The speed of light in a gravitational field

[Passionflower]

I think it's do-able. I just think the tail-end won't be able to maintain the separation after the leader goes through the EH. They will definitely lose radar contact after that time, because no light signal can be sent to outside the EH from inside.

I'll have a go at the calculation when I find a pencil and an old envelope

Actually I am not totally sure they will lose radar contact or not, since both the head and the tail are non stationary. If the signal sent from the head will reach the tail by the time the tail passed the EH it should be fine. I think in fancy terms you call this the "perceived event horizon".

 

[PAllen]

I wrote 'I went a step further'. Do you understand what the plot represents? Do you understand that by using this we can express the measured speed of light as ruler or proper distance divided by the time it takes light to get there?

All I did was to express things with physical distances instead of coordinate values mainly because there are a few individuals here who seems to think that everybody else knows next to nothing and that they are the only ones who know that r2-r1 is not a physical distance.

Well, in return, did you read and understand what I wrote? I did enough of the calculation to see that tangential and radial *non-local* lightspeeds are different. Thus a claimed formula for 'the general case' that does not take direction into account is wrong. All formula's you've shown are clearly valid only radially. Using proper distance/time does not remove real anistropy; it will only remove coordinate anisotropy. My claim, which are free to refute is that nonlocal measurements of lightspeed in this geometry are anisotropic, while local ones are isotropic.

I think my formulas are right, the derivation was done by Maple, perhaps you want to argue that Maple has a bug?

Well, it is a question of Maple+person+copying. The antiderivative of the dr integral is of the form: sqrt(r(r-R)) -R ln(sqrt(r) + sqrt(r-R))
[EDIT: I see my mistake. I should have +R ln(..) instead of -R ln(...).
Now you see why i would prefer to comment and question without calculating. I have no maple or any form of math software. It is a lot
harder for me. Thus Passionflower's main formulas are correct. ]
If you evaluate this from r1 to r2, you get:

sqrt_expr(r2) - sqrt_expr(r1) + R ln_expr(r1) - R ln_exr(r2)

which produces what I claim, not what you claim.

As to your simplification, it looks like somehow, you ended up multiplying where you should have divided.

Perhaps this cannot be settled without someone else doing the calculation from scratch.

We would measure the distance from the head as we use the clock on the head to measure the elapsed time of light. We can use a distance based on Lorentz factoring the integrand of a stationary observer or we can use a distance in Fermi coordinates. Point is we can calculate it and it is a good exercise doing it.

All it takes is a positive attitude, if I make a mistake then help correcting it instead of throwing 'doom' at it.

I raise a real issue, you get mad that I raise it. It is possible to raise an issue, without proposing an answer. There is no law that says I must either know the answer or spend the time to compute an answer, in order to raise a valid issue. If the you think the issue is not really an issue, respond with an argument to the content, rather than a snit.

In no way did I imply it couldn't or shouldn't be done; I just observed that simultaneity for a free falling observer will be different from a stationary one, and that if coordinate t defines simultaneity for the static observer, it cannot also specify simultaneity for the free falling observer. If you think this is not a true statement, argue against it. If you were already planning to consider it, just say so.

There is nothing 'expert' or positive about an attitude that it is all messy, too hard, not defined, meaningless etc. There is nothing 'expert' or positive about implying that people are too uninformed to even attempt to do it.
There is nothing 'expert' about giving only 'baby talk' by repeating that the the speed of light is c locally when clearly the discussion goes beyond that.

Why must you be so rancorous? I implied none of these things. You seem to think discussion is attack.

The comment about local measurement of lightspeed was specifically to pervect, and also yuiop. Pervect expressed doubt about the calculation showing measurement of lightspeed less than c. He had made a specific suggestion about doing a local computation. So I did, and confirmed it had the expected result. yuiop had raised the idea the coordinate lightspeed in these coordinates was locally anisotropic. I wanted to point and compute that using proper distance and proper time successfully removed this coordinate anamoly. Do you think I need you permission to comment?

Contributing means sticking out one's neck and provide things we can calculate, so what if a mistake is made, I make many. But I do not sit back and say 'no, no, no, wrong, you don't understand' without actually doing anything positive except for implying I know it all and the other knows nothing.

I disagree that it is not permissible to raise questions without doing calculations. There is no such forum rule or practice.

Indeed, we can perhaps collectively attempt to calculate it?

And yes GP coordinates would work, would be a nice change from Schw. coordinates.

Can we find the formulas for the 'raindrop'?

e.g.

- Distance to the EH (both in terms of Lorentz factoring the integrand used for a stationary observers and in Fermi coordinates (and if the 'doomsayers' there are many more, please come up with them, a few formulas more or less will not break it) ) for a given r value.
- Time it takes light to go a given physical distance away from a 'raindrop' for a given r value?

The trailing part moves obviously non inertially (a raindrop with a little rocket engine :) ) but I do not see how that matters for calculations at the head, but of course we have to watch for the acceleration at the tail to go to infinity.

Seems that is all we need, then we can start to drive this towards the EH and see what happens.

I completely agree these latter points address the simultaneity issue. My only goal was making sure it was addressed. Sorry if it was obviously going to be addressed. But for the case we computed, you never really made any attempt at all to justify coordinate t as a valid simultaneity for all static observers. It was as if you rejected the question. You left it to others to justify this.

 

[Passionflower]

Sorry PAllen but your formula is wrong.

Consider the coordinate difference versus the proper distance. You will see your formula shows a smaller proper distance than coordinate difference while obviously the opposite should be true.

 

[Mentz114]

I've done some work on the leader-trailer scenario. I found the normalised 4-velocity of the rain-drop

$$\begin{align*} U_\mu=\left[ \begin{array}{c} \frac{\sqrt{2\,r\,M+{r}^{2}}-2\,M}{r} \\\ \sqrt{\frac{2M}{r}}\\\ 0 \\\ 0 \end{array} \right] \end{align*}$$

Using this metric to calculate the norm

$$d\tau^2 = \left(1-\frac{2M}{r} \right) \, dt_r^2-2\sqrt{\frac{2M}{r}} \, dt_r \, dr - dr^2- r^2 \, d\theta^2 - r^2\sin^2\theta \, d\phi^2$$This is a geodesic if the covariant derivative of $U$, in the $U$ direction is zero, i.e. $U_{m;n}U^n=0$, and my calculation finds this is so. Obviously this needs to be checked, but I got good cancellation of terms to get zero.Now, this is the speculative bit. I propose the normalised 4-velocity of the trailng observer is

$$\begin{align*} V_\mu=\left[ \begin{array}{c} \frac{\sqrt{\frac{M}{r-1}}\,\left( 2\,r\,\sqrt{\frac{M}{r}}-2\,\sqrt{\frac{M}{r}}\right) -\sqrt{\frac{-4\,{M}^{2}-4\,r\,M}{2\,M}+2\,r\,M+{r}^{2}+1}}{1-r}\\\ \sqrt{\frac{2M}{r-1}}\\\ 0 \\\ 0 \end{array} \right] \end{align*}$$

which introduces a coordinate singularity at $r=1$.

The acceleration vector is not zero, see further down. Suffice to say, the accelerations become also become infinite at $r=1$.

This is not a problem as long as $1<2M$, because both observers will have gone through the EH by then. It's tidy that when the leader hits $r=0$ the trailer will be at $r=1$.

Anyhow, that's my first attempt. I suspect this can be done more neatly in the local frame of the raindrop, which has simple physics.

