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Free fall and relativity

  1. Nov 12, 2013 #1

    Buckethead

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    Two very simple questions, but ironically I've listened to a number of podcasts and articles that leave me confused as they don't always sound consistent so...

    1. Is a satellite that orbits a planet (i.e. a satellite in free fall) feel the GR effects of the gravity of the planet? In other words, disregarding SR for the moment, is a clock on a satellite ticking at the same rate as a clock in the middle of space. I'm guessing they will because neither feels gravity.

    2. If you are standing on the North Pole and a satellite is orbiting around the equator, and disregarding any gravitational effects and only addressing SR effects, does a clock on the North Pole tick at the same rate as a clock on the satellite. I'm guessing it will since the distance between the satellite and observer never changes.

    Thanks, hope to bury these questions once and for all.
     
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  3. Nov 12, 2013 #2

    PAllen

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    Both your suppositions are incorrect.

    1) It is correct that a satellite in orbit is in free fall, and is following an inertial trajectory in GR. However, it is not correct in GR that all inertial trajectories mutually observe the same time rate. Further, if two events are connected by two different inertial trajectories, they will end up aging differently, producing a pure inertial version of twin differential aging. For example, suppose a clock is passing an orbiting satellite moving radially away from the planet at a speed such that it will fall back and cross the satellite's orbit again after one orbit. Both trajectories, separating and meeting again, are purely inertial. Yet, the radial free fall trajectory will be the one that shows greater elapsed time. You might be thinking of separating this into SR vs. GR effects, but GR is a unified theory including SR. However, to make some attempt to do this separation for your example, you don't get the result you expect because the orbiting satellite is at a different gravitational potential than a far away clock. Where GR can be approximated by a potential, time dilation can be separated into a relative speed component (call it SR) and component due to potential difference.

    2) This is really confused. If you subtract gravity, the satellite around the equator is not in free fall, and is, instead, analogous to circular trajectory with constant centripetal acceleration. This would be analogous to particles in an accelerator ring, which would clearly be time dilated relative to an observer stationary with respect to the ring's center.
     
  4. Nov 12, 2013 #3

    Buckethead

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    How does one determine that the radial unit shows the greater elapsed time? (excellent example BTW) Is it because the radial unit will have gone a lesser distance between the two meetups?

    My mistake. I was trying to think of this in the same way that a diagonal vector can be broken down into an x and a y component. So I assumed that time dilation could be broken down the same way into a "SR axis" and a "GR axis" in a way. As you are perhaps suggesting in the below quote.

    OK so a stationary object and an orbiting object at the same elevation show the same GR effects (same time dilation degree) and in addition the fact that one is moving relative to the stationary one adds even more time dilation. Correct?




    I was not really trying to subtract gravity, only separate it into a gravity component and a relative velocity component, so are you saying that this is not allowed as it is meaningless to think of an object in both a free fall and a circular trajectory at the same time unless a gravity component is preserved (or in the case of the accelerator a mag field)? I think I'm seeing a deeper meaning here so let me ask this question:

    In the case of the accelerator, there is no gravity so there has to be a force (mag field) to keep the particle in line. If I were to sit at the center and spin around so that the particle seemed to stand still, would I have to say the particle is still moving when determining its time dilation because of the required force to keep the status quo and not because of any relative speed between me and the particle? And does this extend to the example of the orbiting satellite as well in that I am not allowed to say the satellite is stationary just because I spin around on the North Pole due to the gravity of the planet keeping the satellite from flying off.

    What I'm getting at with these questions is that I almost feel like there is always some kind of relative speed involved but in the case of GR this relative speed translates into the opposing force (gravity or the mag field in the accelerator) rather than an actual relative speed and because of this it doesn't make sense to think of an orbiting satellite in free fall unless you also include the required force. Am I on the right track?

