How does light slow in the presence of gravity?

[Naty1]

Jonathan: From my post # 5:

Contrary to intuition, the speed of light (properly defined) decreases as the black hole is approached. ...The reason for the qualification 'properly defined' above is that the speed of light depends upon the vantage point (frame of reference) of the observer. When we say that the speed of light is decreased, we mean from the perspective of an observer fixed relative to the black hole and at an essentially infinite distance. On the contrary, to an observer free falling into the black hole, the speed of light, measured locally, would be unaltered from the standard value of c.

http://www.physlink.com/Education/AskExperts/ae13.cfm

Note: Time stops at the event horizon, light appears to have zero speed.

appears to conflict with your post:

From the point of view of the local observer, the velocity and wavelength are unchanged, but from the point of view of the global observer, the speed decreases in a lower potential

and the wavelength shrinks (which is equivalent to an increase in momentum as it gets closer to the central mass, just as for a massive object).
[/QUOTE] (my boldface emphasis)
Can you explain the difference...frames of reference are still not perfectly clear to me. Is your global observer equivalent to properly defined above?? I thought light would be blue shifted approaching a mass and red shifted when radiated from a mass for a distant observer.

 

[spidey]

Sorry for not being clear. The frequency appears to be different (red-shifted or blue-shifted) as seen by local observers at different potentials, but from the point of view of an observer looking at what is happening on a larger scale, the frequency is actually the same everywhere but the local observers' clocks are subject to time dilation effects.

From the point of view of the local observer, the velocity and wavelength are unchanged, but from the point of view of the global observer, the speed decreases in a lower potential and the wavelength shrinks (which is equivalent to an increase in momentum as it gets closer to the central mass, just as for a massive object).

The frequency of a light wave means the frequency of its electric or magnetic components (which have the same frequency but are out of phase with one another). Similar considerations apply to any other form of frequency associated with a signal moving at light speed, such as a light being turned on or off, or the image of a turning wheel.

Thanks for the info..i understand that local observer will see change in frequency and global observer will see change in wavelength and velocity...but isn't this weird? how it is different for different observers?

 

[Jonathan Scott]

Thanks for the info..i understand that local observer will see change in frequency and global observer will see change in wavelength and velocity...but isn't this weird? how it is different for different observers?

Yes, that's relativity!

It's not really true that the local observer will see a "change" in frequency. Each observer, global or local (at least in a static case) sees the signal to have a constant frequency along its path, but that frequency is different for observers at different gravitational potentials, which can be attributed to the fact that their clocks are running at different rates.

In particular, if a signal is emitted by a process which is known to have a specific characteristic frequency in its local frame (such as a spectral line), the frequency of that signal will appear to be shifted compared with a similar process in the observer's frame by the relative difference in the gravitational potential, hence the "red shift" or "blue shift" terminology.

 

[Phrak]

The physically measure speed of light does not slow in the presence of gravity, only the coordinate speed does. Depending on the coordinates, the coordinate speed of light can be anything, even in special relativity.

No way dude! I always measure my velocites using coordinates.

Good grief, you pop-in, drop a bomb, and maybe cite a reference where you're claim is burried somewhere... and disappear.

So in what coordinate system, in special relativity is the velocity of light anything? Just one will do.

 

[DrGreg]

I think it is a point worth making that some of the counterintuitive features of GR in general, and black holes in particular, are not directly due to the curvature of spacetime but are due to the acceleration of the observer making the measurement. Some of these effects just disappear when we consider a free-falling observer near a black hole, and conversely some of these effects appear when we consider an accelerating observer in empty space (i.e. in flat spacetime).

Consider yourself as an observer in a uniformly accelerating rocket in flat spacetime, with a constant upward proper acceleration of g. Your coordinate system is called Rindler coordinates.

This is "Born rigid acceleration". Each part of your rocket above and below you is a fixed distance from you (as measured in a co-moving inertial frame). But the bottom of the rocket accelerates more than g and the top of the rocket accelerates less than g. According to you, a clock at the bottom of the rocket ticks slower than your own clock, and light from that clock is received by you as red-shifted.

