A-wal said:
Because it describes something important. Free-falling without tidal force is the equivalent to moving freely, while tidal force describes acceleration.
I *think* you're saying the same thing that I was saying in what you referred to here:
A-wal said:
You've lost me. "... would feel acceleration, which we would interpret as being caused by the internal forces on each part by the other parts. If such an object were also affected by tidal gravity..." Also affected by tidal gravity? Same thing isn't it?
However, I'm not sure, and it's irrelevant to the main point (whether or not a free-falling observer can cross the horizon). So for the time being, anyway, I'd like to table the whole issue of tidal "force". Let's assume for the time being that we are dealing with a black hole that is so large that the tidal gravity at the horizon is negligible. (We can do this because the tidal gravity at the horizon goes like 1 / M^{2}, so we can make the tidal gravity as small as we like by making M large enough.)
A-wal said:
You can plot a course and say there's enough proper time to reach the horizon before it dies, but I think it would change as you get closer.
The proper time along a particular curve in spacetime *doesn't change*--it's a geometric invariant. Saying it can change as you get closer to the horizon is like saying the distance from New York to Los Angeles can change while you're passing through Las Vegas. But instead of just repeating this, since you asked me to go through the computation, here it is. Once we've gone through it, I think a lot of your other questions will be easily answered.
(BTW, I'm basing this on Misner, Thorne, and Wheeler, Chapter 25, but you'll find similar calculations in just about any textbook on General Relativity.)
The general method we will use is as follows: we define a curve in spacetime along which we want to integrate the metric to obtain a "length"--either proper time elapsed for an observer who has that curve as his worldline (if the curve is timelike), or proper distance seen by an observer moving on a worldline orthogonal to the curve (if the curve is spacelike). We then write the metric in appropriate form, using the equation for the curve to reduce the number of integration variables to one. This will (hopefully) allow us to evaluate the integral.
For the case of proper time elapsed for a freely falling observer to reach the horizon from some radius R > 2M, we first define the curve along which we will do the integral as a purely radial curve, so theta = pi / 2 (this makes the math easiest) and phi = 0 all along the curve. We start at r = R (some value >> 2M) and t = 0. We end at t = infinity. (Minor technical point: I have idealized the curve we're using so that, strictly speaking, it is the worldline of an observer "falling in from rest at infinity"--that is, at any finite value of r, this worldline has a non-zero inward velocity. This doesn't change the essential conclusion--that an observer falling from any finite radius will reach r = 2M in a finite proper time--but it makes the calculation simpler.)
The key thing we need to complete the definition of the curve is a function r(t)--or t(r)--that will let us write everything in terms of a single coordinate. For a freely falling, purely radial curve, this function is implicitly given by the equation for the coordinate velocity:
\frac{dr}{dt} = - \sqrt{\frac{2M}{r}} \left( 1 - \frac{2M}{r} \right)
This let's us write the metric as follows:
d\tau^{2} = dt^{2} \left[ \left( 1 - \frac{2M}{r} \right) - \left( \frac{dr}{dt} \right)^{2} \frac{1}{1 - \frac{2M}{r}} \right] = dt^{2} \left( 1 - \frac{2M}{r} \right)^{2}
which gives, on taking the square root and integrating,
\tau = \int_{0}^{\infty} \left( 1 - \frac{2M}{r} \right) dt
As it stands, this integral can't be evaluated, because we still have r in the integrand. To do the integral directly we would need to first solve the differential equation above in dr/dt (which is not at all straightforward) in order to get an explicit expression for r(t) to substitute into the integrand (which would still leave us with a very messy integral to do).
The usual way of dealing with this is to rewrite the integral in terms of dr, which makes it easy to evaluate. However, before doing this, we need to satisfy ourselves that as t -> infinity, r -> 2M. We can show this by noting two things about the equation above for dr / dt: first, dr / dt is negative for any r > 2M, so r will decrease with t until r = 2M is reached (i.e., for any finite value of t, if r > 2M, r will still be decreasing, so it will be closer to 2M at some larger finite value of t); second, as r -> 2M, dr / dt -> 0, so we don't expect r to actually reach 2M at any finite value of t (because the closer it gets, the slower it approaches), but only asymptotically as t -> infinity. (There's probably a slick mathematically way of proving this formally.)
