Entering a black hole -- I know I must be wrong (Help me understand why)

[Somoth Ergai]

TL;DR Summary

Based on a new explanation I read about what happens from the perspective of an outside observer, It seems like it should take an infinite amount of time to enter a black hole, even from the perspective of the object entering it.

From what I understand about time dilation and the relativity of simultaneity; if we imagine two people near a black hole and one of them begins to approach the black hole on a trajectory that crosses the event horizon. The stationary observer will never see the moving observer enter the black hole. And in fact it will take an infinite amount of time to even simply reach the "edge" of the event horizon. as past this point no future events can be observed from the outside.

the explanation I saw recently says that form the perspective of the moving observer they do in fact see themselves enter the black hole. This doesn't seem right to me. Suppose they are entering the black hole feet first. Since there is a distance from their feet to their head it seems to me that time dilation would mean their feet take an infinite amount of time to reach the edge of the horizon too. and likewise their knees and on up the rest of their body.

Which also implies, to me, that they would sort of get "squashed" as the rest of there body got closer and closer to the horizon without any of it ever actually entering or crossing over. which would also mean the exact opposite of spaghettification would happen. Since every physics source tells me that spaghettification does happen, I know my interpretation must be wrong. I'm just having a hard time understanding why. Any help is appreciated.

 

[Ibix]

Time dilation is a concept that applies between hovering clocks above the event horizon. Someone falling in to the black hole is not hovering. Furthermore, time dilation is a concept whose derivation fails at the event horizon. Trying to analyse a horizon crossing in such terms is a self-contradiction and you will not get coherent sense from it.

Ultimately the reason that a distant observer cannot see someone falling in to a black hole is that light cannot leave the event horizon. If you fall in, you meet the light emitted by your feet at or below the horizon and you can see it then.

 

[Orodruin]

@Ibix just saved me a lot of typing. Let me just point out that “I read somewhere” is not a very good reference. We have no idea where this “somewhere” is or what it says exactly.

 

[PeterDonis]

Based on a new explanation I read

Where? Please give a specific reference.

It seems like it should take an infinite amount of time to enter a black hole, even from the perspective of the object entering it.

This is wrong. The proper time for an object to fall to a black hole's horizon from a finite distance is finite. The calculation is simple and can be found in most GR textbooks.

See this Insights article for some further discussion:

https://www.physicsforums.com/insights/black-holes-really-exist/

 

[PeterDonis]

I know my interpretation must be wrong. I'm just having a hard time understanding why.

You haven't given a reference, but I strongly suspect that wherever you read what you say you read is not a textbook or peer-reviewed paper--which means it's not the kind of source you should be trying to learn from. Learning science from pop science sources does not work. You are trying to reason from the mental model you have of how GR works, which is fine, but only if your mental model is the right one, i.e., one from which you can reason correctly. Pop science sources will not give you such a model. You need to learn from textbooks and peer-reviewed papers.

 

[Somoth Ergai]

Where? Please give a specific reference.

I should not have said read. It was a video from a youtube channel called floatheadphysics



he also gave this post in response:

