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Proper acceleration

  1. May 22, 2012 #1
    I'm all messed up here trying to learn about proper acceleration. As I understand it, an observer with constant acceleration undergoes parabolic motion but one with constant proper acceleration undergoes hyperbolic motion as seen from an inertial frame.Wikipedia says that proper acceleration is the rate of change of proper velocity with respect to coordinate time, but this doesn't agree with the definition of prop acc given in post 14 in this thread, where it depends on the rate of change of rapidity.In Don Koks' book, according to equation 7.6, the proper acc is γ^3 times the acceleration measured in an inertial frame(x dot equals v).Also, I would like to know how to derive the rindler metric
     
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  3. May 22, 2012 #2

    PeterDonis

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    The hyperbolic part is right. I think the parabolic part is right (assuming that by constant acceleration there you mean constant coordinate acceleration, i.e., the second derivative of coordinate position, x, with respect to coordinate time, t), although it's not the way I would normally think about this scenario.

    The Wikipedia page's presentation seems somewhat strange to me, assuming you meant this Wikipedia page:

    http://en.wikipedia.org/wiki/Proper_acceleration

    It's not the normal way of defining proper acceleration that I'm used to, but without taking more time to look at it I can't say for sure whether it is actually inconsistent with this...

    ...which is the (correct) definition that I'm used to seeing.

    This looks OK to me, although Google Books isn't letting me look at the actual book page you linked to, so I can't see the context.

    Try the Wikipedia page here (this one looks OK to me):

    http://en.wikipedia.org/wiki/Rindler_coordinates

    Or Greg Egan's page here:

    http://gregegan.customer.netspace.net.au/SCIENCE/Rindler/RindlerHorizon.html
     
  4. May 22, 2012 #3

    Bill_K

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    vin300, Uniform acceleration is a good example to practice on, but how much you learn from it depends on how you solve it! If you use the rapidity approach you've solved one problem. Better is to use a general approach, and thereby learn how to solve most problems in relativistic kinematics.

    The motion of a relativistic particle can be described in either of two ways: in terms of three-dimensional quantities t, x, v and a, or in terms of four-dimensional ones: τ, x, v and a. Of course dt/dτ ≡ γ = (1 - v2/c2)-1/2.

    The 4-velocity v = dx/dτ is a timelike 4-vector with space and time components v = (γv, γc). Note that v has dimensions of velocity and is constant length: v·v = c2.

    The 4-acceleration a = dv/dτ is a spacelike 4-vector orthogonal to it: v·a = 0. To calculate the components of a you just have to differentiate the components of v. (It helps to do a little algebra on the side first and note that dγ/dt = γ3(v·a)/c2.)

    a = dv/dτ = γ dv/dt = (γ2a + γ dγ/dt v, cγ dγ/dt)
    = (γ2a + γ4(v·a)/c2, γ4(v·a)/c)

    Yes indeed, this is messy! But it's completely general. You can apply this formula to things like circular motion, etc. But to make matters simpler from now on, let's assume the motion is linear. It then reduces to something rather nice:

    a = (γ4 a, γ4 va/c)

    What is the magnitude of a?

    a·a = - γ6 a2, so |a| = γ3 a

    That's the result that you found mentioned in Don Koks' book: the γ3 factor between the 4-acceleration and the 3-acceleration.

    We still have not assumed the acceleration is uniform. If you do that:

    γ3 a = const ≡ A

    you get a differential equation

    dv/dt = A (1 - v2/c2)3/2

    which you can easily integrate to get a hyperbola.
     
  5. May 22, 2012 #4

    stevendaryl

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    I think the Wikipedia article is wrong. Proper acceleration is the derivative of 4-velocity with respect to proper time, not with respect to coordinate time.
     
  6. May 22, 2012 #5
    OK, thank all of you.
    BTW, there's a small error here:
     
  7. May 22, 2012 #6

    Bill_K

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    Thanks, should be: (γ2a + γ4v(v·a)/c2, γ4(v·a)/c)
     
  8. May 23, 2012 #7
    How to prove using the general approach, that the magnitude of proper acceleration is invariant? The coordinate acceleration measured by all observers is the same, and γ is observer dependent, so the magnitude of four acceleration will be different for different observers.
     
  9. May 23, 2012 #8

    PAllen

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    The norm of any vector is invariant (note: norm is defined using the metric, which is Minkowski in SR, not Euclidean). 4-velocity is a vector, proper acceleration (derivative of it by an invariant τ) is thus also vector.
     
  10. May 23, 2012 #9
    And how about the general (not minkowskian) case?
     
  11. May 23, 2012 #10

    stevendaryl

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    The coordinate acceleration is not the same for all observers. If the coordinate acceleration is g in the initial rest frame of a rocket, then it will be g/γ3 in a frame in which the rocket is initially traveling at velocity v.
     
  12. May 23, 2012 #11
    I'm pointing at the less frequently mentioned situation in GR involving non-inertial observers, certainly in this case one cannot affirm so lightly that 4-acceleration magnitude is an invariant quantity, at most one must say that it is undefined. The fact is that as it happens with energy 4-acceleration magnitude in GR is not a conserved quantity (except in static spacetimes), it is also frame-dependent. My question is then how does the 4-vector concept in SR evolve to the GR context? can it still be considered a 4-vector if its magnitude is not a scalar invariant?
     
  13. May 23, 2012 #12

    PAllen

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    The argument is completely general. I mentioned Minkowski metric only to make sure the questioner understood how the norm is defined (contraction of a vector twice with the metric).
     
  14. May 23, 2012 #13
    I qualified my question in my next post, I guess you missed it.
     
  15. May 23, 2012 #14

    PAllen

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    Proper acceleration is defined by the same definition in GR as SR and is a well used concept. There is no requirement of static spacetime. It is a geometric invariant by construction. Part of the physical interpretation of GR is that the mathematically defined proper acceleration norm (invariant) is what is measured by an accelerometer. This physical prediction is either true or false for our universe, but there is no ambiguity in the definition or interpretation of proper acceleration in GR.

    [Edit: To make explicit, again, the argument: 4-velocity is a vector in GR as well as SR; derivative by tau along a timelike world line is also a vector in GR as well as SR. Norm of a vector is defined as a contraction of the metric with the vector (twice). That contractions to scalars are invariant in diff.geometry is an elementary consequence of definitions. Thus, proper acceleration, so defined, is invariant with no caveats in arbitrary GR spacetimes.]
     
    Last edited: May 23, 2012
  16. May 23, 2012 #15
    You mean a non-inertial observer will measure the same four-acceleration magnitude than an inertial one?
     
  17. May 23, 2012 #16

    stevendaryl

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    For any 4-vector Aμ, the quantity gμη Aμ Aη is an invariant---it has the same value to all observers, regardless of whether they are inertial or not.
     
  18. May 23, 2012 #17

    PAllen

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    What? 4-acceleration (with norm of proper acceleration) is a feature of a world line (of some object). The object may be supposed to have an accelerometer. Who observes the measurement being taken is irrelevant. They won't see the instrument read different values. The fact the norm of 4-acceleration is an invariant mathematically captures the irrelevance.
     
  19. May 23, 2012 #18
    Nevermind. I had some sort of silly confusion writing about 4-quantities when I was thinking in terms of 3-quantitities.
     
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