Having shown that, we can invert the equation for dr / dt to obtain:
\frac{dt}{dr} = - \sqrt{\frac{r}{2M}} \frac{1}{1 - \frac{2M}{r}}
We can then rewrite the metric as:
d\tau^{2} = dr^{2} \left[ \frac{r}{2M} \frac{1}{1 - \frac{2M}{r}} - \frac{1}{1 - \frac{2M}{r}} \right] = dr^{2} \frac{1}{1 - \frac{2M}{r}} \left( \frac{r}{2M} - 1 \right) = dr^{2} \frac{r}{2M}
Taking the square root and integrating gives:
\tau = \int_{2M}^{R} \sqrt{\frac{r}{2M}} dr = \frac{2}{3 \sqrt{2M}} \left[ \left( R \right)^{\frac{3}{2}} - \left( 2M \right)^{\frac{3}{2}} \right]
This is often re-written to put all the variables in dimensionless form, which makes things look more elegant:
\frac{\tau}{2M} = \frac{2}{3} \left[ \left( \frac{R}{2M} \right)^{\frac{3}{2}} - 1 \right]
A-wal said:
I know what you said. How can the *circumference* and *area* be invariant if they're not in the radial direction? How is it not true that time dilation shortens the length of time something exists? Length contraction would extend the length between the observer and the singularity, and therefore the distance between the observer and the rainbow horizon. The "experiences it slower" means that time dilation makes everything else speed up. It doesn't matter if an observers proper future time is infinite. I would dispute that, but it's besides the point. Even an observer with infinite potential future time will be unable to reach an event horizon in any given amount of finite time.
Using the general method outlined above, here are the answers to the substantive questions you've raised. I'll take them in a somewhat different order from the order in which you raise them.
First, time dilation. Consider the following two curves: (A) r = some value close to 2M, theta = pi / 2, phi = 0, t = 0 to infinity; (B) r = some value much much larger than 2M, theta = pi / 2, phi = 0, t = 0 to infinity. Both of these curves have infinite "potential future time". However, if I pick any positive finite value of t (say t = 100), the proper time elapsed along curve A up to that value of t will be much less than the proper time elapsed along curve B up to that value of t. This is what we mean when we say that there is "time dilation" close to the horizon, and it's *all* we mean.
In particular, note that, for an infalling observer, the r coordinate is *not* constant, and (as we'll see below) the "lines of simultaneity" for the infalling observer are *not* the same as those for the hovering observer, so there's no straightforward way to say what event on the hovering observer's worldline "corresponds" to a given event on the infalling observer's worldline. This means that there's no straightforward way to assess the "time dilation" of the infalling observer relative to the hovering observer (or to one far away). In particular, as we've seen, the fact that there is a sense in which "time dilation becomes infinite" at the horizon does *not* imply that an infalling observer cannot reach the horizon in a finite proper time--he does.
Now consider a third curve: (C) t = 0, theta = pi / 2, phi = 0, r = R (some value > 2M) to 2M. This is a spacelike curve whose length is the "proper distance" from curve (A) to the horizon (as seen by a "hovering" observer--see below). The integral is similar to the one we did above: the only term in the metric that matters is the dr^{2} term, so we have:
s = \int_{2M}^{R} \frac{1}{\sqrt{1 - \frac{2M}{r}}} dr = \int_{2M}^{R} \sqrt{\frac{r}{r - 2M}} dr
This is not the sort of integral that one can do "by inspection" (at least, I can't), but that's what tables of integrals are for.