From Floatheadphysics

"So, I got a lot of heat on my BH video in the comment section. I think most had a problem with the part of the video that said - "Nothing can ever enter the horizon from an outside perspective". A common perspective seems to be that things do enter, but it's hidden from the outside perspective. I wanted to clarify that that's not the case. Here's a simple way to think about it. If you dropped a stone from height h, it will hit the ground after a time t = root (2h/g) in your wrist watch. So, even if you couldn't see it, you could still predict when exactly it happens. The stone hitting the ground is an event that can be assigned a time co-ordinate. But, if we asked a similar question - 'If we dropped a stone from a radial distance r (r > the event horizon radius) when, according to our wrist watch, will the stone cross the event horizon' - the answer is ... error! We will find that at time t = infinity, the stone will just reach the horizon. And that's it. I can't assign any value of t for the crossing or any events beyond those (hence the name event horizon). However, the stone will cross the horizon (and hit the singularity) at some finite time in it's clock. (It's proper time - tau). This is the description of what we call a Schwarzschild black hole. An ideal black hole which doesn't rotate, has no charge, and is static. So, is it accurate to say event's behind the horizon 'happen' but are just invisible for the outside perspective? I don't think so. If that were true, we should be able to assign a time co-ordinate, and we can't. Saying event's happen but they don't have a 'when' is meaningless! This is relativity in all it's glory - the happening of those events are .. relative. They do happen for folks who are at those events but not for folks outside. And a black hole is ultimately a collection of those set of events! But, o'course there are more nuances to it. In relativity, we know simultaneity far away means nothing! So, a beautiful advice in the book I mention below is to always make local measurements. So, another way to piece together the story of this falling stone from the outside perspective is to place a set of stationary observers near the blackhole and record their observations as the stone just passes them. Then if we piece together all these recordings, we get a different story :). But, o'course we need to take baby steps in understanding all this. If you are interested to dive deeper, do check out the book - 'Exploring Black Holes by Wheeler'. I think it's excellent! It revises the basic ideas of special relativity and immediately dives into Schwarzschild metric. Finally, I did address how can black holes merge or grow in the pinned FAQs. In the hindsight, I think I could have addressed those in the video itself."

Which is the main thing that prompted my question.

I unfortunately do not have access to the textbooks or know where to look for peer reviewed papers to answer my specific questions. I'm not a student either. I'm just someone fascinated by physics and doing the best I can with the resources I have to try and learn and understand more. I only recently even discovered physicsforums which seemed like a pretty good place to start trying to get real answers to the things I see in pop science that don't seem to add up. I'm at least smart enough to notice when something doesn't make sense.

Also thank you for providing the article. I read it to the best of my understanding. I will probably need to read it a few more times to fully understand

 

[Ibix]

If that were true, we should be able to assign a time co-ordinate, and we can't.

Sure we can. Gullstrand-Painleve, Eddington-Finkelstein and Kruskal-Szekeres coordinates can all assign times to events inside the horizon. Schwarzschild coordinates can't, but that doesn't mean that other systems can't.

 

[Dale]

an infinite amount of time

This should be "an infinite amount of coordinate time in the Schwarzschild coordinates".

reach the "edge" of the event horizon

1) the event horizon doesn't have an edge
2) the event horizon is outside of the Schwarzschild coordinate chart, so Schwarzschild coordinates cannot be used to say anything about the horizon. Specifically, it cannot say anything about reaching the horizon.
3) if you want to say things about the horizon you will obviously need to use coordinates that cover it (there are several)

Suppose they are entering the black hole feet first. Since there is a distance from their feet to their head it seems to me that time dilation would mean their feet take an infinite amount of time to reach the edge of the horizon too.

Again, "an infinite amount of coordinate time in the Schwarzschild coordinates". The Schwarzschild time coordinate is adapted to a hovering observer at spatial infinity. They are neither hovering nor at infinity, so the Schwarzschild coordinates are even more irrelevant than coordinates usually are.

that they would sort of get "squashed" as the rest of there body got closer and closer to the horizon without any of it ever actually entering or crossing over.

Again, the correct statement would be "squashed according to the difference in the Schwarzschild radial coordinate". The Schwarzschild radial coordinate has to do with the surface area of a sphere around the horizon, not about distances between different radii. It is irrelevant to any physical notion of squashing, such as material stress or strain.

I know my interpretation must be wrong. I'm just having a hard time understanding why.

The principle issue is that you are attributing physical meaning to coordinates. Coordinates are just labels that identify specific events in spacetime. They don't tell you anything about the physics.

Similarly, my address is just a label for my house's location. Suppose that I live at 1234 Some Street and my next door neighbor lives at 1236 Some Street, while you live at 4321 Different Drive and your next door neighbor lives at 4323 Different Drive. That does not imply that your neighbor lives as close to you as my neighbor lives to me. Similarly with coordinates.

