Widdekind Posted August 24, 2011 Share Posted August 24, 2011 If, in the FRW metric, the 'interval' is: [math]ds^2 = (c \, dt)^2 - R(t)^2 \left( \frac{dr^2}{1 - K r^2} + r^2 \left( d \theta^2 + ( sin(\theta) d\phi )^2 \right) \right)[/math] And if, in GR, that 'interval' is derived from the 'metric' tensor, via: [math]ds^2 = g_{\mu \nu} ds^{\mu} ds^{\nu}[/math] And if [math]ds \equiv \left[ c dt, dr, d\theta, d\phi \right][/math]; then, why wouldn't the 'metric' tensor be: [math] \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & R^2 / 1- K r^2 & 0 & 0 \\ 0 & 0 & R^2 r^2 & 0 \\ 0 & 0 & 0 & R^2 r^2 sin(\theta)^2 \end{array} \right)[/math] or, if [math]ds \equiv \left[ c dt, R dr, R r d\theta, R r sin(\theta) d\phi \right][/math] (to put all the physical distances into the interval); then, why wouldn't the 'metric' tensor be: [math] \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 / 1- K r^2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right)[/math] ? Link to comment Share on other sites More sharing options...
ajb Posted August 24, 2011 Share Posted August 24, 2011 Are you asking about the minus sign difference between the time and space? Link to comment Share on other sites More sharing options...
Widdekind Posted August 26, 2011 Author Share Posted August 26, 2011 (edited) According to Eric V. Linder's First Principles of Cosmology, pg.88, [math]1+z = \frac{\left( \vec{k} \circ \vec{u} \right)_e}{\left( \vec{k} \circ \vec{u} \right)_o}[/math] where [math]\vec{k} \circ \vec{u} = g_{\mu \nu} k^{\mu} u^{\nu}[/math]. Yet, when that four-dot-product is taken, the Minkowski (locally flat-space) metric tensor seems to be utilized, such that (for a radially-inbound [math]e \rightarrow o[/math] photon): [math]1+z = \frac{\lambda_o}{\lambda_e} \frac{\gamma_e \left( 1 + \beta_{r,e} \right)}{\gamma_o \left( 1 - \beta_{r,o} \right)}[/math] What would happen, if you considered cosmologically-distant light-sources, deep down in a gravity well? If, as a first approximation, one were to assume a Schwarzschild space-time, is not the Schwarzschild Metric: [math] \left( \begin{array}{cccc} 1-R_S/r & 0 & 0 & 0 \\ 0 & -1 / (1- R_S/r) & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{array} \right)[/math] ? Then, the four-dot-product time-time components [math]1 \rightarrow 1-R_S/r[/math]; and, the space-space components [math]\beta_r \rightarrow \beta_r / (1-R_S/r)[/math], so that: [math]1+z = \frac{\lambda_o}{\lambda_e} \frac{\gamma_e \left( (1- \frac{R_{S,e}}{r}) + \frac{\beta_{r,e}}{1-R_{S,e}/r} \right)}{\gamma_o \left( (1- \frac{R_{S,o}}{r}) - \frac{\beta_{r,o}}{1-R_{S,o}/r} \right)} \approx \frac{\lambda_o}{\lambda_e} \frac{ \left( (1- \frac{R_{S,e}}{r}) + \beta_{r,e} (1+R_{S,e}/r) \right)}{\left( (1- \frac{R_{S,o}}{r}) - \beta_{r,o} (1+R_{S,o}/r) \right)} = \frac{\lambda_o}{\lambda_e} \frac{ \left( 1 + \beta_{r,e} - \frac{R_{S,e}}{r} (1 - \beta_{r,e}) \right)}{\left( 1 - \beta_{r,o} - \frac{R_{S,o}}{r} (1 + \beta_{r,o}) \right)}[/math] to lowest orders, in velocities and radii terms. For the Schwarzschild terms, the time-originating piece dominates the space-originating piece, and actually reduces the redshift, if the emitter is deep down in a gravity well. Could that be correct ?? Is not the RS of our galaxy, of [math]\approx 10^{12} M_{\odot}[/math], thusly about one-third-light-year ? If so, then for us, out ~30Kly from our galactic core, the effect would be O(10-5). EDIT: To emphasize the 'gravity-well' effect, assume zero velocity; and, assume that some emission line implies [math]\lambda_e[/math], whilst [math]\lambda_o[/math] is known. Then: [math]\frac{1 - \frac{R_{S,o}}{r} }{ 1 - \frac{R_{S,e}}{r}} (1+z) \approx\frac{\lambda_o}{\lambda_e} [/math] showing that light, blue-shifting in-bound at the observer, much