Gravitational Redshift

1. mysearch

522
Based on the quote below and the link to Wikipedia, I believe the implication is that gravitational redshift will occur when a photon climbs out of the gravity well, while blueshift will occur when falling into the gravity well. However, I am trying to clarify whether both time dilation and space expansion have to be taken into consideration when an observer measures the frequency or wavelength of a photon at different positions in the gravity well.

Not sure as to what is intrinsic to frequency as I would have thought frequency is subjective to an observer?

If the observer is collocated with the photon source within a gravity well, does local time define the frequency and therefore the energy E=hf?

To a distant observer, far from the gravity well, time ticks faster, so would the photon, when arriving, appear to have a lower frequency and energy?

However, along the way, space would have contracted as space is said to expand in the direction of increasing gravity. Therefore, would this effect cause the wavelength to shorten?

If $$c=f\lambda$$ and [c] is constant, would the shorten wavelength not cause frequency to increase and cancel the effect of time dilation?​

Clearly theory suggests that this is not the case, but then implies there is a different perception of the energy associated with the photon. Therefore, would welcome any clarification of these issues.

2. Mentz114

4,110
I think it would be the other way round. The wavelength would appear longer.

For light, frequency and wavelength are not independent, and so you can regard a red-shift either as an increase in wave length or a reduction in frequency.

Red-shifted light will deliver less energy. This is not a surprise because energy is not relativistically invariant.

Have you come across the Pound-Rebka experiment ? I guess you have since you quote Prof. Wiki.
en.wikipedia.org/wiki/Pound-Rebka_experiment

Last edited: May 10, 2008
3. pmb_phy

All that is neccesary is that time dilation is present, i.e, that g00 is a function of position. g00 depending on position does not imply that space is altered.

Pete

4. mysearch

522
Mentz114: Thanks for the comments. I have attached a response against each below:

As you approach a large mass [M] doesn’t time tick more slowly and the radial distance [r] expand? If so, I was assuming a photon travelling away from [M] would perceive a contraction in unit space and therefore the wavelength would shorten, although this doesn’t seem to be the case.

I agree based on the assumption that $$c=f\lambda$$ and [c] is constant. This was why I was asking about the combined effects of time dilation and space contraction when the photon travels away from [M]. Pete in post #3 appears to be implying that time dilation alone causes the gravitational shift, which suggests that the frequency is fixed at N cycles per second, when initially emitted, but as the photon moves out of the gravity well, N cycles per second appears lower because time is ticking faster. What I didn’t understand was why the radial contraction per unit length didn’t not shorten the wavelength, i.e. increase frequency, and therefore cancel the effects of time dilation. Clearly, experiments suggest this is not the case, but I couldn’t really explain to myself why both time and space effects didn’t apply.
How does the conservation of energy get resolved if the photon appears to lose energy? In classical physics a physical mass can lose kinetic energy, but its total energy is conserved in its potential energy.

Actually I hadn’t, so thanks for the reference.

Pmb_phy/Pete: Thanks for the feedback, unfortunately I not that familiar with the meaning of $$g_{00}$$ notation. As I have tried to explain above, I don’t really understand why the changes in both time and space don’t cancel out. Most texts seem to simply quote the net effect without reference to the underlying physical cause, e.g.

You seem to suggest that time dilation causes the photon to have a lower frequency, i.e. longer wavelength, and that the effects of radial space contracting do not cause the wavelength to shorten, i.e. higher frequency.

5. Mentz114

4,110
The speed of light would have to change to accomodate a lower frequency with a shorter wavelength.

Pete is right, only $$g(r)_{00}$$ is used to calculate gravitational redshift. The radial component is irrelevant. This must be in the nature of light. A mass thrown up can lose velocity, light cannot.

6. mysearch

522
Request for confirmation

As indicated earlier I have not really studied the vector-tensor maths associated with $$g_{00}$$, however looking around on the web I found one reference that suggested that $$g_{00}$$ is associated with time dilation where:

$$d\tau = \sqrt{g_{00}} dt$$

This suggests that in the context of gravity only, i.e. no relative velocity, that $$g_{00}$$ might be related to the relativity factor $$[\gamma]$$, e.g.

$$g_{00} = 1-Rs/r$$??