The physical meaning comes from the metric and from invariants that you calculate using the metric. The coordinates are only useful insofar as they help make relevant calculations easy with a particular metric. By themselves the coordinates have no meaning.

 

[PeterDonis]

It was a video from a youtube channel called floatheadphysics

So, as I said, not a good source to learn from.

A common perspective seems to be that things do enter, but it's hidden from the outside perspective.

More precisely, that things at or below the horizon cannot be seen by an observer outside, because the light can never reach them.

I wanted to clarify that that's not the case.

All of what follows is wrong. The author is mistaking what physicists call a "coordinate singularity" for an actual physical issue. This is a common error in pop science discussions.

I unfortunately do not have access to the textbooks or know where to look for peer reviewed papers to answer my specific questions.

Yes, you do. Sean Carroll's GR notes are available for free online. Plenty of peer reviewed papers are available on arxiv.org. Other good references are available in threads here on PF.

 

[PeterDonis]

I did address how can black holes merge or grow in the pinned FAQs. In the hindsight, I think I could have addressed those in the video itself.

Apparently the author fails to realize that the fact that black holes can merge and grow contradicts his claim that it is impossible for anything to reach the horizon. Another common issue with pop science videos is failure to do basic critical thinking and analysis.

 

[Nugatory]

Suppose they are entering the black hole feet first. Since there is a distance from their feet to their head it seems to me that time dilation would mean their feet take an infinite amount of time to reach the edge of the horizon too. and likewise their knees and on up the rest of their body.

You mentioned the relativity of simultaneity above, so you should be able to understand how it is connected to time dilation. Let’s attach clock A to the toes and B to the head, and let’s synchronize them while we’re far away from the black hole so there’s no problem getting them both to read the same value at the same time, namely the moment that we synchronize them.

When we say that clock A is running slower than clock B, we’re saying that even though the clocks were once synchronized at some now at the same time that clock B reads $T$, clock A reads something less. Clearly this is going to depend how we’ve defined “at the same time”. In flat spacetime there’s a simple and obvious definition: if light from an event reaches me when my clock reads $T$, and it took the light time $\Delta T$ to get to me, then we’ll say the event happened at the same time that my clock read $T-\Delta T$. But this definition fails when we’re watching things fall through the event horizon because the light never gets from the event to us. The “infinite time dilation” you’ve read about is just saying that using this definition of “at the same time” we will never get to a time on our clock that is the same as the horizon crossing, no matter how long we wait. Clearly this infinite time dilation is an artifact of the broken definition of “at the same time” that we’re using - just because we never see an object falling through the horizon doesn’t it didn’t happen.

All of these “can’t reach the event horizon” paradoxes are resolved by using a different definition of “at the same time” that isn’t broken (“doesn’t have a coordinate singularity at the horizon” in the lingo). There are several good ones, but if you’re going to be reasoning about what’s going on in the vicinity of the horizon you will be best served by Kruskal coordinates (Google is your friend, and even though the math is daunting you don’t need it to get a qualitative understanding from a Kruskal diagram).

When we plot the paths of the head and toe of your infalling observer on a Kruskal diagram we get a clear picture of what’s really going on. The light from the toes crossing the horizon gets stuck at there until the head reaches the horizon - more precisely the light from the toe crossing follows a path through spacetime that reaches the head at the moment that the head passes through the horizon. If the infaller survives the tidal forces as they approach the black hole they don’t even notice anything unusual as they pass through the horizon, there's never a moment when they can't see their toes.

 

[Somoth Ergai]

The light from the toes crossing the horizon gets stuck at there until the head reaches the horizon - more precisely the light from the toe crossing follows a path through spacetime that reaches the head at the moment that the head passes through the horizon. If the infaller survives the tidal forces as they approach the black hole they don’t even notice anything unusual as they pass through the horizon, there's never a moment when they can't see their toes.

