Nonlinear relation between coordinate time and proper time

[craigthone]

No, it isn't. Check your algebra. The answer you should be getting is

$$
d\tau^2 = \frac{\left( r - 2GM \right) \left[ r^2 - \left(2GM\right)^2 \right]}{r \left(2GM\right)^2} dt^2
$$

Start with the above formula
$$
d\tau^2 \approx 2 \left( 1 - \frac{2GM}{r} \right)^2 dt_{ H}^2
$$
This is the relation for the infalling observer near the horizon. Then we need to relate the coordinate time $t_H$ to the coordinate time, and also the proper time of distant observer $t_\infty$. The relation is
$$
dt_\infty = \left( 1 - \frac{2GM}{r} \right)^{1/2} dt_{ H}
$$

This can be obtained as this: Suppose at each position there is an observer with fixed $(r,\theta,\phi)$, (called Schwarzschild observer), carrying two clocks. One is the stand clock which record his proper time. One clock records the coordinate time $t$ which will runs at different proper rate at different position. At infinity, the two clocks run at the same rate. At each position, we have
$$d\tau=\left( 1 - \frac{2GM}{r} \right)^{1/2} dt=dt_{\infty}$$

 

[craigthone]

Why does what not work?

If you mean, why isn't the final answer just some numerical factor times $\sqrt{1 - 2GM / r}$, the latter is $d\tau / dt$ for a hovering observer--an observer who is at a constant altitude above the horizon. You should not expect $d\tau / dt$ for an infalling observer to have the same dependence on $r$; the infalling observer is moving relative to the hovering observer, and their relative speed is itself $r$ dependent.

I mean the approach here using the information of the free falling observer rather than using the out going light ray equation.

 

[PeterDonis]

This is the relation for the infalling observer near the horizon.

No, it isn't. We just spent several posts showing that.

Then we need to relate the coordinate time $t_H$ to the coordinate time, and also the proper time of distant observer $t_\infty$.

Coordinate time is the same whether you're at infinity or near the horizon. So there's nothing to relate; it's just $t_H = t_\infty$. And since coordinate time is the same as proper time for the observer at infinity, there's nothing further to relate.

The rest of your post just builds on these two errors.

 

[PeterDonis]

I mean the approach here using the information of the free falling observer rather than using the out going light ray equation.

It should be obvious why that doesn't work: you're leaving out the process of the light traveling outward from the infalling observer emitting it to the observer at infinity receiving it. How can you expect to get a correct answer if you leave that out?

 

[PeterDonis]

I read the initial condition of a frequency source body is at (t_0, r_0) 0< r_0 - GM << 1 close enough to EH.

I don't. I read the paper as saying the following: an observer at infinity (or at least at a very large value of $r$) throws a source of fixed emitting frequency into a black hole at $t = 0$. He watches the light signals coming back to him from the source as it falls and measures their frequency at reception as a function of $t$ (which is coordinate time = the observer at infinity's proper time). Equation (4) is what he finds (more precisely it's the ratio of the reception frequency to the known fixed emission frequency).

 

[Ibix]

Is the notation in the paper potentially slightly confusing? What it's calling $d\tau$ is the proper time of an infalling emitter between the emission of wave crests. What it's calling $dt$ is the proper time of an observer hovering at infinity between the reception of those crests. The latter is, of course, equal to the coordinate time differential between reception events. But it isn't equal to the coordinate time differential between the emission events.

Am I right so far? If so, observe that light pulses emitted radially outward $\Delta t$ apart at the same r arrive $\Delta t$ apart at infinity. Then draw a little triangle in the Schwarzschild t-r plane and start filling in results.

I admit I don't see immediately where the exponential comes from in this procedure. My guess would be the exact result is a truncated Taylor series for the paper's exponential. Or that I'm going about this the wrong way.

 

[PeterDonis]

Is the notation in the paper potentially slightly confusing?

It certainly seems to have confused the OP, yes.

 

[PeterDonis]

Am I right so far?

No. Your next sentence after the one I just quoted is correct, and contradicts your previous paragraph:

light pulses emitted radially outward $\Delta t$ apart at the same r arrive $\Delta t$ apart at infinity.

This is true, but it means that the coordinate time differential between reception events must be the same as the coordinate time differential between emission events. The only thing that changes is the proper time differential between events (since proper time has a different relationship to coordinate time for the emitter and receiver).

