Energy of a photon

[Sriman Dutta]

According to Einstein's Special Theory of Relativity, time passes slowly for an observer when the observer is near a gravitational source. So, if there are two observers, time passes slowly for the one whose distance from the gravitational source is less as compared to the other. So, time dilates for the lower observer in relation to the upload observer ( considering the case in a gravitational potential well). However, if the two observers are not allowed to communicate or see each other, they would never realize the time difference.

The direct result of the gravitational time dilation is gravitational redshift. For an upper observer, time passes more quickly than the lower one. If a ray of light travels out of the potential well, the ray would be redshifted for the upper observe ( because frequency of light is inversely proportional to time elapsed).

This means that the energy of the released photon decreases because E=hf.

However my intuition tells me that the energy only seems to decrease to the external observer as he is placed outside the photon's frame of reference.

Am I correct. ?

[swansont]

According to Einstein's Special Theory of Relativity, time passes slowly for an observer when the observer is near a gravitational source. So, if there are two observers, time passes slowly for the one whose distance from the gravitational source is less as compared to the other. So, time dilates for the lower observer in relation to the upload observer ( considering the case in a gravitational potential well). However, if the two observers are not allowed to communicate or see each other, they would never realize the time difference.

The direct result of the gravitational time dilation is gravitational redshift. For an upper observer, time passes more quickly than the lower one. If a ray of light travels out of the potential well, the ray would be redshifted for the upper observe ( because frequency of light is inversely proportional to time elapsed).

This means that the energy of the released photon decreases because E=hf.

However my intuition tells me that the energy only seems to decrease to the external observer as he is placed outside the photon's frame of reference.

Am I correct. ?

All observers are outside the photon's frame of reference. Photons travel at c, and anything with mass can't.

[Sriman Dutta]

OK fine but that doesn't work for my question.

[swansont]

OK fine but that doesn't work for my question.

It was a condition of your question. I was pointing out that it's moot: there is no "photon's frame of reference"

[Sriman Dutta]

So does the energy of the photon decreases actually ?

[swansont]

So does the energy of the photon decreases actually ?

The energy of a photon decreases as it moves up in a potential well. Relativity predicts this, but conservation of energy does as well.

[Tim88]

According to Einstein's Special Theory of Relativity, time passes slowly for an observer when the observer is near a gravitational source. So, if there are two observers, time passes slowly for the one whose distance from the gravitational source is less as compared to the other. So, time dilates for the lower observer in relation to the upload observer ( considering the case in a gravitational potential well). However, if the two observers are not allowed to communicate or see each other, they would never realize the time difference.

The direct result of the gravitational time dilation is gravitational redshift. For an upper observer, time passes more quickly than the lower one. If a ray of light travels out of the potential well, the ray would be redshifted for the upper observe ( because frequency of light is inversely proportional to time elapsed).

This means that the energy of the released photon decreases because E=hf.

However my intuition tells me that the energy only seems to decrease to the external observer as he is placed outside the photon's frame of reference.

Am I correct. ?

 

My short answer: yes, you're very right! (apart of the glitch in your last sentence of course.)

 

Long answer: this is one of the trickiest issues in GR with a lot of blurred description in the literature, so here's my attempt to a self consistent analysis (if there are errors, corrections are welcome. And perhaps someone else may give a disagreeing but equally self consistent answer).

 

[edit:] First of all, QM (photons) and GR (waves) speak a different language; technically speaking they should not be mixed. In what follows I'll assume that a number of photons corresponds to a certain number of wave cycles.

 

Next, energy is conserved but also cycles are conserved:

- no energy can "get lost" or "appear from nothing". Essential requirement: we must do the analysis with a single reference system.

- if a source emits a 1 kHz signal for one second locally, 1000 cycles should arrive.

 

When this radio signal of 1000 cycles is sent upward, 1000 cycles arrive at a detector with a faster resonating quartz crystal. The measured frequency is therefore reduced ([edit:] ceteris paribus of course: here we assume that the Earth surface is in rest during the interval, as motion is irrelevant for your question.)

 

Locally f and E seem to be decreased because they are compared with detectors that posses higher gravitational potential energy.

More elaboration to show the self consistency: when an object with a number of atoms falls down to the ground, kinetic energy is released.

