Two watches, one thrown in the air.

[CarlB]

Standing on the North pole of the Earth, I have two watches. I synchronize the watches. I throw watch "A" into the air, for a height gain of 10 meters (I have a good arm). Watch "B", I keep in my hand. I catch watch "A" when it descends to its original height, the same height as "B".

Obviously, the watches are no longer synchronized. Which shows the later time? By how much?

Of course any high school student knows that the difference in time will be too small to distinguish in a practical experiment. What I'm asking for is the theoretical difference.

Carl

 

[whozum]

10 meters isn't high at all.

 

[HallsofIvy]

I don't see why there would be any discernable difference. The speeds are no where near relativistic speeds.

 

[SplinterIon]

The larger a gravitational field (or the closer you get to one), the slower time runs. A clock in the ISS and on Earth differ by about 2E-9 seconds (Earth one is "slower"). So the magnitude of the difference you're talking about is ridiculous.

My $0.02

 

[KataKoniK]

Don't know the answer to your question, but I never knew this could happen in real life.

 

[Dr.Brain]

Don't know the answer to your question, but I never knew this could happen in real life.

Ofcourse, this can happen for velocities approaching the value of 'c', the 'time-dilation' effect can take place.

Time-dilation can take place if the watch A is in relative motion to the watch B, which is the case but the falling watch A will fall down reaching a speed of $\sqrt 2gh$ which is nowhere near c. If the watch A had fell with a velocity of 0.8c ...then time-dilation is observable.Anyways time-dilation takes place as:

$t= \frac{t_o}{\sqrt (1- \frac{v^2}{c^2})}$

where denominator almost equals 1 in the case when watch A falls froma height of 10m and thus no observable time-dilation takes place.

 

[whozum]

Don't know the answer to your question, but I never knew this could happen in real life.

Hehe, yeah man, relativity is about real life, physics explains real life.

Time-dilation can take place if the watch A is in relative motion to the watch B, which is the case but the falling watch A will fall down reaching a speed of [itex] v= \sqrt{2gh} which is nowhere near c

Careful there, it could be for sufficient values of h :)

 

[KingNothing]

He was asking a theoretical question and wanted a theoretical answer. Many people need to stop being rude by saying "there won't be any difference". A theoretical question is used to grasp the concept, even if it is unable to be performed. clearly he is trying to learn something here.

 

[HackaB]

He was asking a theoretical question and wanted a theoretical answer. Many people need to stop being rude by saying "there won't be any difference". A theoretical question is used to grasp the concept, even if it is unable to be performed. ...

Word.

Can one use regular old special relativity to do this? For instance, the elapsed time on the thrown watch would be given by

$$d \tau^2 = dt^2 - \frac{dx^2}{c^2}$$

where x and t are measured in the "stationary" ground frame. x and t are related by

$$x = \sqrt{2gh} \ t - \frac{1}{2}g t^2$$

$$d \tau = \sqrt{1 - \frac{1}{c^2}(gt - \sqrt{2gh})^2} \ dt$$

Then the total elapsed time on the thrown watch would be

$$\Delta \tau = \int d \tau = \int_0^{2\sqrt{\frac{2h}{g}}} \sqrt{1 - \frac{1}{c^2}(gt - \sqrt{2gh})^2} \ dt$$

$$\Delta \tau = \frac{c}{g} \int_{-\frac{\sqrt{2gh}}{c}}^{\frac{\sqrt{2gh}}{c}} \sqrt{1 - x^2} \ dx = 6 \times 10^7 \int_0^\epsilon \sqrt{1 - x^2} \ dx$$

where epsilon = Sqrt(200) / (3 x 10^8)

$$\Delta \tau \approx 6 \times 10^7 \int_0^\epsilon 1 - \frac{x^2}{2} \ dx = 2 \sqrt{2} - 10^7 \epsilon^3$$

So if that's the way to do it, it looks like the difference is of order 10^-18 seconds.

edit: forgot to mention I was using g=10. So the 2 Sqrt(2) should really be 2 Sqrt(20/g). The order of the magnitude of the difference should not be affected much, I think.

 

[Gokul43201]

Word.

Can one use regular old special relativity to do this? ...

So if that's the way to do it, it looks like the difference is of order 10^-18 seconds.

It looks like the time dilation from the variation of the gravitational field (in GR) might be slightly bigger ~ 10^-15 or thereabouts.

