I Nonlinear relation between coordinate time and proper time

[craigthone]

For Schwarzschild geomery
$$ds^2=-(1-\frac{2GM}{r})dt^2+(1-\frac{2GM}{r})^{-1}dr^2+r^2d\Omega^2$$
For a Schwarzschild observer , the proper time and coordinate time are related by
$$d\tau=(1-\frac{2GM}{r})^{1/2}dt$$
There is a often used relation between proper time and coordinate time
$$d\tau \sim \exp(-\frac{t}{4GM}) dt$$
I do not know how to relate the two relation, i.e. representing position $r$ in terms of coordinate time $t$. Are there any suggestions?

 

[PeterDonis]

There is a often used relation between proper time and coordinate time

Where are you getting this relation from?

 

[craigthone]

Where are you getting this relation from?

In this paper https://arxiv.org/abs/1505.08108, formula (1.4).
This is the redshift of signals of the infalling observer through the horizon. I know a messy way to derive this. But I want to know if there is some simple way to do it since the relation is so simple. Maybe it is a exercise of some textbook, but I do not remember it.

 

[PeterDonis]

In this paper https://arxiv.org/abs/1505.08108, formula (1.4).

This is not a relation between proper time and coordinate time. It's a relation between the time at infinity $t$ since the instant you dropped something towards a black hole, and the frequency of light you receive at infinity from that object.

 

[PeterDonis]

This is the redshift of signals of the infalling observer through the horizon.

It's the dependence of the redshift of signals you receive at infinity on the Schwarzschild coordinate time $t$ (which is the same as the observer's time at infinity) since you dropped the object. It's not the redshift of a single signal. Nor is it the relationship between proper time and coordinate time for any Schwarzschild observer, which is what you appeared to be comparing it to in the OP.

In short, the paper you linked to is talking about something completely different from the relationship between proper time and coordinate time for a Schwarzschild observer, which is what this thread is about (per your own thread title). So if you want to talk about what's in the paper you linked to, you should start a separate thread on that separate topic.

 

[craigthone]

This is not a relation between proper time and coordinate time. It's a relation between the time at infinity $t$ since the instant you dropped something towards a black hole, and the frequency of light you receive at infinity from that object.

Thanks! I know my mistake now. It is the relation about proper time of signals and coordinate time of signals. Is there any simple way to get it?

 

[PeterDonis]

It is the relation about proper time of signals and coordinate time of signals.

No, it isn't. You are misunderstanding the entire topic of the paper you linked to; it has nothing whatever to do with the proper time vs. coordinate time of anything.

 

[craigthone]

No, it isn't. You are misunderstanding the entire topic of the paper you linked to; it has nothing whatever to do with the proper time vs. coordinate time of anything.

$d\tau$ is proper time of signals.
$dt$ is the time meaured at infinity. You mean I can not call it coordinate time of signals? why?

 

[PeterDonis]

$d\tau$ is proper time of signals.

No, it isn't. The signals are light signals and have no proper time. To the extent it's the proper time of anything, it's the proper time of the infalling object that is emitting the signal; $d\tau$ is the proper time interval between two successive wave crests of the signal, along the infalling object's worldline, when the signal is emitted.

$dt$ is the time meaured at infinity.

The time measured at infinity between two successive wave crests when the signal is received at infinity.

You mean I can not call it coordinate time of signals?

Not in any meaningful sense. The expression $d\tau / dt$ is just the reciprocal of the frequency of the signal when it's received at infinity.

 

[craigthone]

Thanks for your correction

 

[craigthone]

No, it isn't. The signals are light signals and have no proper time. To the extent it's the proper time of anything, it's the proper time of the infalling object that is emitting the signal; $d\tau$ is the proper time interval between two successive wave crests of the signal, along the infalling object's worldline, when the signal is emitted.
The time measured at infinity between two successive wave crests when the signal is received at infinity.
Not in any meaningful sense. The expression $d\tau / dt$ is just the reciprocal of the frequency of the signal when it's received at infinity.