The acceleration vector of the trailer has 2 non-zero components, the r-component,

$$-\frac{\sqrt{r}\,\left( 2\,r-1\right) \,M\,\sqrt{2\,M+r-1}+\left( 2-2\,r\right) \,{M}^{2}+\left( -2\,{r}^{2}+2\,r-1\right) \,M}{{r}^{4}-2\,{r}^{3}+{r}^{2}}$$

and the t-component
$$-\frac{\sqrt{M}\,\left( \left( \sqrt{2}-\sqrt{2}\,r\right) \,M-2\,\sqrt{2}\,{M}^{2}\right) +\sqrt{2}\,\sqrt{r}\,{M}^{\frac{3}{2}}\,\sqrt{2\,M+r-1}}{\sqrt{r}\,\left( {r}^{3}-{r}^{2}\right) }$$

 

[pervect]

I've done some work on the leader-trailer scenario. I found the normalised 4-velocity of the rain-drop

$$\begin{align*} U_\mu=\left[ \begin{array}{c} \frac{\sqrt{2\,r\,M+{r}^{2}}-2\,M}{r} \\\ \sqrt{\frac{2M}{r}}\\\ 0 \\\ 0 \end{array} \right] \end{align*}$$

I also did some work on this problem, quite a while ago. However, I did it in Schwarzschild coordinates with M set to 1/2. , i.e.

ds^2 = (1-1/r) dt^2 - dr^2 / (1-1/r)

For the 4-velocity of a raindrop falling from infinity, I get (these are the superscripted components, not the subscripted ones!)

$$\frac{dt}{d\tau} = \frac{1}{ \left( 1 - 1/r \right) }}$$

$$\frac{dr}{d\tau} = \frac{1}{\sqrt{r}}$$

Unfortunately, even comparing this first preliminary (and standard) step is a lot of work, due to the differences in the choice of metric.

The coordinates of the following particle I got were given by

r_following = r(s)
t_following = t0 - t(s)

where r0 and t0 are the coordinates of the lead particle, s is the fixed distance of the following particle, and you need to integrate to find the functions r(s) and t(s)

for r(s)
$$\frac{dr}{ds} = \sqrt{1 + \frac{1}{r0} - \frac{1}{r(s)} }$$

with the initial conditon r(0) = r_0, and for t(s)

$$\frac{dt}{ds} = \frac{1}{\sqrt{r0} \, \left(1 - 1/r(s) \right) }$$

with the intial condition t(0) = 0

I went on to find a series expansion in s for r(s) and t(s) of order three (and a higher-order one that I never posted), to give a complete closed form (but approximate) rermi chart.

 

[PAllen]

Thanks for taking the time to explain. Additional comments in line.

Given the worldline of an observer, accelerated or not, in curved space-time or flat, for points sufficiently close to the observer there is a fairly natural notion of simultaneity, and of distance.

This happens because the geometry of space-time is locally Lorentzian - as is described in MTW on pg 19, if you happen to have that textbook. I'll give a short quote:

Yes, I know this, but have no problem with someone telling me something I already know; helps establish what the common level of understanding is. We are all different.

Yes, I have MTW, bought it when it first came out, which was after I last seriously studied physics. I've only read selected sections over the years since. So pointing me to section in it is quite useful to me, as are references to anything online. References to Wald, for example, are not useful to me ( but I hardly expect anyone to know what books I have).

Given a specific metric, it's fairly easy to recover said notion of local distance. What you do is introduce a set of coordinates that make the space-time metric at that point diagonal and unity (assuming that you've set c=1). All you need to do is to find a linear transformation that diagonalizes the metric.

You can transform to new coordinates either by the usual tensor transformation laws, or by simple algebra. It's easiest if you write the old variables in terms of the new, i.e. if you have a metric in (x,t) and you want to change to (x', t') you can write:

x = ax' + bt'
t = cx' + et'

then you can just write dx = a dx'+b dt' and dt = c dx' + e dt and substitute to get the metric in terms of x' and t'.
'
Given such a swath of space-time with a locally Minkowskian metric, the coordinate differences actually represent physical distances (in the small region where space-time is flat), so you can read distances directly from the new coordinates, and you can define the natural notion of local simultaneity for said observer by setting dt' = 0.