    Looks like I won't be burying any questions today. :cry:
     
  5. Nov 12, 2013 #4

    PAllen

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    Well the real answer is to integrate the proper time for the paths using the Schwarzschild metric. You can justify by other arguments, but none covers the general case in GR. For example, note that the radial trajectory spends more time at higher gravitational potential. Posit that since both motions are inertial, this dominates, so the radial inertial trajectory elapses more time due to the higher potential. You can analyze many near spherically symmetric situations this way, and also weak gravity cases that are less symmetric, but it is not really GR - it is an approximation to GR that works in many simple cases.
    This is only possible in very specialized cases. To the extent you do this, you are not really grappling with GR.
    Yes, for a spherically symmetric, nearly static field (that thus can be treated with a potential).
    Yes, I think it is meaningless to talk about free fall circular trajectory without gravity.
    Rotating frames are complex in relativity. However, yes, a clock spinning in the center at the same angular speed as a particle moving at some signficant radius will see the circular moving particle time dilated. Further, the circular moving particle will see the spinning central clock going fast. This is not a case mutual time dilation. This is all purely SR - no gravity inolved.
    Actually, for a polar observer and an equatorial orbiting satellite you would have to balance two effects - the satellite's higher potential would make it go fast per the polar observer, but its speed would make it go slower. The altitude of the satellite would determine which wins. If in a low orbit, it would be dilated relative to the polar clock, if in a high orbit would run faster.
    For simple, symmetric, weak gravity situations you can pursue this direction. But the key to doing it right is to forget force and focus on potential. For a distant observer, a clock on a massive shell of matter is dilated; a clock inside the shell, experience no force, is even more dilated.
     
    Last edited: Nov 12, 2013
  6. Nov 12, 2013 #5

    Buckethead

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    Why is this pure SR? I would have thought it would be pure GR for the reason that 1) There is no relative speed between the particle and the axis (the distances between the two remain constant), 2) There is a constant accelerating force on the particle meaning that its clock will move slower than the axis clock due to the acceleration, 3) There is no mutual time dilation as there is in two inertial ships passing each other (just as there is no mutual time dilation between a clock in and out of a gravity field), and 4) Because there is a preferred frame of reference as in when the axis clock is spinning at the same rate as the particle, the axis knows it by its centriputal force.

    I understand the time dilation due to gravity, but the time dilation due to the orbit still confuses me in that to the observer, whether or not the satellite is actually moving depends only on whether the observer is spinning around in his north pole chair. Since distance never changes between the observer and the satellite and since (if the observer is spinning with the satellite) there isn't even a relative movement of any kind between the two, it seems that gravity is the only thing that has any say in the matter, not SR. Now I understand that there is an underlying asymmetry in that the observer obviously knows he is spinning due to the forces that he feels and also because he knows that the satellite would fall to earth if it were not orbiting (although he could equally just assume that the earth had no mass I guess).

    Part of the difficulty I'm having with some of this is the nagging question of how an object knows when it is spinning or to put it another way, how does it know it's suppose to feel centripetal forces when it rotates relative to the surrounding space when it can't possibly know if it is spinning or if the space around it is spinning and it is stationary. But perhaps that is a question for another thread or a question that can't be answered.

    I'd like to just for a moment go back to the question of why there is a gravity potential for an orbiting satellite when according to Einstein's philosophy there is no gravity in the sense of a force but rather just a mass and along with that mass comes a curvature in spacetime and it is this curvature alone that determines the actions on orbiting objects. So if there is no force and only curvature, then doesn't this imply that a orbiting object and a linearly traveling inertial object should be at the same gravitational potential or is it that the degree of curvature in space is what dictates a "gravitational" potential and not really the massive object (other than of course the massive object is causing the curvature)?
     
  7. Nov 12, 2013 #6

    PAllen

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    If gravity can be ignored, it is pure SR. The central clock and the clock revolving about it could be nearly massless in empty space. There is no significant gravity, so it is purely SR. SR distinguishes inertial frames from accelerating or rotating frames. Acceleration and rotation are absolute, not relative - they can be detected locally, inside a black box. Acceleration and rotation were analyzed using SR before GR existed.