I mention this simply because I am interested in trying to resolved some physical interpretation that explains the effects under discussion. Therefore, I wanted to try to expand on your other comment:

While possibly over simplistic, I have described relativity in terms of its separate effects on space and time. Special relativity describes the effect of velocity, which remains invariant because time dilation and space contraction are equal and maintain constant velocity in both frames of reference. General relativity incorporates the additional effect of gravity, which also leads to time dilation, although space expands in the radial direction. As such, velocity cannot be invariant. As such, velocity is simply a concept of distance travelled in a given time, i.e. v=d/t, and this is all $$c=f\lambda$$ would appear to be saying. However, the assumption that v=d/t may be the issue that has misled me, because in the context of wave propagation, the frequency [f] is defined by the energy [E=hf] and the propagation velocity [v=c] is defined by the media, i.e. vacuum permittivity & permeability. As such, it might be more accurate to say that wavelength results from $$\lambda=c/f$$.

I not sure whether this is an appropriate physical interpretation, but it suggests that the wavelength of a photon, as measured by a distant observer, can only result from changes in [c] and [f]. If [c] is a constant in each frame of reference, then only the time dilation effects on [f] lead to the gravitational redshift observed by experiments. As such, the radial contraction that occurs when moving out of the gravity well would not directly affect wavelength as I had originally assumed. Is this a valid conclusion?

7. MeJennifer

The frequency the distant observer measures depends on the curvature differential between the event of emission and absorption. In other words, f does not change in a gravitational field but the measuring apparatus changes in such a field. A measured blue shifted photon is really a photon observed by a red shifted apparatus.

8. mysearch

522
Response to post #7

I accept that we could describe the gravitational redshift in terms of a differential curvature of spacetime. However, I would have thought this curvature would still correspond to the space and time effects relative to 2 positions in the gravity well. As far as I was aware, time dilation is a real effect against which we then subsequently define frequency. In this context, the real change in the perception of time appears to be more fundamental than the knock-on effects on the measuring apparatus. This is not being forwarded as a statement fact, simply the assumptions to which I was working

9. Mentz114

4,110
mysearch,

going back to an earlier remark,

Well, why should they ?

The 'cause' is the difference in gravitational field strength. This is measured by g_00. Actually it's a bit more subtle but without a full understanding of curved space, hard to explain.

This agrees with experiment. If you're looking for an explanation in terms of underlying structures, then you're into uncharted territory.

In post #6 you're making a mistake by trying to make analogies between the gamma of SR and gravitational effects. SR is formulated in the Minkowski space-time, GR extends this to a general spacetime where curved worldlines can be caused by geometry alone ( gravity) whereas in SR, a curved worldline always implies an external force. The conclusions of SR do not carry over to GR in a simple way.

One more thing. Suppose we have observers A and B, and A sends some light to B who measures the frequency. It is not possible to say whether any difference observed is a result of an 'intrinsic' change in the light or a difference in the measuring apparatus, caused by transporting it to another place. No experiment can distinguish these situations.

Last edited: May 11, 2008
10. mysearch

522
Response to #9

A fully accept that I am on a learning curve and for those who have studied this subject in detail, it must be difficult to simplify some concepts to basic principles. It is true that I was hoping to rationalise, at least to myself, the basic effects of relativity in terms of time dilation and space expansion/contraction.

This wasn’t my intention; I was simple reproducing an inference in
https://www.physicsforums.com/archive/index.php/t-125057.html

This is a very good point and one also argued in post #7. However, I thought it was an axiom of relativity that everything is relative. As such, it seems to be a philosophical point as to whether the photon changes in transit or the observer/measuring apparatus changes with respect to the photon - see rationale below. From the perspective of the receiving observer, the photon has less energy than a photon from a comparable source emitted locally.

I accept that my attempt to rationalise an explanation in #6 may appear naïve, but it appeared to help clarify some important aspects of gravitational redshift. First, $$\lambda=c/f$$ seem to help explain why only time dilation affected frequency. Equally, it appears to provide a practical interpretation of the energy lost associated with a photon moving away from a gravitational mass. Based on the conservation of energy, this lost energy has to be accounted in some other form. In classical physics, the energy associated with a mass [m] can be defined in terms of its kinetic (Ek) and potential (Ep) energy that in-turn is linked to velocity and gravitation:

$$Total Energy (Et) = Ek + Ep =1/2mv^2 + (-GMm/r)$$

At face value, this is more difficult for a photon as it is said to have no rest mass [m0]. However, if we link Planck’s equation (E=hf) with Einstein’s equation $$(E=mc^2)$$, it suggests that photons have kinetic mass $$(mk= hf/c^2)$$. Now if time dilation causes frequency [f] to reduce, in the case of an outgoing photon, it also causes the photon to lose energy. However, the photon now has more potential to gain energy, i.e. increase in frequency, by falling back to a lower position in the gravity well. As such, mass and frequency act as equivalent concepts against which the total energy can be accounted in any frame of reference. It just seems easier to me to assign the changes to the photon coming into the observers frame of reference rather than assume the photon is unchanged, which then requires entire frame of reference to be recalibrated to some other arbitrary, but remote frame of reference.