I have seen all of this before, but your explanation is really nice and appreciated.

This notion of simultaneity will make the speed of light isotropic, as should be obvious (I hope) from the Mikowskian metric, which defines the path light must take by ds^2 = 0.

Right, and I explicitly verified, without such coordinates, that if I carefully used proper distance and time, lightspeed was 1 and isotropic locally, while it could differ from 1 and was not isotropic, non locally. This makes perfect sense to me, as a non-local radial back and forth path traverses completely different geometry than a non-local back and forth path that starts perpendicular to a radius.

In curved space-time, the notion of how to extend the notion of simultaneity beyond a small local region is not clearcut. One possibility, which however, isn't unique, is especially useful. This is to extend the definition of simultaneity by drawing geodesic curves through the locally simultaneous points as described above. This leads to "fermi normal" coordinates. Another way of saying this is that simultaneous points in time are generated by the set of space-like geodesics passing through your observer's worldline at a given point that are orthogonal to said worldline.

While this seems like (and is) a very natural choice for simultaneity, it's not the only one in common use by any means. Cosmologists, for instance, do NOT use fermi normal coordinates when they report on distances within the universe. They use surfaces of constant cosmological time, cosmological time is time elapsed in the comoving frame since the big bang, instead.

The fermi-normal defintion of simultaneity (and of distance) is useful because it's compatible with the notion of Born rigidity. You can construct a family of observers all of whom measure the distance to their neighbors as constant, which is exactly what you need for a notion of distance that's compatible with Born rigidity.

If you try this with the cosmologists notion of distance, you find that it won't work, because observers with constant coordinates don't maintain a constant distance from each other, so the conditions you need for Born rigidity aren't met by the coordinate system.

Could you verify the following that seem to follow for Fermi normal coordinates:
- Especially for a rapidly accelerating observer, or extreme geometries, you can't extend them very far without having the spatial geodesics from one point on the 'central world line' intersect those from another. This defines a basic limit on the size of coordinate patch you can construct this way for extreme observers.

-An alternative definition of simultaneity I have used is what has been referred here as radar or Dobie-Gulls (?sp). This can often be extended much further than Fermi-Normal, while matching Fermi-normal locally. However, there are tradeoffs, of course. These really only define simultaneity within the prior lightcone of a point on a worldline, if you don't define the future progress of the world line. This could be seen as an advantage rather than a limitaion, since the procedure refuses to define simultaneity for events you can't know about yet.

A further limitation of radar simultaneity is that it is useless cosmologically. This isn't because it only applies to the prior lightcone (that's all we can see anyway), but because it also requires that the observer's world line can be extended to the prior light cone of some distant object. For distant galaxies, you might reach the big bang without achieving this.

It's a bit off the topic, but https://www.physicsforums.com/showthread.php?t=435999&highlight=fermi+normal does do a series expansion for fermi-normal coordinates for observers "falling from infinity" into a black hole, which provides one answer to the question about a "constant distance" observer falling into a black hole.

On a more general note, there's some reasonable-looking discussion at the Wikipedia at http://en.wikipedia.org/w/index.php?title=Rindler_coordinates&oldid=392242531, about Rindler coordinates which are the flat space-time analogue of Fermi-normal coordinates which goes into some detail about distances - though it's a bit lacking in references, alas.

I'd recommend getting familiar with Rindler coordinates first before worrying too much about Fermi-normal coordinates. MTW is a good reference, if you have it, and a bit more reliable than the wiki article - not that it helps if you don't have the textbook. This would address some of your concerns about the issue of the static observers accelerating.

I'm familiar with Rindler coordinates to the extent of having read through derivations of them and of their main properties. I have never tried my own calculations using them.

There are apparently exact solutions for Fermi-normal coordinates in the literature for the interior Schwarzschild space-time, unfortunately I don't have access to compare to my series expansion for the exterior region, i.e http://jmp.aip.org/resource/1/jmapaq/v51/i2/p022501_s1?isAuthorized=no .

