You're messing up with simple time dilation. Time intervals are relative quantities. Two observers may not be agree with measured time intervals of an event. You see other moving observer's time dilated. Also, you see other observer's time dilated if she is deep in Gravity well than you are. Meaning, you find other observer's measured time interval more than your own measurement result of the same event. That's it.

Now, come to Radioactive decay: You measured half-life of a substance on Earth. Another observer who is independent of motion of Earth etc and far from any Gravity well (the notion of stationary is irrelevant), would find your measured half-life more than hers.

the Doppler component $1+\beta$ and the time dilation component [...] $\gamma$

That decomposition really makes no sense from a special or general relativistic perspective. The only reason you might write the Doppler shift that way is if you're trying to make a connection with nonrelativistic Doppler shift.

A manifestly covariant formula for relativistic redshift is $$1{+}z = \frac{p_\text{light} \cdot v_\text{detector}}{p_\text{light} \cdot v_\text{emitter}}$$

where the $v$s are four-velocities and the $p$ is the four-momentum of the light, or any nonzero scalar multiple of it (i.e., any four-vector pointing along the path of the light).

This formula works in general relativity too, in *all* cases, if you parallel transport the vectors to a common location along the path of the light before taking the dot products. All light frequency shifts in general relativity are aspects of the same underlying phenomenon. The decompositions into "cosmological redshift", "gravitational redshift" and so on are human inventions.

Some argue that there is no time dilation in this case

What John Rennie is saying there is just that the situation is symmetrical, like two people in Minkowski space who are moving apart. They see each other redshifted forever, but they're not mutually aging more slowly than each other. In some sense, they are aging "at the same rate" if you consider the symmetry. In another sense there's no way to compare the rates. The "universal time coordinate" (FLRW time) that he mentions is the proper time of the receding people, or the proper time of clocks moving with the Hubble flow.

Doppler component due to the galaxies recession speed and the time dilation component due to the space expansion

There is a redshift from the recession speed of the galaxies, and there's a redshift from the expansion of space, but they don't combine with each other, they're just equal to each other, because "recession speed" and "expansion of space" are different names for the same thing.

In the special case of linear expansion, $a(τ) = τ/τ_0$, spacetime is flat and you can actually put a global Minkowski coordinate system on it. The Minkowski coordinates $(x,t)$ are related to the FLRW coordinates $(χ,τ)$ by $$\begin{eqnarray} t &=& τ\,\cosh\,(χ/τ_0) \\ x &=& τ\,\sinh\,(χ/τ_0) \end{eqnarray}$$

and the cosmological recession is just SR relative motion and the cosmological redshift is SR redshift given by the SR formula. (As I said, you can always use the SR formula if you're willing to parallel transport the vectors, but in this case you don't have to transport them if you use Minkowski coordinates.)

You shouldn't attach too much physical significance to this coordinate system (if any at all), but it illustrates that there's no real difference between redshifts that are usually attributed to different physical mechanisms.

## Best Answer

My answer assumes that by "at the center", you mean orbiting the supermassive black hole (SMBH) at the centre.

The amount of time dilation or gravitational redshift (as measured by an observer at infinity) depends on how deeply into a gravitational potential well the clock or light source is.

From the point of view of various observers that are

inthe potential well, what matters is theirrelativegravitational potential.The potential well of our observable Galaxy will reach its deepest point at the event horizon of the supermassive black hole at its centre (or actually there will be local minima at the event horizons of any other black holes in the Galaxy). For an observer orbiting not far from the central SMBH receiving signals from Earth, they would notice that: (A) The signals were gravitationally blue-shifted, but that this overall blue-shift would be modulated by the Doppler shift caused by the orbital motion around the SMBH. (B) Events on Earth would appear to be speeded up by the same factor as the gravitational blue-shift (again, modulated by the orbital motion). These phenomena occur because the Earth is at a higher gravitational potential (less negative) than an object in a close orbit around the SMBH.

For an observer on Earth observing signals from something orbiting the SMBH, the opposite would be the case.

Some detailsGravitational redshift/blueshiftBoth the gravitational redshift and the time dilation are governed by the same mathematical factor (though frequency is decreased, whilst the spacing between ticks on a clock is increased of course); thus: $$ \tau = \tau_{\infty} \left( 1 - \frac{2GM}{rc^2}\right)^{1/2}\ \ \ \ \ f = f_{\infty} \left( 1 - \frac{2GM}{rc^2}\right)^{-1/2},$$ where $\tau$ and $f$ would be the spacing between ticks on a clock or the frequency of a light source at the observer in orbit around the black holes, whilst $\tau_{\infty}$ and $f_{\infty}$ would what those quantities were as deduced by an observer at infinity. 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 is $\sqrt{2/3}\simeq 0.82$. Of course an observer at Earth is not "at infinity", because the Earth has its own little potential well and is also sitting in the relatively weak (compared with the black hole) potential well of the Galaxy. Both of these are comparatively negligible in terms of redshift - e.g. at the surface of the Earth, the redshift factor due to the Earth's potential is $(1-1.4\times10^{-9})^{1/2}$ (small, but important for GPS systems); or if we say there is $\sim 10^{11}$ solar masses inside the Sun's orbit in the galaxy, then the redshift factor due to this would be $\sim (1- 1.1\times 10^{-6})^{1/2}$ (again, small compared with 0.82).So an observer at Earth would see signals redshifted (frequencies decreased) by a factor of 0.82 from a source in orbit around the black hole, whilst an observer in orbit around the black hole would see signals blue-shifted by a factor 1/0.82 from a source on the Earth.

Relativistic Doppler shiftThe basics of the relativistic doppler effect are covered here. 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 a factor $\gamma^{-1}$. 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, there would be a net redshift due to the transverse doppler effect. This effect is fully reciprocal, so an observer in orbit around the black hole would see light from the Earth with a net blueshift of a factor of $\gamma$ 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 transver doppler effect would be a factor of 0.87 and almost the same as the gravitational redshift. At larger orbital radii, the gravitational redshift becomes more dominant.

On top of the net tranverse doppler redshift there will be a periodic modulation of the frequency as the source orbits the black hole. Again this would be reciprocated if the source was on Earth and the observer was orbiting the black hole). The amplitude of this will depend on the inclination of the orbit to the line of sight. At its largest, $\theta = 0$ and the redshift/blueshift will be factors of $\gamma^{-1}(1 \pm v/c)^{-1}$. Thus for a source in orbit at the innermost stable circular orbit, the maximum redshift would be a factor of 0.58, whilst the maximum blueshift would be a factor of 1.74.

NB All the above calculations assume a non-rotating black hole. The details are different for spinning black holes, but are qualitatively similar, and the fact of reciprocity remains.

Sensitivity of detectionIf you are also asking what size of redshift/blueshift is detectable - at present, (optical) astronomical spectrographs are capable of detecting shifts corresponding to speeds of 1 m/s or equivalently, time dilations of 1 part in 300 million. X-ray spectrographs, which now routinely detect both the gravitational redshift and tranverse doppler effect in material circulating around supermassive black holes (though not in our own Milky Way) can detect shifts of about 1 part in a 1000.