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Gravitational time dilation reason

  1. Aug 6, 2013 #1

    I am really bad at physics so bear with me...

    I saw some very interesting 3 hour physics-marathon on TV the other day.

    I have never gotten an actual clue on how things work relativisticly even though I've read a book by Brian Greene some years ago.

    Since then I have tried to deduce the formula for time dilation and failed.

    But seeing this TV-documentary made me actually deduce the equation for time dilation.

    This equation being:


    Where t0 is the traveller.

    I like this version better because it tells that same thing everybody is talking about i.e time stops when you are travelling at the speed of light.

    But this was not my question.

    My question is this:

    What is the reason or theory behind gravity slowing down time?

    I have searched the internet but all I can find is the formula


    rsch being the schwartz-child radius of a black-hole equivalent for the planet.

    And I have always been a curious guy so I ask myself why?

    I can accept that the velocity of light is constant.

    But what more do I have to accept?

    That gravity stops time?

    I am very interested in hearing some kind of physical explanation of why.

    I know about the Doppler-effect. But that works on relative velocities (and not accelerations).

    I do think there is some kind of connection here but I do not see it.

    Thankful for any answer.

    Best regards, Roger
    Please excuse my bad english. I'm just a curious swede.
  2. jcsd
  3. Aug 6, 2013 #2


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    Do you know what the equivalence principle is?
  4. Aug 6, 2013 #3
    There is a lot of food for thought in these equations--especially those involving the Schwarzchild radius. But on a basic level, you can see why gravitational time dilation must occur without getting into advanced topics like the Schwarzchild radius by thinking about the following gedanken experiment (which I have stolen from Schutz's "A First Course in General Relativity").

    First, assume there is a magical machine which can convert a photon of energy into some massive particle according to E=mc2. Recall the energy of a photon: E=hv, where h is Planck's constant and v is the frequency of the photon. Further, assume we're staying close to the earth where the potential energy of a massive particle is V=mgr, where m is the mass of the particle, r is its height above some reference point, and g is the gravitational acceleration in the region close to earth.

    Now imagine we shoot a (massless) photon from the surface of the earth (r=0) to the magical mass-energy converter suspended at a height r=R. If the photon's frequency v did not change between the surface of the earth (r=0) and the magical converter, then we could generate energy out of nothing--starting with a photon with energy hv at the surface, if it arrives at the converter with energy hv, then we could convert it to a massive particle with E=hv=mc2, then drop it down to the original spot, adding an additional energy mgR, so the final energy back at the surface is E'=hv+mgR. We could repeat the process and generate arbitrary amounts of energy out of nothing. This contradicts conservation of energy.

    The answer is that the photon must have a different frequency when it reaches r=R due to gravitational time dilation. The decrease in the frequency is the manifestation of time dilation. This is the way GR avoids contradicting energy conservation.

    Hope this argument for why gravitational time dilation must be there helps you. But there are other ways to see why it must happen.
    Last edited: Aug 6, 2013
  5. Aug 6, 2013 #4


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    Schutz got that argument from MTW by the way, so you aren't really guilty of stealing anything :)!
  6. Aug 6, 2013 #5
    Thank you Jolb!

    I'm sorry, I still don't understand.

    Why would the photon gain mgH just by being dropped to the earth surface?

    This is how I see it:

    Reaching H the photon energy is hf+mgH.

    Dropping down to the surface of earth, the photon energy is hf (h=0).

    Have I missed something?

    Due to your nice example I however think I now understand that my initial thoughts about why gravity can bend light is correct.

    The simple reason is that E=hf=mc^2 and this means that the photon can be given an equivalent mass and thereby be affected by the gravitational pull of large objects.

    This insight also helps in stating the equations above.

    This do however not help me understand why gravity affects time.

    Best regards, Roger
    Last edited: Aug 6, 2013
  7. Aug 6, 2013 #6
    No, please explain.
  8. Aug 6, 2013 #7


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  9. Aug 6, 2013 #8
    Hmmm, I think you have it a little mixed up. The experiment I described would go:
    1) Photon initially at surface, which we will call altitude=0. Energy of photon=hv
    2) Photon travels and ends up at an altitude=R. Here we make the assumption that E=hv. (This will be the assumption that leads to the contradiction.)
    3) Photon enters the magical converter, which generates a particle at rest with mass m=E/c2=hv/c2
    4) Particle of mass m falls freely to the surface. Its kinetic energy when it arrives at the surface is mgR. The total energy is mc2 + mgR = (hv/c2)c2+mgR = hv+mgR > hv. This contradicts the conservation of energy.

    Hence, at step 2), we made the wrong assumption. The correction that is needed has to do with the fact that the frequency v has been incorrectly assumed to be the same between steps 1) and 2). The frequency of the photon after step 2) should really be a smaller frequency, and the difference is due to gravitational time dilation.

    Let me know if you understand now.

