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Questions about acceleration in SR

  1. Jun 29, 2008 #1

    Fredrik

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    I'm going to use this thread to ask questions about acceleration in SR. I'll just ask one or two at at time.

    First question: How do you calculate what the world line of a rocket with constant proper acceleration looks like in an inertial frame? Assume v=0 at t=0 and v>0 after that.
     
  2. jcsd
  3. Jun 29, 2008 #2

    Easy, just plot t on the y axis against d on the x axis for t>0 using this equation:

    [tex]d = \frac{c^2\sqrt{1 + (at/c)^2}}{a} [/tex]

    Generally a=1/d when t=0

    See http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html
     
    Last edited: Jun 29, 2008
  4. Jun 29, 2008 #3

    Fredrik

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    Thanks. I'm going to get some sleep now, but I'll have a look at it tomorrow.
     
  5. Jun 29, 2008 #4

    Hurkyl

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    More geometrically... the curves of constant (proper) acceleration are hyperbolas with lightlike asymptotes. The Minkowski metric bestows upon such hyperbolas (which I will henceforth call H's) many of the properties of Euclidean circles:


    An H is the set of all points a fixed distance from its center. (timelike distance or spacelike distance yield hyperbolas opening in different directions)

    Lines through the center are orthogonal to H's.

    Two cocentric H's are equidistant from each other.

    An H is parametrized by hyperbolic trig functions.

    Equal hyperbolic angles (with vertex at the center) sweep out equal length hyperbolic arcs.


    In higher dimensions, we get hyperboloids and such.
     
  6. Jun 29, 2008 #5
    Hi, I think I understand most of your statements here but I am not clear about the "Lines through the center are orthogonal to H's." property. Could you explain in more detail referring to the diagrams here http://mysite.du.edu/~jcalvert/math/hyperb.htm or here http://mathworld.wolfram.com/Hyperbola.html

    Perhaps I am getting confused between the terms orthogonal, normal and right angles in Euclidean terms?
     
  7. Jun 29, 2008 #6

    Hurkyl

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    I'm not really sure how much there is to say; if you draw a line through the center of an H, it will intersect the H in a (Minkowski) right angle. e.g. that means the (Minkowski) dot product of the vector describing the line with the tangent vector to the hyperbola is zero.

    (Compare with Euclidean geometry: any radius of a circle meets that circle at a Euclidean right angle)

    e.g. in your first link, in the section "Hyperbolic functions", the line segment OP intersects the hyperbola at a (Minkowski) right angle. In fact, you can 'rotate' the diagram (i.e. apply a Lorentz boost centered at O) so that the line segment OP is moved to the x-axis. Then it might be more clear that it's a right angle, since in that drawing, a Euclidean reflection about the x-axis is the same thing as a Minkowski reflection about the x-axis.

    edit: clarified we are to use the Minkowski dot product, not the Euclidean one
     
    Last edited: Jun 30, 2008
  8. Jun 29, 2008 #7

    robphy

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    A better characterization of the hyperbola in dealing with uniform acceleration is that
    the hyperbola is the "curve of constant curvature [in the Minkowski metric]".
    With a Galilean metric, the corresponding "curve of constant curvature" is the parabola.
     
  9. Jun 30, 2008 #8

    Fredrik

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    Thanks guys. Your answers are helpful, but I'm not just looking for a description of the properties of a world line of constant proper acceleration. I was hoping someone would show me the trick that makes it easy to prove that these curves are hyperbolas.

    I'm not even sure I have the right idea about what constant proper acceleration means. I think it means that

    [tex]\frac{d^2x}{d\tau^2}=\begin{pmatrix}0 \\a\end{pmatrix}[/tex]

    in every co-moving inertial frame. Is that right?
     
  10. Jun 30, 2008 #9

    Hans de Vries

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    Well, from the link you provided yourself: :smile: http://en.wikipedia.org/wiki/Rindler_coordinates

    The Rindler coordinates guarantee that the lengths in the restframe of the accelerating
    object stay the same ( otherwise, things would start breaking apart)

    The first figure at the right in link demonstrates this. The number of steps in the changing
    x' axis stay constant which means that the lengths in the rest frame stay constant.



    Regards, Hans
     
  11. Jun 30, 2008 #10
    Thanks Hurkyl, I get it now :smile:
     
  12. Jun 30, 2008 #11

    Fredrik

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    I was a little confused by your answer at first, but now I see that you must have interpreted my question in a different way than I intended. You told me what the rocket would look like, but I was trying to ask what the world line of a rocket (of zero length) would look like (t as a function of x) and why. (The why is the part I'm having problems with). I think the Wikipedia page answers the "what" but not the "why".
     