The only problem I have with the localise perspective is that I seems to suggest that the energy of a photon falling into the event horizon would be infinitely blueshifted, implying infinite energy. Again, I am not making statement of fact here, simply outlining some assumptions on which I am attempting to learning more about this subject, so have much appreciated the help and feedback.

11. yuiop

Consider the following example. There are 3 observers (A,B,C) at various heights in a gravity well. According to an observer at infinity the gravitational gamma factor $$g= 1/ \sqrt{1-Rs/R}$$ is 8, 4 and 2 for A,B and C respectively with A being the deepest in the well. Say a photon is emitted with a frequency of 1 and wavelength of 1 at location A as measured by A. The measurements of (frequency, wavelength, speed of light) made by the observer (D) at infinity at locations A, B, C, and D would be:

A' = (1/8, 1/8, 1/64)
B' = (1/8, 1/2, 1/32)
C' = (1/8, 2, 1/4)
D' = (1/8, 8, 1)

and the local measurements would be:

A = (1, 1, 1)
B = (1/2, 2, 1)
C = (1/4, 4, 1)
D = (1/8, 8, 1)

In other words the coordinate measurements according to the observer at infinity indicate the frequency remains constant and wavelength get longer by a factor of gamma squared and the coordinate speed of light also increases by a factor of gamma squared as the photon climbs out the well. To the observer at infinity the energy of the photon is constant.

To the local observers each observer measures a lower frequency and a longer wavelength locally than the observers further down and the same value for the local speed of light c.

That is my analysis that is consistent with the Schwarzschild metric. As with special relativity, it is not possible to state which observers point of view is the more "fundemental".

12. yuiop

The radial distance effect is the other way round. Approaching a large mass, clocks ticks more slowly by gamma and rulers contract by gamma and the coordinate speed of light slows down by gamma squared, but the combined effects result in the local speed of light being constant for all observers.

13. mysearch

522
Kev, thanks for the comments and especially the analysis in #11, as it is always helpful to look at the issues in terms of a practical example. I was in the process of working through your example, but then saw post #12, which has thrown up some doubts in my mind.

I accept I am no expert, but I was pretty sure that the radial distance expands under gravity as [r] approaches [Rs], i.e. its the opposite to special relativity, in which length contracts as [v->c]. This is why the (1-Rs/r) is the denominator in the [dr] term in the Schwarzschild metric.

It will probably be tomorrow before I can respond properly. Thanks.

14. yuiop

Which of these statements do you disagree with?

a) Gravitational time dilation slows clocks by the gamma factor.
b) Coordinate speed of light slows by gamma squared.
c) Local speed of light is always c.

If you agree with statements a,b and c you have to come to the conclusion that length (rulers) have to contract by gamma aproaching a massive body.

The confusion may be the difference between "radial distance" for which there is no simple conversion factor and the length of local almost infinitesimal ruler.

15. mysearch

522
Response to Kev #11, 12 & 14

Hi Kev

This could well be the case, but I will try to explain my frame of reference then you can tell me if we are talking at cross-purposes. I have no problem with bullets (a) or (c), so maybe the confusion is with respect to bullet (b). I am only providing the following simplified form of the Schwarzschild metric as a reference:

$$c^2 d\tau^2=c^2\left(1-Rs/r\right)dt^2 - \frac{dr^2}{\left(1-Rs/r\right) }$$

Without resorting to any derivation, the placement of the (1-Rs/r) terms suggests that the relativistic effects, under gravity, works in different direction for time and space. Also the equations for proper time $$[d\tau]$$ and proper distance [ds], given in various sources, appears to support this suggestion:

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

$$ds = \frac {dr}{\sqrt{\left(1-Rs/r\right)}}$$

Before continuing, let me just clarify a few assumptions. I am assuming that the local observer/apparatus measures 1 second per second and 1 metre per metre. Therefore, the time and space effects are all relative to a distant observer, e.g. (D) observing (A). I am also assuming that both the local and distant observers are stationary, so no velocity effects. What I not too sure about is that if the local observer (A), in curved space around a black hole, measures the circumference, unaffected by gravity, and then divides by $$2\pi$$, would the radius calculated and the actual radius observed differ and allow the local observer to recognise he was in curved space?

