Well done!
Hi again, Jorrie,
Jorrie said:
I have tried my hand for a Lemaitre observer. Despite expecting hand slaps for not using Painleve coordinates, I worked in a good old Schwarzschild
chart and hence had to stay outside of the event horizon.
Right, you used the method suggested by pervect, which combines the usual gravitational redshift (for a distant static observer and a static observer at some finite Schwarzschild radius) with the usual Doppler shift (valid for two local frames at event E), which is perfectly valid in the exterior region. And good, you used the correct velocity v = \sqrt{2m/r} (taking the positive sign since the Lemaitre observer is falling away from the distant static observer), and then as you say
\frac{ \lambda_{{\rm received}} }{ \lambda_{{\rm emitted}} } = \sqrt{1-2m/r} \, \frac{ \sqrt{1+\sqrt{2m/r}} }{ \sqrt{1-\sqrt{2m/r}} } = 1+\sqrt{2m/r}
and as you observed this approaches a wavelength ratio of two as the Lemaitre observer approaches the horizon. So if our Lemaitre observer has a rear pointing spectrograph, he knows he is near the event horizon when he sees a redshift in his rear view mirror of about two. But if he only performs experiments inside the cabin of his rocket ship, then he won't notice anything in particular as he passes the horizon!
Next, because the interior region belongs to an
real analytic extension of the result you found, by a computation which is valid only the exterior region, and since
the formula you found continues to make sense on 0< r < 2m, we might expect that it is also valid in the interior. To confirm that I suggested using a second method to directly compute the frequency ratio, which is valid on both regions. (More on that below.)
Also, it is worthwhile finding the radius as a function of the proper time of the Lemaitre observer remaining until impact at r=0, and then plugging into the expression you found above, you should find a certain power law expression for the observed redshift as a function of the proper time remaining until impact. So if our Lemaitre observer keeps an eye on his rearpointing spectrograph, he can estimate his remaining lifetime (assuming he doesn't turn on his rocket engine).
Likewise, going the other way, you found the correct redshift for signals sent radially from the Lemaitre observer back up to the distant static observer. It is interesting to express this in the proper time of the latter observer. His proper time agrees with Schwarzschild coordinate time (by definition of the exterior Schwarzschild chart!), so you need only compute r(t) for the Lemaitre observer. The coordinate speed is dt/dr = -1/(1-2m/r)/\sqrt{2m/r}, so integrating we find
t-t_0 = -\sqrt{\frac{2r^3}{9 m}} - \sqrt{8 m r} + 2 m \, \log \left( \frac{1+\sqrt{2m/r}}{1-\sqrt{2m/r}} \right)
This is hard to invert! But notice that the first term dominates when r \gg 2m, while the third term dominates when r \approx 2m. So you should be able to find two approximate expressions from which you can see that the redshift obeys a power law for most of the infall, but very near the horizon obeys an exponential law.
pervect said:
Yes, I've gotten that result in the past. Note that it corresponds to setting E=1 in the more general expression for the velocity (relative to a stationary observer) of a radially infalling observer
<br />
\frac{\sqrt{E^2 - (1 - \frac{2M}{r})}}{E}<br />
and that E=1 corresponds to a LeMaitre observer (someone who's velocity is zero at infinity).
Exactly. This is a third method (valid only in the exterior region, where the standard analysis via effective potentials of the null geodesics is valid).
Before we move on to the second method, note that the first method can be applied to other radially infalling observers, such as the slowfall observers. In the exterior Schwarzschild chart, the slowfall observer has tangent vector
\frac{1-m/r}{1-2m/r} \, \partial_t - \frac{m}{r} \, \partial_r
That is, in the exterior he has a world line of form
t-t_0 = -\frac{r^2}{2 m} - r - 2m \log ( r-2m)
Using the second method, can you find the wavelength ratio for signals sent from this observer back up to the distant static observer? And vice versa!
Here, recall that the slowfall observers, by definition, accelerate radially outward with just the right magnitude of acceleration (as a function of position) which would suffice to maintain their position according to Newton's theory of gravitation, namely m/r^2. But in gtr, the energy of the gravitational field itself gravitates, so the gravitational attraction is a bit stronger (roughly speaking), so these observers slowly fall radially inwards. How does this agree with the result you found for signals sent from the distant static observer to a slowfall observer? Can you find the wavelength ratio measured by the slowfall observer as a function of his proper time, and (suitable approximationate expressions for) the wavelength ratio measured by the distant static observer is
his proper time?