Consequently all consistuents of an atom must be assumed to have reduced energy. An emitted light ray ("photons") will therefore be released at lower frequency, and if it propagates upwards it will be detected at a lower frequency than the emitted frequencies of atoms at that height.

 

Note also that a change of frequency "in flight" results in self contradiction. If for 24h a certain number of cycles is sent upward, that number must also be received in the same time interval. Else cycles would go missing "in mid air"!

Edited October 23, 2016 by Tim88

[Strange]

Note also that a change of frequency "in flight" results in self contradiction. If for 24h a certain number of cycles is sent upward, that number must also be received in the same time interval. Else cycles would go missing "in mid air"!

 

 

That implies that the Doppler effect cannot exist. But both that and gravitational red-shift are well attested.

[Tim88]

That implies that the Doppler effect cannot exist. But both that and gravitational red-shift are well attested.

 

That implies rather that you did not fully read my post, as I clarified redshift in agreement with Sriman. And perhaps you missed my additional precision of ceteris paribus: motion isn't part of the question.

 

Maybe interesting in this context: I know a mechanic teacher who wrote a textbook with a section on GR, and as I recall it (it was a long time ago), when looking through it I was disappointed to see that he included a section in which a photon "loses energy" while climbing up a gravitation well. When I brought up the inconsistency, he defended it as "just a heuristic". He would take it up with his co-author, but I never heard about it again.

See also Okun: ON THE INTERPRETATION OF THE REDSHIFT IN A STATIC GRAVITATIONAL FIELD

http://cds.cern.ch/record/393302/files/9907017.pdf

 

The abstract is very clear if it's separated in two paragraphs (emphasis mine):

 

The classical phenomenon of the redshift of light in a static gravitational potential, usually called the gravitational redshift, is described in the literature essentially in two ways: on the one hand the phenomenon is explained through the behaviour of clocks which run the faster the higher they are located in the potential, whereas the energy and frequency of the propagating photon do not change with height. The light thus appears to be redshifted relative to the frequency of the clock.

 

On the other hand the phenomenon is alternatively discussed (even in some authoritative texts) in terms of an energy loss of a photon as it overcomes the gravitational attraction of the massive body. This second approach operates with notions such as the "gravitational mass" or the "potential energy" of a photon and we assert that it is misleading.

 

Regretfully, while I mostly agree with his arguments, he didn't fully make the case for the correct interpretation there. Therefore I did it here (despite the usage of a huge amount of equations by him, and the total lack of that by me: for the logic it's irrelevant!).

 

PS. In view of the OP's young age, I'll try to answer any follow-up questions in less sophisticated English for improved clarity.

Edited October 23, 2016 by Tim88

[edguy99]

This may help:

 

http://www.perimeterinstitute.ca/videos/david-wineland-keeping-better-time-era-optical-atomic-clocks

 

The video of nobel prize winner David Wineland talking about the different speed of clocks in different gravitational potentials helps with "real life" understanding of Einstein's Special Theory of Relativity where time passes slowly for an observer when the observer is near a gravitational source. I found it very valuable and informative to watch.

[Sriman Dutta]

So now I get it.

Thanks to all of you.

 

Now I understand SR. But still I haven't got a mathematical model for explaining the time dilation due to speeds near to c and the phenomenon of gravitational time dilation. Can anybody provide me that. ?

[Mordred]

Here is the basics of the Lorentz transformations. I included an open source textbook at the end.

 

Lorentz transformation.

 

postulates.

 

1) the results of movement in different frames must be identical

2) light travels by a constant speed c in a vacuum in all frames.

Consider 2 linear axes x (moving with constant velocity and [latex]\acute{x}[/latex](at rest) with x moving in constant velocity v in the positive [latex]\acute{x}[/latex] direction.

Time increments measured as a coordinate as dt and [latex]d\acute{t}[/latex] using two identical clocks. Neither [latex]dt,d\acute{t}[/latex] or [latex]dx,d\acute{x}[/latex] are invariant. They do not obey postulate 1.

A linear transformation between primed and unprimed coordinates above

in space time ds between two events is ( this below doesnt have curvature as SR assumes Euclidean)

[latex]ds^2=c^2t^2=c^2dt-dx^2=c^2\acute{t}^2-d\acute{x}^2[/latex]

Invoking speed of light postulate 2.