 

[Passionflower]

So does everybody agree now that the measured light speed between two radially separated stationary observers depends both on the distance between them and on the distance to the EH in a Schwarzschild solution?

Here is a plot showing this:

[PLAIN]http://img717.imageshack.us/img717/2414/001lightspeed3d.gif

And yes, when we make the distance between the two observers 0, the 'holy mantra' that the light speed is locally c obviously becomes true.

So, in short, the measured speed of light between two radially stationary objects is NEVER c, except when the separation is zero.

[ Edit ] updated plot

 

[PAllen]

So does everybody agree now that the measured light speed between two radially separated stationary observers depends both on the distance between them and on the distance to the EH in a Schwarzschild solution?

Here is a plot showing this:

And yes, when we make the distance between the two observers 0, the 'holy mantra' that the light speed is locally c obviously becomes true.

So, in short, the measured speed of light between two radially stationary objects is NEVER c, except when the separation is zero.

None of my posts ever disagreed with this. First I was asking questions about the basis of the calculations which you didn't want to answer. Once answered by others, I agreed with this result except for an arithmetic sign mistake I made, but thought you made.

Note that I also still claim that this graph applies only to observers with the given radial separation. The results are different for separation by proper distance in non-radial directions.

 

[pervect]

Yes, I know this, but have no problem with someone telling me something I already know; helps establish what the common level of understanding is. We are all different.

:: I'm glad to hear that, on both counts

Could you verify the following that seem to follow for Fermi normal coordinates:
- Especially for a rapidly accelerating observer, or extreme geometries, you can't extend them very far without having the spatial geodesics from one point on the 'central world line' intersect those from another. This defines a basic limit on the size of coordinate patch you can construct this way for extreme observers.

Yes - you are correct. There is a definite size limit on the size of the coordinate patch - though I'm not sure what it is for the black hole case. I"ve done a few plots of the space-like geodesics I generated, my best guess is that you have to be fairly close to the black hole before the issue of geodesic crossing arises and that even then they don't seem to cross in the region between the observer and the event horizon, so the event horizon seems to be inside the coordinate patch.

In flat space-time, the geodesics cross at a distance c^2/g from the accelerating observer, at the Rindler horizon - though you probably knew that already, as you mentioned later on that you were familiar with Rindler coordinates.

-An alternative definition of simultaneity I have used is what has been referred here as radar or Dobie-Gulls (?sp). This can often be extended much further than Fermi-Normal, while matching Fermi-normal locally. However, there are tradeoffs, of course. These really only define simultaneity within the prior lightcone of a point on a worldline, if you don't define the future progress of the world line. This could be seen as an advantage rather than a limitaion, since the procedure refuses to define simultaneity for events you can't know about yet.

Ah - I"ve seen some discussion on radar distance - mostly a brief mention in MTW, I think - though I haven't seen the other name you mentioned (Dobie-Gulls). A quick search didn't turn up any hits - can you check the spelling or provide a reference?

I suspect that one of the reasons I didn't like the radar notion much in what I did read is just what you mentioned, that it depends on the future path of the observer, while the Fermi-normal notion doesn't.

 

[Passionflower]

None of my posts ever disagreed with this.

But you are not the only poster. ;)
Please don't take things too personal and please don't assume that all comments are addressed directly to you.

Note that I also still claim that this graph applies only to observers with the given radial separation. The results are different for separation by theta.

Well I do not disagree with that, it is just a lot harder to calculate if we allow theta to vary.

Feel free to show a formula or a graph with theta separation.

 

[PAllen]

Ah - I"ve seen some discussion on radar distance - mostly a brief mention in MTW, I think - though I haven't seen the other name you mentioned (Dobie-Gulls). A quick search didn't turn up any hits - can you check the spelling or provide a reference?

