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## Main Question or Discussion Point

If we have an infalling observer through the event horizon,will she see the end of the Universe?

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If we have an infalling observer through the event horizon,will she see the end of the Universe?

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This topic is discussed here with great regularity. I suggest a forum search and the best place to start is likely to be the threads listed at the bottom of this thread.

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As phinds says, there is no "end of the universe" in our best current models in cosmology. However, your question could be rephrased as "will an observer who falls through the horizon of a black hole see arbitrarily far into the future of the universe?" The answer to that question is no. An observer who falls through the horizon will hit the singularity fairly quickly by his clock; and he will not see the rest of the universe outside "speeded up" before he does so. In fact, light coming in to him from the outside universe will be increasingly redshifted, so he will actually see the rest of the universe outside "slowing down", and he won't see very much of it at all before he hits the singularity and is destroyed.If we have an infalling observer through the event horizon,will she see the end of the Universe?

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How do you come to the conclusion that light from outside would be increasingly redshifted? Could it be that you confuse the coordinate systems of the infalling observer with the one outside?In fact, light coming in to him from the outside universe will be increasingly redshifted, so he will actually see the rest of the universe outside "slowing down", and he won't see very much of it at all before he hits the singularity and is destroyed.

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By calculating the inner product of the tangent vectors of ingoing null geodesics and infalling timelike geodesics. In other words, by using the mathematical framework of GR as applied to this scenario.How do you come to the conclusion that light from outside would be increasingly redshifted?

The redshift of ingoing light as observed by an infalling observer is an invariant; it is independent of coordinates.Could it be that you confuse the coordinate systems of the infalling observer with the one outside?

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Let's say you have an observer, called Brand, outside, and an infalling observer, called Cooper (with his spaceship Ranger), falling in a black hole called Garguntua.By calculating the inner product of the tangent vectors of ingoing null geodesics and infalling timelike geodesics. In other words, by using the mathematical framework of GR as applied to this scenario.

The four names are invariant under the coordinate transformation. What do you think about the following citation:

"She (Brand) sees light from the Ranger shift to longer and longer wavelengths (...), until the Ranger turns completely black and unobservable. And bits of information that Cooper transmits to Brand one second apart as measured by his time on the Ranger arrive with larger and larger time separation as measured by Brand. After a few hours Brand receives the last bit that she will ever receive from Cooper emitted before piercing the horizon.

Cooper, by contrast, continues receiving signals from Brand even after he crosses the horizon. Brand's signals can't get out to Brand. Einstein's laws are unequivocal.

This is how it must be.

Moreover, those laws tell us that Cooper sees nothing special as he crosses the horizon. He can't know, at least not with any cease, which bit that he transmits is the last one Brand will receive. "...

From:

Kip Thorne, The Science of Interstellar

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It's fine. What's the problem?What do you think about the following citation

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As I understand it, the question is about what Cooper sees. Kip Thorne's remarks are perfectly correct, but don't answer the question. One web reference is http://iopscience.iop.org/article/10.1088/1742-6596/104/1/012008/pdf "Doppler effect in Schwarzschild and Kerr geometries". It's a rather coordinate-dependent approach, not as elegant as the coordinate-free way Peter described earlier, but it may be easier to understand if you're not familiar with the mathematical machinery Peter described, for instance how to affinley parameterize null geodesics (a pre-requisite for normalizing the length of their tangent vectors properly).

[add]Note that while there is no inertial frame of reference stationary at the event horizon, there are plenty of inertial frames that fall through it - at least in the sense that inertial frames exist at all in a curved space-time, i.e. in the sense that they exist as an approximation in the neighborhood of said infalling observers.

Also note that the paper I cited gives the answer for the redshift that Cooper sees in (14). It's just a factor of the received frequency being 1/2 the transmitted frequency, when the infalling observer starts out at rest at infinity. This is consistent with what Peter has said and what many other PF posters have derived by various different approaches.