  13. Jun 30, 2008 #12

    (t as a function of x) would look like any of the curved worldlines in this diagram. http://upload.wikimedia.org/wikipedia/commons/0/0f/RindlerObserversCartesian.png

    The "why" tends to stray into philosophical arguments while the "what" is is more scientific and less disputed. As a rough handle on why the wordline curves the way it does, you could imagine a simple solid fuel rocket with a fixed nozzle that is designed to burn f Kgs of fuel per unit time. Time dilation affects slows down the chemical reactions so the when the rocket is moving it burns less fuel per unit time according to the inertial observer although it appears to burn the same amount of fuel per unit time according to the co-moving observer onboard the rocket. Another way to look at it (but more controvertial) is to consider the inertial mass of the rocket as increasing, and requiring more energy to accelerate so burning a fixed quantity of fuel per unit time causes less increase in velocity per unit time as the rockets relative velocity increases. There are probably other ways to visualise what is "really happening" which is why the "why" is much harder to nail down than the "what".
     
  14. Jun 30, 2008 #13

    DrGreg

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    Proof (part 1)

    I'll restrict myself to one dimension of space, so we can consider an arbitrary particle's worldline defined by the vector (t, x) relative to a given inertial frame. Because of the way proper time [itex]\tau[/itex] is defined we know that

    [tex] c^2 \left( \frac {dt}{d\tau} \right)^2 - \left( \frac {dx}{d\tau} \right)^2 = c^2 [/tex] (1) ​

    The properties of hyperbolic functions mean that it is possible to find a number [itex]\phi[/itex] such that

    [tex] \frac {dt}{d\tau} = \cosh \phi [/tex] (2)
    [tex] \frac {dx}{d\tau} = c \sinh \phi [/tex] (3) ​

    [itex]\phi[/itex] is called the rapidity of the particle relative to the inertial frame. At low speeds [itex]c\phi[/itex] approximates to the velocity v, which can be seen by dividing (3) by (2):

    [tex] v = \frac {dx}{dt} = \frac {dx/d\tau}{dt/d\tau} = c \tanh \phi [/tex] (4) ​

    Note that the Lorentz factor is given by [itex]\gamma = \cosh \phi[/itex], and [itex]\gamma v = c \sinh \phi[/itex].

    The hyperbolic version of the Lorentz transform can be used to transform to another inertial coordinate system (t', x') moving at velocity [itex]v_0 = c \tanh \phi_0[/itex] relative to the first:

    [tex] dt' = dt \, \cosh \phi_0 \, - \, \frac {v_0 \, dx} {c^2} \sinh \phi_0 [/tex] (5)
    [tex] dx' = dx \, \cosh \phi_0 \, - \, c \, dt \, \sinh \phi_0 [/tex] (6) ​

    This forum has problems uploading posts with many equations, so I will continue in the next post...
     
  15. Jun 30, 2008 #14

    DrGreg

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    Proof (part 2)

    This post continues from the immediately prededing post

    Now, from (4)
    Substituting (2) and (3) we obtain

    [tex] \frac {dt'}{d\tau} = \cosh (\phi - \phi_0) [/tex] (7)
    [tex] \frac {dx'}{d\tau} = c \sinh (\phi - \phi_0) [/tex] (8) ​

    Thus it can be seen that the rapidity [itex]\phi'[/itex] relative to the new frame is given by

    [tex] \phi' = \phi - \phi_0 [/tex] (9) ​

    This shows that rapidity is additive (just like Newtonian velocity). Furthermore

    [tex] \alpha = \frac {d\phi'}{d\tau} = \frac {d\phi}{d\tau} [/tex] (10) ​

    so [itex]\alpha = d\phi / d\tau[/itex] is an invariant -- all inertial observers agree on its value.


    This forum has problems uploading posts with many equations, so I will continue in the next post...
     
  16. Jun 30, 2008 #15

    DrGreg

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    Proof (part 3 of 3)

    This post continues from the immediately prededing post

    Now, from (4)

    [tex] \frac {dv}{d\tau} = c \frac {d\phi}{d\tau} \, sech^2 \, \phi = c \alpha \, sech^2 \, \phi [/tex] (11) ​

    and so the coordinate acceleration is given by

    [tex] a = \frac {dv}{dt} = \frac {dv/d\tau}{dt/d\tau} = c \alpha \, sech^3 \, \phi [/tex] (12) ​

    (Note that, although [itex]\alpha[/itex] is invariant, [itex]\phi[/itex] is not, so different inertial frames measure different coordinate accelerations.)

    When the inertial frame is the co-moving frame, i.e. when v = 0, i.e. when [itex]\phi[/itex] = 0, we have

    [tex] a = c \alpha [/tex] (13) ​

    This is, by definition, the proper acceleration of the particle at that event. The invariance of [itex]\alpha[/itex] means that the quantity [itex]c\alpha[/itex] is the proper acceleration at all other events -- choose the appropriate co-moving frame at each event along the particle's worldline.


