# Thread: gravitational redshift and speed of light

1. ## gravitational redshift and speed of light

We know that light which travels perpendicularly to a gravitational field will bend, producing gravitational lensing. It won't speed up or slow down, but only its direction changes, right? But what if it travelled straight outward? We know it will redshift, but will it slow down, even a little bit, due to gravity, similar to that of a massive object, maybe in proportion to the redshift or something, or speed up as it falls inward? Consider it for these scenarios:

1) The Earth. Will light slow down as it leaves the Earth's surface due to Earth's gravity?

2) The sun, which is a few million times more massive than Earth. Will light slow down as it leaves the surface of the sun?

3) The redshift of the sun is about one part in a couple million, so what if the sun were a few million times more massive than it is? Would light slow due to its gravity?

2. Grav,

I thought we'd gone over this before in other threads. In the Schwarzschild coordinatization, light *speeds up* as it climbs out of a well, and slows down as it "falls". While radial and tangential coordinate speeds of different by a square root of the (1 - R/r) factor, both still speed up and slow down accordingly.

And one might expect that a particle moving very close to light speed should behave the same. They do. The "critical speed", the turning point, is c/sqrt(3) (see GEM, c/sqrt(3) thread).

And, to make it clear, all observers (including properly rotating observers, as I've learned) measure the speed to be 'c' locally using their own ruler and clock (whether their ruler is rotating or not. ) It just speeds up or slows down when it moves away from them where they can't actually measure it locally.

-Richard

3. ## Re the catcher

The sun is much larger than the earth so the light must slow more as it travels out of a larger gravity well. On earth in a smaller gravity well it falls in and is caught at the speed of light.

On the ISS where there is negligible gravity it is also caught at the speed of light. So if the difference is not with the transmitter it must therefore be with the receiver.

Second electron orbits change position not at the speed of light but instantly this means the particle is a cathers mit desiged to compensate for the speed the light is travelling and record the variation as a shift of frequency either red or blue.

Regardless of the speed the light is travelling it is caught by the receive the light relative to their local reference frame.

Cheers

I am more than prepared to be wrong but this sounds more logical to me.

4. Originally Posted by publius
Grav,

I thought we'd gone over this before in other threads. In the Schwarzschild coordinatization, light *speeds up* as it climbs out of a well, and slows down as it "falls". While radial and tangential coordinate speeds of different by a square root of the (1 - R/r) factor, both still speed up and slow down accordingly.

And one might expect that a particle moving very close to light speed should behave the same. They do. The "critical speed", the turning point, is c/sqrt(3) (see GEM, c/sqrt(3) thread).

And, to make it clear, all observers (including properly rotating observers, as I've learned) measure the speed to be 'c' locally using their own ruler and clock (whether their ruler is rotating or not. ) It just speeds up or slows down when it moves away from them where they can't actually measure it locally.

-Richard
Thanks, publius. This stuff just flies over my head all too often, so I have to refresh my memory. I saw the c/sqrt(3) in the thread you're referring to, but I have no idea what it means. So let's keep it simple. I'm assuming that if the speed of light is to remain the same in any local reference frame, then any apparent change in speed must be in direct proportion to the time dilation between frames, is that right? Now, if we were to measure the speed of light locally on Earth, it would be c. And if we were to send a pulse into space and split the beam so that we could then measure it between two sensors, it would be c in that frame as well, correct? So if we sent a pulse from Earth to a station is space every second and vice versa, then we would know exactly what any time dilation would be by comparing the rate they are received to their own clocks. So if the speed of light speeds up by some amount, say ten percent, to the people on Earth after they emit it, then that would mean that the clocks on the station are running ten percent faster as well, wouldn't it?

5. Grav,

That is conceptually how it works, but a little different in the details. The radial speed of light comes from both radial legnth contraction, as well as time dilation. There is no difference in tangential rulers, so only time dilation effects the speed in that direction.

But, the way local rulers and clocks work, the only difference in the light received will be seen as a redshift or blueshift depending on which it goes. The gravitational time dilation factor is this:

dtau/dt = sqrt(1 - R/r).

Radial length is scaled by the same factor.

