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Gravitational Redshift 
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#1
May1008, 07:27 AM

PF Gold
P: 515

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.
[URL="http://en.wikipedia.org/wiki/Gravitational_redshift"] 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
May1008, 08:37 AM

PF Gold
P: 4,087

For light, frequency and wavelength are not independent, and so you can regard a redshift either as an increase in wave length or a reduction in frequency. Redshifted light will deliver less energy. This is not a surprise because energy is not relativistically invariant. Have you come across the PoundRebka experiment ? I guess you have since you quote Prof. Wiki. en.wikipedia.org/wiki/PoundRebka_experiment 


#3
May1008, 08:41 AM

P: 2,954

Pete 


#4
May1008, 10:34 AM

PF Gold
P: 515

Gravitational Redshift
Mentz114: Thanks for the comments. I have attached a response against each below:
Pmb_phy/Pete: Thanks for the feedback, unfortunately I not that familiar with the meaning of [tex] g_{00}[/tex] 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. 


#5
May1008, 01:31 PM

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P: 4,087

Pete is right, only [tex]g(r)_{00}[/tex] 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
May1108, 05:10 AM

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P: 515

[tex]d\tau = \sqrt{g_{00}} dt[/tex] This suggests that in the context of gravity only, i.e. no relative velocity, that [tex]g_{00}[/tex] might be related to the relativity factor [tex][\gamma][/tex], e.g. [tex]g_{00} = 1Rs/r[/tex]?? 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: 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
May1108, 05:21 AM

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#8
May1108, 06:56 AM

PF Gold
P: 515

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 knockon effects on the measuring apparatus. This is not being forwarded as a statement fact, simply the assumptions to which I was working



#9
May1108, 07:31 AM

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P: 4,087

mysearch,
going back to an earlier remark, 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 spacetime, 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. 


#10
May1108, 08:47 AM

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P: 515

http://www.physicsforums.com/archive.../t125057.html 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, [tex]\lambda=c/f[/tex] 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 inturn is linked to velocity and gravitation: [tex] Total Energy (Et) = Ek + Ep =1/2mv^2 + (GMm/r)[/tex] 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 [tex] (E=mc^2)[/tex], it suggests that photons have kinetic mass [tex](mk= hf/c^2)[/tex]. 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
May1108, 09:05 AM

P: 3,967

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
May1108, 09:18 AM

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#13
May1108, 11:20 AM

PF Gold
P: 515

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 (1Rs/r) is the denominator in the [dr] term in the Schwarzschild metric. It will probably be tomorrow before I can respond properly. Thanks. 


#14
May1108, 11:37 AM

P: 3,967

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
May1208, 06:28 AM

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P: 515

Hi Kev
Saw your last post, so will start with the last point you raised: [tex]c^2 d\tau^2=c^2\left(1Rs/r\right)dt^2  \frac{dr^2}{\left(1Rs/r\right) }[/tex] Without resorting to any derivation, the placement of the (1Rs/r) terms suggests that the relativistic effects, under gravity, works in different direction for time and space. Also the equations for proper time [tex][d\tau][/tex] and proper distance [ds], given in various sources, appears to support this suggestion: [tex]d\tau = \sqrt{\left(1Rs/r\right)}dt [/tex] [tex] ds = \frac {dr}{\sqrt{\left(1Rs/r\right)}} [/tex] 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 lightsecond. 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 lightseconds, 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, inline 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 inturn would affect the frequency based on [tex]c=f\lambda[/tex]. 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 [tex]\lambda=c/f[/tex] 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
May1208, 07:08 AM

P: 3,967

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. [tex]0 = c^2\left(1Rs/r\right)dt^2  \frac{dr^2}{\left(1Rs/r\right) }[/tex] Rearrange and simplify: [tex]\frac{dr^2}{\left(1Rs/r\right) } = c^2\left(1Rs/r\right)dt^2 [/tex] [tex]\frac{dr^2}{dt^2} = c^2 \left(1Rs/r\right)^2 [/tex] [tex]\frac{dr}{dt} = c \left(1Rs/r\right) [/tex] .. 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. [tex]dt' = dt \sqrt{\left(1v^2/c^2)} [/tex] [tex] dx' = \frac {dx}{\sqrt{\left(1v^2/c^2)}} [/tex] 


#17
May1208, 08:45 AM

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P: 515

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 [tex]\pm[/tex] to reflect the quadratic solution:
Photon: [tex]\frac{dr}{dt} = \pm c \left(1Rs/r\right) [/tex] Simply for reference, here are some other solutions, starting with the observed freefall velocity from (D): FreeFall Mass: [tex]\frac{dr}{dt} = c \left(1Rs/r\right) \sqrt{Rs/r} [/tex] 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 [tex][d\tau][/tex] suggest otherwise: Photon: [tex]\frac{dr}{d\tau} = c \sqrt{Rs/r} \pm c [/tex] FreeFall Mass: [tex]\frac{dr}{d\tau} = c \sqrt{Rs/r} [/tex] Now the speed of light [c] appears to maintain its value [c] with respect to the local observer (A), even when (A) is freefalling 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: [tex]t' = \gamma \left( t  \frac{v x}{c^{2}} \right)\rightarrow \gamma (t) [/tex] when x=0 [tex]x' = \gamma \left( x  v t \right) \rightarrow \gamma (x) [/tex] when t=0 Apologises for raising so many issues, but I would be interested how your position explains gravitational redshift. Again, many thanks. 


#18
May1208, 01:45 PM

P: 3,967

For length: [tex](x_2'  x_1')/\gamma = \left( x_2  v t \right) \right  \left( x_1  v t \right) [/tex] [tex](x_2'  x_1')/\gamma = \left( x_2  x_1 \right) [/tex] [tex](x_2'  x_1') = \left( x_2  x_1 \right)\gamma [/tex] For time: [tex](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) [/tex] Since x2 and x1 are not the same for non zero t we use the reverse transformation: [tex]t = \gamma \left( t' + \frac{v x}{c^{2}} \right) [/tex] [tex](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) [/tex] Now we can say x2' = x1' for the clock in the moving frame and it follows: [tex](t_2 t_1) = \gamma \left( t_2'  t_1' \right) [/tex] [tex](t_2' t_1') = \left( t_2  t_1 \right)/\gamma [/tex] 


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