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How can angular momentum and relativity coexist?

  1. Apr 10, 2013 #1
    I fully expect to be taken out to the intellectual wood shed by any number of candidates (GhWells, DaleSpam come to mind), but I do not understand how you can have relative motion and angular momentum?

    Angular momentum seems to suggest there is some absolute type of motion, i.e. rotation of planets, whereas Einstein talks about the general relativity of all forms of motion.

    If all forms of motion are relative, how do we know we're really rotating around the sun, or that angular momentum is real?
     
  2. jcsd
  3. Apr 10, 2013 #2

    ghwellsjr

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    Proper Accelerations are absolute of which rotations are an example.
     
  4. Apr 10, 2013 #3

    PeterDonis

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    Angular momentum *is* relative; you have to pick a "center" about which to measure it, and different choices of "center" will give different values for angular momentum. The same goes for "rotation".

    What is *not* relative is the proper acceleration of a particular observer following a particular worldline. For example, imagine two space stations far out in deep space. Astronauts at rest relative to the first station are weightless, and measure zero proper acceleration on themselves. Astronauts at rest relative to the second station feel weight: they can stand up within the station just as they could on the surface of the Earth, and they measure a nonzero proper acceleration on themselves.

    The usual way of describing the difference between these two stations is to say that the second station is "rotating", while the first is not. But really it's the observable difference between the two that's primary; attributing it to "rotation" is the result of trying to construct a consistent theory of physics that predicts the observable difference correctly.

    This brings up another point of possible confusion. Consider the two space stations above again, and suppose that in the first station, astronauts looking out a particular window always see the same patch of stars apparently motionless in the sky, while astronauts looking out a particular window of the second station see the starfield continually changing, as stars move across the field of view.

    It's tempting to say that these observations confirm that it's the second station that is "rotating" in some absolute sense. But the only thing the observations really say, definitely, is that the second station is rotating relative to the stars, while the first is not. Again, calling the rotation of the second station "absolute" is the result of constructing a physical theory, it's not something that's given directly by observation, the way the observed proper acceleration is.

    In the same way, we say the Earth revolves around the Sun, but really what we directly observe is that the Sun moves against the background of distant stars with a period of a year. We have to use some physical theory to attribute that observation to the Earth revolving around the Sun.

    One other note: strictly speaking, in GR it would indeed be perfectly possible to describe the entire solar system using a frame of reference in which the Earth, not the Sun, was at rest. However, everything will look much more complicated in such a frame than it does in a frame in which the Sun is at rest. GR says you can use any coordinates you like, but it doesn't say that all coordinates will make things look equally simple.
     
  5. Apr 10, 2013 #4

    Bill_K

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    Wow, I can't believe you said that, Peter. :frown: Rotation is absolute in general relativity, just as acceleration is absolute. And it's something that can be locally determined without looking at the stars. The prototype instrument used to determine one's absolute rotation is a "photon gyroscope" which examines the interference between two counterrotating beams of light.
     
  6. Apr 10, 2013 #5
    Please explain why you think that is the case? :confused:

    Unless we are talking about a rotating point any extended object will fee stress under rotation, this can obviously be measured. The most simple observation I can think of is an equatorial bulge for rotating objects. You do not need stars for that.
     
  7. Apr 10, 2013 #6

    PeterDonis

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    What kind of "rotation" does this measure? It seems to me that it would be measuring something like the rotation of the spatial coordinate axes along the device's worldline, i.e., the failure of the spatial axes to be Fermi-Walker transported? Or is it measuring something more like the Sagnac effect?

    [Edit: Unfortunately, Motorola appears to have a 4G phone out with "photon" in its name that has a gyroscope in it (probably just a fancy name for the accelerometer), so Google isn't coming up with anything useful. :grumpy:]
     
  8. Apr 10, 2013 #7

    Doc Al

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    You and Bill_K are in violent agreement. Did you miss the words "can" and "without"?
     
  9. Apr 10, 2013 #8

    WannabeNewton

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    What is your mathematical definition of absolute rotation for arbitrary spacetimes Bill e.g. akin to ##a^b = u^{a}\nabla_{a} u^{b}##.
     
  10. Apr 10, 2013 #9

    A.T.

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    Rotation and proper acceleration can be determined locally, but can we say for sure that this local effect (inertia) has nothing to do with the rest of the universe? We cannot remove the rest of the universe to test this.
     
  11. Apr 10, 2013 #10
    Oops, I did misread Bill's words.
    Sorry.
     
  12. Apr 10, 2013 #11

    Nugatory

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    The bolded text suggests to me that you two might be talking at cross purposes. PeterDonis is saying that a physical theory is required to interpret the observed physical accelerations as absolute rotation; you're pointing out that GR is such a theory.
     
