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

Timelike geodesics might not be maxima of the proper time

  1. Feb 13, 2004 #1
    A commonly accepted modern SR/Gr statement is "timelike geodesics are maxima of the proper time".

    I really shall not dispute with these experts, who are all professors.
    But this statement is clearly questionable. It bothered me for a long time.

    The H & T experiment and GPS all show the statement is questionable. As I mentioned in another thread, in a gravitational field, an orbiter's world line is at geodesic and a standing person's worldline is not a geodesic, but the clock of the orbiter is slower than the standing person. This contradicts that statement.

    My guess is that the statement shall be altered as "Timelike geeodesics are the maxima of the proper time compared with an interseting worldline that share the same local comoving inertial framet at the initial intersecting event point."
  2. jcsd
  3. Feb 13, 2004 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    1) You can't compare the times elapsed on two clocks except at points where their worldlines cross.

    2) You've got gravitational time dilation backwards. The orbiter's clock would tick faster than the Earthlings.

    3) The definition of a geodesic is done through differential geometry. Rather than comparing the total proper times of two very long worldines, you compare the differential proper time along small intervals of two worldlines. In all cases, the geodesic is the curve with maximal proper time.

    - Warren
  4. Feb 13, 2004 #3

    Thank you for your reply.

    In the document I saw, Eastward airplane has -184 ns Kinematic effect and 275 ns for Westward airplane. My take is that Eastward airplane runs slower than the westward airplane, since it used less time in airtraffic. Is that so? Also, the gravitational effect is 144 positive. As we know, the higher you are, the faster you clock is. Is that so? If that's the case, the positive figure here means faster clock and the negative figures mean slower clock.

    If Eastward is running slower, the airplane's speed is in the direction of catching up with an orbiter if there is any. Eastward is the same direction as Earth's spin. Westward is offseting the Earth's spin then. It took almost two days to get back, so its speed is just about half of the Earth's spin. The Westward airplane is still considered eastward spin in half of Earth's spin speed. So the westward airplane is intended toward to a standing clock.

    The orbiter seems to be slower than a standing clock at the same altitude to me.

    Please correct me where I am wrong.
  5. Feb 14, 2004 #4
    Hi, Warren,

    I know you are a resourceful person. Your postings benefited me a lot.

    My previous response was taken from Hafele-Keating Experiment. I believe you knew already, but just in case you did not know.

    I was concerned whether I interpreted that experiment correctly. That's why my first response is how the interpretation of that experiment shall be and did not explain diretly to your questions.

    Any way, let me my case clearer:

    1. First I can build a very high tower and I can place a clock in it. So the clock is more likely standing than orbiting.

    Or, you think this is not good enough, I can send a spaceship into a orbit and adjust it slowly until it's at rest to the Earth frame. Well it will fall without the escaping speed so I have to keep pumping out fuel toward the Earth to keep it float in some height without any spinning speed.

    2. Second, I can send an orbiter orbiting at that height. The orbiter will repeatedly pass by the stading spaceship for every circle it makes. I can synchronize and compare their clocks as many times as I want. Their world lines will definitely intersect many times.

    The question is, which clock will tick slower?

    From my interpretation of H & K experiment, the orbiter will tick slower. You know the orbiter's worldline is a geodesic and the standing spaceship isn't. That's why I see that contradicts this common accepted statement.

    Even locally, You can image many standing spaceship or tower clocks at that height thru the orbit that orbiter run through. I can't see how we can say locally the geodesic is the maxima of the proper time.

    That's why I think the statement shall be adjusted, the initial clock ticking rates are already different when the two world lines intersected and that invalidated this statement.

    I have no interest in battling anyones here. I am just interested in what are the facts. If I am wrong, please correct me. I will appreciate your help.
  6. Feb 15, 2004 #5
    The correct statement is that timelike geodescics are extremals of proper time. See Exploring Black Holes by Taylor and Wheeler
    Here is a relavent example from Rindler's text
    Last edited: Feb 15, 2004
  7. Feb 15, 2004 #6

    Thanks again. I can't open that pdf, somehow I got a mesaage "unknown format".

    Who is Rindler? Apparently, he knew my confussion here. I was going to try to show that any objects freeing fall thru the same event point with different velocities will have different clock ticking rate and so as different proper time. So, how to interpret this extremal is really an issue.

    He apparently knew my question. He must have already provided an answer.
  8. Feb 15, 2004 #7


    User Avatar
    Staff Emeritus
    Gold Member
    Dearly Missed

    Purists insist that we say not maximum reading but rather extremal
    reading: either maximum or minimum.

    I have always wondered, since the deeper requirement is that the time derivative vanish, whether an inflexion point case exists.
  9. Feb 16, 2004 #8
    I've changed my mind on the derivation. Will get back if/when I'm satisfied with one.
    Last edited: Feb 21, 2004
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook