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Proper acceleration

  1. Jul 18, 2008 #1
    Hi,

    Imagine a rocket accelerating in flat space. Let's say for the sake of argument it measures its acceleration by using an onboard accelerometer as (a).

    (E1)What would a momentarily comoving inertial observer measure the acceleration of the rocket to be using his own local rulers and clocks as the rocket passes him?

    (E2)What would an observer on the rocket measure the acceleration of a mass released from a somewhere near the nose of the rocket falling over a short distance? This measurement should be the same as the apparent acceleration (as in change of velocity per unit time) of the inertial observer in (E1) as measured by the observer in the rocket. This should also be similar to the average acceleration of a ball tossed up into the air by an observer onboard the rocket as measured by that same observer.

    Using the equivalence principle, is the acceleration calculated by placing a mass on some weighing scales in a gravitational field exactly the same (in relativistic terms) as the acceleration measured by timing the fall time over a short distance so that the change in gravitational radius and velocity is infinitesimal?

    Which of the above practical measurements is closest to the formal definition of proper acceleration?
     
  2. jcsd
  3. Jul 18, 2008 #2
    Both sound like measurements of coordinate acceleration instead of proper acceleration to me, since both would measure the second derivative of position wrt time (d^2 x/dt^2, or dv/dt), which is the definition of coordinate acceleration. I think the official definition of proper acceleration is the acceleration "felt" by an observer. (Change in momentum per unit mass/time). But they should be equal unless the coordinate acceleration is measured by an observer with relative velocity a significant fraction of c, wrt the observer being accelerated. In which case they would be related by sqr[1 + (at')^2/c^2].

    Maybe someone else can post the equations in latex. And/or correct me if I got them wrong.

    Al
     
    Last edited: Jul 18, 2008
  4. Jul 18, 2008 #3

    Mentz114

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    If the momentarily comoving inertial observer always shares the proper time of the accelerating frame, then the acceleration will be measured as (a) at all times.

    (a). Same as the icm observer.

    Given all your caveats, I'm inclined to believe ( back of envelope) that they will give the same answer.

    I can't work out how to use scales to measure acceleration. But a spring balance will give the same result as dropping and timing ( over a short enough distance).

    [caveat]There are three ways known to define distance in Rindler space, but they converge for very short distances.
     
    Last edited: Jul 18, 2008
  5. Jul 19, 2008 #4

    Jorrie

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    Hi Kev, interesting questions! Here's my view for what it's worth.
    If one takes the rocket as pushed from the rear (as is usual), then the proper acceleration differs over the length of the rocket (even after transients have damped out), so you have to specify where the acceleration is measured. If you say "a rocket accelerating at a", one may perhaps assume that you mean the acceleration of the center of mass (COM) of the rocket.

    Assuming the COM interpretation, a momentarily comoving inertial observer will measure the proper acceleration of the COM.

    Again, you will have to specify where on the rocket the observer resides. My understanding is that the proper acceleration of the nose (in geometric units) will be a(x) = (1-ax)a, where x is the distance from the COM (or wherever you have measured a) to the nose. So, for an observer residing in the nose, the acceleration of the mass will then be be -a(x).

    Does this make any sense in your scenario?
     
  6. Jul 20, 2008 #5
    Hi Al68, Mentz114 and Jorrie! Thanks for all the thoughtful and helpful answers. :)

    As always I have more questions :P

    I agree that if the relative velocity is a significant fraction of c then the acceleration measured by an inertial observer would be different. I am however trying to analyse the situation where the relative velocity is an insignificant fraction of c and hence the specification of a momentarily comoving inertial observer. An example would be tossing a ball one meter into the air. The initial and final velocity of the ball would be insignificant relative to the the speed of light. The difference in gravity at the surface of the Earth and one meter above the Earth would also be almost insignificant as the radius of the Earth is much larger than one meter. For all practical purposes the kinematic acceleration of a ball thrown one meter into the air is a constant 1g going up and coming back down and even at the apogee where the ball is momentarily almost stationary.


    Usually in descriptions of proper acceleration there is a assumption of a infinite amount of momentarily comoving observers, one for each "moment" and so your observation would seem to apply here.


