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Proper time of accelerated clocks

  1. Feb 18, 2008 #1
    In other threads the topic of how the proper time of a clock under acceleration has come up a number of times so I have decided to analyse this subject in a little more detail.

    For this analysis we have these conditions:

    a)Two clocks are initially at rest separated by a distance [itex]L_o[/itex] along the x axis.
    b) Both clocks are simultaneously born-rigid accelerated along the x axis (maintaining constant proper separation).
    c) The experiment happens in flat space.
    d) When the rear clock reaches velocity v relative to the initial frame it stops accelerating and maintains constant velocity v.
    e) When the front clock reaches velocity v relative to the initial frame it stops accelerating and maintains constant velocity.
    f) In the final state both clocks have constant and equal velocity and are separated by the proper distance [itex]L_o[/itex]

    Referring to the attached diagram:

    [itex] a_F [/itex] is the acceleration of the front clock proportional to [itex]c^2/x_F[/itex]

    [itex] a_R [/itex] is the acceleration of the rear clock proportional to [itex]c^2/x_R[/itex]

    [itex] t_F [/itex] is the coordinate time that the front clock clock takes to get to velocity v.

    [itex] t_R [/itex] is the coordinate time that the rear clock clock takes to get to velocity v.

    The dashed green line from the origin going through [itex]T_{R1}[/itex] and [itex]T_F[/itex] is a line of simultaneity in the accelerated clock frame.

    The red lines are constant velocity phase after the acceleration phase.

    When the front clock stops accelerating the proper time that has elapsed ([itex]t_f[/itex]) is given by the formula:

    (Eq 1) [tex]T_F = { c \over a_F} \sinh^{-1} \left({a_F t_F \over c} \right)[/tex]

    When the rear clock stops accelerating the proper time that has elapsed ([itex]t_R[/itex]) is given by the formula:

    (Eq 2) [tex]T_{R1} = { c \over a_R} \sinh^{-1} \left({a_R t_R \over c} \right)[/tex]

    The rear clock accelerates faster and gets to velocity v quicker than the front clock. The proper time that elapses from the coordinate time the rear clock stops accelerating to coordinate time that the front clock stops accelerating is simply the coordinate time difference adjusted by the Lorentz factor:

    (Eq 3) [tex]T_{R2} - T_{R1} = (t_f - t_r)\sqrt{1-v^2/c^2}[/tex]

    The difference in proper time between the front clock and the rear clock after accelerating is given by

    (Eq 4) = (Eq1) - (Eq 2) - (Eq 3)

    The following equations are generally known:

    (Eq 5) [tex] a=\sqrt{\left({c^4 \over x^2-c^2 t^2} \right)}[/tex]

    (Eq 6) [tex] v={a t \over \sqrt{\left(1+(a t /c)^2\right)}}[/tex]

    and the acceleration of the front clock in terms of acceleration of the rear clock can easily be derived as :

    (Eq 7) [tex] a_F = {a_R c^2 \over (L_o a_R +c^2) [/tex]

    By substituting (Eq 5), (Eq 6) and (Eq 7) into (Eq 4) and solving the time difference between the front and rear clocks as seen in the initial frame is given by:

    (Eq 8) [tex] T_F - T_{R2} = {-L_o v \over c^2} + {L_o \over c} \sinh^{-1} \left({ v \over c \sqrt{1-v^2/c^2}} \right)[/tex]

    When the clocks are synchronised again it is well known that:

    (Eq 9) [tex] T_F - T_{R2} = {-L_o v \over c^2} [/tex]

    It is interesting to note that the difference in the proper times of the clocks after born-rigid acceleration and before synchronisation (Eq 8) is completely determined by the final velocity and the proper spatial separation and is independent of the rate of acceleration or time taken to transfer from one frame to another.

    It is obvious from (Eq 8) that if constant proper separation is maintained the clocks will not remain naturally synchronised.

    It is further conjectured that there is no acceleration scheme that can keep two spatially separated clocks synchronised when transferring the clocks from one inertial reference frame to another.*

    I have left quite a few gaps in the derivation to try and keep it brief, but I will fill the gaps in, if anyone is really interested.

    [EDIT] *I found a counter proof to this conjecture in post #9. (This does not change the validity of (Eq 8) in the above context of born-rigid transportation.

    Attached Files:

    Last edited: Feb 19, 2008
  2. jcsd
  3. Feb 18, 2008 #2
    Interesting. The result is counter-intuitive because of the asymmetry.
    What about a simular situation where the clocks are alined in a perpendicular direction instead of in the direction of acel?
    Any asymmetry in the final time frame there?

