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The Twins Paradox in curved spacetime

  1. Jan 1, 2006 #1
    The Schwarzschild Metric on a spacial plane passing through the center of a spherically symmetric (non-spinning) center of gravitational attraction is:
    [tex] d \tau^2 = (1- \frac{2M}{r})dt^2 - \frac{dr^2}{(1- \frac{2M}{r})} - r^2d \phi^2 [/tex]

    If there are two spaceships at a distance [tex] r_1 [/tex] from a gravitating body, and one moves radially ([tex] d \phi = 0 [/tex]) to a radius [tex] r_2 [/tex], in a time [tex] \tau_1 [/tex] as measured by its own clock, and then immediately returns from [tex] r_2 [/tex] to [tex] r_1 [/tex] in a time [tex] \tau_1 [/tex], how much time will have elapsed on the clock on the spaceship which remains at [tex] r_1 [/tex] ?

    I suppose what I'm looking at is a double integral wrt t and r but being a little unfamiliar with how this is done.
     
    Last edited: Jan 1, 2006
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  3. Jan 1, 2006 #2

    pervect

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    We need a little more information. The path that the radially moving spaceship takes can be expressed as a curve [itex](r(\tau),t(\tau))[/itex]. We know that the spaceship is at r1 at [itex]\tau=0[/itex], and that it's at r2 at [itex]\tau=\tau1[/itex], and that it's back at r1 at [itex]\tau=2*\tau1[/itex].

    What we need to know to compute the elapsed time for the other ship is the exact details of the path. There are an infinite number of ways of going between two points. Two plausible possibilities for the intent of the problem

    1) [itex]dr/d\tau[/itex] stays constant on the outgoing and ingoing path

    2) The spaceship follows a geodesic, so that it does not have to fire its engines on the outgoing and ingoing path (it free-falls).
    BTW, is this a homework problem?
     
    Last edited: Jan 1, 2006
  4. Jan 1, 2006 #3
    There aren't an infinite number of ways, there's just a geodesic. The motion is either towards or away the spherical mass. No tangential motion.

    No it's not homework. I'm trying to get through "Exploring Black Holes" and I asked a question that isn't in the book.
     
  5. Jan 1, 2006 #4

    pervect

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    OK, I think I have all the latex errors fixed, and I've made a few additions to the text...

    You have specified three points in space-time. There are an infinite number of curves connecting those three points - thus it was not clear that your spaceship followed a geodesic curve connecting them until you stated that it was.

    The simple way to solve your problem is to take advantage of the fact that objects following a geodesic in a Schwarzschild geometry have a conserved energy.

    This fact may or may not be in your particular text - it's online at http://www.fourmilab.ch/gravitation/orbits/

    If you look at the section under "gravitational effective potential" on this webpage, and you realize that L=0 because the object is falling radially, you can write

    [tex]
    \left( \frac{dr}{d\tau} \right)^2 = E^2 - 1 + \frac{2M}{r}
    [/tex]

    where E is some constant. I suppose I should note that this webpage is using geometric units, so c=G=1.

    It's possible to derive this from the fact that the geodesic path is the path that maximizes proper time, if you're interested.

    To find the value of E, you need to solve
    [tex]\frac{dr}{\sqrt{E^2 -1 + \frac{2M}{r}}} = d\tau
    [/tex]

    for the proper value of E to meet your boundary conditions. This is unfortunately rather messy, Maple finds that

    [itex]
    \tau = - \left( -\sqrt {r \left( {E}^{2}r-r+2\,M \right) }\sqrt {{E}^{2}-1}+M
    \ln \left( {\frac {M+{E}^{2}r-r+\sqrt {r \left( {E}^{2}r-r+2\,M
    \right) }\sqrt {{E}^{2}-1}}{\sqrt {{E}^{2}-1}}} \right) \right)
    \sqrt {{\frac {{E}^{2}r-r+2\,M}{r}}}r{\frac {1}{\sqrt {r \left( {E}^{2
    }r-r+2\,M \right) }}} \left( {E}^{2}-1 \right) ^{-3/2}
    [/itex]

    (Looks like this gets truncated, well anyway it's a real mess).
    [add] Fortunately George found a simpler way of solving for E, see his post.


    You have already written down

    [tex]
    d \tau^2 = (1- \frac{2M}{r})dt^2 - \frac{dr^2}{(1- \frac{2M}{r})} - r^2d \phi^2
    [/tex]

    We can divde both sides by [itex]d\tau^2[/itex] to get

    [tex]
    1 = (1- \frac{2M}{r}) \left( \frac{dt}{d\tau} \right) ^2 - (1- \frac{2M}{r}) \left( \frac{dr}{d\tau} \right) ^2
    [/tex]

    Substituting for [itex]dr/d\tau[/itex] we get

    [tex]
    1 = (1- \frac{2M}{r}) \left( \frac{dt}{d\tau} \right) ^2 - (1- \frac{2M}{r}) (E^2 - 1 + \frac{2M}{r})
    [/tex]

    This allows you to solve for [itex]t(\tau)[/itex]

    [add]
    Actually there's an easier way - there's already an expression for [itex]dt/d\tau[/itex] on the webpage I cited! But you have to solve the intergal to get the solution, in any case.

    If you don't have Maple or Mathematica, you might try

    http://integrals.wolfram.com/index.jsp
    [end add]

    This solution for [itex]t(\tau)[/itex] will give you the elapsed coordinate time for the object to rise and then fall. You now have to account for gravitational time dilation for the hovering observer to find his elapsed time that he will observe, this is given by

    [tex]
    d \tau_{hover} = \sqrt{(1- \frac{2M}{r})} dt
    [/tex]

    since dr=0 for the hovering observer. Because r is a constant, this is just a constant multiplier for [itex]\Delta t[/itex], which is the change in coordinate time t.
     
    Last edited: Jan 1, 2006
  6. Jan 1, 2006 #5

    George Jones

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    [tex]E[/tex] can be expressed in terms of DiamondGeezer's [tex]r_{2}[/tex]. At the turning point [tex]r_{2}[/tex], [tex]dr/d\tau = 0[/tex]. Using this in the above gives

    [tex]E = \sqrt{1 - \frac{2M}{r_{2}}}[/tex].

    If [tex]r_{2}[/tex] is: the event horizon, then [tex]E = 0[/tex]; infinity, then [tex]E = 1[/tex].

    Regards,
    George
     
  7. Jan 1, 2006 #6

    pervect

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    Ah yes - that's MUCH simpler.
     
    Last edited: Jan 1, 2006
  8. Jan 1, 2006 #7

    George Jones

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    I haven't worked the problem through to the bitter end, but from what pervect has done, and from my own cursory analysis, the algebra still looks rather messy.

    It might help to express r as a multiple of 2M, i.e., to make the change of variable x = r/(2M). Also, using actual numbers for the x's (and thus E) might help.

    Just a couple of thoughts - I don't know how useful they are.

    Regards,
    George
     
  9. Jan 2, 2006 #8

    pervect

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    I've played around with the problem some more too, and it still comes out to a very messy intergal, even with your suggestions.

    If we let u = r/2M, and k = r2/2m, we can, with a bit of messing around, find

    du/dt = f(u), so we can find t by integrating du/f(u). Messing around consists of changing the variables, then taking du/dt = (du/dr) (dr/dtau) / (dt/dtau).

    f(u) is not terribly complex, but du/f(u) is difficult to integrate :-(, the computer doesn't want to come up with a correct closed-form intergal (neither do I :-)).

    [tex]
    f(u) = \frac{ (u-1) \, \sqrt{1-\frac{u}{k}}}{2M u^{1.5} \sqrt{1-\frac{1}{k}}}
    [/tex]
     
    Last edited: Jan 3, 2006
  10. Jan 2, 2006 #9
    Hmmm. I'm going to have to look at this problem a bit more closely. It's apparent that the moving ship won't measure the same distance from r_1 to r_2 as the ship at r_1

    From the viewpoint of the ship at r_1, the moving ship travels (inertially) a distance of
    [tex]\Delta r = \int_{r_1}^{r_2} \frac {dr }{(1- \frac{2M}{r})^ \frac{1}{2}}[/tex]

    This becomes
    [tex]\Delta r = \int_{r_1}^{r_2} \frac { r^ \frac{1}{2} dr}{(r-2M)^ { \frac{1}{2}} }[/tex]

    Substituting [tex]r = z^2[/tex]

    [tex]\Delta r = \int_{z_1}^{z_2} \frac { 2z^2 dz}{(z^2-2M)}[/tex]

    -> [tex]\Delta r = z(z^2 - 2M)^ \frac{1}{2} +2M \log_e \abs {z+(z^2-2M)^ \frac{1}{2})[/tex]

    ....from z1 to z2

    So the round trip should be twice this value (by symmetry).
    And so to the time bit....
     