 

[HallsofIvy]

He was asking a theoretical question and wanted a theoretical answer. Many people need to stop being rude by saying "there won't be any difference". A theoretical question is used to grasp the concept, even if it is unable to be performed. clearly he is trying to learn something here.

He referred to throwing a watch up 30 meters- a completely "non-relativistic" situation- and said nothing about relativity. I said there would be no discernable difference. There is nothing rude about telling the truth.

 

[HackaB]

It looks like the time dilation from the variation of the gravitational field (in GR) might be slightly bigger ~ 10^-15 or thereabouts.

Cool. Is it still the thrown watch that has less time elapsed, though? Seems like I remember reading that the elapsed proper time is maximized in free fall, so I was wondering if in GR it was the thrown watch that read a later time.

He referred to throwing a watch up 30 meters- a completely "non-relativistic" situation- and said nothing about relativity...

He asked for the difference in elapsed time on the two watches. What else could he be talking about?

 

[DaveC426913]

He referred to throwing a watch up 30 meters- a completely "non-relativistic" situation- and said nothing about relativity. I said there would be no discernable difference. There is nothing rude about telling the truth.

Pah, you're rationalizing. You should know what a thought experiment is.

It is not a "non-relativistic situation", since the watches are experiencing gravity - as in: changes in-. And the question is obviously about relativistic effects.

The watch thrown in the air experiences less gravity briefly. It will fall out of sync. Pointing out that the AMOUNT that it is out-of-sync is vanishingly small that it might as well be zero does not shed any light on the poster's question.

 

[CarlB]

Cool. Is it still the thrown watch that has less time elapsed, though? Seems like I remember reading that the elapsed proper time is maximized in free fall, so I was wondering if in GR it was the thrown watch that read a later time.

Okay, here's another thrown watch problem, one that eliminates the gravitational interaction and may get you thinking about your answer to the first question.

I'm in a spaceship somewhere free of significant gravitational effects. As before, I've got two perfect watches, A and B. The ship has its rocket engine running providing a forward acceleration of $$9.8 m/s^2$$.

I throw watch A ahead of the ship at a speed of $$\sqrt{9.8 \times 10 \times 2}\;m/s$$ in a forward direction. Since the rocket is accelerating, I catch up with the thrown watch and catch it. Now which one shows the later time? By how much?

Carl

 

[HackaB]

Okay, here's another thrown watch problem, one that eliminates the gravitational interaction and may get you thinking about your answer to the first question.
Carl

If you are just testing us to see if we can solve the problems you thought up, won't you at least tell us the right answer after we've given it a try?

 

[OlderDan]

Okay, here's another thrown watch problem, one that eliminates the gravitational interaction and may get you thinking about your answer to the first question.

I'm in a spaceship somewhere free of significant gravitational effects. As before, I've got two perfect watches, A and B. The ship has its rocket engine running providing a forward acceleration of $$9.8 m/s^2$$.

I throw watch A ahead of the ship at a speed of $$\sqrt{9.8 \times 10 \times 2}\;m/s$$ in a forward direction. Since the rocket is accelerating, I catch up with the thrown watch and catch it. Now which one shows the later time? By how much?

Carl

When are the clocks synchronized? Before you throw one out or as it leaves the spacecraft with its newly acquired velocity?

 

[CarlB]

When are the clocks synchronized? Before you throw one out or as it leaves the spacecraft with its newly acquired velocity?

It doesn't matter. Under the special theory of relativity, acceleration itself (i.e. the act of throwing the watch) doesn't effect the passage of proper time. Only velocity (i.e. the integral of acceleration) effects the passage of proper time. The proper time discrepancy is proportional to the length of time you spend on the other reference frame, not on how you got there.

If you don't like that answer, then choose either one. I'd love to hear your explanation for the difference.

This is all under the assumption of perfect time keepers and all that. Most watches don't work so well when you put an infinite acceleration on them, even for a brief instant.

Carl

 

[CarlB]

If you are just testing us to see if we can solve the problems you thought up, won't you at least tell us the right answer after we've given it a try?

You do see that the example of the watch thrown from the spaceship eliminates the need to consider the gravitational field of the earth, don't you? (Well, at least to first order, since the gravitational field of the Earth isn't quite linear even over a distance of 10 meters.) It gets back to the principle that an acceleration is equivalent to a gravitational field.