    Has this equation been experimentally verified? Reference?
  4. Feb 18, 2008 #3
    Very interesting, nice work.
    How do you reconcile (eq3) with the fact that, as seen from your Minkowski diagram:

    [tex]T_{R2} = { c \over a_R} \sinh^{-1} \left({a_R t_r \over c} \right)[/tex] ?
  5. Feb 18, 2008 #4
    No, clocks exactly perpendicular to a clock on the x axis and syncronised with that clock, will remain syncronised with that clock when the acceleration occurs purely along the x axis.

    I used equations from the following sources:

    http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken]

    http://en.wikipedia.org/wiki/Hyperbolic_motion_(relativity [Broken])


    The final result obtained in (Eq 8) is my own work. I am hoping the experts on this forum can confirm that the maths and logic is correct.

    P.S. The diagram was created using a free geometry software program (called CaR) that can plot functions and evaluate expressions. That software at least confirms that (Eq 8) = (Eq 1) - (Eq 2) - (Eq 3)

    For software that does not handle hyperbolic functions (Eq 8) can be expressed in the alternative form:

    [tex] T_F - T_{R2} = {-L_o v \over c^2} + L_o c \log \left( \left({ v \over c \sqrt{1-v^2/c^2}} \right) + \sqrt{\left({ v \over c \sqrt{1-v^2/c^2}} \right)^2 +1 \right) [/tex]

    using the general relation

    [tex] \sinh^{-1}(x) = \log\left(x+ \sqrt{x^2+1}\right) [/tex]

    where log is natural logarithm.
    Last edited by a moderator: May 3, 2017
  6. Feb 18, 2008 #5

    The rear clock maintains constant velocity from [itex]T_{R1}[/itex] to [itex]T_{R2}[/itex] (the straight diagonal red line) so the hyperbolic motion equations are not required for that segment and the straight forward Lorentz transformation of the time inteval is sufficient to obtain (Eq 3).

    The total proper time experienced by the front clock is given by (Eq 1)

    The total proper time experienced by the rear clock is (Eq 2) + (Eq 3)

    The rear clock has a combination of hyperbolic motion and constant velocity.
  7. Feb 18, 2008 #6
    OK, I understand, you have some typos in the text and in the diagram that threw me off.
    So, you are trying to calculate [tex]Ttotal_F-Ttotal_R[/tex], right?
    But, in eq(8) you are calculating:

    [tex]Ttotal_F-T_R_2[/tex], unless it is another typo?
  8. Feb 18, 2008 #7

    [tex]Ttotal_F-Ttotal_R = (Eq 1) - (Eq2) - (Eq 3) = T_F - T_{R1} - (T_{R2}-T_{R1}) = T_F - T_{R1} - T_{R2} + T_{R1} = T_F - T_{R2} [/tex]


    [tex]Ttotal_F-Ttotal_R = Ttotal_F-T_R_2 [/tex]
  9. Feb 18, 2008 #8

    Thank you for the previous clarifications. I hope that I am helping you cleaning up the post.

    Now, this is somewhat misleading. The accelerations [tex]a_F[/tex] and [tex]a_R[/tex] do not appear in the final result because of the extra conditions that you set on them (measure the elapsed proper time difference when the clocks have achieved the same speed [tex]v[/tex] while maintaining the same proper separation [tex]L_0[/tex]). This is why [tex]v[/tex] replaces [tex]a_F[/tex] and [tex]a_R[/tex] in the final formula of eq(8)

    This is well known, indeed.

    This is an artifact of the fact that the two clocks will describe different trajectories in space-time, so, when you try to calculate the total elapsed proper time via integration, you may get different results (I don't think proper time is a conservative field :-) )
    Last edited: Feb 18, 2008
  10. Feb 18, 2008 #9
    As always, constructive criticism is always welcome ;)

    Accelerating while maintaining constant proper separation is the definition of born-rigid acceleration which is probably the most used acceleration method in discussions of accelerating rockets. It is admittedly an ideal situation, that is probably not often found in nature except as an approximation. The result that clocks go out of synchronisation under ideal born-rigid acceleration is not obvious to everyone.

