## View of universe from event horizon of black hole

If I flew over to a the nearest black hole with the Hubble scope on a trailer (cough), how would the performance of the scope differ from current, particularly with regards to observing extremely distant objects.

In particular, when time dilation becomes extreme as my orbit of the BH nears the even horizon, how does that effect my observation of far redshifted objects that are also extremely time-dilated compared to the earth frame?
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 Recognitions: Homework Help You'd need a better IR scope then. Maybe radio-astronomy would be more helpful. How are you measuring "performance of the scope"? I imagine there would be some fun with tidal stresses on it for example.
 Recognitions: Gold Member Science Advisor Distant objects would be increasingly blue shifted as you approached the event horizon. Unfortunately, by the time you got close enough to the EH for the effect to become significantly noticeable, you would become distracted by more ... pressing ... issues.

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## View of universe from event horizon of black hole

Incoming light will gain energy - ergo - blue, not red, shift: I'm an idiot :(
The famous redshift is the other way. Hence:
 when time dilation becomes extreme as my orbit of the BH nears the even horizon, how does that effect my observation of far redshifted objects that are also extremely time-dilated compared to the earth frame?
They'd get shifted back wouldn't they? You'd get a gravitational blue shift of a Hubble red shift... there are easier ways...
 Recognitions: Science Advisor Staff Emeritus A static observer would see a gravitational blueshift. But if you're orbiting, you need to consider the doppler shift due to your motion relative to said static observer as well. If you are careful define the velocity correctly, you could just multiply the gravitational doppler shift by the SR dopler shift. THis only works with the correct defintion of velocity, though - a velocity you measure using local clocks and rulers for the static observer, or the same sort of velocity the static observer measures for you. Such velocities will always be lower than 'c'. If you use coordinate based defintions of velocity, this approach will give you WRONG answers. If you're falling straight in, and looking straight outwards, and your fall started at infinity with zero velocity, I believe you'll wind up with a net redshift. But I couldn't find the detailed calculation. If you're orbitting with some radial velocity, the calculations get more complex, and the results will depend on the details of your orbit , and the direction in which you look. Hamilton has a "redshift map" at http://casa.colorado.edu/~ajsh/singularity.html for one particular orbit, which might give you a general idea of how the sky redshifts / blueshifts as you fall in, and he has some computer generated movies on the same page.

 Quote by pervect ... If you're falling straight in, and looking straight outwards, and your fall started at infinity with zero velocity, I believe you'll wind up with a net redshift...
Pervect, Would the net effect be driven by your infalling acceleration at the point were you carry out the observation, or would it always be redshifted?

Regards,

Noel.
 Recognitions: Gold Member Science Advisor If you are in free fall, the external universe will be progressively redshifted as you approach the EH. By the time you reach the EH, the external universe will appear infinitely redshifted [just like you will appear infinitely redshifted to a stationary external observer].

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 Quote by pervect If you're falling straight in, and looking straight outwards, and your fall started at infinity with zero velocity, I believe you'll wind up with a net redshift. But I couldn't find the detailed calculation.
Yes, I have done, but never posted, the calculation in Painleve-Gullstrand coordinates.

If observer A, who hovers at great distance from the black hole, radially emits light of wavelength $\lambda$, then observer C, who falls from rest freely and radially from A, receives light that has wavelength

$$\lambda' = \lambda \left( 1+\sqrt{\frac{2M}{R}}\right).$$
The event horizon is at $R = 2M$, and the formula is valid for all $R$, i.e., for $0 < R < \infty$. In particular, it is valid outside, at, and inside the event horizon.
 Quote by Lino Pervect, Would the net effect be driven by your infalling acceleration at the point were you carry out the observation, or would it always be redshifted?
According to the formula posted above, there is always a redshift.

Suppose observer that A hovers at a great distance from a black hole, and that observer B hovers very close to the event horizon. The light that B receives from A is tremendously blueshifted. Now suppose that observer C falls freely from a great distance. C whizzes by B with great speed, and, just past B, light sent from B to C is tremendously Doppler reshifted. What about light from A to C? The gravitation blueshift from A to B is less that the Doppler redshift from B to C. As C crosses the event horizon, C sees light from distant stars redshifted, not blueshifted.

The argument in the previous paragraph only works for B placed anywhere outside the event horizon, but the formula is valid every.

 Quote by Chronos If you are in free fall, the external universe will be progressively redshifted as you approach the EH. By the time you reach the EH, the external universe will appear infinitely redshifted [just like you will appear infinitely redshifted to a stationary external observer].
No, at the event horizon, light is redshifted by a factyot of 2.

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 Quote by Lino Pervect, Would the net effect be driven by your infalling acceleration at the point were you carry out the observation, or would it always be redshifted? Regards, Noel.
I'm not positive I understand the question. Your acceleration won't affect the redshift, an accelerating observer will measure the same redshift as a comoving inertial observer for a given wavelength.

However, it' important to note that if you look at other directions other than the radial direction, you can see blueshift.

The details of your orbit can also change the amount of redshift, for instance if you start at rest at some height that's less than infinity, it's easy to calculate that you'll initially see a blueshift when looking radially outwards.

Tidal forces will always cause radial light from infinity to redshift more as you get closer to the black hole. The same tidal forces can and will cause light approaching from the sides to blueshift, because radial tidal forces stretch in the radial direction and compress in the traverse direction.
 Thanks all. I think that I (reasonably) understand it - for a free falling observer, the radial view will be redshifted. George, using your model, I appreciate that light from C to (B or) A would be redshifted as it climbs out (before the horizon / as long as it can) of the gravity well and also as a result of the Doppler effect, but that the light (from A to C) and the measures by observer C are impacted differently is less intuitive! I guess that it is to do with Pervect’s comment that an accelerating observer and a commoving inertial observer measure the same redshift in these circumstances, and I just need to read a bit more about it. * Pervect, when you mentioned the impact of tidal forces, are you talking about tidal forces on a photon / beam of light? I didn’t think that there would be any such forces on a point-particle – am I wrong? * Regards, * Noel
 Recognitions: Gold Member Science Advisor There is a paper on this [which I incorrectly recalled] - Exchange of signals around the event horizon in Schwarzschild space-time, http://arxiv.org/abs/1201.4250.
 Thanks Chronos. That's great. Regards, Noel.

 Tags black hole, hubble distance, time dilation