Hi Al68, Mentz114 and Jorrie! Thanks for all the thoughtful and helpful answers. :)
As always I have more questions :P
Al68 said:
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
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.
Mentz114 said:
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.
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.
kev said:
(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?
Mentz114 said:
(a). Same as the icm observer.
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)
Mentz114 said:
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).
Good point. I should have said spring balance rather than scales. ;)
Jorrie said:
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.
Another good point. Assume measurement at the COM or a very short infinitesimal rocket.
Jorrie said:
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?
Again, assume dx is much smaller than x where dx=x2-x1 and x2~x1~x where ~ means aproximately equal.
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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)
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From
http://www.mathpages.com/rr/s7-03/7-03.htm
(Eq A): \frac{d^2 r}{d\tau^2}=\frac{-4m}{R^2(1+cos(\theta))^2}
"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."
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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) \frac{d^2 r}{d\tau^2}=\frac{-m}{R^2}
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.
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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): \frac{d^2 r'}{d\tau^2}=\frac{-m}{r^2\sqrt{1-2m/r}}
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."
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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:
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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): \frac{d^2 r}{dt^2}=\frac{-m}{R^2}\left(1-\frac{2m}{R}\right)
as expected. If R is infinite, the coordinate acceleration reduces to
(Eq D): \frac{d^2 r}{dt^2}=\frac{-m}{R^2}\left(1-\frac{2m}{R}\right)\left(1-\frac{6m}{R}\right)
"
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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): \frac{-m}{r^2\sqrt{1-2m/r}}
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) \frac{-m}{R^2}
and the coordinate acceleration of that same particle according to an observer at infinity is:
(Eq C): \frac{-m}{R^2}\left(1-\frac{2m}{R}\right)
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?
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"I have made this letter longer than usual because I lack the time to make it shorter." Blaise Pascal