At a distance of c²/g underneath you an event horizon forms. If you drop an apple out of the rocket (so it then moves inertially), as it approaches this distance you will see its image red-shifted and slowing down. It will take forever to reach c²/g. This applies not only to the image of the apple that you see, but also to your coordinate calculation of the apple's position relative to you. You will never see light emitted from any object further than c²/g below you. And you will calculate that the upward coordinate speed of light, measured in your own Rindler coordinate system, approaches infinity at a distance of c²/g below you, even though it is c locally.
The rocket's Rindler coordinates (T, X) are related to the apple's inertial coordinates (t, x) by the equations

ct = (X + c²/g) sinh (gT/c)
x + c²/g = (X + c²/g) cosh (gT/c)​

(You can get further details by googling for "Rindler coordinates" DrGreg site:physicsforums.com.)

To summarise, variable coordinate speed of light, event horizons, "gravitational" time dilation and "gravitational" red/blue shift are all phenomena you can study in accelerating coordinate systems in flat spacetime. So, in this thread's title, "the presence of gravity" is, in a way, an irrelevance; it's the acceleration of the observer that leads to the explanation. Gravity is only indirectly responsible, because it forces observers to accelerate to remain "stationary".

 

[DrGreg]

No way dude! I always measure my velocites using coordinates.

Good grief, you pop-in, drop a bomb, and maybe cite a reference where you're claim is burried somewhere... and disappear.

So in what coordinate system, in special relativity is the velocity of light anything? Just one will do.

If George were talking about the standard Einstein-synchronised Minkowski coordinates of an inertial observer, he would be wrong. But I reckon he was talking about:

- either accelerated coordinates in SR (see my previous post in this thread)

- or non-standard coordinates of an inertial observer, such as those at the bottom of this post

And in either of those cases, the coordinate speed of light need not be c.

 

[Naty1]

DRGREG: Greatpost...I even understood it...

Some of these effects just disappear when we consider a free-falling observer near a black hole, and conversely some of these effects appear when we consider an accelerating observer in empty space (i.e. in flat spacetime).

To illustrate a black hole effect, here's an explanation I just came across in THE RIDDLE OF GRAVITATION, Peter Bergmann:

The Schwarzschild radius is that region in which the escape velocity approaches the speed of light...(For an outside stationary observer relative to the black hole)...after a while the speed of a falling particle will no longer increase; on the contrary it will slow down because the motion of the particle is governed by local conditions. As it approaches the Schwarzschild radius the particle must traverse increasingly larger distances...in doing so it is controlled by local time which passes more slowly than the outside observer's time.These two effects reinforce each other; they result in such a slowdown that it takes the particle forever to reach the Schwarzschild radius...(same effect for a light ray)...(yet) free falling objects take only a finite amount of their own time to reach the Schwarzschild region.

 

[spidey]

DRGREG: Greatpost...I even understood it...



To illustrate a black hole effect, here's an explanation I just came across in THE RIDDLE OF GRAVITATION, Peter Bergmann:

So you say that light doesn't reach schwarzchild radius?

 

[Naty1]

So you say that light doesn't reach schwarzchild radius?

yes...I,too, was surprised,That's why I posted the excerpt...it depends on your frame of reference...I assume Bergmann stated it correctly as he was a student of Einstein's...

I read the adjacent 5 or so pages on either side of the quote several times to be sure I had the authors context...that's why I also posted: (For an outside stationary observer relative to the black hole) which was stated several pages away...

I am not quite sure how close to that frame an oberver here on Earth would be...hence we might see it slightly differently due to our motion...but I have repeatedly read that light/mass slows upon it's approach to a black hole, that from an Earth fixed frame at essentially infinite distance. Maybe somebody can clairfy that point.

 

[DrGreg]

So you say that light doesn't reach schwarzchild radius?

yes...I,too, was surprised,That's why I posted the excerpt...it depends on your frame of reference...I assume Bergmann stated it correctly as he was a student of Einstein's...

I read the adjacent 5 or so pages on either side of the quote several times to be sure I had the authors context...that's why I also posted: (For an outside stationary observer relative to the black hole) which was stated several pages away...