Thank you for taking the time to try and walk me through an understanding. Forgive me if I am still misunderstanding though. Because it sounds to me like,

1 if their toes cross the horizon, and
2 they can't see their toes cross the horizon until their eyes do, and
3 if they can always see their toes, then
4 doesn't that imply that from the perspective of their eyeballs their body will be getting squashed more and more toward their feet until the point they actually cross the horizon?

I guess what I'm asking is. Is it correct to interpret this as them seeing their body length contracted more and more until the point of crossing the event horizon?

Also, Are you saying that I'm correctly understanding that they will indeed see the clock at their feet time dilated compared to the clock on their head?

Also also. Suppose they were flipped around so they enter head first. and yes ignoring the complications of the tidal forces for the moment. I know it's been stated elsewhere that it is a misconception that looking outward from the black hole would mean the universe appears to be going by in fast forward but I struggle to understand why.

I am probably misunderstanding a fundamental aspect of this yet again but, as I understand it, the way time dilation works in a simplified way is that it takes light longer and longer to get from point a to point b. and if this is happening to an entire object then from its perspective time is passing normally because all of the processes that rely on the transfer of photons are slowed down. But outside of that area of time dilation processes are occurring normally. which is what leads to time appearing to be slowed down or stopped in extreme time dilation situations.

 

[PeterDonis]

1 if their toes cross the horizon, and
2 they can't see their toes cross the horizon until their eyes do, and
3 if they can always see their toes, then
4 doesn't that imply that from the perspective of their eyeballs their body will be getting squashed more and more toward their feet until the point they actually cross the horizon?

No. They are free-falling; it's no different than if they were just floating in deep space far from all gravitating bodies. The horizon is just a lightlike surface that passes by them at the speed of light. Imagine you are floating in deep space far from all gravitating bodies, and a light ray coming from the direction of your feet passes you. Nothing about that light ray passing you makes any difference to how you see your toes. That is all that is going on in your scenario. You are just confusing yourself by insisting on using coordinates that don't work well at or near the horizon.

Is it correct to interpret this as them seeing their body length contracted more and more until the point of crossing the event horizon?

No. See above.

Are you saying that I'm correctly understanding that they will indeed see the clock at their feet time dilated compared to the clock on their head?

No. See above.

I know it's been stated elsewhere that it is a misconception that looking outward from the black hole would mean the universe appears to be going by in fast forward but I struggle to understand why.

If you are falling into the hole, the rest of the universe looks redshifted to you, not blueshifted.

I am probably misunderstanding a fundamental aspect of this yet again

Nothing in your scenario has anything to do with any concept of "time dilation". So any reasoning you are doing from any concept of "time dilation" will be wrong for your scenario.

 

[Nugatory]

1 if their toes cross the horizon, and
2 they can't see their toes cross the horizon until their eyes do, and
3 if they can always see their toes, then
4 doesn't that imply that from the perspective of their eyeballs their body will be getting squashed more and more toward their feet until the point they actually cross the horizon?

No. Before the head reaches the horizon the it is seeing light that left the toes before they crossed the horizon (so were one body length below the head when the light was emitted). When the head reaches the horizon it is seeing light that left the toes when they reached the horizon (so were still one body length below the head when the light was emitted). Once the head is below the horizon it is seeing light that left the toes when after they were one body length further "in". So there's never any perceived length contraction or squashing, the light that reaches the eyes has always traveled one body length in the appropriate amount of time.

Another way of thinking about it: The light that reaches the head when it is $\Delta t$ microseconds away from reaching the horizon (using the clock attached to the head, of course) is the light that left the toes when they were $\Delta t$ microseconds away from reaching the horizon. Reaching the horizon is just the $\Delta t=0$ case of this statement.

If you can plot this on on a Kruskal diagram it will be way simpler to see what's going on - that's why I so strongly recommend learning to read one.

(Be aware that I have made some simplifying assumptions here, but what I am saying is qualitatively accurate)

 

[Dale]

1 if their toes cross the horizon, and
2 they can't see their toes cross the horizon until their eyes do, and
3 if they can always see their toes, then
4 doesn't that imply that from the perspective of their eyeballs their body will be getting squashed more and more toward their feet until the point they actually cross the horizon?