By the way, by transforming to the accelerated spaceship problem, you eliminate the need to include the mass and radius of the earth, and the gravitational constant G. That makes the problem a lot easier, I think. All that you need is the acceleration 9.8m/sec^2, the distance, 10m, and the speed of light 3x10^8 m/sec.

Carl

 

[OlderDan]

It doesn't matter. Under the special theory of relativity, acceleration itself (i.e. the act of throwing the watch) doesn't effect the passage of proper time. Only velocity (i.e. the integral of acceleration) effects the passage of proper time. The proper time discrepancy is proportional to the length of time you spend on the other reference frame, not on how you got there.

If you don't like that answer, then choose either one. I'd love to hear your explanation for the difference.

This is all under the assumption of perfect time keepers and all that. Most watches don't work so well when you put an infinite acceleration on them, even for a brief instant.

Carl

First let me say that I am not speaking with authority on this subject. But I'm going to respond anyway to present what prompted me to raise the question. I think it does matter, but that does not mean I will be able to do the calculation in either case.

I know almost nothing about general relativity, or how to properly treat accelerating reference frames from a special relativity viewpoint, but I do recall the notion that gravity and acceleration are more or less equivalent in general relativity, so if one affects the way a clock runs, so should the other. I was not thinking in terms of an infinite acceleration when I asked the question. If infinite acceleration avoids the issue, then so be it. Let's just say the clock is locally synchronized when it gets up to speed and leaves the spaceship, however that is accomplished.

Beyond that, I'd also like to see a solution to the modified problem. I do understand the motivtion for eliminating the gravitational effects and focusing on the acceleration alone. I suppose one could do some sort of integral over quasi-inertial reference frames, but I think there is more involved than just time dilation. As in the case of resolving all those special relativity paradoxes like the twins, I expect those synchronization issues will come into play.

 

[CarlB]

Okay, here's the answer.

Dear Dan,

> I do recall the notion that gravity and acceleration
> are more or less equivalent in general relativity,
> so if one affects the way a clock runs, so should the other.

This is true, and it is why we can work the problem out for the accelerated case, and the answer we get will be identical to the situation on the earth.

Rather than work with accelerated frames of reference, don't you think it would be easier if you used an inertial reference frame? There is one that is available, and that is the one attached to the thrown watch.

In that reference frame, the equation for the passage of proper time (using only one spatial dimension for simplicity) is

$$ds^2 = dt^2 - dx^2/c^2$$

The position of the thrown watch in the (accelerated) reference frame of the held watch is given by

$$x_t = V_0\, t -4.9 t^2$$

where "4.9" is half the gravitational acceleration of the earth. Therefore, the position of the held (accelerated) watch, in the frame of reference of the thrown (inertial) watch is the negative of this:

$$x_h = -V_0\,t +4.9 t^2$$

This gives the velocity of the held watch as:

$$\frac{d x_h}{dt} = -V_0 + 9.8 t$$

The duration of the flight is given by $$2 V_0 / 9.8 = V_0/4.9 = \tau$$ seconds, and this is the proper time as experienced by the thrown watch.

The proper time of the accelerated (held) watch is given by integrating $$ds$$ over this time period. That is,

$$\Delta s = \int_0^\tau ds = \int_0^\tau \sqrt{1 - (\frac{dx_h}{dt})^2/c^2}dt < \tau$$

where the final inequality is due to the fact that the integrand is less than one.

So there is no need to know anything about general relativity, other than, as you mentioned, an accelerated reference frame is equivalent to a gravitational field, from the point of view of a small observer. Just choose the right (inertial) reference frame and special relativity will crank out the answer.

The fact that proper time depends on velocity and not on acceleration is encoded in the fact that

$$ds = \sqrt{1 - (v/c)^2} dt$$

The above equation has no acceleration in it, only a velocity.

Now the thing to note is that it is the accelerated (i.e. held) watch that is the one which is slowed down (i.e. the traveling twin has $$\Delta s < \tau$$). When you compare the watches upon catching the thrown one, the one you catch will show the later time.

This is contrary to the intuitive expectation that the thrown watch is the one that corresponds to the traveling twin, which is why I posed the problem.

By the way, if you don't like using the inertial reference frame attached to the watch, you can do the problem in any other inertial reference frame of your choosing, and you will similarly avoid needing to write out a solution to Einstein's field equations. It's just that the equation for the proper time experienced by the thrown watch is particularly simple in the inertial frame of that watch.