I am not quite sure how close to that frame an oberver here on Earth would be...hence we might see it slightly differently due to our motion...but I have repeatedly read that light/mass slows upon it's approach to a black hole, that from an Earth fixed frame at essentially infinite distance. Maybe somebody can clairfy that point.

The light never gets there according a "stationary" (and therefore accelerating) observer's coordinate system. According to a free-falling observer, the light would get there in a finite time (I think). Certainly, a free-falling observer who falls through the event horizon at just the right moment would measure the infalling light overtake him/her, through the horizon, at the speed of light.

The light's non-arrival is, in a sense, an illusion caused by the accelerating coordinate system's constantly changing definition of relative simultaneity.

In the quote from Bergmann in post #42, "speed" means "coordinate speed according to the stationary observer".
Bergmann's quote would also apply to the apple dropped from my rocket in post #40, provided you replace "Schwarzschild radius" with "event horizon" and omit the sentence "As it approaches the Schwarzschild radius the particle must traverse increasingly larger distances", which is true only in curved spacetime.

By the way I made a mistake in post #40. When I said "the upward coordinate speed of light, measured in your own Rindler coordinate system, approaches infinity", I should have said "aproaches zero". It's the proper acceleration of a point that is a fixed distance below the rocket (according to the rocket) that approaches infinity.

 

[treffibug]

I have a question that is a variant of the twin paradox that I am having trouble solving:

Consider a planet that is not rotating and we have a twin at the top of a mountain and a twin at the bottom. Neither twin is moving relative to the other, so special relativity equations do not play a part here.

The two twins are experiencing different gravitational field strengths (ie the twin at the top of the mountain is less bound to the planet). At what rates are the two experiencing the passage of time.

I know that the higher twin is aging faster, but how much fast?

- Trevor.

 

[treffibug]

Follow up:

I have found a formula I think may answer my question, but for a long time I have been trying to link the slowing of the rate of time passage (erm ... "dt/dt"?) to an object's loss in potential energy.

I remember someone telling me (when I was too young to get it) that as you climb out of an energy well, potential energy is transferred into the speeding up of internal processes, ie thermal energy, as you shift into a faster time rate. The converse situation: as you fall into a gravity well, the slowing down of said internal energy processes is transferred into kinetic energy, until you hit something solid of course.

The formula I have found is:

T(surface) = T(infinity) / sqrt ( 1 - (2GM/R.sqr(c) ) )

Given that potential energy is -GM/R per kilo (ie energy lost due to being down a well per kilo, let's say 'W') I can rearrange the formula to yeild:

(sqr(c)/2)*(1 - sqr(T(infinity)/(T(surface))) = W.

Have I made any mistakes? I have been hunting for this formula for a long time :S

 

[Jonathan Scott]

Follow up:

I have found a formula I think may answer my question, but for a long time I have been trying to link the slowing of the rate of time passage (erm ... "dt/dt"?) to an object's loss in potential energy.

I remember someone telling me (when I was too young to get it) that as you climb out of an energy well, potential energy is transferred into the speeding up of internal processes, ie thermal energy, as you shift into a faster time rate. The converse situation: as you fall into a gravity well, the slowing down of said internal energy processes is transferred into kinetic energy, until you hit something solid of course.

The formula I have found is:

T(surface) = T(infinity) / sqrt ( 1 - (2GM/R.sqr(c) ) )

Given that potential energy is -GM/R per kilo (ie energy lost due to being down a well per kilo, let's say 'W') I can rearrange the formula to yeild:

(sqr(c)/2)*(1 - sqr(T(infinity)/(T(surface))) = W.

Have I made any mistakes? I have been hunting for this formula for a long time :S

The slowing of the time rate IS the change in potential energy. Provided that you are not in the vicinity of a neutron star or black hole, you can just assume that the fractional change in time rate is equal to the difference in the Newtonian potential,-GM/rc^2.

 

[Naty1]

According to a free-falling observer, the light would get there in a finite time (I think)

Yes, that is what Bergmann says in my posted quote...it makes sense, actually, because attached to a photon, you'd be freely falling (no acceleration) and all would be "normal" from your frame of reference...

I remember someone telling me (when I was too young to get it) that as you climb out of an energy well, potential energy is transferred into the speeding up of internal processes, ie thermal energy, as you shift into a faster time rate.