Again, you are focusing on coordinates, and a bad choice of coordinates. Coordinates simply don’t matter. What matters is the metric and frame invariant quantities.

In a frame invariant sense, the event horizon crosses their body at the speed of light. So the light from their toes reaches their head at the same time the horizon does. This satisfies conditions 1 to 3 without implying condition 4.

Is it correct to interpret this as them seeing their body length contracted more and more until the point of crossing the event horizon?

Again. No.

Are you saying that I'm correctly understanding that they will indeed see the clock at their feet time dilated compared to the clock on their head?

Again. No. Time dilation simply isn’t relevant here. Time dilation is a coordinate artifact and is only useful is it makes the calculation of some invariant easier. In this scenario it does not. There is no invariant quantity in the scenario that is equal to or proportional to time dilation.

 

[Orodruin]

If you are falling into the hole, the rest of the universe looks redshifted to you, not blueshifted.

Well, that depends. At any event I can find a free falling observer that sees a particular light signal blue shifted, or one that sees it red shifted. It all depends on exactly how it is falling.

Sure we can. Gullstrand-Painleve, Eddington-Finkelstein and Kruskal-Szekeres coordinates can all assign times to events inside the horizon. Schwarzschild coordinates can't, but that doesn't mean that other systems can't.

To add to this, there are also coordinates on regular Minkowski space that have the same type of issues assigning a time coordinate to some events. That obviously does not mean those events suddenly don’t happen.

 

[PeterDonis]

At any event I can find a free falling observer that sees a particular light signal blue shifted, or one that sees it red shifted. It all depends on exactly how it is falling.

I said "free-falling into the hole" since that is what the OP was assuming in their scenario. That means either an infalling Painleve observer (if they are free-falling from rest at infinity) or an infalling Novikov observer (if they are free-falling from rest at a finite altitude). Both of those will see light coming in from the rest of the universe as redshifted--the former always, the latter once they have fallen for a sufficient (not very large) time.

 

[Orodruin]

I said "free-falling into the hole"

I mean, there are no observers free falling out of the hole … I assume we still talk about the horizon crossing here as the meaning of “falling into the black hole”.

 

[PeterDonis]

there are no observers free falling out of the hole

Not outgoing free-fall trajectories crossing the black hole event horizon, no. But there are plenty of free-fall trajectories above the horizon that do not end up crossing the horizon. And among those will be trajectories that never see light from the rest of the universe redshifted. That is the kind of thing I assumed you were referring to in your previous post; I was merely clarifying that in this thread we are not talking about those kinds of free-fall trajectories. We are only talking about the ones that do fall into the hole from outside.

 

[Ibix]

I think it's worth adding that the problem with Schwarzschild coordinates at the horizon is similar to the problem with latitude/longitude at the poles. The coordinates don't work well there. Concluding that you can't cross the horizon because of this is just as misguided as concluding that you can't walk through the north pole because you'd teleport half way along the top edge of the map and teleportation is impossible. The "teleportation" is entirely an artefact of the coordinates going wrong - the pole is actually a point, not the line it is represented as on a Mercator map, and there is no teleportation in reality if you walk through the pole. Any headaching about it comes from treating the map as if it were the territory. The same is true about worrying about crossing the event horizon because Schwarzschild coordinates don't work there.

In both cases you can just pick coordinates that are better suited for the job and the headaches go away.

 

[Orodruin]

Not outgoing free-fall trajectories crossing the black hole event horizon, no. But there are plenty of free-fall trajectories above the horizon that do not end up crossing the horizon. And among those will be trajectories that never see light from the rest of the universe redshifted. That is the kind of thing I assumed you were referring to in your previous post; I was merely clarifying that in this thread we are not talking about those kinds of free-fall trajectories. We are only talking about the ones that do fall into the hole from outside.

Ok, to be clear, I was talking specifically of free falling world lines crossing the horizon at the time of crossing. Some of those will see redshift, some blueshift.