With this example, I'll bet that you are now eager to solve some more problems in general relativity using special relativity mathematics.

Carl

 

[OlderDan]

Now the thing to note is that it is the accelerated (i.e. held) watch that is the one which is slowed down (i.e. the traveling twin has $$\Delta s < \tau$$). When you compare the watches upon catching the thrown one, the one you catch will show the later time.

This is contrary to the intuitive expectation that the thrown watch is the one that corresponds to the traveling twin, which is why I posed the problem.
Carl

OK, you have me thinking that what I referred to as the integral over quasi-inertial reference frames might hold some validity for small distances and times. Fortunately, the result is not at all contrary to what I expected based on analogy to the twin problem, and I recognized the frame of the thrown clock as the inertial frame from which to view the problem. By analogy, the accelerating clock should be the one that records the lesser time interval.

Given that you seem to have a good handle on this stuff, let me pose the following question, which might call into question the symmetry of velocities and accelerations between frames that you spoke of. This is the scenario that prompts me to question the validity of the quasi-inertial approach.

Consider the traditional twin problem. One twin, T_a, accelerates from Earth and travels to a star 20 light years away, reverses direction and returns. Now this is only a thought experiment, so we give this traveling twin the uncanny ability to instantaneously view all the clocks in the rest frame of the inertial twin, T_i, and to locate those clocks in his own reference frame. At the moment he begins his trip, T_a sees the universe exactly as T_i does, so all of T_i's clocks are synchronized.

During the time T_a accelerates to near light speed in T_i's rest frame, the universe he is viewing changes dramatically. That star that was 20 light years away moves to let's say only one light year away (length contraction). The clock on the star becomes unsynchronized to the clock on the earth, leading it by nearly 20 years, because while T_a is making the outbound trip he is going to see that clock running slow (time dilation), but it has to read 20 years when he gets there.

When T_a arrives at the star, he decelerates to a stop and then accelerates for home. Just before this transition, he locates and reads the clock on earth, which has been running slow during his whole trip, so it reads 1/20 of a year and is only one light year away from him. As he slows into the rest frame of T_i, that clock has to move away to a distance of 20 light years, and has to run off almost 20 years of time to become synchronized with the star clock. As he accelerates toward home, the Earth must approach him to within a distance of 1 light year, and must run off an additional interval of almost 20 years so that it will read 40 years after running slow during his return trip.

Now to me this suggests that what T_a is "seeing" during the acceleration phase is vastly different from what a quasi-inertial view would predict. He would have to see the Earth move away from him and back toward him a distance of 38 light years at speeds far in excess of the speed of light as measured by his lengths and his proper time. He would also have to see the Earth clock running forward at super high speed to change from lagging the star clock by about 19 years to leading the star clock by about 19 years.

Now of course T_a does not actually get to see these distant objects and clocks making these transitions because the information cannot get to him faster than the speed of light, but in principle he should be a resident of an extended reference frame that can record these events on his clocks and send him the information to draw conclusions after the fact.

I'm not suggesting this is a proper view of the universe as seen by T_a, but it all seems to be consistent with a special relativistic view of the problem and, to me at least, casts doubt on the validity of any quasi-inertial approach to treating accelerating reference frames. I've always assumed the general theory could account for these intervals of acceleration properly, but I don’t see it happening from a calculation like the one you did.

Feel free to tell me what's wrong with my description of the scenario. That' why I'm throwing it out there.

 

[HackaB]

CarlB:

It does make very good sense that the frame of the thrown watch should be considered the inertial frame and the held watch be should be the accelerated frame. But I don't understand why you insist that acceleration does not affect proper time. Sure, it doesn't appear explicitly in the formula. But if a(t) is the acceleration, a(t) determines v(t) up to a constant. Are you saying that the elapsed proper time on a spaceship going from event A to event B does not depend on how it got there? That is saying that the integral

$$\int_A^B ds = \int_A^B \sqrt{1 - (v/c)^2} dt$$

doesn't depend on the function you choose for v. Please correct me if I misunderstand you.

 

[CarlB]

It does make very good sense that the frame of the thrown watch should be considered the inertial frame and the held watch be should be the accelerated frame. But I don't understand why you insist that acceleration does not affect proper time. Sure, it doesn't appear explicitly in the formula. But if a(t) is the acceleration, a(t) determines v(t) up to a constant. Are you saying that the elapsed proper time on a spaceship going from event A to event B does not depend on how it got there?

