Energy loss of a photon moving against gravity

[Naty1]

so if we speak about curvature instead of potentials (switch from Newton to GR)

inconsistent...no potential effects in Newtonian gravity...

 

[TriKri]

Yes, that's correct.

Ok. Another thing that bugs me, and which really was the original reason for me to start this thread: It is said that a body with sufficient mass can prevent light from escaping from it. But according to my physics book, the gravitational redshift for a photon moving against gravity equals to gh/c² (where g is the gravitational acceleration, h is the distance the light is moving against gravity, and c is the speed of light), thought only for gh/c² << 1. This means that for a distance h that the light has traveled, you will get a factor (1 - gh/c²) which the frequency is multiplied with, and this factor can never < 0, since it's only valid for gh/c² << 1. So how is it possible that gravitational redshift can take the light out completely?

This is how I think it should be (you can tell me where I am going wrong): For large gh/c², you need to divide the distance the photon is traveling into n smaller parts, with distances h_i, 1 < i < n, so that g_ih_i/c² << 1 for each i. Here g_i is the average gravitational acceleration within the i:th distance. When the photon is sent out from the source, an observer O₀ measures the frequency of the photon to be f₀ > 0. It then travels the first distance, and an observer O₁ at the end of the first distance measures the frequency to be

f₁ = (1 - g₁h₁/c²)·f₀​

Then the photon travels the second distance and observer O₂ at the end of that distance measures the frequency to be

f₂ = (1 - g₂h₂/c²)·f₁​

Finally, the photon reaches observer O_n, at the end of the last distance, who measures the frequency to be

f_n = (1 - g_nh_n/c²)·f_n-1 = (1 - g₁h₁/c²)·(1 - g₂h₂/c²)·...·(1 - g_nh_n/c²)·f₀
​

Logarithm both sides:

ln(f_n) = ln(1 - g₁h₁/c²) + ln(1 - g₂h₂/c²) + ... + ln(1 - g_nh_n/c²) + ln(f₀)
​

Since g_ih_i/c² is so small for each i, ln(1 - g_ih_i/c²) is approximately = -g_ih_i/c², so

ln(f_n) = -(g₁h₁/c² + g₂h₂/c² + ... + g_nh_n/c²) + ln(f₀) = ΔV/c² + ln(f₀)
​

where ΔV is the gravitational potential that differs between the source and the destination. Reverse the logarithm again to get

f_n = e^ΔV/c2·f₀ > 0​

This means that no matter the distance or how big the gravitational field is, an observer far away will always be able to see the photon and register a positive frequency for it. This is in contradiction to very many predictions that great masses can prevent light from escaping. So, where do I go wrong?

 

[TrickyDicky]

inconsistent...no potential effects in Newtonian gravity...

This answer is a bit... ,hmmm, inconsistent?

 

[Jonathan Scott]

Hi
Ok, so if we speak about curvature instead of potentials (switch from Newton to GR) and we imagine a spacetime manifold with positive curvature (a hypersphere similar to Einstein model for intance),and events A and B separated by a big enough ds^2, and consider a light signal emmited from A and received by B, would B observe A's photons blueshifted with respect to local photons by the effect of the curvature (different potentials) of the manifold?

Curvature and potential are different things.

Firstly, curvature has multiple meanings anyway.

One type of curvature is analogous to that of a conical surface, which is locally flat but adds up to less than flat around a complete circuit. This is loosely like the way a gravitational field curves space-time in empty space.

Another type is analogous to that of part of a sphere or the top of a hill. This is loosely like the curvature produced by mass and energy which gives rise to fields.

The potential in a static field in Newtonian gravity (when expressed in dimensionless units, that is potential energy per rest energy, rather than potential energy per rest mass) is exactly equivalent to the time-dilation effect in General Relativity, which can effectively be described in term of a position-dependent scale factor between local time and the observer's coordinate time.

If you start considering points which are not a fixed distance apart in a static gravitational field, then comparisons of frequency and energy also have to take into account the effect of relative motion, so you get Doppler effects and time dilation due to relative speed.

 

[Jonathan Scott]

inconsistent...no potential effects in Newtonian gravity...

