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Time dilation and acceleration due to gravity

  1. Feb 24, 2016 #1
    If acceleration and acceleration due to gravity is equivalent , then by equivalence principle if we accelerate through universe at the equal to acceleration due to gravity near black hole , does that mean that there will be same time dilation while we accelerate in the universe like when we stay near black hole ????
     
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  3. Feb 24, 2016 #2

    stevendaryl

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    The equivalence principle only applies locally. An experiment on board a rocket accelerating at rate [itex]g[/itex] will give the same results as a rocket hovering in place over a planet with gravity [itex]g[/itex] only if (in both cases), [itex]g[/itex] is approximately constant, so that we can neglect the difference between [itex]g[/itex] at the top of the rocket and [itex]g[/itex] at the bottom of the rocket. If you do very accurate measurements, then the variation of [itex]g[/itex] between top and bottom will tell which situation you are actually in.

    But given that clarification, what you say is right. If you are in a rocket hovering outside of a black hole's event horizon, then you will observe location-dependent time dilation: clocks at the front of the rocket (away from the black hole) will run slightly faster than clocks at the rear of the rocket (closer tothe black hole). You get the same effect from having a rocket accelerate through empty space.
     
  4. Feb 24, 2016 #3
    In fact, for an accelerated reference frame, there exists the concept of Rindler horizon, which is quite similar to the event or Schwarzschild's horizon of the hypothetical entity which is called black hole.

    Check, for example, this excellent link:[/PLAIN] [Broken] http://gregegan.customer.netspace.net.au/SCIENCE/Rindler/RindlerHorizon.html[/URL] [Broken]
     
    Last edited by a moderator: May 7, 2017
  5. Feb 25, 2016 #4
    So friends is it true that if i say that acceleration of space vehicle and hovering of special vehicle around high gravitational field are the ways to get greater amount of time dilation than moving on constant greater velocity . That means Special theory of relativity is special case of general theory of relativity. That means one can enjoy high magnitude of time dilation doing acceleration or even do it better ( save energy of space shuttle) by sittting in higher gravitational field. So why physicist never popularize the concept that there is greater time dilation in accelation of space shuttle, they only say close to speed of light, why why ????
     
  6. Feb 25, 2016 #5

    jbriggs444

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    Any time dilation that results from the acceleration of the space shuttle is purely illusory. It can be cancelled by using using an inertial coordinate system. That is to say that it goes away if one puts different marks on a piece of paper with a pencil.

    In addition, it is not acceleration that results in time dilation. It is depth in the gravitational potential. Those are not at all the same thing. The one is the integral of the other.
     
  7. Feb 25, 2016 #6

    stevendaryl

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    Well, it depends on what you mean by "time dilation". If you have a long rocket, and clocks in both the front and rear, then accelerating in the same direction for long enough will result in the front clock being ahead of the rear clock, which you can verify by moving them both to the center for side-by-side comparison.
     
  8. Feb 25, 2016 #7

    stevendaryl

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    This effect is almost certainly not measurable aboard the space shuttle--partly because the space shuttle doesn't fly anymore.
     
  9. Feb 25, 2016 #8

    jbriggs444

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    One could quibble that in an inertial frame, neither clock is directly affected at all by the acceleration, but only by its resulting velocity with respect to that frame.
     
  10. Feb 25, 2016 #9

    stevendaryl

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    Yeah, you can understand the reason for clock discrepancy on board a rocket purely in terms of velocity-dependent time dilation, but it's a little surprising, because one wouldn't think that the front of the rocket is traveling any slower than the rear of the rocket. It actually is, because of length contraction.
     
  11. Feb 25, 2016 #10
    So you people mean acceleration of space shuttle contributes no time dilation , whatever time dilation will happen will be due to velocity only. So the equivalence principle doesn't qualify in that case.
     