 

[PeroK]

Also also. Suppose they were flipped around so they enter head first. and yes ignoring the complications of the tidal forces for the moment. I know it's been stated elsewhere that it is a misconception that looking outward from the black hole would mean the universe appears to be going by in fast forward but I struggle to understand why.

I am probably misunderstanding a fundamental aspect of this yet again but, as I understand it, the way time dilation works in a simplified way is that it takes light longer and longer to get from point a to point b. and if this is happening to an entire object then from its perspective time is passing normally because all of the processes that rely on the transfer of photons are slowed down. But outside of that area of time dilation processes are occurring normally. which is what leads to time appearing to be slowed down or stopped in extreme time dilation situations.

Perhaps the best plan if you ever want to cross the event horizon is to keep your eyes closed. Then there's no problem, as you can't see your feet in any case!

 

[Orodruin]

Ok, to be clear, I was talking specifically of free falling world lines crossing the horizon at the time of crossing. Some of those will see redshift, some blueshift.

To put that in numbers. A Novikov observer dropping from rest at $r = r_0$ will observe radially ingoing light signals from infinity to have a frequency shift of
$$
\frac{1}{2}\sqrt{\frac{r_0}{r_0 - r_S}},
$$
where $r_S$ is the Schwarzschild radius. This factor is equal to 1 if $r_0 = 4r_S/3$. If the observer drops from rest below $r = 4r_S/3$, then the radial light signal from infinity will be blueshifted. If the observer drops from above, then it will be redshifted. However, the Novikov observers falling from $r < 4r_S/3$ will be free falling observers entering the black hole (as in - passing through the horizon) while observing light from infinity to be blueshifted.

Edit: Obviously, Painleve observers correspond to $r_0 \to \infty$ and will observe an exact halving of the frequency when passing through the horizon.

 

[PeterDonis]

free falling world lines crossing the horizon at the time of crossing. Some of those will see redshift, some blueshift.

I don't know of any that will see blueshift; AFAIK all such worldlines will be either Painleve or Novikov worldlines, and those all see redshift at the horizon. Can you give a specific example of one that sees blueshift at the horizon?

 

[Orodruin]

I don't know of any that will see blueshift; AFAIK all such worldlines will be either Painleve or Novikov worldlines, and those all see redshift at the horizon. Can you give a specific example of one that sees blueshift at the horizon?

See post #23. A free falling observer dropping from rest at $r < 4r_S/3$ will observe a blueshift on the horizon. This is easy to compute even using Schwarzschild coordinates (the limit of the observation point going towards the horizon is well behaved).

All you need to do the computation is the metric and a couple of constants of motion for the observer and lightlike geodesics.

 

[Orodruin]

To follow that up (I was on phone before so this would have been a lot to write in LaTeX there):

We consider the Schwarzschild metric
$$
ds^2 = \alpha\, dt^2 - \frac{dr^2}{\alpha}
$$
where $\alpha = 1 - 1/r$ (normalising $r_S = 1$ for brevity). This has a Killing field $K = \partial_t$.

Consider a radially ingoing light signal with frequency 1 at infinity. The 4-frequency of such a signal is given by
$$
N = f \partial_t - g \partial_r
$$
with $f$ and $g$ being some positive functions. The 4-frequency is also the tangent of the light-like geodesic followed by the signal. From the 4-frequency being a null vector follows
$$
N^2 = g(N,N) = \alpha f^2 - \frac{g^2}{\alpha} = 0 \quad
\Longrightarrow \quad
g = \alpha f.
$$
From being the tangent of an affinely parametrised geodesic follows that
$$
E = g(K,N) = \alpha f
$$
is a constant of motion and therefore $f = E/\alpha$. At infinity we must have $f = 1$ (frequency at infinity equal to one) and since $\alpha \to 1$ at infinity, $E = 1$. That's it as far as the light signal goes: $N = (1/\alpha)\partial_t - \partial_r$.

