# Proper time for an object falling into a black hole

1. Oct 28, 2006

### MeJennifer

How do you calculate the proper time and the proper distance for an object free falling into a black hole from the event horizon to the singularity?

2. Oct 29, 2006

### pervect

Staff Emeritus
First you have to solve the geodesic equations, which are just 2nd order differential equations. The usual forms for these equations are already paramaterized by proper time, which I will call tau.

The solution of the equations can be greatly eased by the fact that there a couple of constants of motion: ~E and ~L, corresonding roughly to the Newtonian ideas of orbital energy and orbital angular momentum. See for instance http://www.fourmilab.ch/gravitation/orbits/

So you specify ~E and ~L, an easy choice is ~E=1 (zero energy at infnity) and ~L=0 (a straight fall in, no transverse motion).

This gives dr/dtau = sqrt(2m/r), which is easily solved to find r as a function of proper time, giving a result for the above "easy" case which is

r = constant * (tau^2)^(1/3), tau = -infinty ... 0

So r starts out at infinity at tau = -infinty, and decreases to 0 at tau=0 in this solution. The proper time will be infinite if you start at r=infinty, but it will be finite if you start at finite r.

I'm not sure what "proper distance" would be - the "proper time" has a clear interpretation as the time that the falling observer reads on his watch.

3. Oct 29, 2006

### MeJennifer

Well but is the value r that simple? What does r represent here?

Since the closer the object comes to the center the more curvature it will encounter. Distance to the center depends on the curvature right?

It is simply the distance traveled for the free falling object.
So obviously not the distance traveled from the perspective of an observer at a large distance.
So in other words how could we possibly use r for this, unless r is some expression of the curvature?

Last edited: Oct 29, 2006
4. Oct 29, 2006

### pervect

Staff Emeritus
r is the Schwarzschild r coordinate. The set of points of a given, constant value of r can be thought of as a sphereical shell around the black hole, one that has a surface area of 4*pi*r^2. r=0 occurs at the central singularity. r=2M (or in non-geometric units, r = 2GM/c^2) is the "event horizon".

I suppose I should mention that the calculation in my previous post assumes that the interior metric is Schwarzschild, something that strictly speaking isn't likely to be true (it's probalby a BKL metric inside the horizon, though it's Schwarzschild outside the horizon).

The free-falling particle is not in a single inertial frame, and the space-time is curved, so I don't see how one can sensibly define a proper distance. Using "radar" distance shows some of the problems with the idea of "distance to the black hole" - outside the event horion, you'll never get a radar signal return, so there doesn't seem to be any sensible way to define "radar distance" either.

In the particles own frame the distance it travels is of course zero. Thus the time is the only thing that one can measure in the particle's frame.

5. Oct 29, 2006

### MeJennifer

I understand that.
But the issue that I do not understand is how you can use the r coordinate to calculate proper time. Clearly the r coordinate does not translate in some linear measure of distance for a free falling test particle.

Well don't worry everybody seems to do so and then with a straight face declare the value of proper time as if it is a done deal.
To me it is not, but that is probably because I do not understand it

Well exactly, then how can anyone support using r in the calculation for proper time? If we use r we use coordinate distance.

Well since we apparently can use a metric to calculate the proper time of an object in free fall from the event horizon to the singularity what is the argument that we cannot calculate the proper distance traveled between the event horizon and the singularity?

How about calculating the proper distance for an object accelerating away from the black hole, in other words at rest in the gravitational field, just above the event horizon and the singularity?

Last edited: Oct 29, 2006
6. Oct 29, 2006

### pervect

Staff Emeritus
I'm not sure I understand what the difficulty is, but I'll try to go into a little more detail in the hopes it will help.

A parameterized geodesic expresses the worldline of a particle as a function of an arbitrary parameter lambda. It's usually called an "affine parameter" BTW.

So given that the Schwarzschild coordinates are r, theta, phi, and t, one has four functions of lambda:

r(lambda), theta(lambda), phi(lambda), and t(lambda). This specifies an arbitrary worldline in parameterized form. It is a sequence of space-time points that represent the path through space and time that the particle takes.

It turns out that it's convenenient to express the geodesic equations in such a manner that lambda is equal to the proper time.

When this is done one has the r coordinate as a function of proper time - one also has the t coordinate as a function of proper time, etc. When you specify a couple of points on the worldline of the particle, the Lorentz interval integrated over that worldline will be the proper time between those events.

You may be used to thinking of paths expressed as a function of time, i.e rather than expressing

r(lambda)
t(lambda)

we specify only one function, r(t). In that case, one would find

$$d\tau^2 = g_{00} dt^2 + 2 g_{01} dr dt + g_{11} dr^2$$

which can be re-written as

$$d\tau = \sqrt{ g_{00} + 2 g_{01} \frac{dr}{dt} + g_{11} \frac{dr^2}{dt^2} } dt$$

modulo some messing about that's required depending on the sign conventions (the above works as-is if g_00 > 0)

and one finds the proper time by integrating $$\int_d\tau$$. The answer comes out to be the same in either case, whether one has a "traditional" worldline expressed as r(t), or a parameterized worldline which expresses r(lambda) and t(lambda).

If you can draw what you mean by "the proper distance" on a space-time diagram, there's a pretty good chance I can tell you how to calculate it.

It will have to be some space-like curve if the result is going to be a distance, rather than a time. But it's not clear what particular space-like curve you might have in mind when you say "the proper distance".

The most logical curve to integrate the length of is the worldline which the particle follows, but that traces out a time-like curve, not a spacelike one.

A refresher: in SR, the metric coefficients are diagonal (Minkowskian). In such a Minkowskian metric the following defintions apply:

A short segment of a curve dt^2 - dx^2 traces out a time-like interval if dt>dx, and a space-like interval if dt < dx. If dt=dx, it is a null curve.

A curve where any two nearby points are separated by a time-like interval is a time-like curve, if any two nearby points are space-like the curve is space-like, and if all the nearby intervals are zero, the curve is a null curve.

In GR one uses the general formula for the Lorentz interval using the metric coefficients rather than the simpler Minkowskain formulas of SR, i.e in SR one writes dt^2 - dx^2 the equivalent expression in GR is g_ij dx^i dx^j.

Last edited: Oct 29, 2006