I think you've got something muddled up there.

 

[Jonathan Scott]

Ok. Another thing that bugs me, and which really was the original reason for me to start this thread: It is said that a body with sufficient mass can prevent light from escaping from it. But according to my physics book, the gravitational redshift for a photon moving against gravity equals to gh/c² (where g is the gravitational acceleration, h is the distance the light is moving against gravity, and c is the speed of light), thought only for gh/c² << 1.

This approximation only holds where the field is weak enough for Newtonian approximations to be valid, which is definitely not true when you're trying to extrapolate to black holes.

 

[TrickyDicky]

Curvature and potential are different things.

I agree, I guess what I meant to say is that potential has a counterpart in GR that would be the metric tensor r haw wikipedia puts it: "In general relativity, the metric tensor (or simply, the metric) is the fundamental object of study. It may loosely be thought of as a generalization of the gravitational field familiar from Newtonian gravitation. The metric captures all the geometric and causal structure of spacetime, being used to define notions such as distance, volume, curvature, angle, future and past."

Firstly, curvature has multiple meanings anyway.

One type of curvature is analogous to that of a conical surface, which is locally flat but adds up to less than flat around a complete circuit. This is loosely like the way a gravitational field curves space-time in empty space.

Another type is analogous to that of part of a sphere or the top of a hill. This is loosely like the curvature produced by mass and energy which gives rise to fields.

The potential in a static field in Newtonian gravity (when expressed in dimensionless units, that is potential energy per rest energy, rather than potential energy per rest mass) is exactly equivalent to the time-dilation effect in General Relativity, which can effectively be described in term of a position-dependent scale factor between local time and the observer's coordinate time.

If you start considering points which are not a fixed distance apart in a static gravitational field, then comparisons of frequency and energy also have to take into account the effect of relative motion, so you get Doppler effects and time dilation due to relative speed.

Right, so in my imaginary example (that I use as an exercise for my own clarification of concepts of GR, so if the comparison with Newtonian gravity is problematic we can drop it, I just thought it would be ilustrative) I am not taking into account relative motion, since is a geometrical thought experiment not related to reality, just to apply concepts of curvature and null geodesics.
I'm interested in knowing whether I got this right. Do you agree then B would observe light coming from A with a higher frequency than "local" light due to the effect of the positive curvature of the manifold? If not please explain why.

Thanks.

 

[TriKri]

This approximation only holds where the field is weak enough for Newtonian approximations to be valid, which is definitely not true when you're trying to extrapolate to black holes.

I think the book (it's called "Modern Physics" by the way), although not being the most covering book on GR, did a pretty good job in motivating the formula, which it derived starting from an accelerating system; then it used the principle of equivalence. But ok, maybe it left something out from the picture. So, how do you calculate when you are close to a black hole? (Notice that for all steps, h_i is of course still chosen so that g_ih_i/c² << 1 for all i even though the gravity is very big)

 

[Passionflower]

yes, locally the speed of light is always "c". "locally" means over a small enough piece of spacetime that it's flat...once curved spacetime has an effect, that is distances become large enough for the curvature of spacetime to have an effect measurements will detect a different speed of light but that an observational phenomena not a physical phenomena as Dalespam posted...

That is not the complete explanation, in a non-inertial frame, both in flat and curved spacetime the average speed of light is not c, except at the point where the observer is located.

Ok. Another thing that bugs me, and which really was the original reason for me to start this thread: It is said that a body with sufficient mass can prevent light from escaping from it. But according to my physics book, the gravitational redshift for a photon moving against gravity equals to gh/c² (where g is the gravitational acceleration, h is the distance the light is moving against gravity, and c is the speed of light), thought only for gh/c² << 1.

That formula applies to a uniform gravitational field. For the weak field this is compatible with a Schwarzschild gravitational field but only when the locations are stationary. When they fall in the field the locations change and so does the g at each location.

To put it in different terms think of the g in the formula as the g at a particular location x, e.g. g(x), h then is simply g(x+h). But as soon as x changes in time g changes value so this formula no longer applies.

In addition the location in the field with respect to g is different in a uniform gravitational field than that in a Schwarzschild field as the former only depends on the location while the latter depends both on the location and M.