Anyway, returning to your example, albeit in a slight different context, but using your values. The distant observer (D) is conceptually watching a hypothetical experiment in (A). The experiment consists of a photon being fired between 2 points, aligned on the radial axis, separated by 1 light-second. As the photon is fired, the source flashes a light, as does the receiver when the photon is received. In the context of (A), the flashes are separated by 1 second, as the local speed of light is always [c].

However, following your example, time in (D) is ticking 8x faster than (A) and therefore we might conclude the observed frequency was 1/8 as it is inversely proportional to time. Again, this aligns to your example. However, this would imply that the time, as measured by (D), between the flashes was 8 seconds. Of course, if we go with the previous assumption about radial space expanding on approaching [Rs], then observer [D] would measure the distance between source and receiver as 8 light-seconds, which would imply the photon was still travelling at [c]. I fully accept this logic may be flawed, but would like to understand why.

Just a few other points I want to highlight. When I originally raised the question about gravitational redshift, based on the assumptions above, it was because I couldn’t resolve why the gravitational effects on time and space didn’t cancel out. If time slows approaching [Rs], then it was logical to assume that the emitted frequency of a photon would be lower, in-line with experimental observation. However, if radial space contracts as the photon travels from (A) out to (D), then I initially assumed this would affect the wavelength, which in-turn would affect the frequency based on $$c=f\lambda$$. However, I thought this might be resolved based on the fact that the physics of wave propagation only states the frequency [f] is defined by the energy [E=hf], whereas the propagation velocity [v=c] is defined by the media of propagation, i.e. vacuum permittivity and permability. As such, the wavelength $$\lambda=c/f$$ is determined by [c] and [f]. If [c] is constant in all local frames of reference, then the gravitational redshift would correspond to experiment and align to the effects of time dilation only. Again, I fully accept this might all be unsubstantiated speculation on my part, but my goal was to try to provide some physical interpretation of the relativistic effects. Many Thanks

16. yuiop

Hi mysarch,

there is quite a lot in your post so I will address the first few points only for now and come back to the rest as time allows.

For light dtau is zero. So we can say for a falling photon:

$$0 = c^2\left(1-Rs/r\right)dt^2 - \frac{dr^2}{\left(1-Rs/r\right) }$$

Rearrange and simplify:

$$\frac{dr^2}{\left(1-Rs/r\right) } = c^2\left(1-Rs/r\right)dt^2$$

$$\frac{dr^2}{dt^2} = c^2 \left(1-Rs/r\right)^2$$

$$\frac{dr}{dt} = c \left(1-Rs/r\right)$$

.. proving my assertion in bullet point b that the vertical coordinate speed of a falling photon in a gravitational field (as measured by an observer at infinity) is inversely proportional to the gravitational gamma factor squared.

the transformation for time and space in SR are in "different directions" too. Proper time is less than the time measured in the primed frame and proper length is greater than the length measured in the primed frame.

$$dt' = dt \sqrt{\left(1-v^2/c^2)}$$

$$dx' = \frac {dx}{\sqrt{\left(1-v^2/c^2)}}$$

17. mysearch

522
Response to Kev #16

Realise you will need to consider my comments in #15. However, wanted to raise some comments against your initial response. I agree with your basic derivation of the observed photon velocity from (D), although I added the $$\pm$$ to reflect the quadratic solution:

Photon: $$\frac{dr}{dt} = \pm c \left(1-Rs/r\right)$$

Simply for reference, here are some other solutions, starting with the observed free-fall velocity from (D):

Free-Fall Mass: $$\frac{dr}{dt} = c \left(1-Rs/r\right) \sqrt{Rs/r}$$

The implication of this equation is that both velocities perceived by (D) collapse to zero at [Rs], presumably corresponding to time slowing to a halt at [Rs]. Of course, solutions with respect to proper time $$[d\tau]$$ suggest otherwise:

Photon: $$\frac{dr}{d\tau} = -c \sqrt{Rs/r} \pm c$$

Free-Fall Mass: $$\frac{dr}{d\tau} = -c \sqrt{Rs/r}$$

Now the speed of light [c] appears to maintain its value [c] with respect to the local observer (A), even when (A) is free-falling at [c]. To be honest I have never been able to resolve both outcomes from the Schwarzschild metric. At one level, time dilation with respect to (D) appears to be supported by experiment, but leads to a frozen star rather than a black hole. While the other solution, even taking into account arguments concerning coordinate singularity, seems to proceed through the event horizon but cannot seem to relate time back to (D). However, as far as I know, a series of local measurements in (A), (B) and (C) would all measure [c] as [c] and suggest the photon frequency was affected by the initial time dilation.