So, the guys up on the space station will think our frequency is low by that factor. Well, there is a relative speed correction as well that is somewhere around 10-20% -- we think a clock on a LEO satellite is fast, but something like 80 - 90% of what the stationary value would be because of the Lorentz, "SR" factor. Don't hold me to those exact numbers, but it's somewhere around there.

To them, the light is moving at 'c', but is reduced in frequency by that factor, with the local wavelength increased accordingly.

-Richard

6. Thanks again, publius. I'll think about this some more.

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Originally Posted by publius
the way local rulers and clocks work, the only difference in the light received will be seen as a redshift or blueshift depending on which it goes.
Originally Posted by publius
To them, the light is moving at 'c', but is reduced in frequency by that factor, with the local wavelength increased accordingly.
Publius could you answer this question:
lets say that we send light from a satellite to the earth. The light then gets reflected back towards the satellite.
When the light get to the satellite, would the light be redshifted, blueshifted or not-shifted?
Whatever the answer, would this also be true with a black hole?

8. Well, if we were hovering stationary over a non-rotating mass and bounced our light off of a stationary perfect mirror down there, we would we see the frequency as the same when we fired it down. Now, the observer down there at the mirror would say that light was blueshifted. For a non-rotating black hole, if there was a stationary mirror above the event horizon behing held in place, the result would be the same. However, we think the light takes a longer time to make the round trip that it would for the same distance (according to our ruler) in flat space time.

However, one does not hover stationary over masses, rotating or not without doing some serious thrusting, one has to be in orbit, and thus seeing relative velocity with the ground. There will be Doppler shifting see there if we bounce off some mirror on the ground.

Now, if there are any lukers here who wish to exercise their GR skills, consider a geostationary satellite for the *rotating frame* of the earth's surface. Let it be over the equator in a perfect geostationary orbit. The satellite has a mirrored surface. You fire a beam of light at it and it bounces back. What happens there? This homework assignment is due no later than 11PM tonight.

Seriously, get to thinking about that, from the rotating frame, and you'll see why the base GPS frame is the non-rotating, earth-centered
(quasi-) inertial frame. Read those Klauber papers, as well as some GPS material for details.

-Richard

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Originally Posted by publius
Well, if we were hovering stationary over a non-rotating mass and bounced our light off of a stationary perfect mirror down there, we would we see the frequency as the same when we fired it down. Now, the observer down there at the mirror would say that light was blueshifted. For a non-rotating black hole, if there was a stationary mirror above the event horizon behing held in place, the result would be the same. However, we think the light takes a longer time to make the round trip that it would for the same distance (according to our ruler) in flat space time.
thanks

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Originally Posted by publius
Well, if we were hovering stationary over a non-rotating mass and bounced our light off of a stationary perfect mirror down there, we would we see the frequency as the same when we fired it down. Now, the observer down there at the mirror would say that light was blueshifted. For a non-rotating black hole, if there was a stationary mirror above the event horizon behing held in place, the result would be the same. However, we think the light takes a longer time to make the round trip that it would for the same distance (according to our ruler) in flat space time.

However, one does not hover stationary over masses, rotating or not without doing some serious thrusting, one has to be in orbit, and thus seeing relative velocity with the ground. There will be Doppler shifting see there if we bounce off some mirror on the ground.

Now, if there are any lukers here who wish to exercise their GR skills, consider a geostationary satellite for the *rotating frame* of the earth's surface. Let it be over the equator in a perfect geostationary orbit. The satellite has a mirrored surface. You fire a beam of light at it and it bounces back. What happens there? This homework assignment is due no later than 11PM tonight.

Seriously, get to thinking about that, from the rotating frame, and you'll see why the base GPS frame is the non-rotating, earth-centered
(quasi-) inertial frame. Read those Klauber papers, as well as some GPS material for details.

-Richard
(my bold)

They'd have had to work mighty quickly ... the timestamp on your post, in the setting I have, was 10:27 PM!

11. Why I thought everybody knew the correct time was EDT (GMT -4). Yep, my clock is right, and it's up to everybody else to correct their wrong, weird clocks to the correct time, darn it.

Speaking of Daylight Savings time, to perhaps give you an idea of the genetic material that lurks in my cells, I had an uncle (died about 2 years ago), who was absolutely opposed to Daylight time. It made him so mad he would throw hissy fits. He never, ever, changed his clock, and was always getting screwed up because of it, having to remember everyone else was an hour ahead.