increase the inferred red-shift; and, to account for non-cosmological red-shifting out-bound at the emitter, "before the light is loosed-and-let-to-fly out into the Hubble Flow", one must reduce their inferred cosmological red-shift value. Could this be relevant for photons emitted near compact objects, e.g., GRBs ?? However, for GRBs, I understand that [math]\lambda_e[/math] is unknown; and, that known cosmological red-shifts are obtained, separately, from their host galaxy. Thus, [math] \lambda_e \approx \frac{\lambda_o}{1+z} \frac{1 - \frac{R_{S,e}}{r} }{ 1 - \frac{R_{S,o}}{r}}[/math] showing that as [math]r_e \rightarrow R_{S,e}[/math], then [math]\lambda_e \rightarrow 0[/math], i.e., the radiation is "harder than it looks". And, indeed, long-period GRBs typically are observed to have softer radiation, but be BH-generated; whereas short-period "short hard" GRBs typically are observed to have harder radiation, but may be NS-merger generated. That picture appears to be completely self-consistent, since BH-generated l-GRBs emit their light closer to the Schwarzschild radius. Edited August 26, 2011 by Widdekind Link to comment Share on other sites More sharing options...
Widdekind Posted August 27, 2011 Author Share Posted August 27, 2011 (edited) re-EDIT: To emphasize the 'gravity-well' effect, assume zero velocity; and, assume that some emission line implies [math]\lambda_e[/math], whilst [math]\lambda_o[/math] is known. Then: [math]\frac{1 - \frac{R_{S,o}}{r} }{ 1 - \frac{R_{S,e}}{r}} (1+z) \approx\frac{\lambda_o}{\lambda_e} [/math] showing that light, blue-shifting in-bound at the observer, much increase the inferred red-shift; and, to account for non-cosmological red-shifting out-bound at the emitter, "before the light is loosed-and-let-to-fly out into the Hubble Flow", one must reduce their inferred cosmological red-shift value. Could this be relevant for photons emitted near compact objects, e.g., GRBs ?? However, for GRBs, I understand that [math]\lambda_e[/math] is unknown; and, that known cosmological red-shifts are obtained, separately, from their host galaxy. Thus, [math] \lambda_e \approx \left( \frac{\lambda_o}{1+z} \right) \frac{1 - \frac{R_{S,e}}{r} }{ 1 - \frac{R_{S,o}}{r}}[/math] showing that as [math]r_e \rightarrow R_{S,e}[/math], then [math]\lambda_e \rightarrow 0[/math], i.e., the radiation is "harder than it looks". And, indeed, long-period GRBs typically are observed to have softer radiation (~1/3 me c2), but be BH-generated; whereas, short-period "short hard" GRBs typically are observed to have harder radiation (~1/2 me c2), but may be NS-merger generated (Mazure & Basa. Exploding Superstars; Bloom. What are GRBs ?). That picture appears to be completely self-consistent, since BH-generated l-GRBs emit their light closer to the Schwarzschild radius. Note, tho, that equally plausibly, humans on earth only observe GRBs, when looking "down the bore-sight", straight towards a pole, of the central compact object. Thus, matter in-falling towards said central accretor, would be moving radially away from earth -- and, hence, any light emitted "back up-gravity-well", towards earth, would be redshifted. Thus, matter infalling faster, towards a BH, during an l-GRB, could also account for the inferred redshifting of the radiation (assumed to be from pair production, at 511 KeV), relative to a somewhat slower infall, towards the NSs, in an s-GRB. Therefore, matter infalling towards the central compact object, at a considerable fraction of the speed-of-light; or, radiating matter residing, close to said compact object (~1.5-2 RS), could both account, jointly or severally, for the relative redshift, of l-GRBs, compared to s-GRBs. Edited August 27, 2011 by Widdekind Link to comment Share on other sites More sharing options...
