Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

In the realm of physics there is no knowledge, the justification of which does not depend either upon experience or upon reason. In the light of the findings of the preconditions of knowledge made in for instance epistemology, this statement is certainly an oversimplification, but I still think it may very well serve its purpose as far as my actual two questions are concerned.

In order to phrase these two questions, I have first to explain which physical theories I assume as a correct basis for phrasing them. And inasmuch as a question of mine is based on alternative mutually incompatible theories, the issue extends to also asking the answering forum to decide which of the theories that best explains the theoretical logics of physics and the empirical findings that have been made until this date.

When commenting on the topic Gravitation as the source of redshift of light beams, Ted Bunn, among others, have assumed that when light is generated by a light source located at a given distance from a heavy body, and radiated radially from that body, then the energy of the light rays would be decreased by an amount proportional to the difference in gravitational potential between the place where the light source is located and the potential of the place where an observer is detecting the light. Since the phrasing of my questions necessitates a correct model of the gravitational red shift of electromagnetic radiation, I now ask if this explanation of the physics of the gravitational redshift is free from contradictions. The gravitational redshift applying to the case outside of a non-rotating, uncharged mass which is spherically symmetric is described by the known exact equation, where • G is the gravitational constant, • M is the mass of the object creating the gravitational field, • R is the radial coordinate of the point of emission (which is analogous to the classical distance from the center of the object, but is actually a Schwarzschild coordinate), • r is the radial coordinate of the observer (in the formula, this observer is at an infinitely large distance), and • c is the speed of light.

The steady oscillation of a light-emitting source may be used as a clock. A common equation used to determine gravitational time dilation is derived from the Schwarzschild metric, which describes space-time in the vicinity of a non-rotating massive spherically symmetric object. In the known equation

• t zero is the proper time between events A and B for a slow-ticking observer within the gravitational field, • t fast is the coordinate time between events A and B for a fast-ticking observer at an arbitrarily large distance from the massive object (this assumes the fast-ticking observer is using Schwarzschild coordinates, a coordinate system where a clock at infinite distance from the massive sphere would tick at one second per second of coordinate time, while closer clocks would tick at less than that rate), • G is the gravitational constant, • M is the mass of the object creating the gravitational field, • r is the radial coordinate of the observer (which is analogous to the classical distance from the center of the object, but is actually a Schwarzschild coordinate), • c is the speed of light, and • r zero is the Schwarzschild radius of M. • If the distance r from the center of the object M is a fixed, constant distance, we arrive at the result that the ratio between the proper time between events A and B for the slow-ticking observer (or low-frequency observer) and the coordinate time between events A and B for the fast-ticking observer (or high-frequency observer) becomes a fixed, constant ratio.

Thus, the original frequency of the oscillation is not an element affecting the gravitational redshift thereof. From these gravitational red shift equations I must thus conclude that the “effective mass” of the photon is not a factor in the red shift equation. Neither is the “potential energy” of the photon in a gravitational field a factor in the red shift equation.

The reasoning proposed by Ted Bunn is thus incompatible with the exact gravitational red shift equations. Ted Bunn uses the metaphor: "In Newtonian language, if you imagine the source of the light at the center of a spherically symmetric expanding spacetime, then the light travels 'uphill' in a gravitational potential all the way to the observer. The observed redshift is partially due to this redshift and partially due to the observer's motion. Conversely, if you put the observer at the center, the light 'falls downhill' all the way to the observer. This gives a different breakdown of the observed redshift into gravitational and Doppler contributions." This metaphor describes a model according to which the pace of the ticks of two different clocks put beside each other, i.e. a fast-ticking clock and slow-ticking clock, would be affected differently while traveling 'uphill' or ‘downhill’ in a gravitational field. But they are not affected differently in the way proposed by the Ted Bunn model, according to which the high-frequency and low-frequency light would be affected differently by the gravitational potential difference.