    Everything I've said so far applies to an arbitrary particle moving in a straight line in flat spacetime.

    Uniform acceleration occurs when the proper acceleration is constant, i.e.

    [tex] a = c \alpha = c \frac {d\phi}{d\tau} [/tex] (14) ​

    is constant, i.e.

    [tex] \phi = \frac {a\tau}{c} [/tex] (15) ​

    Substitute this into (2) and (3) and integrate with respect to [itex]\tau[/itex] to get

    [tex] t = \frac {c}{a} \sinh \frac {a\tau}{c} [/tex] (16)
    [tex] x = \frac {c^2}{a} \cosh \frac {a\tau}{c} [/tex] (17) ​

    These are the standard equations for constant proper acceleration a.
     
  17. Jun 30, 2008 #16
    Thanks for taking the time to post a derivation Dr greg.

    It is always nice to see a derivation from first principles. I guess in a way it does answer the question "why" in the context that if the first principles are accepted, then the derivation of the equations shows "why it must be so". :smile:
     
  18. Jun 30, 2008 #17

    Demystifier

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  19. Jun 30, 2008 #18

    Hans de Vries

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    The origin of the curves is in fact very elementary and beautifully shown in decent
    treatments of the Poincaré group.

    Repeated application of the operator in the middle where [itex]\epsilon[/itex] is a very
    small number generates the hyperbolic path where [itex]\vartheta[/itex] is the increase
    of the rapidity per unit of time:


    Boost operator:

    [tex]
    \left( \begin{array}{c} x^0 \\ x^1 \\ x^2 \\ x^3 \end{array} \right) ~=~
    \left(
    \mbox{\Huge I} ~+~ \epsilon
    \left(\begin{array}{cccc}
    0 & 0 & 0 & 1 \\
    0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 \\
    1 & 0 & 0 & 0 \\
    \end{array}\right) ~~
    \right)^n
    \left( \begin{array}{c} x^0 \\ x^1 \\ x^2 \\ x^3 \end{array} \right)
    ~=~
    \left(\begin{array}{cccc}
    \cosh(\vartheta t) & 0 & 0 & \sinh(\vartheta t) \\
    0 & 1 & 0 & 0 \\
    0 & 0 & 1 & 0 \\
    \sinh(\vartheta t) & 0 & 0 & \cosh(\vartheta t) \\
    \end{array}\right)
    \left( \begin{array}{c} x^0 \\ x^1 \\ x^2 \\ x^3 \end{array} \right)
    [/tex]


    This is very similar to the rotation operator where [itex]\phi[/itex] is the increase of
    the angle per unit of time. Repeated application of the operator in the
    middle gives the circular path defined by the sines and cosines.

    Rotation operator:

    [tex]
    \left( \begin{array}{c} x^0 \\ x^1 \\ x^2 \\ x^3 \end{array} \right) ~=~
    \left(
    \mbox{\Huge I} ~+~ \epsilon
    \left(\begin{array}{cccc}
    0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 1 \\
    0 & 0 & 0 & 0 \\
    0 & -1 & 0 & 0 \\
    \end{array}\right) ~~
    \right)^n
    \left( \begin{array}{c} x^0 \\ x^1 \\ x^2 \\ x^3 \end{array} \right)
    ~=~
    \left(\begin{array}{cccc}
    1 & 0 & 0 & 0 \\
    0 & \cos(\phi t) & 0 & \sin(\phi t) \\
    0 & 0 & 1 & 0 \\
    0 & -\sin(\phi t) & 0 & \cos(\phi t) \\
    \end{array}\right)
    \left( \begin{array}{c} x^0 \\ x^1 \\ x^2 \\ x^3 \end{array} \right)
    [/tex]

    This beautiful symmetry allows us to combine Lorentz transformations and rotations
    to be handled by a single operator.


    Regards, Hans
     
  20. Jun 30, 2008 #19

    Fredrik

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    Thank you DrGreg. That was awesome, and it completely answers my question. The explanation was very clear. I appreciate that you even took the time to explain rapidity. It's not a difficult concept, but I have never had a reason to learn it before.

    Hans, thanks for explaining that. That's actually pretty cool.

    Demystifier, thanks for the link. I only took a quick look at your article since Greg answered my question completely, but I have a feeling that it can answer some of the other questions I've been thinking about asking.
     
  21. Jul 1, 2008 #20

    Demystifier

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    I guess you might ask similar questions that I asked myself some 10 years ago.
    The answers to some of these questions may be also present here:
    http://xxx.lanl.gov/abs/gr-qc/9904078 [Phys.Rev. A61 (2000) 032109]
    http://xxx.lanl.gov/abs/physics/0004024 [Found.Phys.Lett. 13 (2000) 595]
     
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