    Last edited: Apr 10, 2013
  13. Apr 10, 2013 #12

    Bill_K

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    A locally nonrotating frame is defined along the world line of a freely falling observer ("the space station") by parallel-transport of the basis vectors.
    If you don't have a physical theory, I'm not sure what's left. :confused: But I'm saying that it IS given directly by observation, exactly the way the observed proper acceleration is.
     
  14. Apr 10, 2013 #13
    Wow...I got the big boys all stirred up. I guess I finally asked a question that made some kind of sense! Yay!

    Photon gyroscope?
     
  15. Apr 10, 2013 #14
    The right answer to the OP was given already in post #2, I can't see what else is there to discuss.
     
  16. Apr 10, 2013 #15

    WannabeNewton

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    What basis vectors do you speak of here? There are many different basis vectors I can choose for the tangent space at each point of some appropriate subset of space-time. I can choose for example the coordinate vector fields ##(\partial _{\alpha})^{a}## so that I have the coordinate basis for each tangent space.

    More physically, I can choose the basis field ##(e_{\alpha})^{a}## which will assign to the tangent space at each event an orthonormal basis in which ##(e_0)^{a}## is tangent to the worldline of the observer at that event and the ##(e_i)^{a}## represent the alignment of the observer's measuring apparatus at that event; here your definition becomes ##u^{a}\nabla_{a}(e_{\alpha})^{b} = 0##. I can visualize the statement ##u^{a}\nabla_{a}(e_{i})^{b} = 0## as saying the alignment of the measuring apparatus carried by the observer is unchanged along the worldline of the observer, which for me visually does correspond to being locally non-rotating, however I do not see why ##u^{a}\nabla_{a}(e_0)^{b} = 0## would have to be a requirement because, as per the definition of this basis field, ##(e_0)^{b} ## is tangent to the worldline of the observer so that statement would imply the observer is inertial and this of course won't be true for an arbitrary observer so I don't suppose this is the basis field you were thinking of.

    Also, Wald has an ostensibly different and coordinate dependent definition of locally non-rotating observers for the specific case of stationary axisymmetric space-times. He defines them to be the family who, in the coordinates adapted to the killing vector fields, have 4-velocities ##u^{a} = \zeta \nabla^{a}t##.

    EDIT: I just read that you said freely falling observer so disregard my concerns regarding the statement ##u^{a}\nabla_{a}(e_0)^{b} = 0##.

    EDIT 2: When Wald defines the above basis field ##(e_{\alpha})^{a}## and says that the ##(e_i)^{a}## represent the "alignment" of the observer's carried measuring apparatus he never explains what he means by the "alignment" of the observer's carried measuring apparatus. I personally took it to mean we use a distant fixed star as a reference point and define the "alignment" of the observer's measuring apparatus as pointing at that star so that the statement ##u^{a}\nabla_{a}(e_{i})^{b} = 0## would mean that the measuring apparatus always remains pointing at that star along the worldline of the observer. This however goes back full circle to the original problem of this thread.
     
    Last edited: Apr 10, 2013
  17. Apr 10, 2013 #16

    PeterDonis

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    Ok, I think we agree, but we're using terminology differently. I agree that whether or not an observer's spatial basis vectors are being Fermi-Walker transported is a direct local observable. But you can't tell just from that observation where those basis vectors are pointing globally; for example, you can't tell just from that local observation whether or not the observer's basis vectors are rotating relative to distant stars. For example, the observer could be orbiting a spinning body, so that frame dragging is present. The latter type of thing is what I was referring to.
     
  18. Apr 10, 2013 #17

    PeterDonis

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    I assumed he meant spatial basis vectors which are orthonormal and whose directions are determined by gyroscopes. At least, that's basically the physical realization of Fermi-Walker transport.
     
  19. Apr 10, 2013 #18

    WannabeNewton

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    Would that be similar to what I talked about in my post? Although the basis field I defined, based off of Wald, did leave me with a big concern which I added as "EDIT 2".
     
  20. Apr 10, 2013 #19

    PeterDonis

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    No, because of the concern you brought up, which is the same frame dragging issue I mentioned. Orienting your spatial axes locally, using gyroscopes, is not the same as orienting them with respect to a distant reference. Axes oriented by one method will rotate relative to axes oriented by the other method.
     
  21. Apr 10, 2013 #20

    WannabeNewton

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    Yeah I just saw that you mentioned the same concern I had.

    So in this case do we have a basis field that assigns an orthonormal basis for each tangent space, for points on the observer's worldline, in which the spatial basis vectors' directions at each event are defined by three respective gyroscopes? How do we physically interpret / visualize the statement then that these spatial basis vectors are parallel transported along the observer's worldline if he/she is to be locally non-rotating in terms of the three gyroscopes defining the directions of the respective spatial basis vectors? Thanks Peter (is it just me or is every conservation you and I end up having always involve rotation lol).
     
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