    As far as I can tell the equation I derived in post #1 is correct but unfortunately (for me) I have found a counter example to my own conjecture that I added to the end of the post. “It is further conjectured that there is no acceleration scheme that can keep two spatially separated clocks synchronised when transferring the clocks from one inertial reference frame to another. ( peh! :P) I have found that if we allow for the propagation delay of the acceleration impulse along the length of a rear engined rocket and allow the rocket to compress well beyond its natural length contracted length then it IS possible to find an acceleration pattern that keeps the clocks synchronised. This acceleration pattern allows the clock at the nose of the rocket to follow a longer path through space-time and time dilate to a lesser extent than the rear clock. In the attached diagram the front of the rocket does not start accelerating until coordinate time t2 which is later than the coordinate time t1 when the rear of rocket has finished accelerating. This is because of the minimum time for a signal travelling at light speed to propagate from the rear to the front. In practice this delay would probably be even longer if the impulse wave propagates through the material at the speed of sound.

    Attached Files:

    Last edited: Feb 18, 2008
  11. Feb 18, 2008 #10
    Thank you, I know very well what Born-rigid motion means. I was commenting about why [tex]a_R[/tex] and [tex]a_F[/tex] do not show in eq(8). Speaking of eq(8), I did not redo your calculations but the fraction [tex]\frac{L_0 v}{c^2}[/tex] looks very suspect since the result is made up of irrational expressions (square roots and sinh). The only way you may arrive to the rational term is if you did some approximations.

    You may want to double-check eq (8), it may not be correct.
  12. Feb 19, 2008 #11
    please clarify a few points for me

    Hi Kev, Some readers are clued up, e.g. "
  13. Feb 19, 2008 #12
  14. Feb 20, 2008 #13
    Irrational numbers are simply numbers that cannot be expressed as a ratio of whole natural numbers. I assume you actually meant transcendental expressions which sinh is but square is not. As far as the expression contained within the sinh function is concerned I only did straight substitutions of equivalent values into the function but otherwise performed no mathematical operations on it.

    Square roots are not transcendental functions but they can present problems due to having real or imaginary roots. As long as c is greater than zero and v is less than c there are not too many problems in handling them here.

    The expression [tex]\frac{L_0 v}{c^2}[/tex] comes directly from substitutions into

    (Eq 3) [tex]T_{R2} - T_{R1} = (t_f - t_r)\sqrt{1-v^2/c^2}[/tex]

    which does not contain the transcendental sinh function.

    By re-writing the rhs of (Eq 3) as

    (Eq 11) [tex] \left(t_f\sqrt{1-v^2/c^2} - t_r\sqrt{1-v^2/c^2}\right)[/tex]

    and substituting [tex]t={v \over a\sqrt{1-v^2/c^2}[/tex] into (Eq 11) we get

    (Eq 12) [tex] {v \over a_F} - {v \over a_R} [/tex]

    then substituting [tex] a_F = {c^2a_R \over (L_0 a_R +c^2)}[/tex] into (Eq 12) we get

    (Eq 13) [tex] {v(L_0a_R+c^2) \over c^2a_r} - {v \over a_R} = {vL_0a_r+vc^2-vc^2 \over c^2a_R} = {L_0 v \over c^2}[/tex]

    Nothing too dodgy there I think.
    Last edited: Feb 20, 2008
  15. Feb 20, 2008 #14
    Thank you, this explains very well my point about the solution being independent of [tex]a_F[/tex] and [tex]a_R[/tex] due to imposing that:
    -proper separation [tex]L_0[/tex] is maintained
    -both clocks arrive to the same terminal speed [tex]v[/tex]
    Last edited: Feb 20, 2008
  16. Feb 20, 2008 #15
    reply kev on Proper time of accelerated clocks

    Hi Kev.

    Thank you for your reply.

    Thank you for the link to the article on Born-rigidity. It might even perhaps be said at one point to support Aristotle's much ridiculed notion about inertia! I will have to think about it.

    You write (with a slight editorial tidy-up by me): #If the higher clock sent signals at one second intervals to the lower clock they would appear to be arriving more than once per second to an observer at the bottom referring to the lower clock.# This is the most concisely commonsensical and perspicuously meaningful plain English statement I have read about this. Thank you. Though this comment of yours has a clear meaning, the phenomenon it describes is still baffling. Would it concomitantly be the case that #If the lower clock sent signals at one second intervals to the higher clock they would appear to be arriving less than once per second to an observer at the top referring to the higher clock.#? I think so.