Further, at one point, I find a valid reason to object to Ted Bunn's conclusion that a photon passing a particle, or a larger clump of matter, is blue shifted as it falls in and equally red shifted as it comes out, by way of which Ted Bunn concludes that there's no overall "tired-light-like" redshift due to an effect of this sort. However, wouldn’t it be plausible that the gravitational forces emanating from a light beam passing through the universe would cause a mechanism of action analogous to the mechanism of action causing the interaction between a physical object passing through the space and the surrounding objects?

My suggestion that such an interaction also takes place between a beam of photons and a physical object I claim to be corroborated by the fact that the work performed by a beam of light passing very close to the Sun certainly will make the Sun go for a little ride. Had an object the size of the Sun passed the Sun, the ride would have been very much potent. But the effect of a beam of light upon the Sun’s position in space of course will be impossible to detect by way of current measurement techniques. But the work is nonetheless done. And much smaller objects than the Sun will be noticeably displaced and accelerated. According to the basic laws of physics, the energy that has been used to accomplish the work in question, is taken away from the light beam, which thus becomes red shifted. Qualitatively, it is therefore true that this type of redshift is a real redshift. But it is enough to make a rough calculation in order to establish and conclude that this red shift in our present Universe will be so insignificant that it will entirely drown in the cosmological red shift. However, in the very young Universe this type of red shift was quite a factor to be accounted for.

share|improve this question

closed as unclear what you're asking by Kyle Kanos, Neuneck, Jim, Ali, John Rennie Jul 15 at 15:14

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

    
Is there a question here? –  Kyle Kanos Jul 15 at 11:41
    
What do you hope to achieve by posting this on a question and answer site? Would you like us to say "Yes. Absolutely. He was wrong, by golly!"? Without asking a question, your post is likely to be closed and not viewed again. Furthermore, you have come to professional physicists and said "what this person said is wrong because physics doesn't work that way". We know how physics works. We can put it in math, which is something that is distinctly lacking from your post. If you want to prove him wrong, show him the math proving he's wrong. Nothing else will do and posting it here is pointless. –  Jim Jul 15 at 13:02

2 Answers 2

This is a partial answer for the first part of your question.

You are not completely right.

In the Schwarzschild metrics, for instance, the ratio of the energies (or frequencies) for the photons, is given by (in $c=1$ units) (here no Doppler effect is included, we only check gravitational effect):

$$\frac{E(r_1)}{E(r_2)} = \sqrt{\frac{1 - 2 \Phi(r_2)}{1 - 2 \Phi(r_1)}} \tag{1}$$

where $\Phi(r)$ is the Newtonian gravitational potential.

However, the "Newtonian metaphor" is qualitatively correct, you can check that :

if $r_2>r_1$, then $\Phi(r_2) <\Phi(r_1)$, and so $E(r_2) <E(r_1)$

Thus, irrespective of the wavelength of the light, the gravitational potential difference causes an extra time delay before the light reaches the observer's eye, compared to the time necessary for the light to reach the observer’s eyes when the difference of gravitational potential is zero.

Yes, but this is not the problem. To check the difference of frequencies, you only have to check, for instance, the difference of period of a clock, seen by different observers. So you consider two successive ticks of the clock at position $r=r_1$, separated by $\Delta \tau_1$, and see how the observer at $r=r_2$ see this difference of time between the two ticks ($\Delta \tau_2$). The Schwarzschild metrics is time-invariant in the (non-physical) $t$ coordinate, so the (non-physical) time $\Delta t$ for a photon synchronized with the first tick to go from $r_1$ to $r_2$ is the same that for a photon synchronized with the second tick.

Said differently, the physical period measured in $r=r_1$ is : $$\Delta\tau_1 = \sqrt{1 - 2 \Phi(r_1)} \tag{2} \Delta t_1$$

The physical period measured in $r=r_2$ is : $$\Delta\tau_2 = \sqrt{1 - 2 \Phi(r_2)} \tag{3} \Delta t_2$$

The time-invariance of the Schwarzschild solution (see discussion above) means that : $$\Delta t_1 = \Delta t_2 \tag{4}$$ I insist, equation $(4)$ simply means that the difference $\Delta t$ for the photons synchronized with the first or the second tick of the clock, to go from $r_1$ to $r_2$ is exactly the same.

