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The co-moving frame - acceleration, momentum and velocity

  1. Feb 21, 2015 #1
    Hi all,
    I'm having trouble with the concept of a co-moving frame, specifically in the context of constant proper-acceleration (hyperbolic motion). I feel like I don't "get it"; and perhaps this is related to some deeper misunderstandings I may have regarding the definitions of certain quantities in SR.
    I would really appreciate it if someone could explain this in very plain words, specifically the following points:
    1. When speaking of trajectories in space-time as seen in a certain frame. What exactly do mean by this when relating to the co-moving frame? It is my intuitive understanding that in this frame the particle/observer lays at the origin for all values of tau, there can be no trajectory which isn't identically zero. What are we talking about when saying that in this frame the trajectory is a hyperbola?
    2. How is the acceleration defined in a co-moving frame, if by definition in this frame the velocity is constant? It would seem to me that the derivative of the velocity wrt proper time should vanish in this frame.
    3. Does a body have momentum in it's co-moving frame? That would seem to contradict that fact that the velocity is null, as 4-momentum and 4-velocity are always simply related. On the other hand, if we have proper acceleration, then we have proper force, and the momentum should be changing. How do we settle the definitions of momentum and velocity in this case?
    4. I've seen the relation stating that the 4-velocity and 4-acceleration are always perpendicular. To my understanding, the 4 velocity is just the "velocity of the world line", in other words the velocity of the trajectory wrt to the path parameter - correct? If the world line is a straight line in space time (for example), how can the acceleration be perpendicular to the velocity? That would suggest a change in the path direction. Or am I confusing the definitions here?
    These seem like enough for a start. Thanks a lot for the help!
     
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  3. Feb 21, 2015 #2

    Orodruin

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    The comoving frame is not one singled out frame for all of the world line. However, for each point on the world line, you can find an inertial frame where the tangent vector is (1,0,0,0).
     
  4. Feb 21, 2015 #3

    Nugatory

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    In the comoving frame, velocity and momentum are always zero.

    In the comoving frame, the coordinate acceleration dv/dt is therefore also always zero. It will not be zero in other frames; because v is defined as dx/dt and both x and t are frame-dependent, dv/dt is also frame-dependent. The proper acceleration, which is the the value that an accelerometer would read, is not necessarily zero and will be the same in all frames whether comoving or not.

    The four-velocity is not much like a regular velocity. Its magnitude is ##c## in all frames always no matter the state of motion of the object; its direction is always tangent to the world line of the object. Because the magnitude of the four-velocity is constant the four-acceleration must be perpendicular to the four-velocity; otherwise the magnitude would change.

    And you are right that because the four-acceleration is perpendicular to the four-velocity, a non-zero four acceleration implies a change in direction. But that's the direction in four-dimensional space-time, not three-dimensional space. An accelerating spaceship is moving in a straight line in three dimensional space but a hyperbolic path in four-dimensional space-time.

    By far the easiest way to visualize all this is to practice drawing some space-time diagrams, in which the trajectory of the object is plotted on a graph with the x coordinate on the horizontal axis and the t coordinate on the vertical axis.
     
  5. Feb 21, 2015 #4
    Thanks a lot for taking the time to reply. I would like to delay on each one of your paragraphs if you don't mind:

    By this do you mean the 3-velocity and 3-momentum? How do you settle the following equation?:
    ##f^{\alpha}=m\frac{d^{2}x^{\alpha}}{d\tau^{2}}=ma^{\alpha}=m\frac{du^{\alpha}}{d\tau}##

    If in the co-moving frame I have:
    ##u^\alpha=(1,0,0,0)##
    ##a^\alpha=(0,\vec{a})##

    How do you differentiate the velocity to obtain the acceleration in the co-moving frame?

    Can you please elaborate on this statement? To my understanding, an accelerometer measures force, which is itself a 4-vector, transforming between Lorentz frames. The proper accelerating being the value of this in one specific frame (co-moving/proper frame). What do you mean by being constant in every frame?

    In what frame is this seen to be an Hyperbola? Is it in the laboratory (inertial) frame? In other words, when sketching the world line of the space ship (which is easy enough to do I suppose, if you're measuring it's spatial location at every moment), who sees the hyperbola? Surely it's not in the co-moving frame where the space ship always measures it's location to be the origin?

    Thanks again!
     
  6. Feb 21, 2015 #5

    Nugatory

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    I do indeed mean the three-velocity and the three-momentum, and I do the differentiation the old-fashioned way: I start with the position as a function of time ##x(t)##, differentiate once to get the velocity ##v(t)=x'(t)## as a function of time, then differentiate that to get the coordinate acceleration ##a(t)=v'(t)=x''(t)## as a function of time. If we're using the coordinates of a comoving frame, ##x(t)## is a constant (not necessarily zero, although it is so convenient to choose the coordinate origin that way that you'll seldom see anything else) so all of its derivatives are zero.


    The accelerometer has to produce the same results in all frames (pause now for the mandatory reminder that "in that frame" is a sloppy substitute for the more precise "using coordinates assigned by that frame"). Otherwise we'd have the impossible situation where its output would read differently according who was looking at it.

    A simple model of an accelerometer is a box containg a weight suspended by six springs from the six inside faces of the box. If the box is accelerated, the forces in the springs will change to overcome the inertia of the weight, so measuring these forces is a measurement of the proper acceleration of the box.

    Yes, the world line is a hyperbola of you use x and t coordinates assigned by an inertial frame.
     
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