  12. Feb 25, 2016 #11

    stevendaryl

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    The equivalence principle allows you to make the following deduction:
    1. If a rocket is accelerating through empty space, then the time on a clock at the front of the rocket will get ahead of a clock at the rear of the rocket.
    2. This situation (ignoring the variation of [itex]g[/itex] with location) is approximately equivalent to a rocket hovering over the surface of a planet (or outside a black hole).
    3. Therefore, we can conclude that in a gravitational field, clocks that are "higher up" will get ahead of clocks that are "lower down"
    Step 1 does not require General Relativity, it is derivable from pure Special Relativity. Step 2 is the equivalence principle (weak or strong--I don't remember which).

    So SR + the equivalence principle allows you to derive gravitational time dilation. You don't need full GR to deduce that gravity must affect clocks. (And as a matter of fact, Einstein predicted gravitational time dilation a number of years before he developed GR.)
     
  13. Feb 25, 2016 #12

    stevendaryl

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    Acceleration is not needed to compute the time dilation of any one clock on board a rocket, but the significance of acceleration is that it leads to different time dilations for clocks at different locations within a rocket.
     
  14. Feb 25, 2016 #13
    The clocks run at very very very close to same rate.

    This effect is much bigger:

    As the rocket accelerates, the amount of information that is mid-air on the way from the rear to the front increases.

    If the space between the rear and the front absorbs one seconds worth of information in 100 seconds, then the observer at the front sees a clock at the rear running at 99% of the normal rate.

    Acceleration does not cause time dilation.
     
  15. Feb 25, 2016 #14

    stevendaryl

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    Well, I would say that that way of putting it is misleading. If you understand the discrepancy between the two clocks (at the front and rear of the rocket) as (only) due to the time lag for light signals to travel from front to rear, or vice-versa, then you would be led to the false conclusion that the discrepancy would disappear when the clocks are brought back together. But it doesn't disappear. A clock that spends 10 years at the front of an accelerating rocket will show more elapsed time than a clock that spends 10 years at the rear of the same rocket, even when brought back together.

    And the point (usually) of considering accelerating rockets is to introduce gravitational time dilation, through the equivalence principle. The corresponding comparison on Earth is to keep one clock at the top of a mountain and another clock at the bottom of the mountain. After 10 years, you reunite the clocks and compare their elapsed times. GR predicts that the one on the top of the mountain will show more elapsed time than the one at the bottom. So it's not just a matter of time lag for light signals.

    You're certainly right, that acceleration doesn't cause time dilation, but I made that point clear in my post: It doesn't cause time dilation, but it causes a different amount of time dilation for clocks at different locations within a rocket.

    What is interesting about time discrepancy between clocks at different locations within an accelerating rocket is that there are two effects that work together:
    1. The time lag for light signals gives the appearance of differential time dilation.
    2. The variation in velocity between front and rear (due to length contraction) gives an additional discrepancy.
    The sum of these two effects gives a constant, height-dependent apparent time dilation. For a long voyage, effect number 1 dominates early on, while effect number 2 dominates once the rocket is moving relativistically.
     
  16. Feb 25, 2016 #15
    When a clock is moved from the rear to the front, the clock speeds up its motion and slows down its ticking rate.

    When a clock is moved from the front to the rear, the clock slows down its motion and speeds up its ticking rate.

    I had that same exact idea until this moment. Maybe it's some common misinformation, because now it seems to me that effect 1 dominates always, because even at relativistic speeds the time delay keeps increasing.
     
    Last edited: Feb 25, 2016
  17. Feb 25, 2016 #16

    Janus

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    This has been brought up by other, but it is important enough that I feel that I should make sure that you understand the concept.
    Gravitational time dilation is not due to the local acceleration felt by the clock. It is due to the difference in gravitational potential. Thus a clock on a mountain top does not run faster than one at sea level because gravity in stronger at sea level, but because the Mountain clock is higher in the Earth's gravitational field. This would be true even if gravity did not get weaker with altitude. Thus if you could arrange to have a gravity field that did not change strength over its extent, A clock placed higher in that field would run faster than one lower in the field.
    If we extend this to the accelerating rocket the clock in the nose and the clock in the tail are equivalent to the two clocks separated in the uniform gravity field and also run at different rates. So, yes, the equivalence principle does apply.
     
