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What you're describing is basically the "collapsed star" (Eng) or "frozen star" (Rus) interpretation of black holes that was common prior to the late mid-1960s. It was a mistake.
Suppose you are distant and stationary relative to the black hole. You will observe infalling matter asymptotically approaching the horizon, growing ever fainter as it redshifts. ...
28
Yes, you are absolutely right, from OUR VIEWPOINT it does.
From Kip Thorne's book "Black Holes and Time Warps: Einstein's Outrageous Legacy."
“Like a rock dropped from a rooftop, the star’s surface falls downward (shrinks inward) slowly at first, then more and more rapidly. Had Newton’s laws of gravity been correct, this acceleration of the implosion would ...
22
There are two methods, one more reliable than the other (though both are pretty good.)
Key point: The brighter a star is, the more detail we can see in its spectrum -- you can think of it as being able to magnify the spectrum more so as to be able to see finer details. This also allows us to see fainter lines (not all spectral lines are equally intense.)
...
21
Yes. It can be measured in spectra of the moon.
A paper The solar gravitational redshift from HARPS-LFC
Moon spectra describes the measurment of red-shifts in Iron absorption lines in the spectrum of the moon which result from gravitational redshift from the sun.
The difficulty in measuring these redshifts is not their scale (equivalent to motion of about ...
20
The rate of expansion, measured in the customary units of (km/s)/Megaparsec
is not known with great accuracy. Recent measurements include 67.6 (SDSS-III), 73(HST) 67.8 (Plank) 69.3 (WMAP) [wikipedia]
The Andromeda galaxy is 0.78 Mpc from us, so taking the Hubble constant to be about 70, gives a recession of about 55 km/s. This is not a very great speed: ...
20
When a galaxy recedes from us, the light we see from it is redshifted. For galaxies at cosmological distances, this redshift is fundamentally different from a Doppler shift; whereas the latter is due to a velocity difference between the emitter and the receiver, a cosmological redshift is due to photons traveling through an expanding space$^\dagger$.
Hence, ...
17
You cannot gauge the redshift of a galaxy by looking at a false colour image. The images taken through different filters are stacked and colourised to suit. You can say that the blue galaxies are indeed bluer than the red galaxies, but there is no absolute scale with which to judge redshift by eye.
Secondly, there is no detail in the NASA web page, but the ...
16
Let's start with a quick clarification: Red Shift is not the same as red light.
Red light is just electromagnetic radiation with a 400–484 THz frequency range, the lowest our eyes can see - highest being violet light, with a 668–789 THz frequency.
Red Shift is an observable effect when you analyze the spectroscopic signature of an electromagnetic radiation ...
14
To distinguish galaxies from stars, you can use the spectrum. Roughly, stars have a black-body like spectrum with features depending on the absorption and emission on the line of sight and in the chromosphere of the star.
Galaxies on the other hand of a spectrum that is the composite of tons of stars. The spectrum will for example be much wider (ranging ...
14
tl;dr Because space doesn't contract inside our Solar System.
Wavelength increase is proportional to space expansion
The prediction of general relativity — one of the most thoroughly tested and succesful theories — is that the wavelength of observed light changes in proportion to the factor by which space expands (Lemaître 1927).
If space expands by a ...
13
Even "round" galaxies look different from stars
cphyc's answers the question excellently: Spectroscopy is the answer, although since — as explained below — galaxies are not point sources, the morphology of stars and galaxies is also different: even elliptical galaxies observed along one of their axes look different from stars. Although both are ...
12
We need to think about just where the time dilation effect occurs. By then thinking about the observations from each point of view, that is the free falling object and the external observer, we can come to terms with just what is happening as opposed to what appears to be happening.
The experience of time
We must remember that an object moving at a certain ...
12
Good answers have already been given, but I wanted to provide another way of looking at it. Take a look at the image below, which is the Hubble Extreme Deep Field (XDF) $-$ for those who don't know, this is a small patch of the sky that Hubble has stared at for a total of 23 days over 10 years $-$ and you'll notice something interesting. It is clear to see ...
11
The problem is that conservation of energy is a slippery concept in General Relativity. There are arguments back and forth but most people accept that conservation of energy is only a local law - it applies only to a local inertial frame and cannot be applied to the universe as a whole. However in an expanding universe it is very difficult to identify any ...
10
The Sloan Digital Sky Survey Data Release 15 contains over 4 million spectra of both galactic and extra-galactic origin from the multi-fiber spectrographs. Of these spectra, 0.7 million came from the original spectrographs during the SDSS-I/II Legacy Survey and the remainder from the upgraded spectrographs as part of the BOSS survey during SDSS-III (see SDSS ...