    Last edited: Jan 2, 2006
  11. Jan 2, 2006 #10

    George Jones

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    I'm going to try and see what happens if [tex]r_{1} = 8M[/tex] and [tex]r_{2} = 18M[/tex]. First I'm going to calculate the proper time taken by a radially freely falling observer that moves from [tex]r_{1}[/tex] to [tex]r_{2}[/tex] and back to [tex]r_{1}[/tex], then I'm going to calculate the coordinate time [tex]t[/tex] time taken, and, finally, I'm going to calculate the proper time measured by an observer that hover at constant [tex]r = r_{1}[/tex].

    Please check my work!

    For any observer moving radially, the Schwarzschild metric gives

    [tex]d \tau^{2} = \left( 1 - \frac{2M}{r} \right) dt^{2} - \left( 1 - \frac{2M}{r} \right)^{-1} dr^{2}.[/tex]

    Also, as stated in previous post, the constant

    [tex]E = \left( 1 - \frac{2M}{r} \right) \frac{dt}{d \tau}[/tex]

    is determined by

    [tex]E = \sqrt{1 - \frac{2M}{r_{2}}}.[/tex]

    Taking [tex]r_{2} = 18M[/tex] gives

    [tex]E = \frac{2 \sqrt{2}}{3},[/tex]

    which in turn gives that

    [tex]\frac{dt}{d \tau} = \frac{2 \sqrt{2}}{3} \left( 1 - \frac{2M}{r} \right)^{-1}.[/tex]

    This can be used to eliminate: [tex]dt[/tex] when calculating the freely falling proper time; [tex]d \tau[/tex] when calculating the elapsed coordinate time.

    Calculate proper time for freely falling observer.

    [tex]
    \begin{equation*}
    \begin{split}
    d \tau^{2} &= d \tau^{2} \left( 1 - \frac{2M}{r} \right) \left( \frac{dt}{d \tau} \right)^{2} - \left( 1 - \frac{2M}{r} \right)^{-1} dr^{2}\\
    &= d \tau^{2} \frac{8}{9} \left( 1 - \frac{2M}{r} \right)^{-1} - \left( 1 - \frac{2M}{r} \right)^{-1} dr^{2}\\
    \left( 1 - \frac{2M}{r} \right) d \tau^{2} = \frac{8}{9} d \tau^{2} - dr^{2}\\
    dr^{2} = \left( \frac{2M}{r} - \frac{1}{9} \right) d \tau^{2}
    \end{split}
    \end{equation*}
    [/tex]

    Thus,

    [tex]\tau = 2 \int_{r_{1}}^{r_{2}} \frac{dr}{\sqrt{\frac{2M}{r} - \frac{1}{9}}}.[/tex]

    Making the change of variable [tex]u = r/\left(2M\right)[/tex] gives

    [tex]
    \begin{equation*}
    \begin{split}
    \tau &= 4M \int_{4}^{9} \frac{du}{\sqrt{\frac{1}{u} - \frac{1}{9}}}\\
    &= \left(4M\right) \left(36.1\right).
    \end{split}
    \end{equation*}
    [/tex]

    Calculate the elapsed coordinate time.

    [tex]\frac{9}{8} \left( 1 - \frac{2M}{r} \right)^{2} dt^{2} = \left( 1 - \frac{2M}{r} \right) dt^{2} - \left( 1 - \frac{2M}{r} \right)^{-1} dr^{2}[/tex]

    leads to

    [tex]
    \begin{equation*}
    \begin{split}
    t &= 4M \int_{4}^{9} \frac{du}{\sqrt{ \left( 1 - \frac{1}{u} \right)^2 - \frac{9}{8} \left( 1 - \frac{1}{u} \right)^3}}\\
    &= \left(4M\right) \left(39.7\right).
    \end{split}
    \end{equation*}
    [/tex]

    Finally, the elapsed proper time for the hovering observer is

    [tex]\sqrt{ \left( 1 - \frac{2M}{r_{1}} \right)}t = \left(4M\right) \left(34.3\right).[/tex]

    Regards,
    George
     
    Last edited: Jan 2, 2006
  12. Jan 3, 2006 #11

    pervect

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    For a free-fall from r=18 to r=4 with M=1, I get a proper time of 72.25 (of whatever unit we are using to measure r and M) for the falling observer. This integrates out reasonably well for [itex]\tau[/itex] as a function of r

    [tex]
    \tau = -3\,\sqrt {-r \left( r-18 \right) }+27\,\arcsin \left( 1/9\,r-1
    \right)
    [/tex]

    (I think this is supposed to be a cycloid? I'm not sure anymore, I just read something about that not too long ago, but I'm to tired to look it up at the moment).

    This checks with your result of twice that for the complete trajectory.

    For coordinate time for the fall, I get 79.30, using Maples numeric integration. This also checks with your result.

    I will be correcting some errors in my previous post that I found while doing this integral. (The symbolic intergal I get from maple is still bad even after correcting my errors).

    i.e. the correct intergal in my notation u=r/2M , k=r_max/2M is

    [tex]
    \int_4^9 \, \frac{2M u^{1.5} \sqrt{1-\frac{1}{k}}} { (u-1) \, \sqrt{1-\frac{u}{k}}} du
    [/tex]

    which gives 79.3 (the coordinate time for the fall, double this for the complete rise+fall cycle).

    The correction factor for the proper time of the stationary observer at r=8M is trivial , being sqrt(3)/2 of the coordinate time.
     
    Last edited: Jan 3, 2006
  13. Jan 3, 2006 #12

    pervect

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    This problem gets quite involved. I want to stress one basic point, though, since I am not sure if it's gotten lost in the details.

    This is the fact that the proper time of an object depends on its trajectory. To calculate the proper time, you need to know the path an object follows.

    There are several ways to specify a path, specifying two functions

    r(lambda), t(lambda)

    is often convenient, in this case however specifying r(t) is more convenient as one then needs to specify only one function.

    For the "free-fall" problem, we have to calculate the path from conserved quantites or via some other means, we can then proceed with using the metric to find the elapsed time along the path.

    It turns out to be convenient to find dr/dt as a function of r in the problem as specified.
     
  14. Jan 3, 2006 #13
    What I would probably be looking for would be the path of extremal aging, where one ship moves inertially in both directions, so there is no extra complication of the acceleration of the ship.
     
  15. Jan 3, 2006 #14

    George Jones

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    As pervect said, the details are a bit overwhelming, but pervect and I both calculated this - the inertial case is the freely falling/geodesic case. Freely falling does not necessarily mean falling towards the black hole, it just means zero 4-acceleration.

    Consider the following physical setup. An observer stands on a platform that uses a rocket to hover at constant r = r1 above the black up. The observer throws a ball straight up, the ball reaches a maximum height at r = r2, falls back down, and is caught by the observer at r = r1. The observer wears a wristwatch, and a watch is also attached to the ball. The time elapsed on the ball's watch should should be greater than the time elapsed on the observer's watch, since the ball moves inertially, while the observer (because of the rocket) does not.

    According to calculation performed by both pervect and me, this is indeed the case. My (4M)*36.1 is the elapsed time according to the ball's watch, while (4M)*34.3 is the elapsed time according to the observer's watch.

    Regards,
    George
     
    Last edited: Jan 3, 2006
  16. Jan 3, 2006 #15

    So do we have a final form for the general solution?
     
  17. Jan 3, 2006 #16

    pervect

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    There are three interesting cases, IMO. It is easier to describe their general properties than to work through all the math in detail.

    We want a confluence of observers, one who wind up back at the same point at some distance r1 away from the black hole at the same time.

    The radially falling observer, Observer-1, moving straight up and straight down, spends as much time as possible away from the deep gravity well of the black hole.

    As a free-falling observer (one that is following a geodesic path), his path will be a local extremum of proper time. His path is also the one that globally maximizes his proper time, so he will be the oldest observer when everyone re-unites. This is hard to prove rigorously, but not that hard to argue in general principles. Being near the black hole slows down aging. To age as much as possible, one must get as far away from the black hole as one can. This means moving away from the black hole. The faster one moves, the less one ages. Balancing one's desire to get as far away from the black hole as quickly as possible with the fact that moving quickly retards aging, one finds that the locally optimum path is the free-fall path. This path is locally optimum because of the laws of physics. It's globally optimum, because moving away from the black hole as quickly as possible is the "right" choice of directions in which to move to maximize aging.