I do want to know if there is some simple way to derive the relation.

 

[PeterDonis]

I do want to know if there is some simple way to derive the relation.

I don't know. The paper just states it without proof, and doesn't give any reference.

 

[sweet springs]

I find the trajectory of bodies falling in BH
t=-2m log(r-2m) + const where m=GM in section 19 of General Theory of Relativity, PAM Dirac. I hope this will help you.

PDF is available at http://amarketplaceofideas.com/wp-c...20-%20General%20Theory%20Of%20Relativity1.pdf

 

[PeterDonis]

I find the trajectory of bodies falling in BH

This gives the Schwarzschild coordinate time along the body's trajectory as a function of radial coordinate $r$, assuming that the body is released from infinity at $t = - \infty$.

 

[sweet springs]

assuming that the body is released from infinity at t=−∞t = - \infty.

No, I am afraid. Dirac there does an approximation that r = 2m + $\epsilon$ where more than second order of $\epsilon$ is neglected. This relation is applicable just near above EH.

 

[PeterDonis]

Dirac there does an approximation that r = 2m + ϵ where more than second order of ϵ is neglected.

Ah, you're right. I was confusing what you quoted with a different formula.

 

[craigthone]

I find the trajectory of bodies falling in BH
t=-2m log(r-2m) + const where m=GM in section 19 of General Theory of Relativity, PAM Dirac. I hope this will help you.

PDF is available at http://amarketplaceofideas.com/wp-content/uploads/2014/08/P%20A%20M%20Dirac%20-%20General%20Theory%20Of%20Relativity1.pdf

I think this works.
For the radial infalling observer near the horizon, $r \rightarrow 2GM$ in the Schwarzschild coordinate, so $dr \approx 0$ and we have
$$d\tau \approx (1-\frac{2GM}{r})^{1/2}dt$$
While from formula of Dirac
$$\frac{r}{2GM}-1 \propto \exp(-t/2GM)$$
Combining the above two formulas, we have
$$ d\tau \propto \exp(-t/4GM) dt $$

 

[PeterDonis]

I think this works.
For the radial infalling observer near the horizon

This can't be all there is to it, because the formula in the paper you linked to is not talking just about phenomena near the horizon. 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.

In short, you can't just wave your hands and look for formulas with exponentials in them and say that's a derivation.

$r \rightarrow 2GM$ in the Schwarzschild coordinate, so $dr \approx 0$

This is not correct. The infalling object does not stop at $r = 2GM$, and $dr / d\tau$ does not vanish there.

 

[craigthone]

This can't be all there is to it, because the formula in the paper you linked to is not talking just about phenomena near the horizon. 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.

In short, you can't just wave your hands and look for formulas with exponentials in them and say that's a derivation.
This is not correct. The infalling object does not stop at $r = 2GM$, and $dr / d\tau$ does not vanish there.

For the distant observer, he has $t\rightarrow \infty$, and $r\rightarrow 2GM$. I think $dr=0$ is OK.

 

[PeterDonis]

For the distant observer, he has $t\rightarrow \infty$, and $r\rightarrow 2GM$.

Huh? The distant observer is at $r = \infty$.

I think $dr=0$ is OK.

No, it isn't, because you claimed to be deriving the expression

$$
d\tau \approx \left( 1 - \frac{2GM}{r} \right)^{\frac{1}{2}} dt
$$

for the infalling observer. In order for that to be correct, it would have to be the case that $dr / d\tau \rightarrow 0$ as $r \rightarrow 2GM$. But it isn't. The fact that $dr / dt \rightarrow 0$ as $r \rightarrow 2GM$ is irrelevant.

