# Falling into black hole

I'm a little confused about what an observer would see while watching something fall into a black hole.
From what I've read, an observer would see the object appear to move slower as it approaches the event horizon, due to time dilation, and it woud take an infinite amount of time to reach the horizon (as seen from observer's frame).

Why doesn't the observer see the light from the object shift in spectrum as it loses energy to the black hole's gravity well, until the object gets to the event horizon? And once the object passes the horizon, wouldn't the object disappear because the light loses all it's energy before it escapes the event horizon?

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DaveC426913
Gold Member
The observer will never see it fall into the EH. The light that the observer sees will be the last light leaving the object before it falls below the EH. Those last few photons being emitted will be stretched over time - forever - and because there are fewer and fewer photons, the object will simply appear to get dimmer and dimmer. True, eventually there will be no photons left to observe, but will will be faaaaar in hte future.

Those last few photons being emitted will be stretched over time
I don't know what you mean by this. Don't the photons move at the speed of light from the object, but they are red-shifted by the gravity of the black hole more and more as the object approaches the EH? So wouldn't the observer would see the object's colors shift out of the visible spectrum?

Or do you mean the rate of photons being emitted is time dilated?

Can't quite wrap my head around this.

JesseM
This section of the Usenet Physics FAQ covers it pretty well. Basically, if the falling object emitted electromagnetic radiation in a completely continuous way and an outside observer could detect radiation at arbitrarily large wavelengths, the observer would see the falling object inching towards the horizon increasingly slowly but never quite reaching it; but since light is emitted in discrete photons and really large wavelengths are too difficult to detect in practice, the reality is that the falling object will just seem to fade away and disappear as it gets very close to the horizon.

JesseM

A very detailed description of the entire process.
This one is good on what would be experienced by an infalling observer, but doesn't talk much about the OP's question about what would be seen by an observer watching an infalling object from outside the event horizon.

K^2
The really crazy bit is that to any external observer, the black hole has not even finished collapsing yet, as that would require matter to cross EH, and external observer cannot observe such a thing.

The really crazy bit is that to any external observer, the black hole has not even finished collapsing yet
"Yet" is what reference frame? :)

We wont see the collapse - correct

But regarding the question "has it already collapsed (now) or not" - it really depends on how you define ‘now’, how you define simultaneity in extremely curved spacetimes. I believe that the “it takes infinite time for BH to form” thing comes from a very old misconception of a BH=frozen star (early 20th century) when no solution of the what is inside BH was known.

K^2
Well, sure, you could define a wonky metric where EH singularity isn't even there. Yet, there exists no valid world line from event of collapse to any exterior event. There do exist valid world lines from any time prior to event, however. So an external observer still has to conclude that it has not yet happened.

If Alice can exchange signals with Bob, then they can measure their time dilation and somehow synchronize their clocks. In a one way situation like BH it is not possible. SO the only way is to use some assumptions based on some calculations, based on some coordinate systems.

In another words, I claim that BH is already formed. How can you falsify this statement?

In another words, I claim that BH is already formed. How can you falsify this statement?
I agree. If someone doubts, he can always go there and check.

lavinia
Gold Member
here is another naive question. I am a total beginner at this.

If an object is moving towards the event horizon of a symmetric non-rotating star wouldn't it disappear before it reached the horizon? This seems to be implied in the Schwartzchild metric.

I was thinking of the case where the object is moving radially at half the speed of light (as measured in global Schwartzchild coordinates i.e. using a far distant clock to measure time and the reduced radius for polar space coordinates). Then the Schwartzchild metric becomes

d$$\tau^{2}$$ = (((1 -2M/r)$$^{2}$$ - .25)/(1 - 2M/r))dt$$^{2}$$

so the object's clock appears to stop before the horizon is reached.

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George Jones
Staff Emeritus
Gold Member
In relativity, care must be used when giving physical meaning to coordinate quantities. Suppose your radially moving observer is coincident with a hovering observer at r = 4M. What is the physical relative speed between these two observers when they are coincident?

JesseM
here is another naive question. I am a total beginner at this.

If an object is moving towards the event horizon of a symmetric non-rotating star wouldn't it disappear before it reached the horizon? This seems to be implied in the Schwartzchild metric.

I was thinking of the case where the object is moving radially at half the speed of light (as measured in global Schwartzchild coordinates i.e. using a far distant clock to measure time and the reduced radius for polar space coordinates). Then the Schwartzchild metric becomes

d$$\tau^{2}$$ = (((1 -2M/r)$$^{2}$$ - .25)/(1 - 2M/r))dt$$^{2}$$

so the object's clock appears to stop before the horizon is reached.
As George Jones suggests, the problem is you can't assume dr/dt has the same meaning in Schwarzschild coordinates that it would in an inertial frame, in particular it's not true that for a light ray dr/dt = c in these coordinates (so an object moving at 0.5c in the local inertial frame of a freefalling object instantaneously at rest in Schwarzschild coordinates would not in general have dr/dt=0.5 in Schwarzschild coordinates). All particles, including photons, will have dr/dt approaching zero as they approach the horizon in Schwarzschild coordinates, taking an infinite amount of Schwarzschild coordinate time to reach it. From p. 835 of Misner/Thorne/Wheeler's Gravitation, here's an illustration showing various worldlines in Schwarzschild coordinates (left) and Kruskal-Szekeres coordinates (right), you can see a lightlike worldline shown as a dotted curve marked "B", with the event horizon as the vertical line at position 2 on the horizontal axis of the Schwarzschild diagram (in KS coordinates the event horizon is a diagonal line at 45 degrees, as is the dotted photon worldline...KS coordinates have the nice property that all lightlike worldlines look like straight lines at 45 degrees, just like in inertial frames):