Yes, I've mispelled it. Here is the reference Dalespam provided in another thread here:

http://arxiv.org/abs/gr-qc/0104077

Of course, only stresses the advantages, and I don't recall it even mentioning the issue that it can't be used at all cosmologically. I also disagree with the strength of its critique of other approaches. However, it *is* very intuitive *and* operational, and can be applied in situations where Fermi-normal fails.

 

[PAllen]

Passionflower, can you clear up one point about your calulations?

In your post #65, you have:

$$\sqrt {6}\sqrt {{R}^{2}}-\sqrt {2}\sqrt {{R}^{2}}+R\ln \left( {\frac {\sqrt {3}\sqrt {R}+\sqrt {2}\sqrt {R}}{\sqrt {2}\sqrt {R}+\sqrt {R}}} \right)$$

After simplification this becomes:

$$R \left( \sqrt {2}\sqrt {3}-\sqrt {2}+\ln \left( -\sqrt {3}+\sqrt {2} \sqrt {3}-\sqrt {2}+2 \right) \right)$$

I still don't get this final simplification. Also, if I directly compute from these with numbers, I get a different answers. After correcting my sign mistake, I get exactly your first form, but I can't see how your second is equivalent.

Even more mysterious, if I take your first formula above, combined with (divided by)

$$\tau_{R2} = 1/2\, \left( R+R\ln \left( 2 \right) \right) \sqrt {2}$$

from your post #17 (which is exactly what I got), I now find that measured lightspeed for this case is 1.086, greater than c rather than less. I think this simple division is correct, as the 2R observer thinks the light has gone twice the proper distance, in twice the proper time. The proper time measuered by the 3R observer doesn't seem relevant.

 

[Mentz114]

Pervect, this is a reply to your post #75 about my #74 ( no-one else has anything to say, it seems ).

I understand your calculation and I think producing a Fermi or similar chart is the way it should be done, hence my remark that it must be simpler and more meaningful in local coords.

In my calculation the two observers are separated by a coordinate distance of 1, which is not how a rigid body would behave. But it convinces me that a rigid body could be partly through the EH at r=2m. PassionFlower particularly wanted to 'drive through' the event horizon, so choosing ccords where there was no singularity at r=2m was essential.

I think I'll try and define a meter-rule in local rain coords and see if it can be poked through the EH ( something that has been discussed elesewhere in the forum but not intelligibly for me). I'll post my results elsewhere if I get anywhere, since this thread has slipped into nit-picking.

 

[Passionflower]

Also, if I directly compute from these with numbers, I get a different answers.

Then you must be making a mistake since they give exactly the same results.

I prefer to use Maple, my hand calculations are simply too error prone, but alternatively there are several open source symbolic math programs available, a popular one is Maxima.

I highly recommend anyone to use at least one of those programs.

P
from your post #17 (which is exactly what I got), I now find that measured lightspeed for this case is 1.086, greater than c rather than less. I think this simple division is correct, as the 2R observer thinks the light has gone twice the proper distance, in twice the proper time. The proper time measuered by the 3R observer doesn't seem relevant.

It takes 1.086 times c for light to get there. So the light speed is slower not faster.
[edit] this is not correct.

 

[PAllen]

It takes 1.086 times c for light to get there. So the light speed is slower not faster.

I don't understand this. We are taking a proper distance divided by a proper time, getting a speed of 1.086. This says speed is faster than what is measured locally (which comes out 1).

 

[PAllen]

Then you must be making a mistake since they give exactly the same results.

Ok, yes. But I only used the first form for all work, which I did get right, because now I get the same value using the second form.

 

[Passionflower]

I don't understand this. We are taking a proper distance divided by a proper time, getting a speed of 1.086. This says speed is faster than what is measured locally (which comes out 1).

Now I was getting confused :)

But what you say is correct.

