# Gravitational Redshift in Kerr Spacetime

Well, suppose then Schwarschild black holes. Following the $$[1]$$, we have the redshift factor:

$$d\tau = \sqrt{1-\frac{2M}{r}}dt. \tag{1}$$

This factor have an physical interpretation to be the time a given observer measures on her own clock, i.e., for a stationary observer the proper time relates to the time as measured by a distant observer via $$(1)$$

Conversely, if we take the $$r= cte$$, $$\theta = cte$$ and $$\phi = cte$$ then the metric reduces to:

$$d\tau ^{2} = g_{00}dt ^{2} \tag{2}$$

Which is the gravitational redshift.

Now, this realization seems to be valid for every metric. I mean, take Kerr metric (which is an example of non-diagonal tensor) if we state the same (now Boyer-Lindquist coordinates, of course) $$r= cte$$, $$\theta = cte$$ and $$\phi = cte$$ we get:

$$d\tau = \sqrt{\Bigg(1-\frac{2Mr}{r^{2}+a^{2}cos^{2}\theta}\Bigg)}\hspace{3mm} dt \tag{3}$$

My doubt is:

Can we say that the factor above, using the kerr metric, have an physical interpretation to be the time a given observer measures on her own clock, i.e., for a stationary observer the proper time relates to the time as measured by a distant observer via $$(3)$$?

$$* * *$$

$$[1]$$ Relativity Demystified

## 2 Answers

Yes, this is correct for an observer at rest in Boyer-Lindquist coordinates. The same reasoning applies to any metric. But it isn’t all that interesting because in general observers are more likely to be moving (e.g., orbiting).

For a ZAMO it is

$$\frac{{\rm d} t}{{\rm d} \tau} = \sqrt{g^{t t}}$$

for an object moving with local velocity $$v$$ relative to the ZAMO it is

$$\frac{{\rm d} t}{{\rm d} \bar\tau} = \frac{\sqrt{g^{t t}}}{\sqrt{1-v^2/c^2}}$$

and for an observer stationary with respect to the fixed stars it is

$$\frac{{\rm d} t}{{\rm d} \tilde\tau} = \frac{1}{\sqrt{g_{t t}}} = \frac{\sqrt{g^{t t}}}{\sqrt{1-\tilde v^2/c^2}}$$

where $$\tilde v$$ is the local frame dragging velocity relative to the fixed stars

$$\tilde v = c \sqrt{g_{t \phi} \ g^{t \phi}} = c \sqrt{1 - g_{t t} \ g^{t t}}$$

• Thanks. But this do not invalidate my analysis, right? (also, your work on kerr spacetime simulations are quite cool) – M.N.Raia May 29 '20 at 10:48
• No, it doesn't invalidate it, I just wanted to expand on it. Just keep in mind that the gravitational ratio is absolute while the kinematic component is symmetric, so if you want to know the time dilation of a far away observer in the frame of the orbiting particle instead of vice versa you have to divide by the gamma factor γ=√(1-v²/c²)⁻¹ instead of multiplying with it. – Gendergaga May 29 '20 at 13:58
• If I reproduce these quantities (i.e. do the same calculations as you; for instance the ZAMO) for another rotating metric (a non-kerr black hole, but still a axisymmetric tensor) I will reach similar interpretations? – M.N.Raia May 29 '20 at 15:31
• Yes, but if for example you use raindrop coordinates (Doran) then the $\sqrt{g^{tt}}$ gives you the time dilation relative to a radially infalling and corotating test particle, while in Boyer Lindquist it is a corotating and radially stationary one, it always depends on the local reference of your chosen coordinates. If you want to describe the time dilation of a radially stationary ZAMO, but using Doran coordinates, you have to take the radial velocity of your ZAMO relative to the infalling reference frame into account. – Gendergaga May 29 '20 at 16:43