In any case, all of this doesn't help in figuring out where equation 1.4 in the paper comes from. I have not had a chance to run through the detailed calculation, but here's a sketch of what's required:

(1) Assume a free-falling object dropped from some large radius $r$ at coordinate time $t = 0$. For simplicity, treat the object as though it was dropped from infinity; this makes the object's worldline simpler to express. Thereby obtain a function $r(t)$ giving the radius $r$ to which the object has fallen at any coordinate time $t > 0$.

(2) Consider a light signal from the infalling object received at the same large radius $r$ at some coordinate time $t > 0$. Find the coordinate time $t_e$ at which this signal must have been emitted by the infalling object, in order to arrive at radius $r$ at coordinate time $t$. This means finding the intersection of two worldlines: the light signal, traced back from its arrival at $t$ at radius $r$; and the infalling object, traced forward from its departure from radius $r$ at time $t = 0$. These two worldlines will intersect at coordinate values $t_e$, $r_e$.

(3) Find the redshift that the light signal will experience in its travel outward from $t_e$, $r_e$ to $t$, $r$. The most straightforward way to do this is to find the infalling object's 4-velocity at the emission event; this gives the initial 4-momentum of the outgoing light signal (since the fixed emission frequency of the signal fixes the inner product of the signal's 4-momentum with the object's 4-velocity). Then parallel transport the light signal's 4-momentum along its worldline to the reception event, and find the inner product of the parallel transported 4-momentum with the 4-velocity of the receiver (who is assumed static at radius $r$).

(4) Since the reception time $t$ was left as a free variable, the result of the above will be a function that gives the frequency of the received signal as a function of $t$. If the paper was correct, this should be the same function that appears in its equation 1.4.

 

[craigthone]

No, it isn't. We just spent several posts showing that.

No, it isn't. Check your algebra. The answer you should be getting is

$$
d\tau^2 = \frac{\left( r - 2GM \right) \left[ r^2 - \left(2GM\right)^2 \right]}{r \left(2GM\right)^2} dt^2
$$

You mean this relation is not for infalling observer?
For infalling observer, we have
$$d\tau^2=\left( 1 - \frac{2GM}{r} \right) dt^2-\left( 1 - \frac{2GM}{r} \right)^{-1} dr^2=\frac{\left( r - 2GM \right) \left[ r^2 - \left(2GM\right)^2 \right]}{r \left(2GM\right)^2} dt^2\approx 2\left(1-\frac{2GM}{r}\right)^2dt^2$$
For Schwarzschild observer, we have
$$d\tau^2=\left( 1 - \frac{2GM}{r} \right) dt^2$$

 

[craigthone]

No, it isn't. We just spent several posts showing that.
Coordinate time is the same whether you're at infinity or near the horizon. So there's nothing to relate; it's just $t_H = t_\infty$. And since coordinate time is the same as proper time for the observer at infinity, there's nothing further to relate.

The rest of your post just builds on these two errors.

I should say that the relation between $dt_{H}$ and $dt_{\infty}$. What we need is just the relation
$$
dt_\infty = \left( 1 - \frac{2GM}{r} \right)^{1/2} dt_{ H}
$$
This relation is from the argument

This can be obtained as this: Suppose at each position there is an observer with fixed $(r,\theta,\phi)$, (called Schwarzschild observer), carrying two clocks. One is the stand clock which record his proper time. One clock records the coordinate time $t$ which will runs at different proper rate at different position. At infinity, the two clocks run at the same rate. At each position, we have
$$d\tau=\left( 1 - \frac{2GM}{r} \right)^{1/2} dt=dt_{\infty}$$

 

[craigthone]

Indeed I had a possible solution to the formula (1.4) which I think is messy.
Consider the out-going radial light ray from event $(t_E,r_E)$ to $(t_R,r_R)$ where signal received position is at infity $r_R \rightarrow \infty$. The light ray equation can be written is
$$0=ds^2=-\left(1-\frac{2GM}{r}\right)dt^2+\left(1-\frac{2GM}{r}\right)^{-1}dr^2=-\left(1-\frac{2GM}{r}\right)dv^2+2dvdr$$
where
$$v=t+r^*=t+r+2GM\log|\frac{r}{2GM}-1|$$The solution to the light ray equation is
$$
\left\{
\begin{aligned}
v=const. \\
v-2r^*=const.
\end{aligned}
\right.
$$
For outside black hole, $r>2GM$, $dr^*/dr >0$, so $v=const.$ implies the ingoing light ray and $v-2r^*=const.$ implies the ougoing light ray which we are interested in.