 

[TrickyDicky]

The potential in a static field in Newtonian gravity (when expressed in dimensionless units, that is potential energy per rest energy, rather than potential energy per rest mass) is exactly equivalent to the time-dilation effect in General Relativity, which can effectively be described in term of a position-dependent scale factor between local time and the observer's coordinate time.

I can't make up my mind about this, are you agreeing with me then?

 

[Jonathan Scott]

I can't make up my mind about this, are you agreeing with me then?

No, I just don't understand your description of your "positive curvature" situation with points A and B, and I'm not sure it even makes sense, so I'm telling you how time dilation works so you can try to apply it to whatever you have in mind.

 

[bcrowell]

I agree, I guess what I meant to say is that potential has a counterpart in GR that would be the metric tensor r haw wikipedia puts it: "In general relativity, the metric tensor (or simply, the metric) is the fundamental object of study. It may loosely be thought of as a generalization of the gravitational field familiar from Newtonian gravitation. The metric captures all the geometric and causal structure of spacetime, being used to define notions such as distance, volume, curvature, angle, future and past."

The WP article is using terminology very loosely, and arguably incorrectly. Note that it uses the word "loosely" to describe its own approach. The best GR analog of the gravitational field is the Christoffel symbol, not the metric. In general, the closest analog that GR offers to the Newtonian potential is the metric, but in the special case of a stationary spacetime there is a scalar potential that is more directly analogous.

But ok, maybe it left something out from the picture. So, how do you calculate when you are close to a black hole? (Notice that for all steps, hi is of course still chosen so that gihi/c2 << 1 for all i even though the gravity is very big)

Jonathan Scott is correct that gh/c² is only an approximation. To calculate the time dilation factor, which is also equal to the redshift factor, you can express the metric in coordinates such that it is manifestly stationary and has the Minkowski form at large distances from the center, and then evaluate \sqrt{|g_{tt}|}. In Schwarzschild coordinates, this gives \sqrt{1-2m/r} (or the inverse of that, depending on which way up you want to express the ratio), where r is the Schwarzschild r coordinate, and the units are chosen such that G=1 and c=1.

 

[Austin0]

This was meant as "down".. s= c + ( ( 9.7 m/s)* dt) ,,,"up".. s = c - ( ( 9.7 m/s)* dt)
or something on this order.

If the clock at the bottom of the tower is running slower this would mean less elapsed time for any speed trial , yes??
If the clock at the top of the tower is running faster it would mean greater elapsed time for any speed trial , Yes??
If you make the tower high enough to have significant effects then the dilation and the g acceleration would compound each other , amplifying the difference between the up and down measured coordinate speeds , no??
Is this comprehensible??

The answer is no.
You are still making no sense.

You made a statement regarding the difference in up and down coordinate speeds of light in the tower experiment.
As I am curious about the measured speed of light in a GR context I sought clarification from you.
SO far, beyond a vague reference to theoretical considerations and negative responces to my attempts to figure out the relevant factors, you have provided no explanation of your statement or correction of things I have said or actual information of any kind.
SO what is "no" supposed to mean?? That gravitational acceleration does not effect photons? That dilated clocks do not effect measured velocities?

 

[nonequilibrium]

What happens to conservation of energy in Relativity? Does E = hf only work in flat spacetime or something?

 

[bcrowell]

What happens to conservation of energy in Relativity? Does E = hf only work in flat spacetime or something?

General relativity doesn't have global conservation laws that make sense in all spacetimes. It does have local conservation laws.

In an example like the Pound-Rebka experiment, E=hf works fine, there is no violation of conservation of energy, and the experiment explicitly involves curved spacetime.

 

[Naty1]

My original comment:

inconsistent...no potential effects in Newtonian gravity...

from tricky

This answer is a bit... ,hmmm, inconsistent?

from Scott

I think you've got something muddled up there.

appaently my comment wasn't clear...