Widdekind Posted September 25, 2011 Author Share Posted September 25, 2011 According to Eric V. Linder's First Principles of Cosmology, pg.88, [math]1+z = \frac{\left( \vec{k} \circ \vec{u} \right)_e}{\left( \vec{k} \circ \vec{u} \right)_o}[/math] where [math]\vec{k} \circ \vec{u} = g_{\mu \nu} k^{\mu} u^{\nu}[/math]... What would happen, if you considered cosmologically-distant light-sources, deep down in a rotating gravity well? If, as a first approximation, one were to assume a Kerr space-time, then would -- for calculating the 4-dot product at the emitter -- not the Schwarzschild Metric, for a polar-emitted photon ([math]\theta = d\theta = d\phi \equiv 0[/math]), e.g., long-type-GRBs, look vaguely like... [math] \left( \begin{array}{cccc} 1- r R_S/(r^2 + \alpha^2) & 0 & 0 & 0 \\ 0 & -(r^2 + \alpha^2) / (r^2 - \alpha^2 + r R_S) & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{array} \right)[/math] ? At the emitter, the photon 4-vector is, in natural units (h = c = 1), [math]\lambda_e^{-1} \left( \begin{array}{c} 1 \\ -1 \\ 0 \\ 0 \end{array} \right)[/math]; and, the emitter 4-vector is [math]\gamma_e \left( \begin{array}{c} 1 \\ \beta_{e,r} \\ \beta_{e,\theta} \\ \beta_{e,\phi} \end{array} \right)[/math]. And then, the four-dot-product, at the emitter, becomes: [math]\left( \vec{k} \circ \vec{u} \right)_e = \left( \frac{\gamma_e}{\lambda_e} \right) \left( \left[ 1- \frac{r R_{S,e}}{r^2 + \alpha^2} \right] + \beta_{e,r} \left[ \frac{r^2 + \alpha^2}{r^2 - \alpha^2 + r R_{S,e}} \right] \right)[/math] And more, borrowing from the PP, the four-dot-product, at the observer, assumed-to-be in the observer's rest-frame, where [math]\gamma_o = \beta_o = 0[/math]; and, which rest-frame is further assumed-to-be non-rotating; thusly becomes: [math]\left( \vec{k} \circ \vec{u} \right)_o = \left( \frac{1}{\lambda_o} \right) \left( 1- \frac{R_{S,o}}{r} \right)[/math] And then: [math]1+z = \frac{\left( \vec{k} \circ \vec{u} \right)_e}{\left( \vec{k} \circ \vec{u} \right)_o}[/math] [math]\; \; \; \; = \left( \frac{\lambda_o}{\lambda_e} \right) \left( \frac{\gamma_e}{1- \frac{R_{S,o}}{r}} \right) \times \left( \left[ 1- \frac{r R_{S,e}}{r^2 + \alpha^2} \right] + \beta_{e,r} \left[ \frac{r^2 + \alpha^2}{r^2 - \alpha^2 + r R_{S,e}} \right] \right) [/math] For a maximally-rotating Kerr BH, the "Kerr radius" [math]\alpha = R_S / 2[/math]; and, emission can occur, from this "rotationally reduced" radius [math]r = \alpha[/math]. Assuming such, the above formula simplifies to: [math]1+z = \left( \frac{\lambda_o}{\lambda_e} \right) \left( \frac{\gamma_e}{1- \frac{R_{S,o}}{r}} \right) \times \beta_{e,r} [/math] Ignoring the O(10-5) effect, of our own assumed-to-be-the-observer gravity well, w.h.t.: [math]1+z = \left( \frac{\lambda_o}{\lambda_e} \right) \times \left( \gamma_e \, \beta_{e,r} \right) \rightarrow \infty [/math] What is the formula, for the "fall-back radius", for a given (rotating) BH, and a given "initial launch velocity", i.e., given [math]\beta_r[/math], how close can you get to a given (rotating) BH, and still escape ? Assuming that, per PP, [math]\lambda_o \approx 1/3 \lambda_e[/math]; and, assuming an average red-shift for long-GRBs of [math]<z> \approx 0.7[/math] (Mazure. Exploding Super-Stars); and, assuming essentially totally-radial (electron) emitter velocity; then, [math]\gamma_e \beta_e \approx 5[/math], i.e., ~0.98 c. Presumed-to-have-fallen-back electrons, acted, "long long ago, far far away", like "launch-vehicle rocket sleds for barely-escaping gamma-ray photons" ?? ADDENDA: This source seems partially poignant, and particularly interesting. Link to comment Share on other sites More sharing options...
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