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I How is SR applied to circular motion?

  1. Aug 9, 2017 #1
    Hi all,
    I have a problem to fully understand how we can apply Special Relativity to a system where one observer is still in the center, and other one is moving in a circle around. For example, like a satellite orbiting Earth. In case of GPS, the clocks carried by satellite are running slower than clocks located on the ground - as a result of time dilation predicted by Special relativity due to their relative motion. (I know there is opposite and even more significant influence due to General relativity, but let's ignore this one).

    Now to my question... My understanding is that SR is correct only for observes within inertial frames of reference, i.e. non-accelerating frames of reference. However, in case of satellite orbiting Earth, isn't there a permanent centripetal acceleration produced by gravitational force of the Earth (causing the curved/orbital path of the satellite)? Then how can we use SR to predict time dilation as described above?

    I have two ideas in my mind, how can that be, but not sure whether any of them is correct to answer my question:

    1) The orbital path is so large, so at any given infinitesimal time interval it can be considered as straight path. And consequently the relative motion can be seen as straight uniform motion.

    2) Even there is the centripetal force acting on the satellite, it's frame of reference can be considered as inertial - it is in a free fall, and there won't be any net force acting on it.

    Thanks in advance for your help!
     
  2. jcsd
  3. Aug 9, 2017 #2

    Ibix

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    Special relativity is correct in flat spacetime - anywhere gravity is not important. It can perfectly well handle accelerating objects. It wouldn't be much of a theory of motion if it couldn't.

    Einstein did originally construct special relativity in inertial reference frames, but it's perfectly possible to work in non-inertial frames. Just as it is in Newtonian physics, it's usually much easier to work in inertial frames but sometimes it's worth the pain. It's just a change of coordinates, after all.
     
  4. Aug 9, 2017 #3

    A.T.

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    That is a bad example, if you are interested in circular motion within Special Relativity, because it involves gravity and thus requires General Relativity. You cannot combine Special Relativity with Newtonian Gravity consistently.
     
  5. Aug 9, 2017 #4

    vanhees71

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    No, circular motion in SR has nothing to do with gravity, and you can formulate special relativity also in rotating reference frames (although only for part of the spacetime, i.e., in the sense of a local reference frame). In Newtonian mechanics it's analogous: The inertial reference frames are special, and usually the theory takes the most simple form when described in coordinates relative to an inertial frame, but you can also describe Newtonian physics in rotating frames, which you sometimes do to, e.g., treat the Foucault pendulum.

    It's of course true, if you use arbitrary generalized spacetime coordinates in SR, you are already pretty close in the mathematics to general relativity.

    You find examples on the following problem set for our recent GR lecture (it's even in English :-)):

    http://th.physik.uni-frankfurt.de/~hees/art-ws16/sheet02.pdf

    The solutions are here:

    http://th.physik.uni-frankfurt.de/~hees/art-ws16/lsg02.pdf
     
  6. Aug 9, 2017 #5
    Thanks Ibix, could you pls provide me an example of such change of coordinates, so I can better understand what do you mean.
     
  7. Aug 9, 2017 #6

    Ibix

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    Rindler coordinates describe the frame of reference of an observer eternally accelerating at a constant proper acceleration. Dolby and Gull wrote a paper on radar coordinates that is a more flexible system: https://arxiv.org/abs/gr-qc/0104077
     
  8. Aug 9, 2017 #7
    OK, what about this example: LHC collider, with a particle accelerated to a constant speed close to c (bearing it's virtual clocks). And a scientist sitting in the middle of LHC with it's own clocks. Did we get rid of gravity?
     
  9. Aug 9, 2017 #8
    I will need some time to go through it, but thanks :)
     
  10. Aug 9, 2017 #9

    A.T.

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    Yes, here you can use SR to predict the time dilation of the particle clock, based on its speed in the inertial frame.

    If you chose a rotating frame where the center and particle are both at rest, then you also introduce a centrifugal potential (similar to gravity). So in that frame the time dilation is explained by different positions in that potential.
     