 

[sweet springs]

Dirac

t=-2m\ log(r-2m) + const

const is explisitly written as

\frac{exp(-t/2m)}{exp(-t_0/2m)}=\frac{1-2m/r}{1-2m/r_0}
where initial position of the falling body m/r_0 <<1 at initial time t=t_0

d\tau=(1-m/r_0)\frac{exp(-t/4m)}{exp(-t_0/4m)}dt

 

[PeterDonis]

Dirac

Is not relevant for the calculation we are trying to do in this thread.

 

[craigthone]

Huh? The distant observer is at $r = \infty$.
No, it isn't, because you claimed to be deriving the expression

$$
d\tau \approx \left( 1 - \frac{2GM}{r} \right)^{\frac{1}{2}} dt
$$

for the infalling observer. In order for that to be correct, it would have to be the case that $dr / d\tau \rightarrow 0$ as $r \rightarrow 2GM$. But it isn't. The fact that $dr / dt \rightarrow 0$ as $r \rightarrow 2GM$ is irrelevant.

Thanks for your corrections again.
How about the following argument, for the infalling observer near the horizon we have
$$\frac{dt}{dr}\approx -\frac{2GM}{r-2GM}=-\frac{2GM}{\varepsilon}$$
This is the formula from Dirac (page 33). Then the proper time for the infalling observer
$$d\tau^2=\left( 1 - \frac{2GM}{r} \right) dt^2-\left( 1 - \frac{2GM}{r} \right)^{-1} dr^2$$
Then from the $dt$ and $dr$ relation, we know that the $dr^2$ term in the proper time can be neglected compared with the $dt^2$ term.

 

[PeterDonis]

from the $dt$ and $dr$ relation, we know that the $dr^2$ term in the proper time can be neglected compared with the dt2dt2dt^2 term.

No, we don't. Look at the coefficient of the $dr^2$ term; it goes to infinity as $r \rightarrow 2GM$.

Or, more rigorously, plug the following into the metric:

$$
dr^2 = \frac{ \left( r - 2GM \right)^2 }{ \left( 2GM \right)^2 } dt^2
$$

You get:

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

Once you've canceled common factors in the second term, both terms go like $r - 2GM$, so both are of similar magnitude.

 

[craigthone]

No, we don't. Look at the coefficient of the $dr^2$ term; it goes to infinity as $r \rightarrow 2GM$.

Or, more rigorously, plug the following into the metric:

$$
dr^2 = \frac{ \left( r - 2GM \right)^2 }{ \left( 2GM \right)^2 } dt^2
$$

You get:

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

Once you've canceled common factors in the second term, both terms go like $r - 2GM$, so both are of similar magnitude.

Yes, you are right and you give the answer. Though $dr^2$ terms can not be ignored, it has the same form as $dt^2$ term. In the end we get the relation between $d\tau$ and $dt$. Thanks for all your posts.

 

[craigthone]

$$d\tau^2=\left( 1 - \frac{2GM}{r} \right) dt^2-\left( 1 - \frac{2GM}{r} \right)^{-1} dr^2$$
$$d\tau^2=\left( 1 - \frac{2GM}{r} \right) dt^2-\left( 1 - \frac{2GM}{r} \right)^{-1}\left(\frac{r-2GM}{2GM}\right)^2dt^2\approx 2\left(1-\frac{2GM}{r}\right)dt^2$$
$$d\tau\approx \sqrt{2}\left(1-\frac{2GM}{r}\right)^{1/2}dt$$

This is what we want.

 

[PeterDonis]

This is what we want.

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

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

 

[craigthone]

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

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

You are right again. Why does it not work? This is strange to me.

 

[PeterDonis]

Why does it not work?

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.

 

[sweet springs]

The paper says
----------------
In a black hole, there are two notable exponential behaviors. If a source of fixed frequency is thrown into a black hole, the frequency seen at infinity will be exponentially reshifted,
(1.4)
----------------

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. Proper time of the source will be slowed down exponentially approaching to zero in comparison with the world time t.

Even if it was released more far away, it takes only finite proper time to come the initial position above so it does not matter in the discussion of the author who is focusing on Exponential Behaviors.

 

[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.