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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 [Broken]
     
    Last edited by a moderator: May 3, 2017
  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?
     
    Last edited by a moderator: Apr 23, 2017
  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]
     
  22. Jul 1, 2008 #21

    Fredrik

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    Thanks. It's good to have all these links in the same thread, so that I'll be able to find them easily at any time. I'm not sure how useful these two will be right now because I feel that I understand the twin paradox extremely well and the rotating disc well enough too, but I'm sure you have considered some aspects of these problems that I haven't even thought about yet, so I'll definitely have a look at them.
     
  23. Jul 1, 2008 #22

    Fredrik

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    New question: Suppose we have a laser beam bouncing up and down between two mirrors located on the floor and the ceiling of a rigid accelerating rocket that has a constant proper acceleration. Suppose also that there are clocks next to the mirrors and a device that records the time (as measured by the nearest clock) at each reflection event. Will the measured times between consecutive reflection events on (say) the floor be constant?

    (I don't think the answer is obviously yes. The "obvious" argument is that if it isn't, then the proper length of the rocket must be changing, and that contradicts the rigidity assumption. But is it obvious that we can measure "proper length" by timing a light signal even when the rocket is accelerating?)

    As usual I'm looking for some sort of derivation, not just the answer, but feel free to just tell me the answer if you know what it is.
     
  24. Jul 1, 2008 #23

    Hurkyl

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    Actually, I was thinking the obvious argument was more like "symmetry under boosts", but I suppose that amounts to the same thing -- the entire setup is the same, it just occurs at a different place and orientation in space-time.

    I'd be mildly surprised if this procedure turns out to be the proper length! But we can compute:

    Let's fix a concrete trajectory for the front and back of the rocket. I'll choose hyperbolas centered at the origin, and choose units so that c = 1 and everything works out. Let the tail of the rocket be given by the worldline [itex]x^2 - t^2 = 1[/itex] and the head of the rocket be given by [itex]x^2 - t^2 = 4[/itex]. (So the length of the rocket is 1)

    Suppose we emit a photon from the tail at the event (0, 1). Its worldline is given by the equation x = 1 + t, and it arrives at the head at the event (3/2, 5/2). When it's reflected back, its worldline is x = 4 - t, which arrives at the tail at the event (15/8, 17/8).

    Since the parametrization by proper time of the tail's worldline is [itex](t, x) = (\sinh \tau, \cosh \tau)[/itex], we find that the round trip time is [itex]\tau = \tanh^{-1}(15/17) \approx 1.39[/itex], as measured by the tail of the rocket.


    Let's continue. :smile: The photon reflects again, with worldline x = (1/4) + t. This arrives at the head at the event (63/8, 65/8). The head's worldline is parametrized by [itex](t, x) = (2 \sinh (\tau / 2), 2 \cosh (\tau / 2))[/itex]. The times at the two photon strikes are:

    [itex]\tau = 2 \tanh^{-1}(63/65) \approx 4.16[/itex]
    [itex]\tau = 2 \tanh^{-1}(3/5) \approx 1.39[/itex]

    for a round trip time of approximately 2.77.


    Upon reflection (ha ha), I would have been very surprised if this procedure could measure the length of the rocket, since I knew the front clock would be running faster than the rear clock, so there's no way they could have agreed on the length of the rocket if they were measuring the round trip time experienced by a photon!
     
  25. Jul 1, 2008 #24
    Under the equivalence principle the time interval between consequative reflection events in the accelerating rocket will be constant because for example the measurements in a rcocket experiencing constant proper acceleration of 1g are no different to the measurements you would make in your living room. The measurement of the height of your ceiling in your living room (generally speaking) does not change over time and the same applies to the rocket. I am off course speaking about "proper" masurements made by an observer that is co-accelerating. An inertial observer in free fall would see the rocket or your living room as getting progressively shorter.

    The "rigid" part of the constant proper acceleration is that the proper length remains constant within a relatively localised region according to the co-accelerating observer. An inertial observer on the other hand, measures length contraction of the height and time dilation of the accelerating clocks (coordinate measurements) and he sees then as continuously changing over time. The combined effect is that the length appears constant to the accelerating observer, but he will however see the ceiling clock as blue shifted relative to the floor clock just as you could in your living room with sensitive enough equipment.

    P.S. I just noticed this comment by Hurkle
    and just wanted to add this. If you measured the height of your living room by timing the round trip of a photon sent from the floor to the ceiling and back you would calculate a shorter height than that obtained by timing the round trip of a photon from the ceiling to the floor and back again, if you assumed the speed of light is c in both cases. The average of the measurements would be pretty close to that given by measuring rod.
     
    Last edited: Jul 1, 2008
  26. Jul 2, 2008 #25

    Fredrik

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    Good answers, both of you. Kev, that's a nice way to quickly find the correct answer.

    Hurkyl, that was awesome. I wanted to do that exact calculation but I thought it would be more complicated, probably because I was considering two arbitrary hyperbolas with the same center instead of two specific ones. I checked all your calculations and I'm getting the same results. I also calculated the coordinates and the corresponding proper time of one more reflection event: t=255/32, x=257/32 (on the tail's world line), [itex]\tau\approx 2.77[/itex].

    Let's call the emission event A, and the reflection events B, C, D and E (with higher coordinates in the original rest frame corresponding to letters later in the alphabet). The measured time between A and C is indeed exactly the same as the measured time between C and E.

    [tex]\tanh^{-1}(15/17)=\tanh^{-1}(255/257)-\tanh^{-1}(15/17)\approx 1.39[/tex]

    I also noticed that the measured time between B and D is exactly the same as the measured time between A and E. This must be a consequence of the fact that you chose the proper acceleration of the head to be exactly half the proper acceleration of the tail.

    I didn't really mean that we could measure the proper length just by timing the light signals. I was thinking something similar to what Kev said, that if I do this in my house, the time for light to go from the ceiling to the floor and back to the ceiling again is always the same. If it isn't, my house is shrinking. It's also clear from the calculations that if we don't keep the proper length constant (i.e. if we change the shape of the world line of the head of the rocket) the time it takes for light to go tail-head-tail isn't always going to be the same. So the fact that the time between consecutive reflections at the tail is always the same is a consequence of the rigidity assumption (the requirement that the proper length stays the same in co-moving frames).
     
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