    Presumably we are talking about light pulses. A light pulse can vary in its frequency, its wavelength, its speed, its duration, and its energy, its polarisation, and its coherence. Whence come the extra pulses, and whither go the missing ones? To or from the past, the future, a storage place part way up the tower, the very matter at the top or bottom of the tower? How do the pulses change on their way? If this went on for long enough, what would happen: would it mean that the light pulses on the way up were supplied by actually eroding the earth at the bottom and sending it up to the top where their energy was stored, till there was no more earth left at the bottom, and the tower got very very top heavy? Or that the top of the tower would be eroded by the need to supply energy for the downgoing pulses until there was no more top of the tower? Does conservation of mass-energy apply here? Does this fit with the notion that photons are particles that can be counted as they travel? Would the height of the tower gradually change because of all this? In answers to these questions, what is a priori or ex cathedra postulate, and what is observed or observable fact?

    How about sending not light pulses, but rather atoms of ponderable matter, at speeds less than light speed? Same questions, except for different degrees of freedom for atoms.

    Let's leave the clock design and construction problem to another day.


  17. Feb 21, 2008 #16
    Hi Christopher.


    Observable facts:

    (a) The Pound Rebka Harvard Tower experiment. The frequency of waves sent up a tower was observer to be reduced by a detector at the top of the tower. Conversely signals sent down the tower were seen to increase in frequency.

    (b) Atmospheric muon observations. The half life of muons is known to be too short for muons created at the top of the Earth's atmosphere to live long enough to make to it to sea level. However far too many muons are detected at sea level. This is accounted for by time dilation extending the effective life of the muons long enough for them to make the complete journey.

    (c) Many particle laboratories routinely measure the half life of transient particles as extended by the time dilation factor when the particles are moving at relativistic speeds or when stored in cyclotrons that keep the particles accelerating in circles.

    Time dilation is routinely observed in laboratories every day all over the world. I would be surprised if there was a single employed physicist alive today that doubted the validity of time dilation.

    However, some would argue that experiment (a) does not directly show that the lower clock is running slower than the top clock. They say the wavelength of the photon is continually stretching as the photon climbs up the gravity well (and the frequency of the photon is slowing down and therefore the photon is losing energy). Where has this energy gone that the photon has lost? In classical physics an object thrown upwards loses kinetic energy and gains potential energy, and this potential energy is stored in the gravitational field. In general relativity there isn't a gravitational field that can store potential energy. A distant observer watching the photon climbing up the gravitational well would see the coordinate speed of the photon as apparently increasing so he could argue that the loss of the photon's energy as calculated by Energy=h(frequency) is compensated by the apparent increase in the photon's velocity. Energy=h(frequency) can expressed as Energy = hc/(wavelength). In this view the loss of energy due to increased wavelength of the photon, is compensating by the apparent increase in c as measured by the distant observer. However this is a very controversial view and it should always be stressed that c is always measured as 299792458 m/s by any local observer.

    Personally, I have no doubt that atoms sent down the tower at a frequency of once per second (as measured by a clock at the top) will be seen to be arriving at intervals of less than one second, as measured by a clock at the base (and vice versa going the other way).

    Others may differ.

    P.S. A red light shone from the top of a sufficiently high tower would appear red to an observer at the base and a blue light shone from the base would appear red to an observer at the top of the tower. I think most people would agree with that.
    Last edited: Feb 21, 2008
  18. Feb 21, 2008 #17
    You mean the other way around, right? :-)
  19. Feb 21, 2008 #18
    Quote from that link: "The effect is very small but measurable on Earth using the Mossbauer effect and was first observed in the Pound-Rebka experiment.[32] However, it is significant near a black hole, and as an object approaches the event horizon the red shift becomes infinite." Increasing red shift is synonomous with increasing wavelength which is the same thing as decreasing frequency.

    Quote from this link http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/gratim.html#c2

    "the gain in energy for a photon which falls distance h is ..." gain in energy for a falling photon is the same as saying the frequency of the falling photon is increasing.

    From this wikipedia article on gravitational redshift: http://en.wikipedia.org/wiki/Gravitational_redshift

    " In physics, light or other forms of electromagnetic radiation of a certain wavelength originating from a source placed in a region of stronger gravitational field (and which could be said to have climbed "uphill" out of a gravity well) will be found to be of longer wavelength when received by an observer in a region of weaker gravitational field. "

    Longer wavelength = reduced frequency.

    So no, not the other way round.
  20. Feb 21, 2008 #19

    "Likewise, gravitational blue shifts are associated with light emitted from a source residing within a weaker gravitational field (i.e. , top of the tower) observed within a stronger gravitational field (i.e. bottom of the tower), while gravitational redshifting implies the opposite conditions."

    You are right, we are saying the same thing, just in a reverse order. My bad.
    Last edited: Feb 21, 2008
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