  18. Feb 25, 2016 #17
    In accelerated frame the proper time ## \tau ## of an accelerated particle relative to the inertial reference frame time ## t ## (frame where the "acceleration started") is given by

    ## \tau = \int \sqrt{1-v^2/c^2} dt = \int dt/\sqrt{1+(at/c)^2} ##

    Which is the time given by time dilation due to constantly changing speed. If two particles are in the same accelerated frame their proper time is related by

    ## \tau' = (1 + ar/c^2) d\tau ##

    Where ## a ## is the proper acceleration, ## r ## is the distance between the particles and ## \tau' ## the proper time of the particle "above" or "in front" of the other particle in the direction of acceleration. So the clock "in front" runs faster.

    Edit: to avoid confusion as how two particles can have have different proper times relative to the original inertial frame of reference in the same accelerated frame, one must take into account the Doppler effect between the particles - the "in front" particle's frequency is observed as higher for the other particle and vice versa, so they observe their clocks running at different rates. When the acceleration stops the clocks are synchronized again and if you want to bring the particles to rest relative to the original inertial frame you must accelerate the same amount in the opposite direction (ie. deccelerate) so the final state is symmetric and the total time dilation due to radial difference in the accelerated frames is zero.
     
    Last edited: Feb 25, 2016
  19. Feb 25, 2016 #18

    stevendaryl

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    At most, that effect can make a discrepancy that is equal to the time it takes to bring the clocks back together. Which might be a second. The discrepancy can grow arbitrarily large. So that can't explain the discrepancy when they are brought back together.

    No, that's not true. Effect number 2 can be calculated in the "launch" frame as follows:

    Let [itex]\tau_r[/itex] be the time shown on the rear clock, and let [itex]\tau_f[/itex] on the front clock. These are given (for constant proper acceleration) by (I'll skip the derivation):

    [itex]\tau_r = (X sinh^{-1}(\frac{ct}{X}))/c[/itex]
    [itex]\tau_f = ((X+L) sinh^{-1}(\frac{ct}{X+L}))/c[/itex]

    where [itex]L[/itex] is the length of the rocket, and [itex]X = \frac{c^2}{g}[/itex] and where [itex]g[/itex] is the acceleration felt by the rear of the rocket, and [itex]sinh^{-1}[/itex] is the inverse hyperbolic sine.

    Using properties of the hyperbolic sine, and using the assumption that [itex]L \ll X[/itex], we can (again, I'm skipping the derivation) get the asymptotic discrepancy between the times shown on the two clocks as approximately:

    [itex]\tau_f - \tau_r \sim L log(\frac{ct}{X})/c[/itex]

    So the difference grows logarithmically with time, as viewed in the "launch" frame.

    The first effect that you mentioned is that, assuming that the clock in the rear sends a signal toward the front once per second, the number of signals "in flight" that have been sent but not yet received gives a false impression of the discrepancy between the two clocks. But I'm pretty sure that the number of signals "in flight" is bounded, so it does not make an ever-increasing apparent difference between the clocks. I will try to compute the asymptotic form of this number, but it seems a little tricky...
     
  20. Feb 25, 2016 #19
    Time difference of clocks grows slower and slower? Well that's not surprising, as both clocks are running slower and slower.

    Or what does the result mean, physically?


    Here's my calculation, a numerical one:
    Code (Text):

    #python program simulating two clocks connected by a string, as the front clock is accelerated

    from math import sqrt

    def gamma(v):
      return  1.0/sqrt(1-v**2)

    def invgamma(v):
      return  sqrt(1-v**2)

    def runningdifference(x):
      u=[]
      old=0.0
      for i in x:  u.append(i-old);  old=i
      return u

    #"speeds" will be a list of coordinate speeds of a pushed clock pair when proper force is constant, coordinate time step is constant, and mass increases as gamma to third power, and coordinate force decreases as inverse gamma

    speeds=[0.01]
    timestep=0.002
    while 1:
      if speeds[-1] > 0.98: break
      g=gamma(speeds[-1])
      deltav= timestep / g**4
      speeds.append(speeds[-1] + deltav )

    #list of string lengths at various speeds
    lengths = []
    for s in speeds:
      lengths.append(0.5*invgamma(s))

    lengthchanges = runningdifference(lengths)

    #list of speeds of the rear clock, speed of rear clock is speed of front clock + string's contraction speed
    speeds2=[]
    for a,b in zip(speeds,lengthchanges):
      speeds2.append(a+b)

    #as we know speeds of clocks, we can calculate rates of clocks
    clockrate1=map(gamma,speeds)
    clockrate2=map(gamma,speeds2)

    #now we compare clock rates, and then we print the more or less interesting results
    clockratios=[]
    for a,b in zip(clockrate1,clockrate2):
      clockratios.append(a/b)


    clockratiochanges=runningdifference(clockratios)

    for a,b,c in zip(clockratios,speeds,clockratiochanges):
      if c<0:  change="decreases"
      if c>=0:  change="increases"
      print a,b,change

     


    From speed 0 c to 0.82 c clocks become more different, from 0.82 c to 0.999 c clocks become less different. That's what that simulation says. IOW different rates of clocks observed in the clocks frame is mostly not real difference in the frame where the clocks are moving at very high speed.
     
    Last edited: Feb 25, 2016
  21. Feb 25, 2016 #20

    stevendaryl

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    I'm not sure I understand your simulation, but let me just try to clarify the sources of the time discrepancies on the clocks:
    • Let [itex]F[/itex] be the "launch frame" of the rocket.
    • Let [itex]F'[/itex] be the instantaneous rest frame of the rocket, some time later.
    • Let [itex]e_1[/itex] be the event in which the rear of the rocket comes to rest in frame [itex]F'[/itex]
    • Let [itex]e_2[/itex] be an event at the front of the rocket that is simultaneous with [itex]e_1[/itex], according to frame [itex]F[/itex].
    • Let [itex]e_3[/itex] be the event at the front of the rocket that is simultaneous with [itex]e_1[/itex], according to frame [itex]F'[/itex].
    • Let [itex]T_1[/itex] be the time on the rear clock at event [itex]e_1[/itex]
    • Let [itex]T_2[/itex] be the time on the front clock at event [itex]e_2[/itex]
    • Let [itex]T_3[/itex] be the time on the front clock at event [itex]e_3[/itex]
    Let's define some time delta's
    1. [itex]\delta T_{21} = T_2 - T_1[/itex], the discrepancy as viewed from frame [itex]F[/itex].
    2. [itex]\delta T_{32} = T_3 - T_2[/itex], an additional contribution to the discrepancy due to relativity of simultaneity.
    3. [itex]\delta T_{31} = T_3 - T_1 = \delta T_{32} + \delta T_{21}[/itex], the discrepancy as viewed from frame [itex]F'[/itex].
    My claim is that:
    1. Initially, the biggest contribution to [itex]\delta T_{31}[/itex] is [itex]\delta T_{32}[/itex], and [itex]\delta T_{21}[/itex] is neglible.
    2. Long after the rocket has been accelerating, when it's moving relativistically, the biggest contribution to [itex]\delta T_{31}[/itex] is due to [itex]T_{21}[/itex].
    3. The sum of the two contributions has a simple form: [itex]\delta T_{31} = T_1 \frac{gL}{c^2}[/itex]
     
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