An interesting point: Yet another aspect I had not considered!

Potential energy IS independent of frame of reference, so all observers see that increase; at the lower gravitational potental in a local frame, time stays the same while energy increases...from Earth frame looking out of the well, time speeds up so, processes appear to as well. This must be more than offset by counter velocity effects when moving at high speed because people age more slowly at high speed viewed from an Earth frame.

But I'm not clear on thermal energy increase...anybody comment/explain??

Does radioactive half life appear shorter (due to faster relative time) when looking from Earth to a lower gravitational point, say high altitude? I guess so!

 

[Jonathan Scott]

But I'm not clear on thermal energy increase...anybody comment/explain??

I think the idea of "thermal energy increase" is misleading. If something is in free fall (which includes the case of going up as well as down, and the case of orbit), nothing is changing locally (unless it hits something else).

The total energy (kinetic plus rest energy) remains constant; potential energy is effectively an adjustment to the rest energy caused by the difference in time rate at different gravitational potentials, so the total energy can be considered for example as the sum of kinetic energy plus rest energy at infinity plus potential energy (which is negative).

Any thermal energy as seen locally will be unchanged, but as seen by a fixed observer it will vary with the time rate due to the potential and velocity in the same way as anything else, for example the frequency of a blinking light on the object.

The case where thermal energy is significant is when a free-falling object hits the ground and comes to an abrupt halt, in which case its kinetic energy is converted to a mixture of thermal energy and mechanical/chemical energy (for example in deformed materials).

 

[DrGreg]

Yes, that is what Bergmann says in my posted quote...it makes sense, actually, because attached to a photon, you'd be freely falling (no acceleration) and all would be "normal" from your frame of reference...

Sorry, but you can't attach yourself to a photon, not if you have non-zero mass. By a "free-falling observer" I mean a free-falling massive observer, not quite the same thing as (free-falling) photon. A photon does not define a local inertial reference frame.

Potential energy IS independent of frame of reference,

In my rocket example, the apple has zero kinetic energy and zero potential energy in the apple's own inertial frame. In the rocket's accelerating frame, the apple initially has increasing kinetic energy, offset by decreasing potential energy. So, potential energy is dependent on reference frame.

 

[spidey]

so if we say light slows down near massive object then would that mean the value of \epsilon_0 \ and \mu_0 \ changes ?

 

[Phrak]

If George were talking about the standard Einstein-synchronised Minkowski coordinates of an inertial observer, he would be wrong. But I reckon he was talking about:

- either accelerated coordinates in SR (see my previous post in this thread)

- or non-standard coordinates of an inertial observer, such as those at the bottom of this post

And in either of those cases, the coordinate speed of light need not be c.

So the moral of the story is, don't use stock quotes for coordinates or don't confuse temperature with time. Which begs the question, what constitutes 'good' coordinates.

 

[DrGreg]

So the moral of the story is, don't use stock quotes for coordinates or don't confuse temperature with time. Which begs the question, what constitutes 'good' coordinates.

If an observer A wants to set up a "physical" coordinate system at event E.

1. Find an inertial observer B who is momentarily stationary, relative to A, at event E. An observer is inertial if they are free-falling i.e. they carry an accelerometer that measures a constant proper acceleration of zero. B is the "co-moving inertial observer" at E.

2. Observer B sets up a (local) coordinate system using his proper-time clock and a lattice of relatively-stationary clocks all synchronised to his using Einstein's synchronisation convention; and measuring distance by radar.*

3. Observer A makes local measurements near event E by asking B to make the measurement on her behalf. "Near" means over ranges where spacetime curvature can be ignored and any relative acceleration between A and B can be ignored. This rough statement can be made mathematically precise through calculus.

Under these conditions, A will measure the speed of light at E to be c. However, if she extrapolates her coordinate system to "non-local" events, she may calculate that the speed of light somewhere other than E takes a different value. The extrapolated coordinate system is no longer a "physical" coordinate system (except at E).

It's possible to set up coordinate systems (using time, distance or related quantities such as angle, but not unrelated quantities such as temperature) that don't coincide with "physical" coordinates anywhere.