  11. Aug 9, 2017 #10

    Thank you, I will look at the examples
     
  12. Aug 9, 2017 #11

    PeterDonis

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    No, because in relativity, "acceleration" doesn't mean the same thing it does in Newtonian mechanics. In relativity, "acceleration" is best viewed as proper acceleration, i.e., acceleration actually felt by a body (and measured with an accelerometer). A body moving solely under gravity, like a satellite orbiting the Earth, is in free fall, feeling zero acceleration. So, as others have pointed out, this is not a good case to use if you want to understand how SR treats circular motion; you need to have the circular motion produced by some non-gravitational force, like a centrifuge, or a ball being swung in a circle using a rope. (To put this another way, in relativity, gravity is not a force, and can't be treated the same way as non-gravitational forces, as is done in Newtonian mechanics.)
     
  13. Aug 10, 2017 #12
    Actually this is very interesting (to me), that we can find a frame in which both observers are in rest (no relative motion between them). And still their clocks will run at different rates. It somehow resembles to me the time dilation predicted by GR due to different position in gravitational field.
     
  14. Aug 10, 2017 #13

    PeterDonis

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    But this "frame" is not an inertial frame, so you can't reason about it as if it were.
     
  15. Aug 10, 2017 #14
    Actually this is what I meant by my idea Nr. 2 in the original post. Thanks for the explanation.


    Thanks for clarification, it make sense to me now
     
  16. Aug 11, 2017 #15
    Hi. SR gives you more topics of interest, e.g.
    - New ##\pi## is greater than 3.141592....
    - Clocks still in the coordinate farther from the center, slower they tick.
    - Center observes that right going light and left going light at distance have different speed.
    Best.
     
  17. Aug 11, 2017 #16

    pervect

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    When looked at in a specific way, that's reasonably close to being true. Some formulations of SR require one to do the analysis in an inertial frame of reference, just as some formulations of Newtonian mechanics are formulated to require that the problem be analyzed in an inertial frame of reference.

    From these formulations in inertial frames of reference, one can find more general formulations of SR that aren't so restricted. The same comment can be said of Newtonian physics.

    Not really. You can certainly analyze an object forced to follow a circular orbit by a rocket, or a strong string in an inertial frame of reference, and when you do said analysis, you get the prediction of time dilation.

    A point that you seem to be partially aware of is that this analysis won't apply to a satellite orbiting the Earth due to gravity, because gravity is not a force. To sucessfully incorporate gravity into special relativity requires general relativity. In order to proceed , we need to focus on the simplest problem that you can potentially analyze, and guide you to an analysis of this simplest possible problem. If you insist on running before you can walk, you won't get anywhere :(.

    Now, I gather from your post that you don't understand what equations you need to analyze the simpler problem of a body forced to orbit by a string or rocket, so I'll present the equation you need to carry out this analysis. Here it is:

    $$d\tau^2 = dt^2 - dx^2 - dy^2 - dz^2$$

    Here (t,x,y, and z) are coordinates in an inertial frame of reference, and ##\tau## is the proper time. The meaning of the term "proper time" can be understood to be just what a clock reads.

    That's what you actually need to understand to work the problem. You seem to be going off in some directions in an attempt to solve the problem, but the directions you are heading in are probably not going to work. I can point you in a direction that will work, but that will only help if you actually follow where I'm pointing, if you, attracted by your own ideas, insist on following them, it'll be entirely up to your own efforts whether or not you succeed. From the fact that you are posting, you presumably want help, the help will come in the form of new ideas that may not occur to you naturally as to how to approach the problem.

    I suppose you also have to understand that dx is a differential operator, which is part of calculus. I tend to take that for granted. I really don't know how to explain this without calculus, alas.

    There are other problems that you probably want to understand, but they're more complicated than the above. One other problems that come to mind are the issues of how one goes about constructing an "accelerated frame of reference" in special relativity, and given such an accelerating frame, how does one go about modifying the solution process above?

