I Vary acceleration, maximize dτ, Schwarzschild metric

1. Feb 24, 2017

Spinnor

Two masses, m and M, are a fixed distance R apart. One of the masses is much larger then the other. At time t the masses start to fall towards each other. Using Newton's Law of Gravitation we can determine the acceleration of the small mass. Can one use the Schwarzschild metric in the following way to find the acceleration of the small mass?

Assume the small mass initially has some acceleration a. Treat acceleration a as a variable. After falling for some small time dt, using the acceleration a we can determine dr. We can now plug both dt and dr into the Schwarzschild metric and get a value for dτ, the proper time interval? If we now plot dτ as a function of initial acceleration a we should get a peak for dτ when the acceleration is that value given by Newton's Law?

Thanks.

2. Feb 25, 2017

pervect

Staff Emeritus
It sounds like the approach E.F. Taylor used in "A Call to Action". <<link>>. As long as one is careful to consider a small enough time duration, the answer is basically yes, if I am understand you correctly. Taylor has referred to this approach as the "principle of maximal aging", though in some later papers he is a bit more formal and suggests that "the principle of extremal aging" or even "the principle of stationary aging" as a more technically accurate description of the fundamental principle. For more on this point and where it makes a difference, see for instance "When Action is not Least" <<link>>, co-authored with C.G Gray.

I believe Taylor uses a similar approach in his book "Exploring black holes", an interesting book that tries to present much of GR at an undergraduate level. But this is from memory, not something I've double-checked by referring to the source.

Last edited: Feb 25, 2017
3. Feb 27, 2017

Spinnor

I think there is a problem with my method. What value do we use for r in the Schwarzschild metric? Can anyone see where I went wrong with my idea?

Thanks!

4. Feb 27, 2017

PAllen

There is a problem with your method, and it is instructive to explain what it is. It has nothing to do with choice of r. Assuming your method was ok, you just choose whatever r you want, because newtonian equivalent acceleration will vary with r, so you pick an r and determine it for that r.

If I understand correctly, you are proposing to use metric values at one specific r, some chosen small dt, with dr determined from dt via 'a'. The fundamental problem with this is that you are only using metric coefficients at one point (I guess you could say you are varying time, but for a static metric, this still prevents you from getting change in metric). However, only changes in metric values determine acceleration. By using values only at one point, you are assuming constant metric over small but finite deltas. No matter what metric values you use, constancy equals no curvature (any metric values - up to sign - are compatible with flat spacetime of some suitably strange units). As a result, if you apply your method, you will find that a=0 always maximizes proper time, which is correct for a metric with no curvature over a small finite region. Note that rigorously, Newtonian equivalent acceleration is tied to what are called connection coefficients or Christoffel symbols. These are expressed in term of metric derivatives. They thus vanish for any constant metric, so no acceleration will occur without metric change.

I take the spirit of your idea, which is very instructive, as : deduce Newtonian acceleration from the Schwarzschild metric without recourse to any tensor methods, or even Euler-Lagrange equations, while still using the notion of maximal proper time as characteristic of free fall. Then you must proceed as follows:

- evaluate the metric coefficients at some very close values of r, e.g. r0 and r1. This brings in change of metric, effectively doing finite difference approximation of a derivative.

- integrate d tau from r0 to r1 e.g. via trapezoid rule. Here you can express dt in terms of dr and 'a'. This becomes a discrete approximation of the action.

- Then you can maximize the result over 'a'. This should give a good approximation given close enough r values.

Last edited: Feb 27, 2017
5. Feb 27, 2017

Staff: Mentor

Not really, because that approach does not start with the acceleration, not even as a free variable. It starts with the action. See below.

There is, but it's not picking a value for $r$.

The goal you are trying to reach is fine: you are trying to figure out what worldline--path through spacetime--maximizes the proper time $\tau$. But the approach you are taking to reach that goal is backwards. In order to maximize $\tau$, you first have to carefully define what you are maximizing it with respect to, i.e., what you are going to hold fixed and what you are going to vary. Or, to put it another way, if you say you have found the worldline that maximizes $\tau$, what set of worldlines did you pick it from?

The standard way of doing this is to pick two events in spacetime and compare all possible worldlines that start at one event and end at the other. In other words, we hold the endpoints of the worldline fixed and allow the worldline to vary. Then we can write an equation for the elapsed proper time along a given worldline, formally, as follows:

$$\tau = \int d\tau = \int \sqrt{ \left( 1 - \frac{2M}{r} \right) dt^2 - \frac{1}{1 - 2M / r} dr^2} = \int_{t_0}^{t_1} dt \sqrt{ 1 - \frac{2M}{r(t)} - \frac{1}{1 - 2M / r(t)} \left( \frac{dr}{dt} \right)^2}$$

Here I have assumed that we can express the worldline as a function $r(t)$ so we can use $t$ as the integration variable. Then we use the Euler-Lagrange equation associated with the integral to find the function $r(t)$ that maximizes $\tau$.

The point of all this is that the acceleration, $d^2 r / dt^2$, doesn't appear anywhere in the above--it only appears when we compute the Euler-Lagrange equation. That is when we find out that to maximize $\tau$, we must have $d^2 r / dt^2$ equal to the Newtonian value (with corrections due to the Schwarzschild metric coefficients).

6. Feb 27, 2017

Spinnor

Long day. Will try and digest your help tonight. Thanks Peter and PAllen for your help!

7. Feb 28, 2017

PAllen

Hoping not to confuse the issue, but I played with carrying out my prior suggestion to arrive at an acceleration without using the Euler-Lagrange equations. It bumps into some interesting technical problems, that are messy to fix.

First, one can clarify my correction of the OP method as simply computing the approximate proper time integral of functions of form r(t) = r0 + 1/2 a t2, treated as an approximation of any function with the constraint of derivative being zero at t=0, and using only one further Taylor term. One has then reduced the variation to a single parameter 'a' as desired by the OP. Such simplifications often work for simple, well behaved, variational problems (though there is obviously little point to such a method when you know the correct method). In this case, there is a hiccup. To create a simple numeric integration, we assume we are going from r0 to r1, with t bound determined from 'a', not assumed a-priori. Then one finds that (even if a few integration steps are used), a=0 always maximized proper time, which is not the desired result. Further, even if you put this limited r(t) function into the correct integral, with the bound of integration of time being specified as √(2(r1-r0)/a), and seek the maximizing 'a', you find that 'a' approaching 0 is maximizing.

The problem is that this simple way of setting the problem up maximizes of the wrong set of trajectories. Instead of considering all paths between two events, we are considering paths with all different end events. Lo and behold, for this family of trajectories, acceleration approaching zero is always maximizing. It is, in fact true, and obvious in hindsight, that the proper time maximizing way to get an object from one radius to another with no other constraint, is simply to move it as slowly possible. To get the intended free fall acceleration, one would have to specify both radii and time delta, and use at least r(t) = r0 + v0t + 1/2 a t2, where v0 is determined in terms of 'a' and time delta by the requirement that the second radius is reached after chosen time delta, giving an overall result in terms of 'a'. Then, a single trapezoid integration step in time can be set up, and maximized over 'a'. By making the radii very close and delta t very small, you should finally get close to the correct answer. Using the Euler-Lagrange equation is far simpler.

Last edited: Feb 28, 2017
8. Feb 28, 2017

Staff: Mentor

Exactly; this is the key point I was referring to in my previous post.