in proper rates - if observers on each spaceship take the measure of the other as S2 passes S1, will an observer in S1 arrive at the same conclusion about contraction and clock rate in the S2 frame as S1 determines about contraction and clock rate in the S2 frame?
     
    Last edited: Mar 26, 2006
  2. jcsd
  3. Mar 27, 2006 #2

    Ich

    User Avatar
    Science Advisor

    The proper clock rate in S1 and S2 is 1s/s. Proper (German: "eigen") means belonging to itself, contrary to "with respect to". Try to remember this before posting.
     
  4. Mar 27, 2006 #3

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    S2 is in at least two different inertial systems, because it accelerates. It's unclear if S1 breaks to a stop or not - if S1 does break to a stop, it accelerates as well.

    Accelearting clocks are not inertial clocks by defintion.
     
  5. Mar 27, 2006 #4

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Is this a fair rephrasing of the problem?

    The earth and alpha are inertial and at rest with respect to each other (so, they have parallel worldlines, vertical in their frame).

    Ignoring the time when S1 was on the earth awaiting launch... at event P, S1 travels inertially toward alpha at speed c/2 and arrives at alpha at event P', after 20 years on S1's clock.

    Ignoring the time when S2 was on the earth awaiting launch... at event Q, one year after P according to the earth's clocks, S2 travels inertially at toward alpha some speed greater than c/2 so as to overtake S1. That is, their worldlines cross before alpha. S2 arrives on alpha at event Q', after 10 years on S2's clock.

    Is this the situation?

    If so, then
    true
    false, as Ich says. The "proper clock rate of S1" is the clock rate of S1 measured by clock S1. Likewise, for S2.
    What it all boils down to is that you have two intersecting inertial worldlines.

    S1 and S2 are inertial observers, between P and P' and between Q and Q', respectively. So, identical experiments they perform on each other (e.g. comparing doppler effects or time dilation effects) will agree.

    In fact, their relative-velocity can be computed from the data you've given. I get 0.41c. (S2's velocity must be 0.755c and alpha is 11.5 light years away.) I'm too lazy right now to calculate the intersection event... but according to the earth, S1 will arrive after 23 years, and S2 after 16 years (including the 1 year wait). I hope I didn't make any mistakes.

    You can figure out when and where they intersect according to the earth frame and what each clock reads at that meeting.
     
  6. Mar 27, 2006 #5
    Robphy:

    "Is this a fair rephrasing of the problem?

    The earth and alpha are inertial and at rest with respect to each other (so, they have parallel worldlines, vertical in their frame).

    Ignoring the time when S1 was on the earth awaiting launch... at event P, S1 travels inertially toward alpha at speed c/2 and arrives at alpha at event P', after 20 years on S1's clock.

    Ignoring the time when S2 was on the earth awaiting launch... at event Q, one year after P according to the earth's clocks, S2 travels inertially at toward alpha some speed greater than c/2 so as to overtake S1. That is, their worldlines cross before alpha. S2 arrives on alpha at event Q', after 10 years on S2's clock.

    Is this the situation?"

    Yes - that is a correct. We have two intersecting world lines, and S1 and S2 are both inertial observers whose local clocks are running at different rates. And yes the relative velocity can be computed - but the actual numbers are immaterial, as is the location where S2 overtakes S1. The issue is whether there is any experiment that can be performed by the spaceships themselves to detect the difference between the intrinsic rate of S1 and S2. Between the two spatial points earth and alpha the S1 clock records 20 years, the S2 clock records 10 years. The proper time interval is the time recorded by a clock attached to the observed body. While it is true that an observer in SI will always measure the S1 clock to be runiing at one minute per minute, and likewise an observer in S2 will measure the S2 rate as one minute per minute, there is nontheless a difference in how the proper rates get established ...the ratio of 20 years per 20 years and 10 years per 10 years. Is there a way to recover by experiment the difference between the numbers that formed the ratios in the two systems?
     
    Last edited: Mar 28, 2006
  7. Mar 28, 2006 #6
    Assuming “quickly accelerates” = instant acceleration to use straight SR no GR.
    I get pretty much what robphy gets:

    S1 v= 0.5c
    S2 v= 0.756
    Distance to travel 11.58 LY
    S1 23.16 earth time (20 S1 time)
    S2 16.32 earth time (10 S2 time 11 including wait)

    S1 time .866 of earth
    S2 time .655 of earth

    Relative S1-S2 v= 0.411
    S2 time .912 of S1 (& S2 time .912 of S1 of course)

    They meet at 1.475 LY from earth
    2.95 earth time
    2.6 s1 time
    2.27 S2 time (including the 1 y wait on earth)
    1.27 S2 elapsed time
    (note: S2 had two different speeds to build 2.27 time since being together with S1)

    Classical SR math.
     
