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Gravitational Redshift for Photon around Kerr Black Hole

  1. Jul 26, 2009 #1
    I have a hopefully straightforward question. It is well known that in the Schwarzschild metric the gravitational redshift is given by [tex]1+z=(1-r_{s}/r)^{-1/2}[/tex]. Clearly this is just the ratio of observed to emitted frequencies (or energies). I understand this so far. However, for the case of the Kerr spacetime, in Boyer-Lindquist coordinates
    [tex]ds^{2}=\bigg(1-\frac{2Mr}{\Sigma}\bigg)dt^{2}+\frac{4aMr \sin^{2}\theta}{\Sigma}dt d\phi-\frac{\Sigma}{\Delta}dr^{2}-\Sigma d\theta^{2}-\bigg(r^{2}+a^{2}+\frac{2a^{2}Mr \sin^{2}\theta}{\Sigma} \bigg)\sin^{2}\theta d\phi^{2},[/tex]
    where

    [tex]\Sigma \equiv r^{2}+a^{2}\cos^{2}\theta[/tex] and [tex]\Delta \equiv r^{2}-2Mr+a^{2}.[/tex]This asymptotes to the Schwarzschild case in the limit [tex]a\rightarrow 0[/tex]

    For the Schwarzschild black hole [tex]1+z=(g_{tt})^{-1/2}[/tex]. I believe this is not the case for the Kerr spacetime (because of frame-dragging in the cross-term?).

    What is the expression for the gravitational redshift in the Kerr spacetime for a photon (I can list the geodesic equations of motion if needed)? Or, how would one go about deriving such a formula? Presumably there would be some [tex]r[/tex] as well as [tex]\theta[/tex] -dependence in said expression (as well as spin, a)?

    Thank you.
     
  2. jcsd
  3. Jul 31, 2009 #2
    Any help, please?
     
  4. Jul 31, 2009 #3
    Gravitational redshift for the Kerr metric is-

    [tex]\alpha=\frac{\rho}{\Sigma}\sqrt{\Delta}[/tex]

    where

    [tex]\rho=\sqrt{r^2+a^2 cos^2\theta}[/tex]

    [tex]\Sigma=\sqrt{(r^2+a^2)^2-a^2\Delta sin^2\theta}[/tex]

    [tex]\Delta= r^{2}+a^{2}-2Mr[/tex]

    (note the above is a slightly different version of expressing Kerr metric to the one in your first post but the equations/maths are the same)

    The above reduces to the Schwarzschild solution when a=0 [itex](\alpha=\sqrt{1 - 2M/r})[/itex]
     
    Last edited: Jul 31, 2009
  5. Aug 4, 2009 #4
    Thanks stevebd1,

    I take it that [tex]\alpha \equiv (1+z)^{-1}[/tex], where z is redshift?
     
  6. Aug 4, 2009 #5

    George Jones

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    In order to calculate gravitational redshift, you have to specify which observer emits the photon and which observer receives the photon.
     
  7. Aug 5, 2009 #6
    This is correct for an observer at infinity where z is the fractional shift in a spectral wavelength ([itex]\alpha[/itex] is sometimes referred to as the reduction factor). You can also say for a rotating object-

    [tex]z=\Sigma\left(\rho\sqrt{\Delta}\right)^{-1}-1[/tex]

    which reduces to the Schwarzschild solution when a=0-

    [tex]z=\left(1-\frac{2M}{r}\right)^{-1/2}-1[/tex]
     
  8. Aug 5, 2009 #7

    George Jones

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    Yes, this is the expression for Kerr gravitational redshift.
    Yes, this is the expression for Kerr gravitational redshift.

    How can both of these expressions be the expression for Kerr gravitational redshift? :biggrin: They relate exchange of photons between different pairs of observers.

    Given the thread

    https://www.physicsforums.com/showthread.php?t=309048,

    I suspect that neither of these expressions is the expression in which you're (Ruf30) really interested, i.e., you're really interested in exchange of photons between a completely different pair of observers.

    It might be fun to work through the math, and the physics behind the math, of the derivations of above expressions. Maybe we should start with
    How is this expression derived?
     