From $(2),(3),(4)$, and noting that the ratio of energies/frequencies is the inverse of the radio of periods, you will easily get $(1)$

share|improve this answer
    
Oh, hello! Great minds think alike. –  rob Jul 15 at 12:54

I'm not entirely sure that I understand your question, but what leaps out at me is

it's wrong to suggest that the energy of the light rays are decreased by an amount proportional to the difference in gravitational potential between the place where the light source is located and the potential of the place where an observer is detecting the light. The reasoning is wrong since it assumes that the high-frequency and low-frequency light would be affected differently by the potential difference. … [T]he observer will note that any specific size of difference in gravitational potential will result in a specific size of red shift which is totally independent of the wavelength of the light.

The gravitational redshift for a spherical, non-rotating mass distribution (such as the earth, to good approximation) is given by \begin{align} 1+z &= \frac{f_\text{emit}}{f_\text{observe}} = \frac{E_\text{emit}}{E_\text{observe}} \\ &= \sqrt\frac{1-r_s/r_\text{observer}}{1-r_s/r_\text{emitter}} = \sqrt\frac{\frac1{r_s} - \frac1{r_\text{observer}}}{\frac1{r_s} - \frac1{r_\text{emitter}}} \end{align} where $r_s = 2GM/c^2$ is the Schwarzchild radius for the mass distribution. If both of the other $r$ are much larger than $r_s$, we can use the binomial expasion to approximate \begin{align} 1+z &\approx \left( 1 - \frac12\frac{r_s}{r_\text{observer}}{}\right) \left( 1 + \frac12\frac{r_s}{r_\text{emitter}}{}\right) \\ &\approx 1 + \frac{r_s}2 \left( \frac1{r_\text{emitter}}-\frac1{r_\text{observer}} \right) \end{align} So you can see that, in this limit where $r_s$ is very small, the photon's energy difference $E_\text{emitted}-E_\text{observed} = zE_\text{observed}$ is exactly equal to the change in the Newtonian potentials $\Phi=-GM/r$ for an object with equivalent mass $E_\text{observed}/c^2$.

While you are correct that the size of the redshift $z$ is independent of the initial wavelength of the light, the change in the energy of photon depends on the photon's "effective mass," which does depend on its wavelength $\lambda = hc/E$.

share|improve this answer
    
Thank you rob for drawing my attention to the following fact: "While you are correct that the size of the redshift z is independent of the initial wavelength of the light, the change in the energy of photon depends on the photon's 'effective mass', which does depend on its wavelength λ=hc/E." Please, present the expression for the photon's "effective mass"! –  Carl Johan Soderquist Jul 16 at 18:45
    
OK, I understand –  Carl Johan Soderquist Jul 16 at 19:04
    
And Rob, in this context it's really important to discern that GR stipulates the following direction of the chain of causalities involved. The difference in the gravitational potential provides the real, concrete cause of the change in energy of the photon. This change of the photon's energy provides in the next step the primary premise of the change of the photon's wavelength or frequency. Then, arriving at the last link of the causalities involved, it's easily understood that this change manifests itself in a change of the photon's "effective mass". Thus, it’s not the other way round. –  Carl Johan Soderquist Jul 20 at 20:44
    
The classic approach is that the "pace" of the flow of time will manifest itself in the tick of a clock or in the frequency of a beam of light. And that pace is slowed down in a gravitational field due to the "energy level" or potential energy of that field. Normally no attention is paid to the fact that at any such potential energy level, the space-time is caused to flow inwards into the object generating the gravitational field. I suggest that this flow could be characterized as the cause of the slowing down of the pace of time. Which in turn manifests itself in a reduced frequency. –  Carl Johan Soderquist Jul 20 at 21:21

Not the answer you're looking for? Browse other questions tagged or ask your own question.