Earlier I remarked that for signals sent radially from a distant static observer down to a radially falling observer, we should expect the usual gravitational blue shift to oppose the Doppler red shift. We could have guessed which would win out in the case of Lemaitre observers from your answer to the question about slowfall observers. This should help to explain why we found a redshift for both ingoing and outgoing signals in the case of a Lemaitre observer and a distant static observer.
OK, on to method two:
The ingoing Eddington chart is
defined so that the
ingoing radial null geodesics appears as horizontal coordinate lines. We can ensure this by setting du = dt + dr/(1-2m/r), which integrates to u = t + r + 2m \, \log(r-2m). The line element in the new chart is
ds^2 = -(1-2m/r) \, du^2 + 2 \, du \, dr + r^2 \, \left( d\theta^2 + \sin(\theta)^2 \, d\phi^2 \right),
-\infty < u < \infty, \; \; 0 < r < \infty, \; \; 0 < \theta < \pi, \; \; -\pi < \phi < \pi
This chart is well defined on the right exterior and future interior regions, the same places where congruences of infalling observers are defined. It is perfectly suited for analyzing wavelength ratios for pairs of observers in which a more distant observer is sending signals radially downward to a closer observer. Can you figure out how to draw the light cones in this chart? The frame fields for static, Lemaitre, and slowfall observers? (Hint: if you know how to transform a vector field into a new chart, that's all you need since the frame fields are made up of unit vector fields.) Can you draw a diagram from which, using similar triangles, you can confirm our guess above about the general expressions for signals sent from the distant static observer down to the Lemaitre or slowfall observers? (The article "Frame fields in general relativity" in the version listed at
http://en.wikipedia.org/wiki/User:Hillman/Archive should help if any of this seems confusing.)
The outgoing Eddington chart is defined so that the
outgoing null geodesics appear as straight lines, by an expression very similar to the above. This chart is perfectly suited for analyzing wavelength ratios for pairs of observers in which a closer observer is sending signals radially outward to a more distant observer. Can you use the second method with this chart to confirm the expressions we found earlier?
Note: the Lemaitre and slowfall
frame fields are defined on the right exterior region and future interior region, whereas the outgoing Eddington chart is defined on the past interior and right exterior region (referring to the usual block diagram exhibiting the global causal structure of the maximal analytic extension of the exterior Schwarzschild vacuum). So, our results here only make sense for the exterior region. Of course, that is just what we expect since infalling observers cannot send signals outside the horizon if they have fallen into the future interior region!
You may be familiar with the (past)
interior Schwarzschild chart in which the line element becomes
ds^2 = \frac{-1}{2m/t-1} \, dt^2 + \left( 2m/t-1 \right) \, dz^2 + t^2 \, \left( d\theta^2 + \sin(\theta)^2 \, d\phi^2 \right),
0 < t < 2m, \; \; -\infty < z < \infty, \; \; 0 < \theta < \pi, \; \; -\pi < \phi < \pi
Observers who maintain constant spatial coordinates in this chart are called Frolov observers and they are geodesic observers who never emerge into either exterior region, although the Frolov congruence is also defined in the future interior region (it has a "caustic" at "the center of the X" in the usual block diagram, i.e. some pairs of world lines of Frolov observers intersect at that two-sphere). If you have seen the fine embedding diagrams of various spatial hyperslices inside the Schwarzschild vacuum as discussed in MTW,
Gravitation, then you might recognize the spatial hyperslices above as the expanding (shrinking) cylinders {\mathbold R} \times S^2; in the Kruskal-Szekeres chart they appear inside the past (future) interior regions as hyperbolic arcs nested between the horizon and the past (future) curvature singularities.
Can you find the Frolov frame (in the future interior region) in terms of the ingoing Eddington chart? Can you then extend your computations above to obtain the wavelength ratio observed by a Frolov observer in the future interior region, for signals sent radially inward by a distant static observer? Can you re-express your answer as a function of the proper time of our Frolov observer? You should a blue shift evolving into a red shift! (Where does the transition occur?) Can you find the world lines of the Frolov observers in terms of the ingoing Eddington and Painleve charts? If you plot the world lines Lemaitre, slowfall and Frolov observers in the ingoing Eddington chart, can you "see" why we obtain redshift or blueshift?
My previous (post #52) was strictly aimed at Schwarzschild coordinates, where I tried to correct my poor definitions of observers outside the horizon. 
)