Consulting one, we find that the indefinite integral is:
\left[ \sqrt{r \left( r - 2M \right)} + 2M ln \left( \sqrt{r} + \sqrt{r - 2M} \right) \right]
which we then have to evaluate from r = 2M to r = R. The 2M endpoint is simple since everything but one term is zero, and we end up with:
s = \sqrt{R \left( R - 2M \right)} + 2M ln \left( \sqrt{\frac{R}{2M}} + \sqrt{\frac{R}{2M} - 1} \right)
where we have used the fact that subtracting logarithms is the same as dividing the arguments. In dimensionless form, this becomes:
\frac{s}{2M} = \sqrt{\frac{R}{2M} \left( \frac{R}{2M} - 1 \right)} + ln \left( \sqrt{\frac{R}{2M}} + \sqrt{\frac{R}{2M} - 1} \right)
Armed with this, we can now investigate "length contraction". The answer we just derived was for a "line of simultaneity" in the frame of a hovering observer (which will also be a line of simultaneity for an observer very far away--in fact, for any observer who remains forever at a constant r coordinate; the only difference will be the specific value of R that we plug into the above formula). The line of simultaneity of an infalling observer will be different because that observer is changing his radial coordinate with time, and so the "proper distance" to the horizon seen by such an observer might also be different.
To calculate this, we need an equation for the line of simultaneity of the infalling observer. This will be a spacelike geodesic that is orthogonal to the infalling observer's worldline. We could laboriously write out an equation for that geodesic in Schwarzschild coordinates, but it's easier to transform to Painleve coordinates, in which the metric is:
ds^{2} = - \left( 1 - \frac{2M}{r} \right) dT^{2} + 2 \sqrt{\frac{2M}{r}} dT dr + dr^{2} + r^{2} \left( d\theta^{2} + sin^{2} \theta d\varphi^{2} \right)
where T is the "Painleve time" defined by
dT = dt - \sqrt{\frac{2M}{r}} \frac{1}{1 - \frac{2M}{r}} dr
In these coordinates, lines of constant T are the "lines of simultaneity" for the infalling observer, who moves on a worldline defined by
\frac{dr}{dT} = - \sqrt{\frac{2M}{r}}
This makes it easy to calculate the proper distance to the horizon from a radius R for the infalling observer; if dT = 0 (and also theta and phi are constant, as above), the metric is simply ds = dr[/tex], and we have<br />
<br />
s = R - 2M<br />
<br />
or, in dimensionless form,<br />
<br />
\frac{s}{2M} = \frac{R}{2M} - 1<br />
<br />
for the proper distance from radius R to the horizon at r = 2M. Note that this number is *smaller* than the proper distance for a hovering observer that we derived above. So the proper distance to the horizon *is* "length contracted" for an infalling observer. But, as we've seen, the "length contraction" and "time dilation" do *not* imply that the infalling observer can't reach the horizon. (By the way, you can also use the Painleve metric to check that I did the proper time integral correctly above; just integrate the metric from r = R to r = 2M, using the equation for dr / dT, in inverted form, to write the integral in terms of dr. You should get the same answer I got above.)<br />
<br />
Finally, let's look at the circumference and area of the horizon. The circumference of the horizon, as seen by a "hovering" observer (any observer with a constant r coordinate), is simply the invariant length of the curve: r = 2M, t = constant (any value will do), theta = pi / 2 (as above, this value makes the math easiest since it makes a "great circle" around the horizon a curve where only phi varies), phi = 0 to 2 pi. Since only phi changes, the integral is easy: we can ignore every term in the metric except the d phi term, so we get:<br />
<br />
s = \int_{0}^{2 \pi} r sin \theta d\varphi = 4 \pi M.<br />
<br />
The area calculation is similar; we just need to vary theta from 0 to pi as well as phi over the range just given.<br />
<br />
What will this calculation look like for the infalling observer? It will be the same except that we look at a constant "Painleve time" T instead of a constant "Schwarzschild time" t. But, as we saw from the Painleve metric above, the r-coordinate is still a constant, 2M, in both cases. So the integral for the infalling observer *looks exactly the same* as the one for the hovering observer. The circumference of the horizon does *not* "length contract". (Neither does the area, for the same reason.) Fundamentally, the circumference and area are the same for all observers because the r-coordinate of the horizon is the same for all observers.
It was just a comment.