I meant to comment that "no potential effects in Newtonian Gravity" result in red shift and blue shift of light...which is the context of the original post...Newton assumed instantaneous effects, no finite speed "c".
/////////////////

my post

yes, locally the speed of light is always "c". "locally" means over a small enough piece of spacetime that it's flat...once curved spacetime has an effect, that is distances become large enough for the curvature of spacetime to have an effect measurements will detect a different speed of light but that an observational phenomena not a physical phenomena as Dalespam posted...

passionflower comment:

That is not the complete explanation, in a non-inertial frame, both in flat and curved spacetime the average speed of light is not c, except at the point where the observer is located.

...both statements seem ok to me...

["mr Vodka" signature yesterday reminded me I "deserved" a vodka...perhaps that was more than I could handle!]

 

[Naty1]

Scott:

Curvature and potential are different things.

and

The potential in a static field in Newtonian gravity (when expressed in dimensionless units, that is potential energy per rest energy, rather than potential energy per rest mass) is exactly equivalent to the time-dilation effect in General Relativity, which can effectively be described in term of a position-dependent scale factor between local time and the observer's coordinate time.

No problem with the above...can we explore that a bit more:

Is it better to say: "In general curvature and potential are different things, but can have identical effects?"

I thought curvature and potential COULD have identical effects...Scott describes one in the second quote above...in other words, doesn't the Einstein formulation of curvature (Ricci,Weyl,etc) lead to the curvature of geodesics...and that in turn changes distance and time parameters for remote observers?? in fact isn't that a lot of what the "equivalence" principle of Einstein is all about...seems like acceleration in flat space (no gravity) has essentially the same affects as uniform motion with a gravitational field (curved space)...

next day edit: What I was trying to ask above is " Aren't gravitational curvature and gravitatonal potential two sides of the same coin,,,,aren't they "equivalent"??

 

[Naty1]

Austin: Try your post #43 as a new thread (rephrase/clarify it for those who have not read here) if you don't get a straightforward answer...

I started to reply, got myself confused, and believe there may be a number of subtlies involved...or ...maybe I don't understand the principles as well as I should...

I don't want to further derail the OP's post replies here ...

 

[Jonathan Scott]

Is it better to say: "In general curvature and potential are different things, but can have identical effects?"

Curvature is just one geometric aspect of the shape of space time, as described by the metric. Potential corresponds to a different aspect, which is like a scale factor.

If by curvature you simply mean "the fact that space-time is curved" that might be sufficiently vague to be just about acceptable, but it wouldn't be very accurate.

 

[bcrowell]

Curvature is just one geometric aspect of the shape of space time, as described by the metric. Potential corresponds to a different aspect, which is like a scale factor.

The way I would put it is that curvature is the only locally observable geometrical fact about the shape of the spacetime. A gravitational potential \phi isn't a separate thing you could observe about the spacetime. It may not be defined at all, if the spacetime isn't stationary. If the spacetime is stationary, then the potential isn't observable, because it's arbitrary up to an additive constant (similar to how electric potential isn't observable). Differences in gravitational potential are observable, but they don't give extra information about the shape of the spacetime that wasn't available from measurements of curvature.

 

[nonequilibrium]

General relativity doesn't have global conservation laws that make sense in all spacetimes. It does have local conservation laws.

Hm, I'm just reminded of the fact that in Newtonian Mechanics non-inertial reference frames don't obey the principle of conservation of energy. So in relativistic mechanics we lose even more? In relativistic mechanics the old adagium "In an inertial frame of reference the total amount of energy is constant" fails? Might there be something else accounting for the loss of energy? If not, are there also situations in which we gain energy instead of lose it (like with redshift)?

EDIT: strange thought experiment: let a photon near a galaxy, it will experience redshift. Now turn back time, you see a "blueshift". Yet if you'd shoot a photon away from a galaxy, you'd experience redshift, right? Huh?

 

[bcrowell]

Hm, I'm just reminded of the fact that in Newtonian Mechanics non-inertial reference frames don't obey the principle of conservation of energy. So in relativistic mechanics we lose even more?

You actually gain something, because you still have a local conservation law, even if you choose a noninertial frame. All you lose is the ability to make global conservation laws: http://www.lightandmatter.com/html_books/genrel/ch04/ch04.html#Section4.5

In relativistic mechanics the old adagium "In an inertial frame of reference the total amount of energy is constant" fails? Might there be something else accounting for the loss of energy? If not, are there also situations in which we gain energy instead of lose it (like with redshift)?