If you use standard Schwarzschild coordinates, the gravitational blueshift approaches infinity as ##g_{tt}## approaches zero, but the special relativistic doppler shift for the received signal also approaches zero. The classical doppler shift is well behaved and has a value of 1/2. One can take the appropriate limit, but it's really better to use better behaved coordinates. Regarless, one need either to work it out, or find a source that one trust, because ones intuition will typically not work correctly. It helps to realize that the event horizon is not a place that any material object can "stop at".. Light always travels at "c" in any inertial frame, but at the event horizon, light has a constant coordinate value. Thinking that there is an inertial frame of reference stationary at the event horizon is tantamount to think their is an inertial referece frame where the velocity of light is zero. This line of thinking is well known to cause immense confusion.Radosz et al said:This establishes an expression for the frequency shift, the total Doppler effect, in the case of an infalling observer in a Schwarzschild geometry. The frequency of light signals ##\omega_{IF}## received by IF observer compared to the frequency ##\omega^•## emitted by a distant inertial observer IO is modified by a product of three factors. The total Doppler effect is composed of a classical kinematical Doppler redshif ##1/(1+V_{IF}## a special relativistic red-shift (time dilation-term) ##\sqrt{1-V_{IF}}^2##, and a gravitational blue-shift ##1/\sqrt{g_{tt}}##

[add]Note that while there is no inertial frame of reference stationary at the event horizon, there are plenty of inertial frames that fall through it - at least in the sense that inertial frames exist at all in a curved space-time, i.e. in the sense that they exist as an approximation in the neighborhood of said infalling observers.

Also note that the paper I cited gives the answer for the redshift that Cooper sees in (14). It's just a factor of the received frequency being 1/2 the transmitted frequency, when the infalling observer starts out at rest at infinity. This is consistent with what Peter has said and what many other PF posters have derived by various different approaches.

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Thanks for the link, nice paper. Nevertheless, in the end it was my fault twice, sorry about that: Don't ask me why, no clue, but yesterday I read instead of "increasingly redshifted" this: "infinitely redshifted". Additionally I thougt Peter meansAs I understand it, the question is about what Cooper sees. Kip Thorne's remarks are perfectly correct, but don't answer the question. .

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No, she will not. See post #3. Your understanding of how black holes work is incorrect.We'll never see her crossing the event horizon,but she will see our Sun turning to a white dwarf.

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stevebd1

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If a person falls into a large enough black hole from rest at infinity and the person viewed outwards, they would see the universe slow down and redshift (there is a fairly thorough thread on this subject somewhere on this forum). If a person was to

[tex]d\tau=dt\sqrt{1-\frac{2Gm}{rc^2}}[/tex]

where G is the gravitational constant, m is the mass of the black hole, r is the radius time is spent at, c is the speed of light, dt is the coordinate time spent at r and [itex]d\tau[/itex] is the proper time. For instance, spending 10 minutes hovering at 1km from the event horizon of a static 4x10

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IIRC, for an observer falling freely from rest at infinity, there is already a redshift factor of 2 at the horizon; so strictly speaking, the light from the outside universe would "begin to redshift" before the horizon was reached.If you were to fall past the event horizon at this stage, the information you receive would still be in a very high wavelength but it would begin to redshift as you fall towards the singularity.

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stevebd1

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This is in reference to someone hovering as close to the BH as possible then falling in.IIRC, for an observer falling freely from rest at infinity, there is already a redshift factor of 2 at the horizon; so strictly speaking, the light from the outside universe would "begin to redshift" before the horizon was reached.If you were to fall past the event horizon at this stage, the information you receive would still be in a very high wavelength but it would begin to redshift as you fall towards the singularity.

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No, it isn't; it's in reference to someone free-falling in from rest at infinity. That means that, when they pass an observer "hovering" just above the horizon, they will be falling inward at almost the speed of light. Someone hovering as close to the hole as possible and then falling in would only be moving very slowly relative to someone who keeps hovering at that altitude.This is in reference to someone hovering as close to the BH as possible then falling in.

For reference, the calculation of the Doppler redshift factor is as follows: a "hovering" observer at radius ##r## sees light coming in from infinity with a Doppler blueshift factor of ##\sqrt{1 - 2M / r}## (meaning, this is the ratio of observed wavelength to emitted wavelength; for frequency the factor would be inverted). An observer falling in from rest at infinity is falling inward, relative to a "hovering" observer at radius ##r##, with velocity ##\beta = \sqrt{2M / r}##. So the blueshift factor of the "hovering" observer is ##\sqrt{1 - \beta^2}##, and the Doppler factor for the infalling observer, looking at the same light at radius ##r## as the "hovering" observer, is ##\sqrt{1 - \beta^2} \sqrt{(1 + \beta) / (1 - \beta)} = 1 + \beta##, which is a redshift. In the limit as ##r \rightarrow 2M##, we have ##\beta \rightarrow 1## and therefore the redshift factor approaches 2.

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stevebd1

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I meant "one plank-length".NOT 1km.For instance, spending 10 minutes hovering at 1km from the event horizon of a static 4x106 solar mass black hole, you would see ~24 days pass outside th