    The first part, how one defines an "accelerated frame of reference" is rather difficult, but I can address the second part somewhat. When you specify an accelerated frame of reference one of the things that comes out of the specification process is a metric.

    Given a metric tensor, ##g_{ij}## we can re-write the previous solution as something like what appears below.

    $$d\tau^2 = g_{00} dt^2 - g_{11} dx^2 - g_{22} dy^2 - g_{33} dz^2$$

    Here all the ##g_{ii}## terms are the new addition, the metric tensor.

    This isn't quite right though. At this point it likely becomes convenient to replace the x,y,z coordinates of the inertial frame with something more general. Cylindrical coordinates ##r, \theta, z## fit this particular problem very well, but we might as well make the leap to completely and totally general coordinates. We'll call these coordinates ##x^i##, and there will be four of them, one coordinate that replaces "t", and three coordinates that replace "x,y,z". We'll number these coordinates, starting at 0, so our general coordinates will be ##x^0, x^1, x^2, x^3##. Given these general coordinates, we can then write:

    $$d\tau^2 = \sum_{i,j} \sum_{i,j} g_{ij} \, dx^i \, dx^j$$

    and ##g_{ij}## is an array of 16 numbers - basically a matrix , though we call it a tensor. The matrix/tensor is symmetric, by and by the way, so that ##g_{ij}## turns out to be equal to ##g_{ji}##.
     
  18. Aug 15, 2017 at 1:01 AM #17
    thank you pervect for your extensive comment.

    Actually that's the initial bug in my head and reason for my original post. According to few sources (e.g. this one, sorry for the wikipedia), the overall time dilation in case of GPS satellite, consist of two components: "due to GR" and "due to SR". The component "due to SR" is calculated via Lorentz transformation similarly as it would be calculated for an observers in a uniform relative motion.
    Thinking about that, maybe what is really confusing me, is that we can split the problem and calculate the two components separately, whereas the gravity plays role only in the component "due to GR".

    I am definitely open to a new (verified) ideas, and I am not insisting on any of my previous ideas. That's why I am here on this forum - I'd like to understand how the things are really working :)
    I know the calculus. And also I understand what is the proper time, and that it is an invariant in SR. If you wish to know, my current understanding of SR is within the scope of the Feynman Lectures, vol. I.


    It means that for constructing an "accelerated frame of reference" (and getting the metric tensor) the GR is required, correct? I haven't studied the general relativity and metric tensors yet, but it is on my TODO list. Some day, I will start this long journey :)
     
  19. Aug 15, 2017 at 1:42 AM #18

    Ibix

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    That's not quite right, although the Wikipedia article is certainly open to that interpretation.

    The time dilation between a ground station and a GPS satellite is completely described by general relativity. However, you can write it as a product of the time dilation between the ground station and a clock in a rocket hovering (not orbitting!) at the satellite altitude, and of the time dilation between that clock and a satellite passing its window. The former is purely gravitational. The latter is purely kinematic, and is exactly as predicted by SR.

    So (in this case) you can decompose the time dilation into a component due to gravitational potential and a component due to kinematics. That doesn't really mean that "there's a component due to SR and a component due to GR". That makes them sound like two different theories, which they're not. It just means that once you've established the time dilation factor between the ground and the rocket, you only need the (much simpler) maths of special relativity to get the time dilation between the ground and the satellite.
     
  20. Aug 15, 2017 at 3:02 AM #19

    vanhees71

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    GR is so much simpler, if you just take it seriously and don't try to split kinematical properties like time dilation into an SR and a GR part. GR, as the name "general" suggests, includes all effects in an ingenious way, making it to the most esthetical fundamental theory of physics ever.
     
  21. Aug 15, 2017 at 3:48 AM #20
    The conclusion seems to be clear: only GR can provide fully satisfactory answers (not only for this topic). I should start to study it - the sooner the better :)

    Thank you all for your comments and feedback! I appreciate it. It gave me at least some clue how should I look at problems like this.
     
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