Really, what he objected to was "someone trying to fool him". At least I think that was his thinking. He didn't like "authority" trying to fool him into thinking it was an hour later than it "really was", to force him to save energy. They tell me when they first did it, he actually left lights on all night, so he would specifically use more energy, thus negating the effect of Daylight time.

However, besides his other ornery characteristics he was also a tightwad extraordinare, and so that didn't last.

-Richard

12. I was born in February, and the clocks go forward in March, and go back again in October, so my first experience is to lose an hour, and then I get it back six months later.

For somebody born between March and October, they first experience the clocks going back and so they gain an hour, and then the clocks go forward six months later and they have that hour taken away again.

So for me, I am always either owed an hour, or working on normal time.

For people born between March and October, they are always either given an extra hour, or working on normal time.

So half the population never experiences getting that extra hour over their lives, they only experience losing it! Unfair!

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Well, except for those who were born in a country (or region) which, at the time, didn't have daylight saving; or for those who ...

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Quote:
Originally Posted by publius
Well, if we were hovering stationary over a non-rotating mass and bounced our light off of a stationary perfect mirror down there, we would we see the frequency as the same when we fired it down. Now, the observer down there at the mirror would say that light was blueshifted. For a non-rotating black hole, if there was a stationary mirror above the event horizon behing held in place, the result would be the same. However, we think the light takes a longer time to make the round trip that it would for the same distance (according to our ruler) in flat space time.

However, one does not hover stationary over masses, rotating or not without doing some serious thrusting, one has to be in orbit, and thus seeing relative velocity with the ground. There will be Doppler shifting see there if we bounce off some mirror on the ground.

Now, if there are any lukers here who wish to exercise their GR skills, consider a geostationary satellite for the *rotating frame* of the earth's surface. Let it be over the equator in a perfect geostationary orbit. The satellite has a mirrored surface. You fire a beam of light at it and it bounces back. What happens there? This homework assignment is due no later than 11PM tonight.

Seriously, get to thinking about that, from the rotating frame, and you'll see why the base GPS frame is the non-rotating, earth-centered
(quasi-) inertial frame. Read those Klauber papers, as well as some GPS material for details.

-Richard

Originally Posted by Nereid
(my bold)

They'd have had to work mighty quickly ... the timestamp on your post, in the setting I have, was 10:27 PM!
And, if they were in the same time zone, one at the top of Mount Everest and one at the bottom, the top ones clock would be running faster, and it wouldn't have ANYTHING to do with 'light travel time' or 'gravity'........simple rotation at different altitudes from the center of the sphere!

15. Originally Posted by publius
(SNIP)
Now, if there are any lukers here who wish to exercise their GR skills, consider a geostationary satellite for the *rotating frame* of the earth's surface. Let it be over the equator in a perfect geostationary orbit. The satellite has a mirrored surface. You fire a beam of light at it and it bounces back. What happens there? This homework assignment is due no later than 11PM tonight.
-Richard
NOW = 10PM at my location. I would have expected that, for this homework, SR would be good enough. And I would also have expected that GPS velocity corrections would be the same, whether you chose the earth centered rotating or earth centered non rotating frame. The SR-correction should depend only on the velocity differences between the geostationary and the earth surface clocks!?

16. I'm going to hunt you all down a good paper on this. Things are a good deal more complex than you think for the level of required accuracy. The rotating frame is avoided specifically because things are so complex there.

Actually, the earth's gravitational effects, as seen from a non-rotating frame are really insignificant -- they only affect clock rates. The actual Shapiro time delay, and the effects of spatial curvature are insigificant for the required accuracy.

However, in the rotating frame, the psuedo-centrifugal and coriolis "gravity" does have an effect on light in that frame -- the path of light is indeed "curved" for the level of required accuracy, and the actual calculations required are done from the non-rotating frame to avoid that.

Think of it this way. You have a large, rotating disc in free space. Observers on that disc wish to know their location relative to a grid on that disc. You would think it would be the easiest to do it from the disc frame because of that.