    Roughly stated, the equivalence principle says that inertial mass is equivalent to gravitational mass. From a different perspective, one might think of the gravitational force as equivalent to the force a body feels when it is accelerated--that gravity is a fictitious force that arises from the curvature of spacetime. This concept is key to GR.
  10. Aug 6, 2013 #9
    I may be missing something, but initially the photon is "massless" with E=hv. It is then sent in the air into the magical energy to mass converter machine, where it is converted into a mass m=hv/c2. Since it is no longer a photon it no longer has the energy hv, as it was converted to a mass, rather it has an energy mgR=hv (or it should be equal to this, provided the machine is 100% efficient). Using the above, hv/c2*gR=hv, thus, c^2=gR. This is not possible as c is constant, and g remains relatively constant. Therefore, v2 must decrease proportional to R, such that hv2/c2*gR=hv1 may hold.

    Hmm, I hope that is accurate?
  11. Aug 6, 2013 #10


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  12. Aug 6, 2013 #11
  13. Aug 6, 2013 #12


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    Oh only the passages I posted are of relevance. You don't need to worry about the next section with regards to the above discussion.
  14. Aug 7, 2013 #13
    Seems like the gedanken experiment is giving people trouble. I'm at a loss on how to explain it more clearly.

    The section in MTW is actually extremely terse, so here is the original section from Schutz, who goes a little more in-depth. MTW and Schutz are consistent in how they spell out the experiment, but my version changes the order around a little. (I start with a photon at the bottom, they start with a mass at the top. But it is the same argument.)

    Attached Files:

    Last edited: Aug 7, 2013
  15. Aug 7, 2013 #14
    Hi Jolb!

    You have been a little too fast...

    You explained it so well that I now think I finally understand. The experiment anyway.

    My interpretation:

    As the photon "leaves" the gravitaional field the gain of mgR has to be reduced from hv due to conservation of energy.

    Because otherwise the photon would gain energy simply by being reflected back from the moon, so to speak.

    How close am I to the truth?

    I do however still not understand how gravitation can affect time.

    Unless the frequency of the photon itself can be related to time. Then strong gravitational fields obviously increses frequency and thereby reduces time.

    How far fetched am I now?

    Finally, your nice answer regarding the equivalence principle:

    First, I read WannabeNewton's Wikipedia suggestion (History of Gravitational Redshift).

    Guess what?

    I read it twice but didn't understand a single thing :)

    But your short explanation made me understand some things.

    Most importantly that accelleration and gravitation are two sides of the same thing.

    And I think the description of the curvature of space* is a beutiful description of how gravity of large planets affects moons.

    Best Regards, Roger
  16. Aug 7, 2013 #15


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    Hi Roger! Let me try to explain to you the equivalence principle in a simple way. If you can get that down then the rest is just taking the kinematical time dilation between the bottom and top of the accelerating box in the accelerating box thought experiment explained in the wiki link and applying the equivalence principle to then state that there must be time dilation between different heights in a gravitational field.

    Say we have an observer inside an elevator and say the observer is holding a small ball in his hand. First consider the scenario in which this elevator is at rest in a downwards uniform gravitational field ##\vec{g}##. If the observer were to drop the ball, he would see it fall to the floor of the elevator at a rate ##g##. Fair enough? Now consider the scenario in which the elevator is accelerating with ##\vec{a} = - \vec{g}##. If the observer were to drop the ball, he would again see it fall to the floor of the elevator at a rate ##g##. So operationally, the observer inside the elevator cannot discern between being at rest in a uniform gravitational field and accelerating in a direction opposite to that of the gravitational field. The two scenarios are physically equivalent.
  17. Aug 7, 2013 #16
    I find this very interesting. Hope you will get an expert's opinion soon.

    Best regards, Roger
  18. Aug 7, 2013 #17
    Hi WannabeNewton!

    Thank you for educating me!

    I started to read it a third time but I couldn't even get over this initial statement:

    "Once it became accepted that light is an electromagnetic wave, it was clear that the frequency of light should not change from place to place, since waves from a source with a fixed frequency keep the same frequency everywhere. One way around this conclusion would be if time itself were altered—if clocks at different points had different rates."

    Please explain the bold part.

    Regarding the rest, I give up.

    The math is simply to hard.

    But let's use what I do understand.

    You pointed out that being in rest in a gravitational field is equivalent to accellerating even opposit that same gravitational field.

    This I find very interesting.

    My quest for understanding gravitational time dilation do however seem to just have begun :)

    The question is, how do I get any wiser?

    I really want to understand this.

    But tensors, give me a break :)

    Best regards, Roger
  19. Aug 7, 2013 #18
    That is absolutely true.

    Fairly close. Instead of reflecting it from the moon back to earth, (which would be completely symmetrical), to make the argument work, you would want to first convert it to a massive particle and then drop it back to earth. Converting between massive particles and massless photons is a key ingredient in the proof by contradiction. But more or less you have it.