Go back and look at the context of my statement. It was in respect to whether the thrown watch was to be synchronized before or after it was thrown. The difference between those two states has to do with the acceleration, or velocity change of the watch. In that context, it doesn't matter at which point the watch was synchronized.

That is saying that the integral

$$\int_A^B ds = \int_A^B \sqrt{1 - (v/c)^2} dt$$

doesn't depend on the function you choose for v. Please correct me if I misunderstand you.

Sometimes I wonder why it is I keep getting asked questions that amount to "are you so stupid that you believe X", when any high school student wouldn't believe X. Is it something about this forum? Look, it was you who twice posted the wrong answer to the problem, not me. The thrown watch is the inertial one, the one with the larger interval of proper time, not the held one.

As it turns out, back when I was in grad school, I taught a "recitation" section for a quarter that included relativity, to undergraduates. It is a common misconception among students that acceleration causes time dilation. The question I was answering was in the context of this misconception.

By the way, during the recitation, the students wouldn't believe me when I insisted on an answer to another relativity question. In fact, they insisted that the professor had told them otherwise. I'll put the question up as another homework problem.

My guess is that the professor did tell them something that wasn't true, but with relativity, there is so much that is unintuitive, that it is easy to do. Of course as a graduate student, I couldn't tell them that a professor was wrong; so I laughed and told them that there was no way that he had said that, but that they must have misheard what he really said. I'm just glad nobody had a tape recorder.

This is very very basic stuff to physicists. No one has the slightest doubt as to what causes time dilation. It's velocity. Sure acceleration causes changes in velocity, but when you write your equations down, the change in proper time will end up proportional to the velocity, not the acceleration (which has to do with changes in velocity and not velocity per se).

Sure, acceleration causes velocity changes, and velocity changes cause time dilation, so in a certain sense, acceleration causes time dilation. You can say this, but your logic is iffy.

Electric fields cause forces. Forces cause acceleration. Acceleration causes velocity. So according to your logic, electric fields cause time dilation.

Proper time gets dilated even in situations where there is no acceleration at all. For example, in the case where the traveling twin doesn't turn around at Alpha Centauri, but instead gives the time to a third astronaut who is going back towards earth.

Carl

By the way, a plastic surgeon wrote an article on this very subject. (No physicist would touch it because it is too simple.) I can't vouch for it. I haven't read it. I haven't read the journal it was published in, etc., but here it is:
http://adsabs.harvard.edu/abs/1997AmJPh..65..979G

 

[HackaB]

Wow! Where did that come from? I'm sorry I participated in your little experiment. I'll let you continue trying to prove that everyone else besides you is an idiot. Good luck.

edit: Why are you so angry?

 

[OlderDan]

It is a common misconception among students that acceleration causes time dilation.

I'm a bit perplexed by this statement. If the equivalence principle says that gravity and acceleration are indistinguishable, and if it is true and has been verified by measurement that time is dilated by a gravitational field, how can acceleration not not cause time dilation? What am I missing?

 

[CarlB]

Dear Dan, Great question.

I'm a bit perplexed by this statement. If the equivalence principle says that gravity and acceleration are indistinguishable, and if it is true and has been verified by measurement that time is dilated by a gravitational field, how can acceleration not cause time dilation?

Let's work out an example in a gravitational field. I know that this is a bit beyond the scope of the usual undergraduate introduction to relativity, but the mathematics involved, if you restrict yourself to metrics and avoid tensors, is not at all difficult.

A man is standing on a planet of mass M at a radius R. He takes out his watch, ties it onto a string, and makes its path follow, in Schwarzschild coordinates, a path defined by $$(r(t),\phi(t),\theta(t))$$, for $$t$$ between zero and one. The equation for the Schwarzschild metric is:

$$ds^2 = ( 1 - r_s / r ) dt^2 - ( 1 - r_s / r )^{-1} dr^2 - r^2 d_o^2$$
http://casa.colorado.edu/~ajsh/schwp.html

Note I switched the above metric from being "space-like" to being "time-like". This is a matter of convention. Then the proper time experienced by the watch will be:

$$\int_0^1 ds^2 = \int_0^1 [(1 - r_s/r) - (1 - r_s/r)^{-1}(dr/dt)^2 - r^2 (d_o/dt)^2] dt$$

Note that the above includes no acceleration. It does include, however, velocity: $$dr/dt$$ and $$d_o/dt$$.