  8. Mar 28, 2006 #7

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    [tex]
    \]
    \unitlength 1mm
    \begin{picture}(45,80)(0,0)
    \linethickness{0.3mm}
    \put(20,20){\line(0,1){60}}
    \linethickness{0.3mm}
    \put(45,20){\line(0,1){60}}
    \linethickness{0.3mm}
    \multiput(20,20)(0.12,0.29){208}{\line(0,1){0.29}}
    \linethickness{0.3mm}
    \multiput(20,35)(0.12,0.17){208}{\line(0,1){0.17}}
    \linethickness{0.3mm}
    \put(20,70){\line(1,0){25}}
    \linethickness{0.3mm}
    \put(20,80){\line(1,0){25}}
    \put(17,30){\makebox(0,0)[cc]{1}}
    \put(35,45){\makebox(0,0)[cc]{20}}
    \put(27,52){\makebox(0,0)[cc]{10}}
    \put(22,30){\makebox(0,0)[cc]{$\theta$}}
    \put(22,45){\makebox(0,0)[cc]{$\phi$}}
    \end{picture}
    \[
    [/tex]

    Here's my calculation using rapidities... analogous to angles in Euclidean trigonometry.... in Maple code.
    Code (Text):

    restart;
    theta:=arctanh(1/2.);
                            theta := 0.5493061443
    d:=20*sinh(theta);
                               d := 11.54700538
    phi:=arcsinh(d/10.);
                             phi := 0.9866469608
    tanh(phi);
                                 0.7559289459
    tanh(phi-theta);
                                 0.4114378276
    20*cosh(theta);
                                 23.09401076
    1+10*cosh(phi);
                                 16.27525231
    cosh(phi-theta);
                                 1.097167541
     
    with a little more trigonometry, you can get everything else you would want. (beta=(v/c)=tanh(rapidity), gamma=cosh(rapidity))

    One can use radar methods to operationally measure all of the kinematic quantities.
     
  9. Mar 29, 2006 #8
    RandallB/robphy - More than I expected - good tutorial(s). Thanks. My interest in posing the question relates the information which is lost (or apparently lost) if we simply consider the two spaceships S! and S2 passing each other. Without the the benefit of knowing the initial conditions, SR predicts that the S1 and S2 frames are equivalent (equally valid inertial frames as they pass one another with relative velocity v) - and accordingly each would arrrive at the same conclusion(s) regarding contraction and clock rate in the other frame (actually Einstein never really said this in his 1905 paper, but SR is generally given this interpretation). So even though the S1 clock and the S2 clock are running at their own proper rate of one second per second, a second in S1 is different than a second in S2 when both are observed in the earth-Alpha frame. So when an observer in S2 uses his watch to make a measurement of the apparent length of S1, he will be using a different time base than the S1 observer uses his watch to measure the apparent length of S2.

    There is a common Gamma factor, but we have different times T1 and T2 to use as a bases for dilation. So is it correct to conclude that the S1 observer figures the T2 time dilation based upon the value of a second in the T1 frame, and the S2 observer figures the T1 time dilation based upon the value of a second in the S2 frame? If they transmit their results to the earth, their conclusions will not be the same because each is arriving at the value of a dilated second in the other frame in terms of his own time.
     
  10. Mar 30, 2006 #9

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I'm not sure what the times T1 and T2 mean precisely.

    Are they the elapsed times for each observer to complete their trips?
    If so, just realize that these a proper times between two different sets of events... (it just so happens that the launch events occurred on the earth's worldline and the arrival events occurred on alpha's worldline). So, there is nothing to be found in comparing T1 and T2.

    If not,...
    Are they the observed time-dilations in studying the duration of one tick of the clock? That is, is T1 the duration of 1 tick of S1's clock , as determined by S2? More precisely, let A and B be events on S1's worldline corresponding to two consecutive ticks of his clock... so, S1 says the "time-difference between A and B" is 1 tick. Is "T1 ticks" what S2 says is the "time-difference between A and B"?
    If so, then (by the relativistic symmetry of these inertial observers.. which can be borne out by a radar experiment) T1=T2, which is essentially gamma, which is a function of the relative-speed between these inertial observers.