    It can be! This is another sort of crucial point that sometimes confuses people. If photons are generated on the surface with a frequency v=1 cycle per second (Hz), then if we sat a clock right next to the source of the photons, we'd see the second hand of the clock tick once for each cycle of the photon. The local correspondence between one tick per photon cycle is a local fact that cannot depend on reference frame.

    But now we look at the photons from a higher altitude, and from there they are measured to have an energy hv-mgR and thus a new frequency v-mgR/h = 1-mgR/h, a slower frequency. The clock next to the source of the photons must be in synchronization with the photons! Thus looking down at the clock from a height R, we see the clock ticking once per 1/(1-mgR/h) seconds, which is longer than one second. So for a clock on the surface, then you will see one tick per second if you view it from the surface, but if you view the clock on the surface from a height R above it, you need to wait a slightly longer time for each tick. This is time dilation.
    Last edited: Aug 7, 2013
  20. Aug 7, 2013 #19


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    Roger let me be a little more quantitative. Consider a rocket ship of length ##h## at rest in the downwards uniform gravitational field ##g## of the Earth, an observer ##O## at the very top of the ship, and an observer ##O'## at the very bottom. Say ##O## has an ideal clock next to him and he uses it to send regularly time pulses of light down to ##O'##, with the regular pulses having an interval ##\Delta t_{O}## as measured by ##O##s clock. We want to know at what intervals ##\Delta t_{O'}## ##O'## receives these pulses as measured by an ideal clock next to him.

    In the spirit of the above, consider now the situation in which the same ship is instead accelerating upwards in free space with magnitude ##g##. Now intuitively we can see that the light pulses sent by ##O## will reach ##O'## faster and faster because the ship is accelerating upwards so ##O'## will be catching up faster and faster with the light signals being sent down from the top of the ship by ##O##. Now, following the exposition in Hartle "Gravity: An Introduction to Einstein's General Relativity", assume for simplicity that the second order terms ##(V/c)^2## and ##(gh/c^2)^2## are negligible so that we can essentially use Newtonian mechanics. Then, choose an inertial frame that coincides with the bottom of the ship at ##t = 0## (call this the origin ##z = 0##) and in which the rocket accelerates upwards along the ##z##-axis of the inertial frame. The position of ##O'## will be given by ##z_{O'} = \frac{1}{2}gt^{2}## and the position of ##O## will be given by ##z_{O} = h + gt^{2}##.

    Now imagine that at ##t = 0##, ##O## sends down a light pulse to ##O'## who receives it at a time ##t_1##. Then ##O## sends a second signal after the aforementioned interval ##\Delta t_{O}## and ##O'## receives it at a time ##t_1 + \Delta t_{O'}## because ##\Delta t_{O'}## represents the time between the first reception and the second reception as defined above hence the total time elapsed from the emission of the first pulse to the reception of the second is just the time ##t_1## from the emission to the first reception plus the time ##\Delta t_{O'}## from the first reception to the second. Now the first pulse travels a distance ##z_O (0) - z_{O'} (t_1) = ct_1## and the second pulse travels a distance ##z_O (\Delta t_O) - z_{O'}(t_1 + \Delta t_{O'}) = c(t_1 + \Delta t_{O'} - \Delta t_{O})## because ##\Delta t_{O} - t_1## gives us the time between the first reception and the second emission so ##\Delta t_{O'} - (\Delta t_{O} - t_1)## gives us the time between the second emission and second reception.

    Plugging in our expressions for ##z_O## and ##z_{O'}## and discarding the negligible second order terms, we find that ##h - \frac{1}{2}gt_1^2 = ct^1, h - \frac{1}{2}gt_1^2 - gt_1 \Delta t_{O'} = t_1 + \Delta t_{O'} - \Delta t_{O}##. Solving this system of equations, we get ##\Delta t_{O'} = \Delta t_{O}(1 - \frac{gh}{c^{2}})##. Hence the observer at the bottom of the accelerating ship receives pulses from the observer at the top of the accelerating ship between intervals that are a fractional rate of the intervals between emissions by the observer at the top. Now the equivalence principle says that the observers inside the ship can equally well consider the ship as being at rest in the uniform gravitational field of the Earth in which case we interpret this as gravitational time dilation between different heights in the gravitational potential of the Earth's uniform gravitational field.

    I hope this helps!
  21. Aug 7, 2013 #20
    I think I've never felt so honored in my whole life!

    Here I am, knowing extremely little about physics.

    And yet, you put so much energy and effort into helping me understand.

    It is amazing.

    Thank you!

    This first read did however leave several questions.

    But the equations are fairly simple so if you give me a week I might be able to fully understand them.

    Strike might, I will fully understand them. I am sure about it.

    Otherwise, you will see me here asking more stupid questions :)

    Take care!

    Best regards, Roger
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