    That would be my intuition too. At the bottom of this post are some quotes from the mathpages website that seem to contradict this view (or more likely... I have misinterpreted what mathpages is saying)

    Good point. I should have said spring balance rather than scales. ;)

    Another good point. Assume measurement at the COM or a very short infinitesimal rocket.

    Again, assume dx is much smaller than x where dx=x2-x1 and x2~x1~x where ~ means aproximately equal.
    .
    .
    .
    The real reason behind the questions in the OP is some statements and equations in the mathpages website which seem to contradict the assumption that acceleration measured by a momentarily comoving acceleration (Method E1) would not be the same as the proper acceleration (as measured by a spring balance type accelerometer onboard the rocket)

    ---------------------------------------------------------
    From http://www.mathpages.com/rr/s7-03/7-03.htm

    (Eq A): [tex]\frac{d^2 r}{d\tau^2}=\frac{-4m}{R^2(1+cos(\theta))^2}[/tex]

    "At q = 0 the path is tangent to the hovering worldline at radius R, and so the local gravitational acceleration in the neighborhood of a stationary observer at that radius equals -m/R^2, which implies that if R is approximately 2m the acceleration of gravity is about -1/(4m). Thus the acceleration of gravity in terms of the coordinates r and tau is finite at the event horizon, and can be made arbitrarily small by increasing m."

    ----------------------------------------------------------

    It should be noted that he using a mixture of coordinate distance and proper time (tau) here. This method of measuring acceleration is interesting because it gives answers that are finite and real below the event horizon and suggests a particle can be momentarily stationary (apogee) at or even below the event horizon.

    At the apogee the above equation can be expressed as:

    (Eq A) [tex]\frac{d^2 r}{d\tau^2}=\frac{-m}{R^2}[/tex]

    It is not clear to me who would actually measure the acceleration to be -m/R^2 because it is not the proper acceleration measured by the hovering observer and it is not the measurent of a coordinate inertial observer at infinity who would measure the hovering rocket to be stationary.
    ---------------------------------------------------------
    From the same page http://www.mathpages.com/rr/s7-03/7-03.htm

    "However, this acceleration is expressed in terms of the Schwarzschild radial parameter r, whereas the hovering observer’s radial distance r' must be scaled by the “gravitational boost” factor, i.e., we have dr' = dr/sqrt(1-2m/r). Substituting this expression for dr into the above formula gives the proper local acceleration of a stationary observer

    (EQ B): [tex]\frac{d^2 r'}{d\tau^2}=\frac{-m}{r^2\sqrt{1-2m/r}}[/tex]

    This value of acceleration corresponds to the amount of rocket thrust an observer would need in order to hold position, and we see that it goes to infinity as r goes to 2m."

    ---------------------------------------------------------

    This last equation is what I would normally assume is meant by proper acceleration and is what would assume a spring balance type accelerometer would indicate. For values of R that are less than the event horizon radius the answers are infinite or imaginary suggesting the impossibility of having a stationary particle hovering below the horizon although it ws shown above that a particle can be momentarily stationary at an apogee, below the event horizon.

    Ordinarily we would normally expect the acceleration indicated by a mass on a spring balance at the surface of the Earth to be the same as the acceleration indicated by dropping a mass a short distance of less than a meter and the equivalence principle indicates similar measurements would be made in an rocket accelerating with a constant proper acceleration of 1g. Mathpages suggests otherwise:

    ------------------------------------------------
    From http://www.mathpages.com/rr/s6-07/6-07.htm

    "At the apogee of the trajectory, when r = R, this reduces to

    (Eq C): [tex]\frac{d^2 r}{dt^2}=\frac{-m}{R^2}\left(1-\frac{2m}{R}\right)[/tex]

    as expected. If R is infinite, the coordinate acceleration reduces to

    (Eq D): [tex]\frac{d^2 r}{dt^2}=\frac{-m}{R^2}\left(1-\frac{2m}{R}\right)\left(1-\frac{6m}{R}\right)[/tex]

    "

    ------------------------------------------------

    Both the above equations are the coordinate measurements of acceleration according to an observer at infinity (because they are derived directly from the Schwarzschild metric) and R is the radius where a projectile initially launched upwards away from the gravitational source momentarily comes to a stop, before falling back towards the gravitational source.