In short, time dilation is caused by velocity, not by acceleration, and this follows into the gravitational theory as well. It is not the fact that you are holding the watch (i.e. accelerating it against the curvature of space-time) that causes its time to dilate, but instead it is velocities (and gravitational wells) that dilate time.

By the way, if you plug the right numbers into the above integral, you will have another way of computing the difference in proper time experienced by the thrown watches. But you'll have to work harder.

Carl

Note: General relativity is pretty arbitrary about what it uses for coordinate systems. Check out the "Free-fall" metric from the above link.

It is a common misconception to think of "time dilation" as if it were an absolute thing. When one discusses time dilation, it is always with respect to something else. With velocity, the "something else" is an inertial (or "stationary") reference frame. With gravitation, it is with respect to stuff so far away that space is flat, and one furthermore chooses a position that is, again, stationary with respect to the massive body. Of course in the context of cosmology, there may be no region where space can be considered flat.

The best way of approaching relativity is in the spirit with which it was defined, that is, as a calculational technique. As with quantum mechanics, don't let the words confuse you, the equations are what is important.

 

[pervect]

This is a variation of a problem Feynman liked to ask graduate students in GR. A variant of this problem is mentioned on one of Feynman's poopular books (I forget which one offhand).

The second form of the problem is essentially the same as the first, but a lot easier to talk about.. The answer in both cases is that the watch thrown upwards shows the longest elapsed time.

This is most easily seen in the second case. The thrown watch is following a geodesic in sapce-time. The held watch is not, it is accelerating. In flat space-time, a geodesic connecting any two points will always have a longer proper time than a non-geodesic path. This same simple statement cannot be made in non-flat space-time (i.e. the case with actual gravity). But we can still use the equivalence principle to convert the the first problem into the second one.

If we look at the second problem from the point of view of the thrown watch, we can see that the problem is equivalent to the standard "twin's paradox". The rocketship starts out with some velocity -v, accelerates to a stop, then continues accelerating until it catches up with the thrown watch. So the rocketship is the accelerating twin, the thrown watch is the "stay-behind" twin, and the rocketship's clock has less elapsed time.

We can use the relativistic rocket equations at

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

to solve the problem without calculus. I'm going to simplify the problem by using geometric units, where c=1.

We will set our reference framw to be the frame of the rocket at t=0. Then the motion of the thrown object is described by d=vt, and the motion of the rocket is described by d=(1/a)*sqrt(1+a^2t^2)-1) (from the above link).

Solving, we find t=0 and t= 2*v/(a*(1-v^2)) are the times when the two objects coincide.

The elapsed time on the thrown object will be simply
*CORRECTION*

tau1 = d/(v*sqrt(1-v^2)) =( 2*v )/( a*sqrt(1-v^2))
$$\tau_1 = \frac{2v}{a\sqrt{1-v^2}}$$

(the second result obtainted by elimianting d from the equation in terms of a and v, d = 2*v^2 / (a*(1-v^2))

The elapsed time on the rocket will be

tau2 = arcsinh(a*t)/a = arcsinh(2*v / (1-v^2)) / a
$$\tau_2 = \frac{arcsinh(\frac{2v}{1-v^2})}{a}$$

It's not obvious which is greater from the analytic solutions, alas. We can see that a*tau1 is a function only of v, as is a*tau2. A computer-generated series analysis shows clearly which is greater for small v, which was the original problem statement.

*CORRECTION*

tau1*a ~ 2*v + v^3 + (3/4)v^5
tau2*a ~ 2*v + (2/3)*v^3 + (2/5)*v^5

so clearly tau1 > tau2 for small v, and plots confirm this for large v (the arcsinh function increases very slowly, it's essentially logarithmic).

[add]
Note that this analysis did not assume that s=.5*a*t^2 for an accelerating object, but used the full relativistic equation. Let's compare the result with the previous analysis that does. First let's find tau2/tau1 as a series in v, the result is

1-(1/6)v^2 - (11/120)v^4

Let's compare with the previously derived result.

$$\tau_1 = \frac{2v}{a}$$
$$\tau_2 = \int_{t=0}^{\frac{2v}{a}} \sqrt{1-(-v+at)^2} \, dt$$

The results are messy, but maple gives
$$\tau_2 = \frac {\left( v\sqrt {1-{v}^{2}}+\arctan \left( {\frac {v}{\sqrt {1-{v}^{2}}}} \right) \right)}{a}$$

The series expansion of tau2/(2*v/a) is

1-(1/6)v^2 - (1/40)v^4

Thus the results are comparable, the relativistic departure from parabolic motion makes a difference only in the fourth-order term in (v/c).

