rjbeery said:
Are you sure there would be a "last flash"? I'd be curious to see this analyzed mathematically. Reason being, if there were such a flash calculable by Bob then he could announce definitively "when" Alice has crossed the EH, which contradicts my understanding.
One can find a web reference by Hamilton (and some diagrams in Eddington-Finklestein coordinates) that show the existence of a last flash.
Answer to the quiz question 5: False. You do NOT see all the future history of the world played out. Once inside the horizon, you are doomed to hit the singularity in a finite time, and you witness only a finite (in practice rather short) time pass in the outside Universe.
Eddingtion Finklstein coordinates. Yellow lines are light, white line is infalling observer. You can see there is a last flash.
http://casa.colorado.edu/~ajsh/collapse.html
For some detailed calculations:
phase 1. Show that in Schwarzschild coordinates for a black hole of mass m=2, the geodesic is given by
-\infty < \tau < 0
r = {3}^{2/3} \left( -\tau \right) ^{2/3}
t = \tau-4\,\sqrt [3]{3}\sqrt [3]{-\tau}+4\,\ln \left( \sqrt [3]{3}\sqrt <br />
[3]{-\tau}+2 \right) -4\,\ln \left( \sqrt [3]{3}\sqrt [3]{-\tau}-2<br />
\right)
By showing that it satisfies
\frac{dr}{d\tau} = \sqrt \frac {2m}{r}
\frac{dt}{d\tau} =\frac{1}{1-2m/r}
See for instance
http://www.fourmilab.ch/gravitation/orbits/, or your favorite GR textbook. m=2 was chosen to make the expressions more tractable, you may choose to repeat without this attempt at simplification if you prefer
phase 2: convert to ingoing Eddington Finklestein coordinates by the transformation
v = t + r + 4\,\ln \left| \frac{r}{2m} - 1 \right|
(recall that we set m=2 in phase 1).
Phase 2a: recall, or derive, that for infalling light, v=constant. Therefore v(tau) gives you the "flash number" you are viewing at time tau.
Get
v =<br />
\tau-4\,\sqrt [3]{3}\sqrt [3]{-\tau}+4\,\ln \left( \sqrt [3]{3}\sqrt <br />
[3]{-\tau}+2 \right) -4\,\ln \left( \sqrt [3]{3}\sqrt [3]{-\tau}-2<br />
\right) +{3}^{2/3} \left( -\tau \right) ^{2/3}-8\,\ln \left( 2<br />
\right) +4\,\ln \left( \left| \left( \sqrt [3]{3}\sqrt [3]{-\tau}+<br />
2 \right) \left( \sqrt [3]{3}\sqrt [3]{-\tau}-2 \right) \right| <br />
\right) <br />
Use the fact that ln(a*b) = ln(a)+ln(b) to rewrite this and cancel out the apparent singularity in v
<br />
v = \tau-4\,\sqrt [3]{3}\sqrt [3]{-\tau}+8\,\ln \left( \sqrt [3]{3}\sqrt <br />
[3]{-\tau}+2 \right) +{3}^{2/3} \left( -\tau \right) ^{2/3}-8\,\ln <br />
\left( 2 \right)
Observe that v is finite (zero) when tau -> 0, so that you do not in fact see the entire history of the universe before you reach the event horizon, furthermore that you don't see the entire history of the universe before you reach the singularity.
If you don't like the formal cancelllation of the divergent terms note that you can compute the limit of v as you approach the event horizon, and show that the limit exists to answer the original question. (You won't see any results inside the event horizon that way though).
Option: recompute the geodesic equations in EF coordinates and show that the modified solution satisfies them to justify the formal cancelation of the divergent terms.If this seems like waaaaay too much work, just study Hamilton's EF plot, or find the plot of an infalling observer in Eddington Finklestein coordinates in your favorite textbook.