    To try and summerise all the above as I see it (and I welcome clarifications) the proper acceleration measured by the accelerometer of a hovering observer is proportional to

    (EQ B): [tex]\frac{-m}{r^2\sqrt{1-2m/r}}[/tex]

    and the local acceleration of a free falling particle that happens to be at apogee at the same radius as the hovering observer as measured by the hovering observer is:

    (Eq A) [tex]\frac{-m}{R^2}[/tex]

    and the coordinate acceleration of that same particle according to an observer at infinity is:

    (Eq C): [tex]\frac{-m}{R^2}\left(1-\frac{2m}{R}\right)[/tex]

    The problem is that the gist of the replies to this thread do not seem to agree that (Eq A) should be any different to (Eq B). Any ideas?


    ---------------------------------------------------------
    "I have made this letter longer than usual because I lack the time to make it shorter." Blaise Pascal
     
  7. Jul 20, 2008 #6

    Jorrie

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    As I see it, Eqs. A and C are coordinate transforms of the proper acceleration (Eq. B) in curved spacetime. This does not quite agree with a rocket that accelerates in flat spacetime (Rindler coordinates)

    Eq. A uses the local (static) observer's time, but the distant (infinity) observer's (Schwarzschild) radial parameter. Eg. B uses local time and radial parameters, while Eq. C uses distant time and radial parameters.
     
    Last edited: Jul 20, 2008
  8. Jul 20, 2008 #7
    I think I just realized the point of your post, maybe. The object released in (E2) would actually be at rest wrt a momentarily comoving inertial observer (E1), and the proper acceleration of that object would equal zero, but the coordinate acceleration of the object (and the inertial observer in E2) as measured by the ship's observer should equal the proper acceleration of the ship.

    But it seems from your last post that the coordinate acceleration of the ship as measured by an inertial observer may not equal the coordinate acceleration of that inertial observer (and a released object at rest with same observer) as measured by an observer on the ship. That seems very strange to me, too, unless I misinterpreted your point.

    I think I should look closer at that page on mathpages and see what others have to say about it.

    Al
     
  9. Jul 20, 2008 #8

    Jorrie

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    I think it must be a "locally static" observer, somehow using Schwarzschild coordinate measuring rods, but neither said observer nor rods can exist at or inside the event horizon.
     
  10. Jul 20, 2008 #9

    Mentz114

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    Kev,
    any remarks I made may only apply to a uniformly accelerating frame. I wasn't thinking about curved spacetime. Rethink needed.

    M
     
  11. Jul 20, 2008 #10

    Dale

    Staff: Mentor

    Hi kev,

    As you know, I am not a GR expert, so take the following with a grain or two of salt.
    a

    a assuming that the length of the rocket is small enough to neglect time dilation etc. between the front and back. (or equivalently that the accelerometer is also in the nose next to the "falling" mass.

    Yes.

    I think that E1, the instantaneous acceleration in the momentarily co-moving inertial frame, is the definition of proper acceleration.
     
  12. Jul 21, 2008 #11
    Hi Al, Mentz and Jorrie :smile:

    It took me a while to figure out your point about the released object being at at rest with the momentarily comoving observer but it appears you are absolutely right. I am also coming to the same conclusion as you, that observers measuring the acceleration using methods E1 and E2 would obtain the same result as the proper acceleration indicated by the accelerometer in the accelerating rocket (or by a stationary hovering observer in a gravitational field as per the equivalence principle).

    I think the confusion (in my head anyway) came from this equation from mathpages:

    (Eq A) [tex]\frac{d^2 r}{d\tau^2}=\frac{-m}{R^2}[tex]

    but because it a measurement made from a mix of proper time of the falling object and coordinate length it is not the actual sensible measurement of acceleration measured by any single observer and is probably best ignored.

    As I mentioned above I do not think the measurement made by a mix of proper and coordinate measuring devices is a sensible measurement.

    As mentioned above, I tend to agree, after further reflection, that proper acceleration measured by an onboard accelerometer and by methods E1 and E2 all yield the same answer (a).
     
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