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lavinia
Gold Member
In relativity, care must be used when giving physical meaning to coordinate quantities. Suppose your radially moving observer is coincident with a hovering observer at r = 4M. What is the physical relative speed between these two observers when they are coincident?
A second passes on the far away clock. The hovering observer records

d$$\tau^{2}$$ = 1/2

in this time he records a shell displacement of

d$$\sigma^{2}$$ = 1/2

So what did I do wrong?

George Jones
Staff Emeritus
Gold Member
A second passes on the far away clock. The hovering observer records

d$$\tau^{2}$$ = 1/2

in this time he records a shell displacement of

d$$\sigma^{2}$$ = 1/2
So, according to the hovering observer, with what physical speed does the radially moving observer whiz by?

lavinia
Gold Member
So, according to the hovering observer, with what physical speed does the radially moving observer whiz by?
sadly i get 1, the speed of light.

George Jones
Staff Emeritus
Gold Member
sadly i get 1, the speed of light.
Why sadly? You have given a coordinate condition, dr/dt = 1/2, for a radially moving observer. This coordinate condition is physically unrealizable for r <= 4M. It is possible to give lots of physically unrealizable coordinate conditions. This is why care is needed, particularly in general realtivity, but also in special relativity, when imposing coordinate conditions.

lavinia
Gold Member
Why sadly? You have given a coordinate condition, dr/dt = 1/2, for a radially moving observer. This coordinate condition is physically unrealizable for r <= 4M. It is possible to give lots of physically unrealizable coordinate conditions. This is why care is needed, particularly in general realtivity, but also in special relativity, when imposing coordinate conditions.
I was just going to ask if the condition is impossible. Very cool.

But now I am worried. I don't understand why physically the distant observer can not observe an object moving at half the speed of light when the object is too close to the event horizon.

George Jones
Staff Emeritus
Gold Member
What is is 4-velocity

$$u = \frac{dt}{d \tau} \frac{\partial}{\partial t} + \frac{dr}{d \tau} \frac{\partial}{\partial r}$$

for the radially moving observer? This is the tangent vector to the worldline (integral curve) of the observer. This vector is timelike only when r > 4M.

It might also be physically interesting to calculate

$$g \left( \nabla_u u , \nabla_u u\right)$$

using the standard metric-compatible, torsion-free connection.

My wife and daughter are in the process of dragging me away from the computer and out the door to do some shopping. I hope to get back to this.

JesseM
But now I am worried. I don't understand why physically the distant observer can not observe an object moving at half the speed of light when the object is too close to the event horizon.
As I said in post 16, you can't treat dr/dt=0.5 as equivalent to "half the speed of light" in Schwarzschild coordinates, light itself doesn't move with dr/dt=1 in these coordinates.

Yeah, one must be careful.
Otherwise, to some, E=MC^2 "proves" that photons must be capable of c^2, when in fact it is not implied at all.

JesseM
Yeah, one must be careful.
Otherwise, to some, E=MC^2 "proves" that photons must be capable of c^2, when in fact it is not implied at all.
Of course, c^2 doesn't even have units of velocity, so it's pretty easy to explain why this one is wrong!

lavinia
Gold Member
As I said in post 16, you can't treat dr/dt=0.5 as equivalent to "half the speed of light" in Schwarzschild coordinates, light itself doesn't move with dr/dt=1 in these coordinates.
Your post was too advanced for me but I see what you are saying. Not only is dr/dt = 1/2 impossible for r < 4M but in Schwarzchild coordinates dr/dt is not in units of the speed of light.
So a dr/dt value of 1/2 does not mean 1/2 the speed of light.

I guess if one just realizes that if dr/dt were 1/2 when r = 4M then

d$$\tau$$

would equal zero which is impossible.

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lavinia
Gold Member
What is is 4-velocity

$$u = \frac{dt}{d \tau} \frac{\partial}{\partial t} + \frac{dr}{d \tau} \frac{\partial}{\partial r}$$

for the radially moving observer? This is the tangent vector to the worldline (integral curve) of the observer. This vector is timelike only when r > 4M.

It might also be physically interesting to calculate

$$g \left( \nabla_u u , \nabla_u u\right)$$

using the standard metric-compatible, torsion-free connection.
- I get that for r = 4M the proper time is zero and for r < 4M that it is negative. So for these values of r the four velocity is not time like as you said.

- I will have to derive the covariant derivative from the metric. Stay tuned.