At the event $E$, i.e. the signal is emitted, $r_E \approx 2GM$, the log term dominates and we have
$$-4GM\log\left(\frac{r}{2GM}-1\right)\approx const.$$
At the event $R$, i.e. the signal is received, $r_R \gg 2GM$, the log term can be ignored and we have
$$v_R-2r_R=t_R-r_R \approx const.$$

Equalizing the constant gives
$$t_R-r_R=-4GM\log\left(\frac{r_E}{2GM}-1\right)$$
$$\frac{r_E}{2GM}-1=\exp\left[ -\frac{1}{4GM}(t_R-r_R)\right]$$
Differentiate it
$$\frac{\Delta r_E}{2GM}=-\frac{\Delta t_R}{4GM}\exp\left[ -\frac{1}{4GM}(t_R-r_R)\right]$$
where $\Delta r_E=|u^r|\Delta \tau$, and $u^r$ is the speed of infalling observer when crossing horizon, something finite. So we have
$$d \tau \propto dt_R \exp\left[ -\frac{1}{4GM}t_R\right]$$

 

[craigthone]

From the discussion above, there seem to be another possible solution.

(1)
Suppose at each position there is an observer with fixed $(r,\theta,\phi)$, (called Schwarzschild observer), carrying two clocks. One is the standard clock which record his proper time. One Schwarzscild clock records the coordinate time $t$ which will runs at different proper rate at different position. At infinity, the two clocks run at the same rate. At each position, we have
$$d\tau=\left( 1 - \frac{2GM}{r} \right)^{1/2} dt=dt_{\infty}$$

(2)
For the infalling observer near the horizon, geodesic equations and velocity square relation gives
$$ \frac{dt}{dr} \approx -\frac{2GM}{r-2GM} \Longrightarrow 1-\frac{2GM}{r} \propto \exp (-t/2GM)$$

(3)
The proper time of infalling observer
$$d\tau^2=\left( 1 - \frac{2GM}{r} \right) dt^2-\left( 1 - \frac{2GM}{r} \right)^{-1} dr^2=\frac{\left( r - 2GM \right) \left[ r^2 - \left(2GM\right)^2 \right]}{r \left(2GM\right)^2} dt^2\approx 2\left(1-\frac{2GM}{r}\right)^2dt^2$$

(4)
Using the relation between $dt$ and $dt_\infty$ in (1), we have
$$d\tau^2\approx 2\left(1-\frac{2GM}{r}\right)dt_\infty^2 \Longrightarrow d\tau \approx \sqrt{2}\left(1-\frac{2GM}{r}\right)^{1/2}dt_\infty $$
Using the relation in (2), we have
$$d\tau \propto \exp(-t/4GM) dt_\infty$$

 

[sweet springs]

It's talking about the frequency of signals received at infinity. Dirac's discussion says nothing about what happens to outgoing light signals after they are emitted from the infalling object; but that has to be included in any derivation of the formula in the paper you linked to.

We get $d\tau/dt$ in the vicinity of EH. Comparison of proper time of falling body with coordinate time,i.e. proper time of an infinite observer, was given. That's all, isn't it? Why do you care how the signal travel to an infinite observer? It may take a long or infinite time for the signal to reach the observer, but we are sure that he observe interval dt given by the formula that has nothing to do with travel between.

 

[PeterDonis]

Why do you care how the signal travel to an infinite observer?

Because equation 1.4 in the paper the OP linked to is the equation for the frequency observed at infinity as a function of time at infinity. If you don't take into account how the signal travels out to infinity, you will not get the right answer for that function. I've already explained this and I don't understand why it's hard to grasp.

 

[PeterDonis]

Consider the out-going radial light ray from event $(t_E,r_E)$ to $(t_R,r_R)$ where signal received position is at infity $r_R \rightarrow \infty$.

At last, you're actually looking at this. However, you're leaving out some things. My post #38 gives a sketch of the full analysis required.

A key thing you are leaving out is: the infalling object is dropped from a very large radius $r$. It starts sending signals back immediately. So the signals the observer at the very large radius (which we can let go to infinity in the limit) receives from the infalling object will not start out being emitted from close to the horizon. So your analysis can't just consider the case of emission close to the horizon; it has to include the entire process of infall and show the exponential decrease of the frequency as received by the faraway observer for that entire process.

This is particularly important when you realize that, if we assume that the infalling object is dropped at $t = 0$, then the light emitted by the infalling object close to the horizon won't be received until a very, very late time $t$. But equation 1.4 in the paper you linked to does not just say the frequency detected at infinity declines exponentially at very late times; it says that for all times after the infalling object is dropped. So for a large proportion of those times, the light being received was not emitted close to the horizon.

 

[craigthone]

Thank all of you for discussions, especially PeterDonis's clear physical picture.