What happens in this scenario? (Poll)

[yuiop]

Consider these two experimetal set ups:

1) An electric battery low down in a gravitational field connected by long wires to a light bulb much higher up in the gravitational field.

2) An electric generator connected by long wires to a light bulb higher up.

The electric power source is at the same height for both experiments and so are the light bulbs. The lengths of the wires are also the same. Assume Schwarzschild geometry. Also assume that both experiments are insulated from temperature changes and that the generator has its own fuel and oxygen supplies in rigid pressure controlled containers so as to eliminate any considerations of changing local atmospheric pressure conditions. Initially both light bulbs have the same brightness before the set ups are lowered.

Both experiments are slowly lowered closer and closer to the event horizon. An observer next to the light bulbs travels with the experimental set ups and remains at rest with the light bulbs. Assume the lower end is not actually lowered exactly to or below the event horizon.

Will the observer see the light bulbs both get brighter / dimmer / no change or he will he see different results for the two experiments?

If you can, try and explain how you reach your conclusions.

Vote now!

 

[Q-reeus]

Silly not to have seen the obvious - only one question need be asked: Is gravity getting stronger as the horizon is approached? Clearly YES! So given that local relative radial displacement is assumed maintained (against increasing tidal forces!) there must be a perceived dimming of the bulbs - equal for both systems. So my belated choice is, after all, not 3 but 2 - both lights get dimmer. Interesting though that no-one else has so far tippy-toed into this 'innocent' pole.
Earlier and wrong stuff:
"Gone with 3rd option - no change. Reasoning: In effect it's a variation of gravitational red-shift, that expression being:

, taken from http://en.wikipedia.org/wiki/Redshift" . On the assumption that r(power source)/r(bulb) remains constant, the red-shift expression predicts no change, the r's being measured by a distant observer. Neglecting effects of elastic stress and strain, a 'small' system shrinks by the red-shift factor, as determined by that distant observer. Therefore to the extent tidal stresses have been factored out of this scenario, everything remains in proportion and indeed there is no locally observed change. Is this wrong? Hope this is not 'trick' question where higher order derivatives of metric curvature become important!"

 

[yuiop]

Gone with 3rd option - no change. Reasoning: In effect it's a variation of gravitational red-shift, that expression being:

, taken from http://en.wikipedia.org/wiki/Redshift" . On the assumption that r(power source)/r(bulb) remains constant, the red-shift expression predicts no change, the r's being measured by a distant observer. Neglecting effects of elastic stress and strain, a 'small' system shrinks by the red-shift factor, as determined by that distant observer. Therefore to the extent tidal stresses have been factored out of this scenario, everything remains in proportion and indeed there is no locally observed change. Is this wrong? Hope this is not 'trick' question where higher order derivatives of metric curvature become important!

It's not a trick question and thanks for your input that has given me some food for thought. I am genuinely not sure of the answer and that is why I put it to a poll. Your post points to some factors I had not previously thought of which I will try to address here.

Let us say the bottom of the apparatus is initially at coordinate radius r1 = 1.01010101 and the top is at coordinate radius r2 = r1+10 = 11.01010101 using units of 2GM/c^2=1. The red shift factor would then be (using your equation):

$$\sqrt{\frac{(1-1/10.01010101)}{(1-1/1.01010101)}} = 9.535$$

Now assuming the apparatus is reasonably rigid and retains its proper length (d) to a reasonable degree of accuracy we have to calculate what it's proper length is. This is done using the formula:

$$d = \int^{r=r2}_{r=r1} \frac{1}{\sqrt{1-1/r}} \, dr = \left[\sqrt{r(r-1)}+\ln(2\sqrt{r}+2\sqrt{r-1}) \right]^{r=11.01010101}_{r=1.01010101} = 12.1659$$

Now the apparatus is transported slowly away from the gravitational source to a great distance where the gravitational time dilation factor is insignificant. At this great distance the proper length and the coordinate length of the apparatus should be approximately equal. Let us say we transport the bottom section to r3 = 1000 then the gravitational time dilation factor would be 1/sqrt(1-1/1000) = 1.00050 which is close enough to unity for our purposes so it would be reasonable to assume a coordinate length of 12.1659 and assume the top of the apparatus is at r4 = r3+12.1659 = 1012.1659. Now we can use your redshift equation again and calculate the redshift factor when far away as:

$$\sqrt{\frac{(1-1/1012.1659)}{(1-1/1000)}} = 1.0000060$$

which is close to unity which we probably could have guessed anyway. So the final result is that far away from the source the the redshift factor is approximately 1.0 and when lowered close to the event horizon (assuming reasonable rigidity and proper length maintained) the redshift factor between the top and bottom is approximately 9.5.