9
Looks like you can measure the cosmological redshift of quasars using that equipment and an 14" reflector:
http://www.rspec-astro.com/sample-projects/
(halfway down the page)
So the answer appears to be yes, you can do it, assuming that page isn't a total fabrication. Seems plausible to me: the redshift of that quasar is sizable at 0.15 (though that's ...
9
Reason 1:
Let's look at the Friedmann equations without the cosmological constant.
$$ \frac{\dot{a}^2 }{a^2} = \frac{8 \pi G \rho}{3}-\frac{kc^2}{a^2}$$
The term on the LHS is just the Hubble constant squared $H^2$ which can be measured the direct measurement of recession velocity of galaxies
The density term can be said to be a combination of $\rho_{...
9
If you had a simple slit spectroscope, and looked at an incandescent light, you'd see a smear of light with red on one end and blue on the other. This is because the filament is producing light by glowing from being heated.If you looked at one of those orange colored sodium vapor street lamps, instead of a smear of color, you'd see a group of lines. This ...
9
Yes, there is direct, non-red-shift evidence of expansion.
The past temperature of the Cosmic Microwave Background Radiation (CMBR) has been directly measured and found to be substantially higher than it is today. Its reduction in temperature over time is direct evidence of expansion. Here are the details:
According to this paper, the CMBR was measurably ...
9
After considering @benrg's comments, I realize that my first answer contained too strong statements about the relation between the two redshifts. I try here to moderate my answer, but you might want to accept their answer instead.
It is common to think of the two redshifts as having nothing to do with each other. Doppler shifts arise when the observer and/...
8
I will make a small calculation here, but please proceed to the results if you may like to.
Calculation
Stars are spherical and static, so metric near their surface (photosphere) and outside on is Schwarzschild. Hence time-time metric component on the surface is:
$$g_{44}=1-\dfrac{R_{grav,*}}{R_*}$$,
where $R_*$ is the radius of the star and $R_{grav,*}$ ...
8
Aside from being, I suspect, totally incompatible with GR, the model fails to match observations.
The universe appears to be homogeneous and isotropic. In the model you suggest, Hubble's law would be different depending on which direction you were looking.
In particular, looking away from the black hole, galaxy redshifts would increase to some asymptotic ...
8
The Hubble law gives the velocity of a distant galaxy right now. A galaxy at a distance $d$ recedes at a velocity $v = H_0\,d$ right now$^\dagger$.
However, the relation between $d$ and the redshift — which is the quantity that we observe — is a non-trivial function of the expansion history of the Universe, obtained by integrating the (inverse) scale factor ...
8
Red shift is used for measuring the distance to very distant stars (galaxies mostly, in fact). The secret is to use spectral lines. Specific elements when very hot emit light at very specific colours and you can spot the pattern of those colours when the whole pattern has been shifted towards the red and see how far it has shifted.
For closer stars, there ...
8
At a distance of $d = 87\,\mathrm{Mpc}$, with a Hubble constant of roughly $H_0 = 70\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$ cosmological expansion should make the host galaxy UGC 11723 recede at $v=H_0 \,d\simeq6100 \,\mathrm{km}\,\mathrm{s}^{-1}$.
However, galaxies also move through space, at typical velocities from several
$100\,\mathrm{km}\,\...
7
Several wonderful yet technical answers have been given, and I cannot add anything to those very nice answers that explain why it is not useful to think black holes get "frozen" at their event horizons. But I can give an answer with a more essentially useful philosophical perspective, which is that the central lesson of relativity is that reality involves a ...
7
Yes, of course. Many, many examples. Telescopes work in the infrared, far-infrared and there are even samples of galaxies that are selected on the basis of their mm emission.
The most distant galaxies detected now have redshifts of 7 or more. This means the wavelength of their light has been stretched by a factor $1+z$ - i.e. by a factor of 8. Thus light in ...
7
Stars consist of similar atoms and molecules that we can find here on Earth.
Radiation from those atoms and molecules often occurs at discrete, identifiable wavelengths that form a unique pattern. This is a consequence of quantum physics - the energies of atomic and molecular states are quantised, and radiation arises from transitions between these. These ...
7
There are two factors here.
First, to construct an image such as this, several optical filters are used, and then these are then merged to give an image. The colours that you see are therefore not the "true" colours (they couldn't be, the light you see is far too faint for the human eye) but are chosen to bring out details in the image. A galaxy that is ...
7
Yes. This is is actually something that is done in the analysis of redshift data. If no correction to the redshifts are made, then the redshifts are known as "geocentric." When you correct for the motion of the Earth around the Sun, the redshifts are known as "heliocentric" (see, for example, the description of data in the 6dFGS database).
Less common is ...
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