[latex]d\acute{x}=\gamma(dx-vdt), cd\acute{t}=\gamma cdt-\frac{dx}{c}[/latex]

Where[latex] \gamma=\frac{1}{\sqrt{1-(\frac{v}{c})^2}}[/latex]

Time dilation

dt=proper time ds=line element

since[latex] d\acute{t}^2=dt^2[/latex] is invariant.

an observer at rest records consecutive clock ticks seperated by space time interval [latex]dt=d\acute{t}[/latex] she receives clock ticks from the x direction separated by the time interval dt and the space interval dx=vdt.

[latex]dt=d\acute{t}^2=\sqrt{dt^2-\frac{dx^2}{c^2}}=\sqrt{1-(\frac{v}{c})^2}dt[/latex]

so the two inertial coordinate systems are related by the lorentz transformation

[latex]dt=\frac{d\acute{t}}{\sqrt{1-(\frac{v}{c})^2}}=\gamma d\acute{t}[/latex]

 

 

Here is relativity of simultaneaty coordinate transformation in Lorentz.

[latex]\acute{t}=\frac{t-vx/c^2}{\sqrt{1-v^2/c^2}}[/latex]

[latex]\acute{x}=\frac{x-vt}{\sqrt{1-v^2/c^2}}[/latex]

[latex]\acute{y}=y[/latex]

[latex]\acute{z}=z[/latex]

 

This free open source textbook is also good to get started.

 

 

http://www.lightandmatter.com/sr/

 

It will help fill in the blanks in the above

Edited October 24, 2016 by Mordred

[Tim88]

The formula's for gravitational time dilation are given in Wikipedia, in the first section of https://en.wikipedia.org/wiki/Gravitational_time_dilation

 

For experiments on Earth, the last, simple equation in that section is usually good enough:

 

time dilation factor = 1 + gh /c²

[Sriman Dutta]

Thanks for that.

 

Now I give here a case to consider.

 

Suppose there is an observer who is travelling at a very high velocity v, v = 80% of c. He finds another light ray travelling parallel to him. According to special relativity he must find the ray travelling at c, as opposed to the classical view that he would see the ray travelling at the velocity of c-v. Now can anybody prove that he will find the light ray travelling at c and not at c-v ?

[swansont]

Thanks for that.

 

Now I give here a case to consider.

 

Suppose there is an observer who is travelling at a very high velocity v, v = 80% of c. He finds another light ray travelling parallel to him. According to special relativity he must find the ray travelling at c, as opposed to the classical view that he would see the ray travelling at the velocity of c-v. Now can anybody prove that he will find the light ray travelling at c and not at c-v ?

 

 

 

The light will still be a wave, and satisfy the wave equation that you get from Maxwell's equations. That would not happen if the wave weren't traveling at c. Since EM communication still works when there is relative motion, we can conclude that the light is moving at c.

[Sriman Dutta]

 

 

 

The light will still be a wave, and satisfy the wave equation that you get from Maxwell's equations. That would not happen if the wave weren't traveling at c. Since EM communication still works when there is relative motion, we can conclude that the light is moving at c.

Swansont can you please explain it deeply (showing the equation and how it fits right) ?

Sorry for troubling you.

[swansont]

Swansont can you please explain it deeply (showing the equation and how it fits right) ?

Sorry for troubling you.

 

 

 

https://en.wikipedia.org/wiki/Electromagnetic_wave_equation

 

In the wave equation, the speed of propagation is always c. Regardless of relative motion between source and receiver. We still observe waves when there is such motion, thus, one can conclude that the wave equation is valid.

[Strange]

 

That implies rather that you did not fully read my post, as I clarified redshift in agreement with Sriman. And perhaps you missed my additional precision of ceteris paribus: motion isn't part of the question.

 

I did read it. And I have just read it again a couple of times. I find it confusing and contradictory. But as it has satisfied the OP, I'll leave it at that.

[Tim88]

 

I did read it. And I have just read it again a couple of times. I find it confusing and contradictory. But as it has satisfied the OP, I'll leave it at that.

I'm sorry if you found it confusing, even in combination with the elaborations by Okun. But I remember that many years ago I was puzzled by it too, due to the conflicting explanations. The OP was satisfied because I agreed with his logical deduction and he had nothing to "unlearn".

[DimaMazin]

Thanks for that.

 

Now I give here a case to consider.