Sorry, I didn’t understand where your last 2 equations come from, as they do not seem to correspond to the Lorentz transforms:

$$t' = \gamma \left( t - \frac{v x}{c^{2}} \right)\rightarrow \gamma (t)$$ when x=0

$$x' = \gamma \left( x - v t \right) \rightarrow \gamma (x)$$ when t=0

Apologises for raising so many issues, but I would be interested how your position explains gravitational redshift. Again, many thanks.

18. yuiop

One difficulty with using the Schwarzchild solution at exactly r=R_s is that no physical observer could remain stationary at the event horizon so the equztion is not physical at that exact point. However, there are other issues. In another thread we showed that the interior Schwarzschild solution shows that time reverses at certain coordinates below the Schwarzschild radius even before a black hole is fully formed (collapsed to a singularity) suggesting that physical process (time related) prevents a singularity forming and all the mass accumulates in the thin shell just outside the event horizon. This is consistent with the frozen star viewpoint. It is also consistent with the Hawking entropy formula that shows the entropy is proportional to the surface area of the black hole event horizon. Charge and mass accumulated on such a shell would appear to anyone outside the shell to be from a point source just as the gravitational effect of the Earth above the surface is mathematically the same as having all the mass located at the centre even though we know it is not really all located there.

That is exactly what I stated the local observers at A, B and C would measure. Is there measurement the only valid viewpoint? I think not. For example the local observers in Special Relativity can not measure length contraction and time dilation of there own rulers and clocks but that does not mean length contraction and time dilation do not exist. Similarly, the two observers with relative velocity each measure the others clock to ticking slower than their own clock and we know that each has a valid viewpoint even thought they seem contradictory.

x is the coordinate of a point. Length would be measured as the difference of two coordinates such as x2-x1.

For length:

$$(x_2' - x_1')/\gamma = \left( x_2 - v t \right) \right - \left( x_1 - v t \right)$$

$$(x_2' - x_1')/\gamma = \left( x_2 - x_1 \right)$$

$$(x_2' - x_1') = \left( x_2 - x_1 \right)\gamma$$

For time:

$$(t_2' -t_1') = \gamma \left( t_2 - \frac{v x_2}{c^{2}}\right) -\gamma \left( t_1 - \frac{v x_1}{c^{2}}\right)$$

Since x2 and x1 are not the same for non zero t we use the reverse transformation:

$$t = \gamma \left( t' + \frac{v x}{c^{2}} \right)$$

$$(t_2 -t_1) = \gamma \left( t_2' + \frac{v x_2'}{c^{2}}\right) -\gamma \left( t_1' + \frac{v x_1'}{c^{2}}\right)$$

Now we can say x2' = x1' for the clock in the moving frame and it follows:

$$(t_2 -t_1) = \gamma \left( t_2' - t_1' \right)$$

$$(t_2' -t_1') = \left( t_2 - t_1 \right)/\gamma$$

Well, my position is that there is no single explanation. From the point of view of local observers frequency of the photon changes due to time dilation and the speed of light is constant. From the point of view of an observer at infinity (or at any constant non local location) the frequency remains constant, the wavelength is stretched as the photon rises and the speed of light increases. Both points of view are equally valid.

19. sirzerp

32
I have a different question related to Gravitational Redshift.

We know that the universe is loosing mass as the stars produce photons. We also know that photons are red shifted when moving away from a star, and blue shifted when moving to a star.

Let’s do a little thought experiment. If a photon leaves the sun at year zero and is red shifted as it leaves, and then travels a 100 million years and is reflected back; and then travels another 100 million years. When the same photon that left 200 million years returns to the sun, at what wavelength is it at?

That sounds contrived, but I’m trying to ask, how is mass lost by a star, related to red shifting of same photon created by the star, when reflected and lack of blue shifting when returning to the star.

As the solar mass decreases, does that mean all reflected photons back to that star system have to return red shifted?

Anyone have the wavelength math for a one percent mass decrease and 200 million year round trip?

Last edited: May 12, 2008
20. yuiop

Hi sirxerp,
ignoring cosmological expansion the photon will red shift on its way out and blue shift on the way back so that it has exactly the same wavelength when it returns as when it it left. I am not sure what you are getting at here unless you are enquiring about the mathematical equations for what happens when cosmological expansion of the universe, during the travel time, is taken into account? In that case the photon will return redshifted.