# How Does Proper Time in Radial Free Fall Differ from Newtonian Predictions?

• Stoonroon
In summary, Foster and Nightingale give an equation for the proper time of a radially falling clock between any two distances, which is the same as its Newtonian counterpart. This raises the question of whether there is any time dilation for such a clock, but further details and calculations are needed for a definitive answer. The Vessot-Levine experiment suggests that the natural clock frequency is not altered, and the first order Doppler cancelation is achieved through a separate out-back link.
Stoonroon
Foster and Nightingale give an equation for the proper time of a radially falling clock between any two distances. (Short Course in General Relativity, 1979, p. 128.) As they state, the equation is the "same as its Newtonian counterpart." So there are no factors representing any slowing of the clocks. This is confusing to me because it seemingly implies that such a clock suffers no time dilation. But that doesn't seem right.

Consider an experiment such as that of Vessot-Levine, but suppose the clock is simply dropped from rest at "apogee" over a large non-rotating planet with no atmosphere. Suppose the time on the clock is set to zero at the moment of release. At rigidly built observing stations alongside the path and at the surface the elapsed time on the falling clock is recorded. The clock's frequency is affected by its speed and its location in the gravitational field. (The frequency of the Vessot-Levine clock was monitored from a distance using a clever first-order Doppler canceling system. But they did not measure elapsed proper time.) My question is, how would these idealized proper time observations differ from the predictions of Newton's theory, in which all clocks always tick at the same rate? Or is it that the combined effects of acquired falling speed and spacetime curvature (i.e, the difference between proper and coordinate distance) "conspire" to make the Newtonian equation match the General Relativistic prediction?

Stoonroon said:
Foster and Nightingale give an equation for the proper time of a radially falling clock between any two distances. (Short Course in General Relativity, 1979, p. 128.) As they state, the equation is the "same as its Newtonian counterpart." So there are no factors representing any slowing of the clocks. This is confusing to me because it seemingly implies that such a clock suffers no time dilation. But that doesn't seem right.

Hi Stoonroon,

This is an interesting question. Could you post the Foster and Nightingale equation so that we can discuss it?

Stoonroon said:
Consider an experiment such as that of Vessot-Levine, but suppose the clock is simply dropped from rest at "apogee" over a large non-rotating planet with no atmosphere. Suppose the time on the clock is set to zero at the moment of release. At rigidly built observing stations alongside the path and at the surface the elapsed time on the falling clock is recorded. The clock's frequency is affected by its speed and its location in the gravitational field. ...

I am not sure but I think the frequency of the falling clock is affected by its frequency but not its location in the gravitational field. This is because the clock is experiencing no proper acceleration as it falls. As I said, I'm not sure and I hope the experts on this forum can give a more definitive answer.

There seems to scant information on the details of the Vessot-Levine experiment (Gravity Probe A) freely available on the internet. It seems ironic that the billions of dollars to fund scientific research on a big scale is ultimately paid for by the general public and yet we have to pay more to be able to read the original papers. :(

Getting off my soap box and back to the original question, there are a number of obvious factors to take into account (and I am sure a greater number of not obvious factors). First, if the observing stations are spaced at regular intervals on a sort of ladder built in space, then when the ladder is lowered into the gravitational field it will be length contracted with the lower parts being contracted to a greater extent. The clocks on the ladder will each be time dilated according to their location in the gravitational field. The falling clock will tick like a clock in flat space far from any gravitational field but it will also be knematically time dilated according to its falling velocity which will obviously be increasing as it falls. Signals sent from the falling clock to a ground station will be doppler shifted and this shift will have to be corrected to obtain the proper time of the falling clock. I think the first order doppler self correction that you mention applies to the functioning of the maser clock itself rather than doppler shift of the signals sent from it, as far as I can ascertain.

I have not done the calculations, but I think the Schwarzschild equation tells us something about the proper time of a falling clock compared to coordinate time, but I am not sure if there is an unambiguous solution in this case.

The Foster and Nightingale equation is attached as a jpg. The initial distance is given by r_0 and the final distance by r.

The General Relativity equation for the clock frequency has in its argument one term for the location ("potential") and one for the speed (squared). They are simply added to give the total effect.

Although I'd still like to see an explicit authoritative assessment, as I contemplate the problem, it seems that use of the Newtonian equation makes sense at least insofar as it assures that both the falling clock and ground-based observers should both measure the same speed. Whereas, if the clock were falling from infinity (and I think also from 2R) the coordinate speed would be diminished by the metric coefficient, 1-2GM/rc^2. (As per Hobson, Efstathiou and Lasenby, General Relativity, p. 210.)

In the Vessot-Levine experiment, the natural clock frequency was not altered; the first order Doppler cancelation was achieved by having a separate out-back link that facilitated the subtraction. (I recommend their original papers, which you may have to go to a university physics library to find.)

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## 1. What is radial free fall proper time?

Radial free fall proper time is the time experienced by an observer who is falling towards a massive object along a radial trajectory, without any external forces acting on them.

## 2. How is radial free fall proper time different from regular time?

Radial free fall proper time is different from regular time because it takes into account the effects of gravity on the passage of time. In regular time, the rate of time passage is constant, but in radial free fall proper time, it is affected by the strength of the gravitational field.

## 3. Is radial free fall proper time affected by the mass of the falling object?

No, the mass of the falling object does not affect the radial free fall proper time. It is only affected by the strength of the gravitational field.

## 4. How is radial free fall proper time calculated?

Radial free fall proper time is calculated using the formula T = ∫√(1-2GM/r) dt, where T is the proper time, G is the gravitational constant, M is the mass of the central object, and r is the distance from the center of the object.

## 5. What is the significance of radial free fall proper time?

Radial free fall proper time is significant because it helps us understand the effects of gravity on the passage of time. It also plays a crucial role in Einstein's theory of general relativity and has practical applications in fields such as space travel and GPS technology.

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