    The easy way to analyze this with pure SR is to pick any inertial frame for the analysis.
    An object always 'knows' if it is spinning. Close your eyes and ask someone to spin you on a chair without warning. You have no trouble knowing when you are spinning. This is an absolute, not relative, effect in both SR and GR.

    However, the spin of a north pole clock is wholly insignificant - one rotation per day. Even if it were more rapid spin, for a sufficiently small, ideal, clock it makes no difference in time dilation between such a clock and a non-spinning inertial clock. What is important is that the clock knows it is spinning, and can measure this even inside a black box. Thus, it knows that something stationary at some distance is moving relative to an inertial frame. For emphasis: there is no principle of relativity in SR except between inertial frames.

    Anyway, I prefer to integrate metrics for such problems. You wanted to try to factor effects. If you want to do that, you have to start with the notion of static world lines in the SC geometry. These define potential as a function of position. These world lines are not inertial - they experience proper acceleration. Then, any object moving relative to static world lines experiences relative motion time dilation compared to an adjacent static observer. For some other static observer, there is then a potential difference effect and a motion effect. The north pole observer, whether you worry about spin or not, is effectively a static observer at a fixed potential.
    Gravitational potential is the energy to get a static object from its current position to 'infinity'. The fact that potential difference for static observers corresponds to relative time dilation between them is derived special case of GR. You don't have to ever use this concept, and it isn't useful in the general case. The general method is to have the metric represented in some coordinates, and compute the proper time along world lines by integrating. This is how you derive the special features of static or stationary metrics, or of the weak field approximation.
     
  8. Nov 12, 2013 #7

    WannabeNewton

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    It can certainly be answered: rotation is absolute not relative. I don't need to pay any attention to the distant stars to tell if I'm rotating or not. There are a myriad of ways I can locally measure rotation. For example I can carry a pronged sphere with beads inserted through the prongs and see if the beads are flung outwards by inertial forces, or see if local gyroscopes rotate relative to me since gyroscopes define what it means to be locally non-rotating.
     
  9. Nov 13, 2013 #8

    Buckethead

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    So if you are in an inertial frame in space (a space station) and you shoot off a rocket with a constant acceleration you can use SR alone to calculate the time dilation on the rocket during its acceleration? And the answer would be the same as if you had a stationary rocket sitting on a planet that had the same acceleration and used GR to determine its time dilation?

    Yes, I recognize that it is a simple matter to know when something is rotating. What I was really asking is, if rotation or acceleration is absolute, what is it absolute to? Einstein said no to the stars (Mach's principle) and everyone says no to absolute space, so what's left? You can't be absolute relative to nothing at all. (although if it was it would be no stranger than quantum mechanics I suppose)

    I'm sorry to say this is a bit over my head but I think I get the general idea.
     
  10. Nov 13, 2013 #9

    WannabeNewton

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    As noted, rotation of frames is with respect to locally non-rotating frames (i.e. Fermi transported frames). This is analogous to how acceleration of frames is with respect to locally inertial frames (i.e. freely falling frames).

    So, in other words, the local standard of freely falling of a Lorentz frame ##\{e_{\alpha}\}##, where ##e_0 = u## is the 4-velocity, is ##\nabla_{e_0}e_0 = 0## and the local standard of non-rotation is ##\nabla_{e_0}e_i = g(e_i,a)e_0## where ##a = \nabla_{e_0}e_0##.

    Therefore if we have an arbitrary Lorentz frame, we can say it is locally rotating if it rotates relative to a coincident Fermi transported frame and accelerating if it does so relative to a coincident freely falling frame. Rotation and acceleration as defined in this way are absolute in the sense that the above equations are covariant.
     
    Last edited: Nov 13, 2013
  11. Nov 13, 2013 #10

    Buckethead

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    I'm sorry to say I don't understand physics math, but I suspect the answer lies in my understanding of what you mean by a Fermi Transported frame (I don't know what that is either). For the case of 2 simple coincident frames that are rotating relative to each other it is of course impossible to know which is rotating and which is free falling in determining which would feel a rotation and which would not. Apparently a Fermi Transported frame is a special case of a frame that is distinguishable from a simple rotating (or non rotating ) frame using some other piece of information other than relative rotations?
     