 

[pervect]

Dear Dan, Great question.

$$ds^2 = [( 1 - r_s / r ) dt^2 - ( 1 - r_s / r )^{-1} dr^2 - r^2 d_o^2$$
http://casa.colorado.edu/~ajsh/schwp.html

Note I switched the above metric from being "space-like" to being "time-like". This is a matter of convention. Then the proper time experienced by the watch will be:

$$\int_0^1 ds^2 = \int_0^1 [(1 - r_s/r) - (1 - r_s/r)^{-1}(dr/dt)^2 - r^2 (d_o/dt)^2] dt$$

Whoa - major typos are not going to help the confused people out there.

We don't want to integrate ds^2, we want to integrate ds as you did in an earlier post. Since ds is proper time, let's use $$d\tau$$. While we are at it, let's get rid of the "solid angle" do, it's not essential and it's just confusing the issue.

So
$$\int_0^1 d\tau = \int_0^1 dt\sqrt{(1 - r_s/r(t)) - \frac{(dr/dt)^2}{1-r_s/r(t)}}$$


The effect of gravitational time dilation can be seen by holding r constant, so that dr/dt = 0.

Then
$$d\tau = dt\sqrt{1-r_s/r(t)}$$

We can see that the rate at which a clock ticks does depend on its height as well as it's velocity in a gravitational field. The acceleration does not appear in the formula directly, it's been subsumed into the parameter r_s.

It is quite true that we can synchronize two clocks at the same point in space when one clock is accelerating. Acceleration does not affect our ability to compare two clocks when they are at the same position. Large accelerations may, as you remark, break a physical clock. It is also true that accleration does not affect the rate at which two clocks at the same point in space run. We can go further, two clocks with the same gravitational potential tick at the same rate, they don't have to be at exactly the same point in space.


However, two clocks at a different gravitational potential do not tick at the same rate.

[add]
Also note that it IS quite true that dr^2/dt^2 does not appear in the expression for calculating the proper time.

 

[PhilG]

Large accelerations may, as you remark, break a physical clock.

I've heard that before. What does that actually mean? Why does it happen?

 

[pervect]

There is a good but longish discussion of this on pg 396 of MTW's "Gravitation", with respect to atomic clocks of the 1972 era, specifically.

The atoms themselves will keep good time as long as tidal forces don't distort their shape. (This is possible under truly extreme conditions, such as a clock or atom near the singularity of a black hole, but not likely to happen in the lab).

In the laboratory, the weak link is the phase-lock loop that keeps a crystal oscillator locked to the resonant frequency of the atoms.

The phase lock loop with proper design can maintain it's lock under steady-state accelerations of up to 50 g's, but a sudden change in acceleration will momentarily unlock the loop. Relatively large frequency errors (1 part in a billion, as compared to the stability of 1 part in 10^12 - remember this is 1972 tech, we can do better nowadays) can occur during the time it takes the loop to re-lock. MTW state that the time it takes the loop to re-lock is proportional to the velocity change, but the constant of proportionality is not given.

 

[mtong]

Throwing a watch fast enough upwards to undergo time dilation would easily exceed the escape velocity for earth.

 

[pervect]

Throwing a watch fast enough upwards to undergo time dilation would easily exceed the escape velocity for earth.

How did you calculate that? Did you include gravitational time dilation? How much of the past (rather long) thread have you actually read?

 

[CarlB]

Whoa - major typos are not going to help the confused people out there.

Dear pervect;

Thanks for correcting the $$\sqrt$$ I left out of the proper time calculation.

[add] Also note that it IS quite true that dr^2/dt^2 does not appear in the expression for calculating the proper time.

Clearly you meant to write "d^2r/dt^2", as in there is no acceleration in the equation for proper time.

Carl

 

[mtong]

Throwing a watch fast enough upwards to undergo time dilation would easily exceed the escape velocity for earth.

I should have made myself more clear, this was in response to CarlB's solution, using special relativity.

 

[pervect]

Dear pervect;

Clearly you meant to write "d^2r/dt^2", as in there is no acceleration in the equation for proper time.

Carl

Yep - I'd go back and fix it, but it's too late to edit it.