This leads me to conclude that the light bulbs would appear to get dimmer to the co-moving local observer and so I am voting for option 2.

 

[Q-reeus]

It's not a trick question and thanks for your input that has given me some food for thought. I am genuinely not sure of the answer and that is why I put it to a poll.

Please pardon any suspicion on my part yuiop - paranoia comes naturally to me
I had earlier made a lazy assumption that length contraction would cancel out red-shift (anyone following the thread would have noticed my several edits!), but as per my revised posting and your calculations show, that is not really so. We agree then - it's now all '2' obvious.
Another problem along these lines I've been toying with addressing is this:
Many popularizing articles treat gravitational red-shift of a photon as one of it 'running out of steam' as it 'struggles' against the 'pull of gravity'. An alternate viewpoint, one that a distant observer would deduce, is quite different. Here the photon is emitted by a source running slower further down the potential well. At all times the frequency is constant - red-shift/blue-shift is a perception of differing clock rates only. On that basis, how to show that a single photon bouncing up and down within a perfectly reflecting box exerts the proper, GR = twice Newtonian gravitational force. The correct 'currency exchange rate' has to be derived such that the box as a whole moves as expected. One could throw in circumferential bouncing to get an averaged 'photon gas' result, but the latter is conceptually much easier - basically just 'angles'.

 

[Ich]

There is energy conservation in static spacetimes, but the answer is still tricky.
Say you have some amount of energy at the bottom. To lift it to a higher position (say, infinity), you have to do work. So not the total energy will arrive, but only a fraction of $\sqrt{1-2M/r}$. (Example: Red shift of photon energy.)
Additionally, the power you generate will be diluted by time dilation. The energy you generate in a year will be spent in $1/\sqrt{1-2M/r}$ years. (Example: Red shift of photon rate.)
That's a relative power at the bulbs of $1-2M/r$.

 

[Zarqon]

Your four options are not mutually exclusive, as option 4 can be true at the same time as either of 1 or 2 :P

I answered option 4, they will change with respect to each other. The reason is I think, that the generator, which runs on mechanical parts, will have its friction between the parts increased drastically as the gravity is increased. This would lead to the generator producing less electric power leading to its corresponding bulb giving less light.

That being said, it's of course still possible that both bulbs descrease overall as well, due to other reasons.

 

[Q-reeus]

...Say you have some amount of energy at the bottom. To lift it to a higher position (say, infinity), you have to do work. So not the total energy will arrive, but only a fraction of $\sqrt{1-2M/r}$. (Example: Red shift of photon energy.)
Additionally, the power you generate will be diluted by time dilation. The energy you generate in a year will be spent in $1/\sqrt{1-2M/r}$ years. (Example: Red shift of photon rate.)
That's a relative power at the bulbs of $1-2M/r$.

The above refers to not how things change relatively when the setup is moved radially as a whole, but just net power transfer between source and load for a fixed relative setting. Given that context, the first red-shift formula surely gives the correct net result for continuous power transfer, unless EM radiation is special in some way. Just think of a third optional arrangement - transmitting and receiving dish antennas instead of wires as link. Unless red-shift and only red-shift applies, some kind of signal chirping (signal pile-up or thinning at a steady rate) is implied, which is unphysical. Power dilution is then part and parcel of just the red-shift factor. Think that's right.

 

[Q-reeus]

..I answered option 4, they will change with respect to each other. The reason is I think, that the generator, which runs on mechanical parts, will have its friction between the parts increased drastically as the gravity is increased. This would lead to the generator producing less electric power leading to its corresponding bulb giving less light...

And here's another factor - the coal-stoker, being dangerously close to the horizon, is just paste on the boiler room floor. That should really throw a spanner in the works!

 

[Ich]

The above refers to not how things change relatively when the setup is moved radially as a whole

There is obviously no effect for r->inf, so I described what changes if you go to small r.

Given that context, the first red-shift formula surely gives the correct net result for continuous power transfer, unless EM radiation is special in some way

No, it gives energy balance. The transmitted power is less.

 

[Q-reeus]

There is obviously no effect for r->inf, so I described what changes if you go to small r.

All in all the original question implies four radii are involved, but given your original comments I take it r here refers to just the generator (bulb being held at 'infinity').

No, it gives energy balance. The transmitted power is less.

In hindsight you are correct - thanks. The time over which the generator works is indeed larger by the inverse redshift factor as seen by the bulb. Which means that for continuous EM wave transmission, it is wave amplitude that drops by the redshift factor, intensity (power) therefore going as the square. Interesting.