 

Suppose there is an observer who is travelling at a very high velocity v, v = 80% of c. He finds another light ray travelling parallel to him. According to special relativity he must find the ray travelling at c, as opposed to the classical view that he would see the ray travelling at the velocity of c-v. Now can anybody prove that he will find the light ray travelling at c and not at c-v ?

v_rel=v₁v₂ / (1+v₁v₂/c²)

v₁=c

v₂=0.8c

v_rel=(c-0.8c)/(1- c _* 0.8c/c²)=0.2c/0.2=c

Edited October 24, 2016 by DimaMazin

[Strange]

I'm sorry if you found it confusing, even in combination with the elaborations by Okun. But I remember that many years ago I was puzzled by it too, due to the conflicting explanations. The OP was satisfied because I agreed with his logical deduction and he had nothing to "unlearn".

 

I don't think it is material I find confusing but your presentation. For example you say:

• When this radio signal of 1000 cycles is sent upward, 1000 cycles arrive at a detector with a faster resonating quartz crystal. The measured frequency is therefore reduced
• An emitted light ray ("photons") will therefore be released at lower frequency, and if it propagates upwards it will be detected at a lower frequency than the emitted frequencies of atoms at that height.

Followed by:

• If for 24h a certain number of cycles is sent upward, that number must also be received in the same time interval.

If the measured frequency is reduced (as you say) then the same number of cycles will not be measured in 24 hours.

 

Unless, of course, you mean that the number transmitted in the receiving clock's 24 hour period must be the same as the number received in the receiving clock's 24 hour period. But as I can't imagine any reason why you would say that (as it is trivially true and therefore meaningless) I assume it is not what you meant.

[Tim88]

Thanks for that.

 

Now I give here a case to consider.

 

Suppose there is an observer who is travelling at a very high velocity v, v = 80% of c. He finds another light ray travelling parallel to him. According to special relativity he must find the ray travelling at c, as opposed to the classical view that he would see the ray travelling at the velocity of c-v. Now can anybody prove that he will find the light ray travelling at c and not at c-v ?

 

First of all, it's not exactly so. If he thinks that he is not traveling at 0.8c, but in rest, then he should find c.

How that is calculated is shown in the post just above by DimaZin.

But if as you say, he thinks that he is traveling at 0.8c, and a light ray is according to him "traveling parallel to him", that means that he assumes that he is not in rest - he actually thinks that he is traveling at 0.8c (relative to the ground?). Right?

 

Then he should make relativistic corrections for his speed. Next he should find that the ray is traveling at the velocity c-v "relative" to him (but note that not everyone uses the same definition of "relative velocity"). Effectively he puts himself in the shoes of a "third party observer". The third party observer method is used in the GPS system: the radio signals that GPS satellites exchange travel at c-v relative to the satellites, which themselves travel at v.

Edited October 24, 2016 by Tim88

[Sriman Dutta]

DimaMazin from where do you get v1 and v2 ? Can you explain the context?

[DimaMazin]

DimaMazin from where do you get v1 and v2 ? Can you explain the context?

v₁ and v₂ are any velosities.

[Tim88]

 

I don't think it is material I find confusing but your presentation. For example you say:

• When this radio signal of 1000 cycles is sent upward, 1000 cycles arrive at a detector with a faster resonating quartz crystal. The measured frequency is therefore reduced
• An emitted light ray ("photons") will therefore be released at lower frequency, and if it propagates upwards it will be detected at a lower frequency than the emitted frequencies of atoms at that height.

Followed by:

• If for 24h a certain number of cycles is sent upward, that number must also be received in the same time interval.

If the measured frequency is reduced (as you say) then the same number of cycles will not be measured in 24 hours.

[..]

 

Thanks - sorry for that, indeed it may be ambiguous out of context.

 

The context: "Essential requirement [for consistency]: we must do the analysis with a single reference system."

I did not mix reference systems without specification in a single sentence (in fact, I try to never do that). In more elaborated phrasing, including the things that I had stated separately:

 

If for 24h, as measured with a system on the ground (which is assumed to be in rest), a certain number of cycles is sent upward to a detector (which is also in rest), then - according to the ground system - that number of cycles must also be received by the detector in the same time interval. Else cycles would go missing "in mid air"!

 

That excludes a change of light frequency "in transit".

 

I hope that now it's clearer.

Edited October 24, 2016 by Tim88