It's not that there's a loss or gain of energy, it's that it's not even possible to define the amount of energy in any useful way.

 

[Naty1]

Ben..so glad you posted

http://www.lightandmatter.com/html_b...tml#Section4.5

to remind me I had promised myself I would read the whole thing...that will be my next "project" for myself...

can you put that link in your signature so when you make posts it appears at the bottom...or did the lords of the forum removed all signatures?? I see mine no longer appears...

 

[Naty1]

Post # 12: TriKri...since this is your thread...

I'm confused ... does the speed of light vary at other locations than the local or not? Jonathan says it does, but Naty1 says it doesn't ... or have I misunderstood anything?

I hope I did not say that. Read my post #8 and Jonathan's post #12 ...I interpret them to say the same thing...distant observers may observe different speeds of light...Only inertial observers in flat spacetime (no gravity) see both local and distant light as "c"...

 

[Austin0]

This is incorrect.

Neither photons nor even massive objects change in frequency or energy when moving in a static gravitational field as observed by anyone observer. When we talk about redshift increasing with potential, what we actually mean is that a series of different local observers at different potentials would observe a photon passing their own location to have different redshift relative to a local reference photon. However, from the point of view of any single observer, the photon has constant frequency and energy when moving freely in a static gravitational field.

In a static gravitational field, using an isotropic coordinate system, the momentum of a free-falling particle increases downwards with time, even if it is a photon, but the energy is constant. This is related to the effect that in such coordinates the speed of light at a location other than the observer's own varies a bit depending on the gravitational potential.

Hi 1) I have searched for a definition of an isometric coordinate system without result.
Intuitively I can picture what this would mean ;essentially a Cartesian system but am not sure if that's what your talking about in this context.
2) In this circumstance how do you differentiate between momentum and energy wrt a photon?
I can see why the energy would apparently increase (blueshift) relative to local electron frequencies but wouldn't this also be equivalent to an increase in momentum??
Also; viewed from a locale of higher potential wouldn't the speed of light decrease at lower levels?
If this is the case, then how would this be related to an increase in momentum?
Thanks

 

[Jonathan Scott]

Hi 1) I have searched for a definition of an isometric coordinate system without result.
Intuitively I can picture what this would mean ;essentially a Cartesian system but am not sure if that's what your talking about in this context.

Your searching may be more successful if you look for isotropic coordinates.

It does not necessarily mean flat (Minkowski) space. "Isotropic" means for example that if you observe a light speed signal expanding from a point (creating a sphere expanding at c according to a local observer) then in the coordinate system the light also expands at the same rate in all directions, still creating a sphere, but at a coordinate speed which is not necessarily exactly equal to c.

In more general coordinate systems, the speed of light is not necessarily the same in different directions, so you can't even talk about the coordinate speed of light unless you also specify a direction.

2) In this circumstance how do you differentiate between momentum and energy wrt a photon?
I can see why the energy would apparently increase (blueshift) relative to local electron frequencies but wouldn't this also be equivalent to an increase in momentum??
Also; viewed from a locale of higher potential wouldn't the speed of light decrease at lower levels?
If this is the case, then how would this be related to an increase in momentum?
Thanks

The magnitude of the momentum of an object is Ev/c² where v is the speed of the object and c is the speed of light. For a photon the speed is equal to c, so the magnitude of the momentum is E/c. Relative to an isotropic coordinate system, c decreases with potential, so the momentum increases if the photon falls down to a lower potential, or decreases if the photon rises to a higher potential. If the photon moves sideways along a line of constant potential, it gets deflected a bit downwards by the field.

The interesting thing is that regardless of the direction in which the photon is travelling, the coordinate rate of change of momentum with time at a given point is always the same downwards pointing vector, equal to exactly twice the Newtonian force that would be experienced on an object of the same total energy at that point in the field. In general, for an object traveling at speed v, it is simply (1+v²/c²) times the Newtonian force, at least in fields where the weak approximation holds (that is, nowhere near a neutron star or worse).