    In any case, consider the following experiment.
    One tick after their meeting (when S1 and S2 meet, as S2 overtakes S1), each observer sends off a light signal to the other. Each observer will measure the same elapsed time on his clock when that signal arrives from the other. So, their identical procedures yields identical measurements (which they can write down in their log books for later comparison)... reflecting their SR-symmetry. (In fact, if each observer understands that the received signal was sent after one tick according to the emitter, the ratio of these proper time-intervals (1 tick for the emitter, T1 ticks for the receiver) gives the Doppler factor, k, which can be used to determine their relative-speed.)
     
  11. Mar 30, 2006 #10
    Thanks robphy - i want to ponder your response before replying

    Yogi
     
  12. Mar 31, 2006 #11
    T1 is simply an increment measured in the S1 frame - for example one second as measured by the clock in S1. Likewsie T2 is an increment measured in the S2 frame by the S2 clock - e.g., also one second as measured by the clock S2. If S1 sends out a light pulse every second as measured by his own clock, and S2 does the same as measured by his clock, an observer on earth, knowing the two recessional velocities, would be able to determine that one second as measured by the S1 clock is different than one second as measured by the S2 clock. In other words one second of proper time in S1 is different than one second of proper time in S2.

    As I read your post 9 you are saying that, in accordance with SR, if S1 and S2 transmit back and forth radio signals they will arrive at reciprocal results - that is, as between S1 and S2 there is no significance to the fact that T1 seconds are not the same as T2 seconds, or in any event it cannot be determined by signals sent between S1 and S2 that there is a difference between T1 and T2 seconds, but only the relative velocity will be revealed.

    But if S1 transmits a signal every second according to the S1 clock and S2 transmits a signal every second according to the S2 clock, then, since the relative velocity v between S1 and S2 is known, each can determine the extra delay caused by their relative recession velocity v (after S2 overtakes S1). Once the relative velocity is determined, either observer can compare his own one second clock ticks with the rate at which pulses are arriving from the other spaceship...After compensating for the changing distance, either observer will see that the received pulses are being sent at a different rate than one per second as measured by the clock in the receiving spaceship, and each will agree upon which spaceship has the longer seconds!
     
  13. Mar 31, 2006 #12

    jtbell

    User Avatar

    Staff: Mentor

    This is the relativistic Doppler effect. Both S1 and S2 observe the other's signals arriving at the rate

    [tex]f_{observed} = f_{source}\sqrt{\frac{1+v/c}{1-v/c}}[/itex]

    where v is the relative velocity of the source and observer, and is positive if the source is approaching the observer.

    If S1 and S2 send signals at the same rate, they each receive the other's signals at the same rate. By the formula above, they each compute the same source frequency.
     
  14. Mar 31, 2006 #13

    Ich

    User Avatar
    Science Advisor

    Yes. Always the other one.
     
  15. Mar 31, 2006 #14

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Just to clarify...
    Technically speaking, "proper-time" is akin to an arc-length... it is a number (as the result of an integral or sum)... and it is invariant (all observers will agree on it). So, one second of proper time in S1 is THE SAME AS one second of proper time in S2. What is different, however, is that each clock has a different spacetime displacement-vector between successive ticks of his clock [i.e., each has a different unit timelike vector tangent to his inertial clock worldline]. In general, an observer will say that the time-components of these vectors (i.e. the apparent durations between ticks for each clock) are unequal. This is time-dilation,

    yes... regarding T1 and T2 to be the vectors descrbed above.

    Conerning that last part of your post, jtbell and Ich addressed that. (I was too slow working on the first part :tongue: )
     
    Last edited: Mar 31, 2006
  16. Mar 31, 2006 #15

    jtbell

    User Avatar

    Staff: Mentor

    Call it parallel processing on a small scale. :cool:
     
  17. Mar 31, 2006 #16
    Granted, if the S1 signals are being sent at the same frequency as the S2 signals, each will draw the same conclusion. But the pulse rate frequencies are not the same. Each is using his own clock to measure one second; the S1 clock is running at a faster rate than the S2 clock.

    Let me analogize to a GPS situation. One clock is considered to be synchronized with a point at the center of the earth (the non rotating earth centered reference frame) we call it E clock. A second clock C2 is first synchronized with E and then put into circular earth orbit.

