# Quasar redshift

The quasar 3C 273 has a redshift z=0.158. A question in a textbook: could this be a gravitational redshift instead of cosmological (=resulting from the expansion of space)?

My answer: no. Firstly, the visible light from the quasar is produced by the accreting gas at a distance from the central supermassive black hole, not by its "surface" (which doesn't physically exist). In the regions where the light is produced, the gravity gradient (?) or the curvature of space isn't large enough to produce z=0.158. Such a redshift could only be produced very near the Schwarzschild radius.

Secondly, if the redshift were gravitational, we wouldn't see a narrow Balmer line H$\beta$ with z=0.158 but instead a broad line because some light originates from near the "surface" (large gravity gradient) and some light from further away.

Am I right?

Your case is not quite watertight - it hinges in your assertion that the optical light that is seen comes from some way out from the black hole (SMBH). The thing is that gravitational redshift can be larger than 0.2 and it is also aided by the relativistic transverse Doppler effect in the orbiting material.

Some details:

Gravitational redshift around a black hole of mass $M$ is governed by $$z = \left( 1 - \frac{2GM}{rc^2}\right)^{-1/2} - 1,$$ where $r$ is the radial coordinate of a light source in orbit around the black holes. This formula would apply for any spherically symmetric mass distribution.

The last stable, possible circular orbit around a non-rotating black hole is at $r = 6GM/c^2$, where $M$ is the black hole mass. This means the gravitational redshift factor could be as large as $0.22$.

On top of this you must consider the Relativistic Doppler shift.

The relativistic doppler shift for a source moving at a speed $v$ at an angle $\theta$ (in the reference frame of the observer), then the emitted and observed frequencies are related by $$f_o = \frac{f_e}{\gamma\left[ 1 + (v/c)\cos\theta\right]},$$ where $\gamma = (1 - v^2/c^2)^{-1/2}$. This means that even when $\theta=90^{\circ}$ and the source orbiting the black hole is moving neither towards or away from an observer on Earth, that there is a "transverse doppler redshift" of $$z = \gamma - 1$$ So although an observer on Earth would see the frequency of a source in orbit around a black hole go up and down due to the doppler shift (a spectral line would be broadened by $\sim \pm v/c$), there would be a net redshift due to the transverse doppler effect.

Material at the innermost circular orbit would have a speed of half the speed of light and $\gamma = 1.15$. Thus the redshift due to the transverse doppler effect would be $z=0.15$ and almost the same as the gravitational redshift. At larger orbital radii, the gravitational redshift becomes smaller but more dominant.

On top of the net tranverse doppler redshift there will be a doppler broadening as the gas orbits the black hole. The amplitude of this will depend on the inclination of the orbit to the line of sight. At its largest, $\theta = 0$, the redshift/blueshift will be factors of $$z = \gamma(1 \pm v/c) -1$$. Thus for a source in orbit at the innermost stable circular orbit, this would lead to a factor of two broadening of any spectral line.

Thus if the gas orbits close enough to the SMBH at 3C273 you probably could produce the redshift seen. The maximum broadening that you should see is much broader than the observed spectral lines, but this could be accounted for by the disc being at a low inclination and the emission coming from a narrow range of radii.

In fact, as you say, the optical emission you see from the broad line region comes from considerably further out than a few Schwarzschild radii, so the effects above, whilst very important for hot X-ray emitting gas in quasars, are probably not a big deal in the optical. So you need a further step in your answer to prove/argue that the optical emission does not arise that close to the SMBH.

• This helps a lot. In 1964ApJ...140....1G Greenstein and Schmidt report narrow hydrogen lines, width 5 nm. Excluding SR doppler shift (as the textbook does) gravitational redshift requires gas in a stable, nearly circular orbit with a specific radius. But probably there are several clouds of gas spiraling towards the SMBH from varying distances. Including an SR doppler shift we also need the orbit to be almost perpendicular to the line of sight, which is even more improbable. Gravitational redshift may exceed z=0.158 but the odds are against a broadening of only 5 nm, if I'm right. – gamma1954 Jan 22 '16 at 22:01
• In 1964ApJ...140....1G Greenstein and Schmidt give $z=GM/(Rc^2)$ (page 6, equation 1). Is this the redshift at a distance R, not in orbit, as opposed to your expression for z in orbit? – gamma1954 Jan 22 '16 at 22:02
• @gamma1954 I'm not arguing that it is gravitational redshift! Quite the reverse. But you need something to argue that the optical emission does not come from very close in and/or comes from a wide range of radii, otherwise GR and SR redshifts could be candidates. $(1-2GM/Rc^2)^{-1/2} -1 \sim GM/Rc^2$ when $GM/Rc^2 \ll 1$. My expression is precise. – ProfRob Jan 22 '16 at 22:20
• I understood you're not arguing that it is gravitational redshift -- but the textbook does argue it is. I'm still thinking about the extra argument that's necessary to make the case against gravitational redshift more convincing, as you have explained. – gamma1954 Jan 22 '16 at 22:33
• How to disprove GR plus SR redshift in 3C273 with only an optical spectrum? The spectrum has a continuum, probably also in UV and X-rays. If this radiation is intense enough to ionise all hydrogen near the SMBH, the same region could not (directly) produce the balmer emission lines that are superimposed on the continuum, because balmer lines are from atomic hydrogen. So the emission lines would have to originate from a region further away. That's an argument against GR and SR redshift, but not very hard. Is there a more convincing argument against GR and SR redshift in this case? – gamma1954 Jan 23 '16 at 13:35