Now consider a radially freely infalling observer with 4-velocity $V$. Such an observer follows a timelike geodesic and therefore
$$
E' = g(K,V) = \alpha V^0
$$
is a constant of motion. Consequently, we can write the 4-velocity
$$
V = \frac{E'}{\alpha}(\partial_t - h \partial_r).
$$
Furthermore, we normalise $V^2 = 1$ which leads to
$$
V^2 = g(V,V) = \frac{E'^2}{\alpha} (1 - h^2/\alpha^2) = 1.
$$
Assuming a drop from rest at $r = r_0$, we would have $h_0 = h(r_0) = 0$ and therefore
$$
\frac{E'^2}{\alpha_0} = 1 \quad \Longrightarrow \quad
E' = \sqrt{\alpha_0}
$$
where $\alpha_0 = \alpha(r_0)$. We can now insert this above to find the general expression for $h$:
$$
\frac{\alpha_0}{\alpha} ( 1 - h^2/\alpha^2) = 1 \quad \Longrightarrow \quad
h = \alpha\sqrt{1 - \frac{\alpha}{\alpha_0}} = \frac{\alpha}{\sqrt{\alpha_0}}\sqrt{\alpha_0 - \alpha}.
$$
Consequently
$$
V = \frac{\sqrt{\alpha_0}}{\alpha} \partial_t - \sqrt{\alpha_0 - \alpha} \partial_r.
$$
The frequency that the observer measures if crossing the light signal at Schwartzschild coordinate $r$ is therefore given by
$$
g(N,V) = \frac{\sqrt{\alpha_0}}{\alpha} - \frac{1}{\alpha} \sqrt{\alpha_0 - \alpha}
= \frac{\sqrt{\alpha_0} - \sqrt{\alpha_0 - \alpha}}{\alpha} = \frac{1}{\sqrt{\alpha_0} + \sqrt{\alpha_0 - \alpha}}.
$$
If this is larger than one the signal is blue shifted and if it is smaller than one it is red shifted. The horizon crossing is given by the limit $r \to 1$, i.e., $\alpha \to 0$. In other words, the frequency observed by the observer when crossing the horizon is
$$
\frac{1}{2\sqrt{\alpha_0}}.
$$
This is larger than one if $\alpha_0 = 1 - 1/r_0 < 1/4$, which works out to $r_0 < 4/3$.

It is therefore simply not the case that a free falling observer crossing the horizon will necessarily see a redshift.

Edit: Note in particular that we recover the well known redshift of a factor of 2 at the horizon for the observer free falling from infinity as then $\alpha_0 \to 1$.

 

[Orodruin]

Come to think of it, this (well, actually the fact that the shift is finite) is a direct counter argument to the idea presented in the OP of the infalling observer never passing the horizon. As the shift is finite, the infalling observer will only receive a finite number of pulses sent at regular intervals from a faraway source before passing through the horizon. Note again that this was concluded using the Schwarzschild metric only. No need to even go to other coordinates.

 

[Somoth Ergai]

If you can plot this on on a Kruskal diagram it will be way simpler to see what's going on - that's why I so strongly recommend learning to read one.

thank you. I will try to learn about this before i continue to make a fool of myself

Yes, you do. Sean Carroll's GR notes are available for free online. Plenty of peer reviewed papers are available on arxiv.org. Other good references are available in threads here on PF.

Also thank you. I was not aware of those resources. I will try to learn more from them before I continue

 

[PeterDonis]

To follow that up

Thanks for the calculations! I think I was misremembering something. Now that I am looking at the math, it's easy to see where I went wrong. Being at rest relative to the timelike KVF, even momentarily, means seeing the same blueshift that a stationary observer there sees; and if you're close enough to the horizon there simply isn't enough free-fall time to turn that blueshift into a redshift (since the blueshift seen while you are stationary also gets larger as you get closer to the horizon).

 

[Orodruin]

Thanks for the calculations! I think I was misremembering something.

One more in the column of math being better than ordinary language to communicate science! Even between people who know the theory.