  12. Nov 13, 2013 #11

    WannabeNewton

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    A Fermi transported frame is the same thing as a locally non-rotating frame. When we say a given frame is rotating, we mean that it is rotating relative to a coincident non-rotating frame. Physically, a locally non-rotating frame is one in which the spatial axes are gyroscopes.
     
  13. Nov 13, 2013 #12

    Buckethead

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    So a Fermi t. frame has no special properties other than its definition as a non rotating frame. This in my mind makes it no different than any other coincident frame so my original question still stands. How does one know when 2 frames are rotating relative to one another which one should be endowed with acceleration and which one should not. The obvious answer of course is that the one that feels acceleration is the one that should be endowed with acceleration but as you can see that's a circular answer.
     
  14. Nov 13, 2013 #13

    WannabeNewton

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    Again, local non-rotation can be measured absolutely and rotation can be measured relative to locally non-rotating frames. The same goes for free fall vs. acceleration. The equations I wrote in post #9 do exactly this.
     
  15. Nov 13, 2013 #14

    Dale

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    No that isn't a circular answer, it is an experimental answer. You have a device for measuring proper acceleration, an accelerometer (or more generally an inertial guidance unit). You can use that to experimentally test whether or not a given coordinate system is inertial.

    Coordinate systems have no inherent physical meaning. They are only labels for events in spacetime. The physical meaning is contained in the metric. To calculate the proper time along some worldline you simply integrate the metric. Because the metric contains the physics, that procedure works regardless of whether you are using an inertial frame in flat spacetime, some non-inertial frame in flat spacetime, or an arbitrary frame in curved spacetime.

    The components of the metric are different in a Fermi frame than in other frames. That is why they are physically identifiable and meaningful. Due to the different metrics, there is no confusion about symmetry between rotating and non-rotating frames.
     
  16. Nov 13, 2013 #15

    PAllen

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    Yes, you can use pure SR to get all possible observations for an accelerating rocket far from a gravitating body. Then, by the principle of equivalence you can conclude that a static rocket in gravitational field should have the same observations to the extent tidal gravity can be ignored. The principle of equivalence is thus always a local and approximate statement (tidal gravity is never exactly zero). It allows one to derive approximate predictions of GR using SR scenarios, and can be used to motivate equations of GR. It never requires that GR be used for a situation without significant gravity.
     
    Last edited: Nov 13, 2013
  17. Nov 13, 2013 #16

    Buckethead

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    I'm sorry but after a very long walk in the woods today thinking about this I have to address it again as I find it contradictory or maybe I just don't understand your answer. Let's take the example of a man jumping off a tall building and half way down passes a woman watering her window flower pots. At the moment the man passes the woman, their wrist watches are ticking at different rates even though they are at the same elevation. The woman is under acceleration so her clock will be ticking more slowly (she is the particle in the accelerator) and the man's watch is going faster because he is not being subject to acceleration. So the answer to my above question has to be no. I still recognize though that this doesn't mean the satellite clock is ticking at the same rate as a distant free floating object.) I guess I'm just still confused why an object in free fall can have a different clock rate than one in orbit when from inside the objects in question there is no experiment that can be done to differentiate between a free floating object in space and a free falling object near a planet (according to Einstein). If there is a gravity potential difference then this can be used to create an experiment to differentiate between the actual state of these two objects simply by doing a communication and comparing clock rates.
     
  18. Nov 13, 2013 #17

    Buckethead

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    I recognize that an accelerometer can be used to measure acceleration, but my question relates not to what has acceleration, but why one has the acceleration and not the other, but I'll explain that more below.

    My brain lacks the ability to understand metrics at this point but the fact that a Fermi frame has a uniquely identifiable metric leads me to believe this is a dominant feature of a frame that can elevate it to the status of a preferred frame of reference. I don't exactly have a problem with that, but I thought preferred frames of reference (with regard to rotating bodies) went out with Newton. I really am fond of the idea of everything being relative even with regard to rotation and acceleration and although this may seem like a pipe dream, I can actually give an example where it is workable at least in a thought experiment. I had this thought several months ago and I almost let it go as being nonsense, but I love that it abolishes preferred frames of references.