    C2 will run at a different rate because of two factors (the height and the velocity). If C2 is corrected for altitude, what is left is an orbiting clock C2 that runs a slower rate than E clock.
    In this case the distance does not vary - so the only correction would be a transverse Doppler - What is observed is C2 running slower than E. Seconds take longer in the Satellite frame than in the ECRF.. If both C2 and E are used to transmit pulses at one second intervals, the pulses received from C2 will have a slower frequency than one pulse per second when received by E and the pulses received from E will have a higher frequency than one pulse per second when measured by C2. C2 must be compensated to make the seconds equal. There is no ambiguity as to which clock is running slower and which clock must be compensated.
     
  18. Apr 1, 2006 #17
    One more example using a linear experiment rather than a satellite clock in free fall - a pion created in the lab and qucikly accelerated to near c velocity. Here the pion clock can be considered to have sent two pulses, one at the time of creation and one at the time of disintegration. If the lifetime of an at rest pion (approx 0.02 usec) is extended by a factor of 20 because of its high velocity wrt the lab, the lab detector will receive 2 pulses in say 20 x 0.02usec and the lab observer will conclude that the pion clock is running slower than the lab clock. But if a lab transmitter is sending signals every 0.02 usec, then during the extended lifetime of the high velocity pion, 20 pulses would have been transmitted. If there is a hypothetical observer P attached to the pion, how could P reach the conclusion that the earth clock is running slower than the pion clock?
     
    Last edited: Apr 1, 2006
  19. Apr 1, 2006 #18

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Using your data...
    T=pion lifetime (0.02 us)
    gamma=20
    k=Doppler factor= gamma+sqrt(gamma+1)*sqrt(gamma-1)=39.97
    (another formula: k=exp(arccosh(gamma)))

    Note that gamma*T="20 x 0.02usec" is the "lab's apparent -duration of the period of the pion's signal-emissions". It is NOT the "lab's reception-period of the pion's signal-emissions", which is kT="39.97T".

    Note further that, while the lab may emit 20 regular signals (at 0.02 us intervals according to the lab) in the lab-time gamma*T, the pion receives only the signal at the meeting... it does not receive any other. (To reach the pion before it dies, the lab must have sent a signal at T/k=0.02usec/39.97. Waiting for 0.02us is too long.) If the pion lived long enough, the pion would have received the lab's signal [sent at 0.02 us after meeting] when the pion's clock read kT=39.97*0.02 us.... just like the lab's result.


    Using a notational variant of the diagram from https://www.physicsforums.com/showthread.php?t=113915

    [tex]
    \]

    \unitlength 1mm
    \begin{picture}(55,90)(0,0)
    \linethickness{0.3mm}
    \put(20,10){\line(0,1){80}}
    \linethickness{0.3mm}
    \multiput(20,90)(0.6,-0.6){50}{\line(1,0){0.12}}
    \linethickness{0.3mm}
    \multiput(20,30)(0.6,0.6){50}{\line(1,0){0.12}}
    \linethickness{0.3mm}
    \multiput(20,10)(0.12,0.2){250}{\line(0,1){0.2}}
    \put(15,30){\makebox(0,0)[cc]{T/k}}
    \put(14,60){\makebox(0,0)[cc]{\gamma T}}
    \put(20,60){\circle*{2}}
    \put(15,90){\makebox(0,0)[cc]{kT}}

    \put(55,60){\makebox(0,0)[cc]{T}}

    \end{picture}
    \[
    [/tex]
     
    Last edited: Apr 1, 2006
  20. Apr 2, 2006 #19
    robphy - You are assuming the lab transmitter and receiver to be at the starting point. We can place the lab transmitter and receiver midway between the starting point of the pion and the point of disintegration (for v=0.99c the disintegration length is only about 6 meters). With the lab transmitter and recorder so located, the pion will receive more pulses during the experiment than the lab recorder.
     
  21. Apr 2, 2006 #20

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I'll try to produce a diagram of what I think you are describing.
    However, after a sketch I made on paper, the following issue presented itself to me.

    The pion's worldline is a segment with a finite proper-time of 0.02us. With the lab "midway between the starting point of the pion and the point of disintegration", the lab worldline bisects the pion's worldline-segment. In order to make their experiments "identical", we should only consider the analogously bisected 0.02us-segment of the lab's worldline. Without that concession, the symmetry between the inertial observers is broken... one is allowed to live a longer proper-time than the other.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?