But it's not. What you do is do it from a non-rotating frame, and locate a receiver in that frame at the instant a signal was received. Once you have that position, you then figure out where the grid on the disc was at that instant, and thus know where the receiver was relative to the grid. But the actual light-travel-time location is done from the non-rotating frame.

ETA: Think of it this way. Consider doing a triangulation calculation where the sides of your triangle were curves, not straight lines. That makes it much harder. Not impossible, just harder. So you do it from a frame where the legs are straight lines. For the level of required accuracy, light has significant curvature in the earth rotating frame, but does not in the non-rotating, earth centered frame.

-Richard

17. I was thinking that if the the Earth was turning underneath the "stationary" station, then a small amount of rotation (when the source is directly below) would be like the source moving tangent to the station, which would be a form of relativistic aberration. But if it is aberrated by one, and then reflected back, would it cancel itself out or would it double?

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Originally Posted by publius
I'm going to hunt you all down a good paper on this. Things are a good deal more complex than you think for the level of required accuracy. The rotating frame is avoided specifically because things are so complex there.

Actually, the earth's gravitational effects, as seen from a non-rotating frame are really insignificant -- they only affect clock rates. The actual Shapiro time delay, and the effects of spatial curvature are insigificant for the required accuracy.

However, in the rotating frame, the psuedo-centrifugal and coriolis "gravity" does have an effect on light in that frame -- the path of light is indeed "curved" for the level of required accuracy, and the actual calculations required are done from the non-rotating frame to avoid that.

Think of it this way. You have a large, rotating disc in free space. Observers on that disc wish to know their location relative to a grid on that disc. You would think it would be the easiest to do it from the disc frame because of that.

But it's not. What you do is do it from a non-rotating frame, and locate a receiver in that frame at the instant a signal was received. Once you have that position, you then figure out where the grid on the disc was at that instant, and thus know where the receiver was relative to the grid. But the actual light-travel-time location is done from the non-rotating frame.

ETA: Think of it this way. Consider doing a triangulation calculation where the sides of your triangle were curves, not straight lines. That makes it much harder. Not impossible, just harder. So you do it from a frame where the legs are straight lines. For the level of required accuracy, light has significant curvature in the earth rotating frame, but does not in the non-rotating, earth centered frame.

-Richard
And beyond GR etc, there's plain ordinary physics - the signals pass through the atmosphere (if light) and ionosphere (if radio), at varying angles; the detectors and transmitters have their own response times and pulse-to-pulse variations (and other quirks); etc, etc, etc.

19. Originally Posted by publius
I'm going to hunt you all down a good paper on this. Things are a good deal more complex than you think for the level of required accuracy. The rotating frame is avoided specifically because things are so complex there. (SNIP) -Richard
Youīre right. Shortly after my response I was thinking about this. But it was too late (10:27 at my location).

20. Apparently, the most oft cited paper on this is "Relativistic effects in the global positioning system" by Neil Ashby which appeared in the "General Relativity and Gravitation" journal. I cannot find an online source (haven't looked that hard, but I did try a bit). However, is a less technical piece by Ashby here that basically summarizes things:

http://www.phys.lsu.edu/mog/mog9/node9.html

-Richard

21. Originally Posted by publius
Apparently, the most oft cited paper on this is "Relativistic effects in the global positioning system" by Neil Ashby which appeared in the "General Relativity and Gravitation" journal. I cannot find an online source (haven't looked that hard, but I did try a bit). However, is a less technical piece by Ashby here that basically summarizes things:

http://www.phys.lsu.edu/mog/mog9/node9.html

-Richard
For those who are interested in more details (and will understand more than I do): An other good one (but very technical piece) is by Robert D. Klauber: arXiv:gr-qc/0108036 v2 24 Jul 2002 (sorry, I donīt know how to post links)

22. MooCow ( )

We've been discussing Klauber and relativsitic rotation in the latter part "Copernicus and Heliocentrism" thread right here.

-Richard

23. Originally Posted by publius
MooCow ( )

We've been discussing Klauber and relativsitic rotation in the latter part "Copernicus and Heliocentrism" thread right here.

-Richard
Do you expect me to follow all threads, Publius Quinctilius Varus ? Or Publius Cornelius Tacitus? If so, si tacuisses MooCows donīt follow all alleys

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