When studying black holes, it's traditional to use a spherical polar coordinate system extrapolated from a "stationary" observer notionally "at infinity". Such a coordinate system never exactly coincides with a "physical" coordinate system as defined above, but at very large distances the difference is negligible.

___
*It has just occurred to me that by specifying in step 2 that distance is measured by radar, the "physical" speed of light is c by definition. This corresponds to the modern-day definition of the metre. However we could instead say that B uses stationary rulers to measure local distance. It's an experimentally verified hypothesis that ruler distance and radar distance are locally the same for inertial observers.

 

[Naty1]

DrGreg...

I posted

Yes, that is what Bergmann says in my posted quote...it makes sense, actually, because attached to a photon, you'd be freely falling (no acceleration) and all would be "normal" from your frame of reference...

You posted

Sorry, but you can't attach yourself to a photon, not if you have non-zero mass. By a "free-falling observer" I mean a free-falling massive observer, not quite the same thing as (free-falling) photon. A photon does not define a local inertial reference frame.

I agree, my wording "attached to a photon" was extremely sloppy, unnecessary...

but what distinction do you make between a free falling mass and a photon...if mass energy equivalence holds, do they move along different curves in free fall??

 

[DrGreg]

but what distinction do you make between a free falling mass and a photon...if mass energy equivalence holds, do they move along different curves in free fall??

The big difference is that non-zero masses always travel slower than light, whereas the photons travel at, ..er.., the speed of light (as measured by any local inertial observer's local coordinate system). So they're not identical in every respect. In 4D spacetime they follow different curves. In 3D space it depends which direction they travel. Anything can travel in a spatially-straight line radially, massive particle or massless photon. But these would be different 4D curves (actually, from the point of view of a local inertial observer, they would be different locally-straight 4D lines).

Here, by "mass" I mean "rest mass" a.k.a. "invariant mass" (not "relativistic mass" which includes kinetic energy). In this terminology, inertial particles have mass ("are massive"), photons are massless.

"Mass energy equivalence" means that mass can be considered as one form of energy (just like kinetic energy, potential energy, heat energy etc are different forms). It doesn't mean all energy "is" mass. It does mean that when you add up all forms of energy to see what has been conserved, you have to include mc² to get the equations to balance.

 

[granpa]

why was my post with the link to a pdf of 'the relativity deflection of light' deleted?
its a pdf of an article in the journal of the royal astronomical society of Canada by the professional astronomer Charles Lane Poor.

wikipedia says this about him:

Charles Lane Poor (1866-01-18 – 1951-09-27)[1][2] was born in Hackensack, New Jersey, the son of Edward Erie Poor. He graduated from the City College of New York and received a Ph.D. in 1892 from Johns Hopkins University. Poor became an American astronomer and professor of celestial mechanics at Columbia University from 1903 to 1944, when he was named Professor Emeritus. He published a monograph disputing the evidence for Einstein's theory of relativity in the pre-war years before the theory became firmly established.[3][4] CL Poor published a series of papers (see bibliography) that reflect his lack of understanding for the theory of relativity.

For 25 years, Poor was chairman of the admissions committee of the New York Yacht Club. In addition, he was a fellow of the Royal Astronomical Society and an associate fellow of the American Academy of Arts and Sciences. He served several terms as mayor of Dering Harbor on Long Island, New York, and invented a "line of position computers" for yachting navigation. At Columbia University, Poor was a teacher of the astronomer Samuel A. Mitchell, who went on to become director of the Leander McCormick Observatory at the University of Virginia.[5]

its one thing to call me a crackpot and delete my ideas but it another thing entirely to delete a reference to an article by a professional astronomer in a professional journal. if there is anything wrong in his calculations then I would be the first to want to know about it. the math in it is simple and straighforward and it should be quiet easy for you to point out where the error lies.

except for the 'historical' section at the bottom of the frontpage most of the other papers in the website where the pdf was located are garbage. I stated so myself. yet you then give me an infraction for posting that link (to a professional paper in a professional journal) as though my only purpose in doing so was to somehow promote all the other crackpot ideas (that I myself called garbage) on that site which I did not link to (and never would)