    Relying heavily on the equivalency principle imagine abolishing the notion of inertia and instead replace all inertial references with the equivalent gravitational potential. There are two scenarios here, the first is a ship dead in space turning on its engines and accelerating and the second is a rotating Earth experiencing rotational forces. Replacing inertia with gravity it could be imagined that when the ship attempts to accelerate, the act of accelerating causes a curvature in space directly behind the ship acting as a gravitational field to resist the ships attemps. This is the inertia that is felt and why the ship requires fuel to proceed. When the ship cuts its engines and coasts, the space curvature dissolves. Remember this is just a thought experiment so cut me some slack.

    Now with regard to the rotating earth (or a rotating skater for that matter). Instead of the rotational forces being due to centripetal force, and because we are replacing inertia with gravity, we can think instead that what is occurring is similar to what is happening in the ship namely when the Earth accelerates in its rotation, at the points of intersection between the rotating Earth and the non rotating space around the Earth a gravitational field is created due to the acceleration. This field will take the shape of a ring around the equator and will resist the acceleration. In other words it will be a gravitational force pulling the equator outward.

    The beauty of this abstract thought is that in both the case of the accelerating ship and the case of the rotating Earth, there need not be any preferred frame of reference. It doesn't matter if the ship accelerates or the space behind it is accelerating instead as the gravitational field being generated will simply pull the space and the ship together and the effect (on the astronauts for example) will be the same even with this perfect symmetry. With regard to the Earth it doesn't matter if you think of the Earth as rotating or the Earth being stationary and the spaced around it rotating as in either case the gravitational ring will ensure that the equator will feel an outward acceleration resulting in a bulge there. Centripetal force and absolute rotation are easily dismissed. Again, not proposing anything here, just kind of a mind game to try an eliiminate a preferred frame of reference when addressing any type of acceleration.
     
  19. Nov 13, 2013 #18

    PAllen

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    This is absurd on its face. There is no equivalence between inertial frames and gravitational potential. What the equivalence principle says about inertial frames is the local physics is the same for all of them. Comparing clocks between different inertial frames (either separated, or adjacent in relative motion) is not local physics, and the equivalence principle does not apply.
     
  20. Nov 13, 2013 #19

    Buckethead

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    Comparing an accelerating rocket with a rocket that has its engines on but is not moving because it is being held by a gravitational field is very much the equivalency principle.
     
  21. Nov 13, 2013 #20

    TumblingDice

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    @Buckethead
    You keep adding more variables to a situation that, if anything, could benefit from simplifying to better understand.

    One problem appears to be that you feel, since gravity offsets the orbital acceleration, there should be no time dilation due to gravity. Let's look at this from an inertial reference frame - you standing outside. Imagine you can see a geostationary satellite above you. Same as if it was on a very, VERY tall mountain, right? Clock on the sat will tick faster than yours because of the component of the Earth's gravitational field, which is weaker.

    Even though the sat appears to be stationary relative to our reference frame, we also know it has to travel at high velocity to maintain its geostationary position, and that is going to add another component, slowing down the tick of the sat clock. All of this has is fact, and a part of the clock adjustments on GPS satellite clocks. They tick 45 microseconds faster due to lesser gravitation, and 7 microseconds slower due to relative motion, per day, so are adjusted to tick 38 microseconds slower to keep synchronized w/Earth time. I think I got that right... Here's a link to check:
    http://www.astronomy.ohio-state.edu/~pogge/Ast162/Unit5/gps.html

    Also, as was mentioned previously, there is 'absolute' acceleration, and rotational acceleration is one type of this, and all can be measured with an accelerometer, even at times when you may feel weightless. The equivalence principle is only in effect if/when you cannot tell the difference with measurements. An accelerometer - even a simple gyroscope - can tell the difference in situations you've posed.
     
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