Here is an example, suppose R=1, Observer 1 is at R=2 and Observer 2 is at R=3 then we get:

Proper Distance R2 to EH: 2.295587149
Proper Distance R3 to EH: 3.595705578
Proper Distance between R2 and R3: 1.300118429

Time for light to go from R2 to R3 in coordinate time: 1.693147181
Time for light to go from R2 to R3 on R2's clock: 1.197235853
Time for light to go from R2 to R3 on R3's clock: 1.382448884

So:

The coordinate light speed between R2 and R3 is: 0.7678708878
The light speed between R2 and R3 for R2 is 1.085933424
The light speed between R2 and R3 for R3 is 0.9404459319

But you made me realize I plotted the wrong graph, I updated the posting with the correct graph.

 

[yuiop]

Actually I am not totally sure they will lose radar contact or not, since both the head and the tail are non stationary. If the signal sent from the head will reach the tail by the time the tail passed the EH it should be fine. I think in fancy terms you call this the "perceived event horizon".

I agree with this observation. Two free falling observers (one above the other) should never lose radar contact with each other as they pass through the horizon (so they do not observe an horizon). This can easily be seen in plots of light rays from the free falling observer's point of view. Eg, see a plot of light rays in Fermi Coordinates here https://www.physicsforums.com/showthread.php?t=443937 and in Gullstrand-Painleve coordinates here https://www.physicsforums.com/showpost.php?p=2948982&postcount=78

If the observers are falling in such a way that one primary observer is free falling and all the others are accelerating in such a way that they maintain constant proper distance then it is still true that they will not lose radar contact.

An interesting exercise would be to do the same thing for an free falling observer at escape velocity with a (non inertial!) observer tagged along in his 'tail' a constant physical distance of 1 removed. If I am not mistaken we can drive this example all the way through passed the EH. Any takers?

I would really like to see this done, if it can be. I do not have much spare time at the moment, so someone please PM me if a result is achieved!

 

[Passionflower]

I agree with this observation. Two free falling observers (one above the other) should never lose radar contact with each other as they pass through the horizon (so they do not observe an horizon). This can easily be seen in plots of light rays from the free falling observer's point of view. Eg, see a plot of light rays in Fermi Coordinates here https://www.physicsforums.com/showthread.php?t=443937 and in Gullstrand-Painleve coordinates here https://www.physicsforums.com/showpos...2&postcount=78

If the observers are falling in such a way that one primary observer is free falling and all the others are accelerating in such a way that they maintain constant proper distance then it is still true that they will not lose radar contact.

Exactly, and since the trailing observer must have a proper acceleration in the direction of the leading observer he would even be ahead compared to a free falling trailing observer.

I do not have much spare time at the moment

Currently I am in a similar position.

 

[unchained1978]

The photon gains energy, but not in the form of velocity. If you remember the energy of a photon is given by E=hf, so the increase in energy is given by an increase in frequency. The light is "blueshifted" as it falls into a gravitational field and "redshifted" as it escapes out of one. A good way to see this is by a simple classical derivation.
[url]http://upload.wikimedia.org/math/d/f/d/dfde118fc179f16b52be2ca7772e2e91.png[/url]
So the work done on a photon for an infinitesimal advance or retreat into or out of the field is given by is h*dv=GMhv/rc^2. Solving this differential equation yields the ratio of light frequencies in a gravitational field in the form [url]http://upload.wikimedia.org/math/d/6/7/d672ae664ef448fc1f03074923929b60.png[/url] where T is the ratio of frequencies and gh is just the Newtonian potential at that point.

 

[amos carine]

I agree with Pallen's approach and questioning, even if he nor I have answers. As the saying of good ole Al- "the important thing is to not stop questioning."

 

[Chronos]

BAO and WMAP results strongly 'constrain' [as in refute] these conclusions. All I can say is you need to get on board with modern science. Show the math.

 

[amos carine]

I agree in my inadequacy and being off the radar, so to speak. Chronos, your words mean that light is not blue shifted when going 'down' and red shifted upon escape of a gravity well or field. Or that the frequency or energy of the photon isn't added to (or subtracted by) the energy it gets from it's acceleration upon this falling? Do you have math to illustrate the different conclusion about falling photons? Seeing it would help understanding.