    The hovering observer, Observer-2, hovering on rocket thrust at a constant distance r1 away from the black hole, is not following a path which locally extremizes proper time. The hovering observer will have a shorter elapsed time on his watch when he re-unites with the radially free-falling observer, as was illustrated numerically in the specific example George and I computed.


    The orbiting observer,Observer-3, orbiting the black hole due to his angular velocity at some constant distance r1 away from the black hole will be following a path which locally extremizes his proper time. This observer has not been analyzed in our example. It will take some "fine-tuning" to make sure that the orbiting observer makes an intergal number of orbits to wind up in the same location as Obserer-1 at the same time, but it's possible with the right choice of initial conditions.

    Globally, though, Observer-3 he has chosen a poor path, as his proper time will be even shorter than Obserer-2! This illustrates the important difference between a local extremum, and a global extemum.

    To see that the proper time of observer 3 is shorter than that of the hovering observer,, it is only necessary to see that the [itex]d\theta^2[/itex] terms in the metric subtract from the proper time intergal, thus the orbiting observer will have a lower value for this intergal than the hovering observer.

    As far as the exact numbers go, I've corrected (I think) the intergal for the coordiante time of the radially falling observer. Wolfram's mathematica apparently is able to handle this intergal correctly in terms of hypergeometric functions (unlike maple). (I'm not actually positive that Mathematica fixes the problem Maple has, but it looks promising). Thus, if you go to Wolfram's integrator website,

    http://integrals.wolfram.com/index.jsp

    and paste in the expression I posted earlier, you'll get a closed form solution in terms of hypergeometric functions. This probably won't actually be all that useful, though, unless you have something that can calculate hypergeometric functions for you. You also may not be able to even read the result unless you switch to "standard form" on the output tab (rather than "traditional form", the default).

    I can't cut and paste the result from the website, as the designers have cleverly made it impossible (I guess it would be too useful, the idea is to sell product, not provide a service :-().
     
  18. Jan 3, 2006 #17

    pervect

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    quiz time: I've probably written too much, so let me ask DiamondGeezer a question to try and get a feel for what is understood and what isn't.

    We've noted that the proper time interval between two points depends on the curve connecting them.

    There are several ways of specifying a curve C between two points, one of the more general is the parametric specification . If we restrict ourselves to one spatial dimension for simplicity, we need only specify x(lambda) and t(lambda) to describe the path the object takes through space (x) and time (t).

    Now here is the question:

    Given such a parametric specification of a curve, how do we calculate the elapsed proper time dtau along the curve, given the functions x(lambda), y(lambda), and the metric?

    (The intergal will be of the form [tex]\int f(\lambda) d\lambda[/tex])
     
  19. Jan 4, 2006 #18

    pervect

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    OK, I'll add a few hints (dunno if DG is still here). Maybe he's already figured it out (it's very simple), but without any feedback it's hard to tell.

    The proper time along a curve (i.e. the time a clock will read if it is moved along the curve) is given by integrating the derivative of the proper time along the curve, i.e. [itex]\tau = \int d\tau[/itex].

    As DG has already written the equation down for [itex]d\tau^2[/itex] in terms of the metric, the only step necessary is to take the square root of this expression and integrate it.

    The only remaining problem is that [itex]d\tau[/itex] is written in terms of dx and dt instead of [itex]d\lambda[/itex]. Our curve is specified by a parameter [itex]\lambda[/itex], which "traces out" the curve as it varies, and we've asked for the lenth as some intergal of [itex]d\lambda[/itex].

    There is a very simple relationship between [itex]d\lambda[/itex] and dx and dt, which can be derived from the definition of the curve itself.
     
  20. Nov 11, 2008 #19
    Pervect,

    Apologies for not replying to this thread sooner. I dropped my attention on this problem for reasons I can't remember (some life stuff getting in the way of important stuff like this - things like emigration and having kids).

    Can I just say that I immensely appreciate your and George's help and insight into my questions which help reduce my ignorance substantially.

    What I'm going to do is re-read the entire thread, do some calculations and see if it all makes sense again.
     
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