  9. Aug 7, 2009 #8
    Thank you George and stevebd1,

    It is true that these expressions are only valid for an observer at infinity. You are right George, neither is what I'm truly interested in. As you know from my previous thread, I'm imaging accretion discs (and tori) around Kerr black holes. It's easier if I explain what I am doing before I ask a question, so as to have some idea of context. I apologize for the inevitable length of this post.

    First of all, redshift is defined by ([tex]E=h\nu[/tex]):
    [tex](1+z)=\frac{\nu_{0}}{\nu}=\frac{(p_{\beta}u^{\beta})_{\lambda}}{(p_{\alpha}u^{\alpha})_{\infty}}[/tex].

    For an (stationary) observer at infinity, or in the case I am working with, at say 200 gravitational radii from the black hole centre, [tex]u^{r},u^{\theta}[/tex] and [tex]u^{\phi}[/tex] are all zero, hence the only contribution is [tex]p_{t}\dot{t}[/tex]. Also, [tex]\dot{t}_{obs}[/tex] is determined in the initial conditions, and [tex]p_{t}\equiv -E[/tex], which is also computed in the initial conditions (and is constant).

    I have no problem in finding the redshift of an observer relative to emission from a material particle on the disc surface: We simply find the components of the four-velocity of a point on the disc. The flow is rotationally supported in the [tex]\hat{r}[/tex] and [tex]\hat{\theta}[/tex] directions, thus [tex]\dot{r}[/tex] and [tex]\dot{\theta}[/tex] are zero. One also finds expressions for [tex]\dot{t}[/tex] and [tex]\dot{\phi}[/tex], but I wont list them here (quite long).

    I take the disc as lying in the equatorial plane, so setting [tex]\theta=\pi/2[/tex] simplifies the expressions considerably. With [tex]E_{em(disc)}=p_{t}u^{t}+p_{\phi}u^{\phi}=p_{t}u^{t}\Big(1+\Omega_{k}\frac{p_{\phi}}{p_{t}} \Big)=p_{t}u^{t}\Big(1-\Omega_{k}\frac{L}{E} \Big)=p_{t}\dot{t}_{em(disc)}\Big(1-\Omega_{k}\frac{L}{E} \Big)[/tex] ,

    [tex]E_{obs}=p_{t}\dot{t}_{obs}[/tex],
    we find the redshift of the observer relative to disc emission as

    [tex](1+z)=\frac{\dot{t}_{em}}{\dot{t}_{obs}}\Big(1-\Omega_{k}\frac{L}{E} \Big)[/tex] , where

    [tex]\Omega_{k}=(r^{3/2}+a)^{-1}[/tex] and [tex]\dot{t}_{em}=\bigg(\Omega_{k}\sqrt{r}\sqrt{r^{2}-3r+2a\sqrt{r}}\bigg)^{-1}[/tex].

    I've included a redshift map of a Schwarzschild accretion disc, viewed at an angle of 75 degrees from the vertical. You can see the features I mention in my previous post.

    In reality I'm beyond this and am now looking at solving the local G.R. radiative transfer equation:
    [tex]\frac{d\textit{I}}{ds}=-\chi \textit{I}+\eta \Big(\frac{\nu_{0}}{\nu}\Big)^{3}[/tex], where

    [tex]\textit{I}=I/\nu^{3}[/tex], [tex]\chi = \frac{\nu_{0}}{\nu}\chi_{0}[/tex] and [tex]\eta=(\nu/\nu_{0})^{2}\eta_{0}[/tex]; observer - no subscript, local frame - sunscript '0'.

    I must solve the transfer equation once I have computed the geodesic for a photon (I'm looking to plot intensity maps and eventually spectra). My problem is how do I find the redshift [tex]\nu_{0}/\nu[/tex] of a photon at each point along its path ([tex]\small{ds}[/tex]) after disc emission relative to the observer, who is not an infinite distance away.

    I'm having problems with this because, photons are null geodesics, so [tex]p_{\beta}u^{\beta}\equiv 2\textsl{H}[/tex], the Hamiltonian, which is zero for photons. This means the redshift after emission from the disc to the observer, at each point along the geodesic is zero (well, (1+z)=0, due to energy conservation)?

    How does one find the redshift of the photon at each point along its geodesic path back to the observer, relative to this observer, if said observer is a finite distance from the black hole?
     

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