Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: james fairclear on 05/04/2020 12:09:41
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Is red shifted light travelling at a speed less than c?
THANK YOU TO EVERYONE FOR ALL YOUR FEEDBACK ON THIS PROPOSITION. I HAVE NOW CONCLUDED THAT MY THINKING WAS FLAWED. ::)
Light emitted from a light source moving away from an observer at a speed v would intuitively be expected to be travelling at a speed (c – v) but is in fact still measured to be moving at a speed of c. The measurement of speed is based on the time interval between the light being emitted and the light being initially detected at the destination.
The difference between light detected from a stationary source and light detected from a receding source is that the latter is red shifted which means that it is less energetic. But what does that really mean?
Measured over a period of T seconds the light received is quantitatively and qualitatively different from the light that has been transmitted over a period of T seconds.
We can characterise light as a continuous waveform whereby there will be less peaks of the waveform detected per second by a measuring device at the destination than the number of peaks per second detected by a measuring device at the receding light source that is co-moving with the source.
Quantitatively there is less energy per second arriving at the destination from a receding light source than light from a relatively stationary light source.
Energy emitted per second = E
Energy received per second = (E – e)
Eventually the total quantity of energy emitted in T seconds will arrive at the destination but with a portion of that energy (e) delayed by t seconds.
Energy emitted in T seconds = ET
Energy received in T seconds = (E – e) x (T)
Energy received in T + t seconds = ET
Energy from the emitted light will start to arrive at the destination at a point in time T1 commensurate with a speed of c from the receding source. However a quantity of Energy E emitted cannot accurately be said to have arrived at the destination until the same quantity of energy E has been absorbed at the destination at time T2 resulting in (by this definition) an effective speed of light less than c.
As an analogy a locomotive leaves station A and collects one mile of carriages in front of it on its way to station B. The first carriage being pushed by the locomotive may arrive at a station B at 09:00 but the locomotive doesn’t arrive until 09:03.
In conclusion red shifted light from a receding light source can be characterised in terms of the respective rates of energy transmitted and received as travelling at a speed less than c.
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Is red shifted light travelling at a speed less than c?
In a vacuum, no. Still c. So I can reflect a beam of light with say a receding mirror and the light comes back to me at c, but lower energy (red-shifted).
Light emitted from a light source moving away from an observer at a speed v would intuitively be expected to be travelling at a speed c – v
This isn't even true of waves in a medium like sound or water waves. You're correct that in fact it is still measured at c.
One can visualise it as follows:
A Quanta of light (Photon) is released from moving light source. The next quanta (Photon) is released at a distance d from the first. Thus a relatively stationary observer will observe a greater distance between each quanta than an observer in the same inertial frame of reference as the moving light source; this is manifested as an increase in wavelength or decrease in frequency.
This makes it sound like the frequency of light is the rate at which the quanta arrive. Not so. Each photon has a frame dependent frequency, and a light source emitting photons at 10x the rate of another is just brighter, not shifted to a different frequency.
If the wave from a stationary light source has a length L then the wave from a moving light source has a length L + n.
If we consider that the full energy of the photon only arrives at the crest of the wave then the amount of energy arriving per second from the moving light source is less than that from the stationary light source. It takes longer for a FULL quanta of light to reach a point A where the light source is moving in a direction away from A than light from a relatively stationary source.
This is a way of looking at it, yes. It works when you do the moving mirror thing I mentioned, but the bit about partial-quanta of light makes no sense. It wouldn't be quanta if it could be emitted and detected over a space of time.
Although energy from each quanta of light will arrive in a continuous stream as its waveform unfolds it cannot accurately be said to have arrived until the whole packet of energy has been absorbed at the destination point. As an analogy a locomotive leaves station A and collects one mile of carriages in front of it on its way to station B. The first carriage being pushed by the locomotive may arrive at a station B at 09:00 but the locomotive doesn’t arrive until 09:03.
Photons are detected as a single event, not some spread out stream. They behave like particles when absorbed. The train analogy doesn't really work here.
The speed of each quanta should be more accurately calculated as distance/time where time is the interval between the FULL quanta being discharged at source and the FULL quanta being fully absorbed at the destination. As its wavelength increases there can be a considerable interval between the arrival of the front of the wave and the back of the wave.
Umm... no.
In conclusion red shifted light from a receding light source can be measured in terms of the quantity of energy transmitted and received per second as travelling at a speed less than c.
Yes except for the speed less than c bit. Yes, it is slowed by lack of a vacuum between source and detection, but there is no slowing because of spread-out emission and detection.
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Energy from the emitted light will start to arrive at the destination at a point in time T1 commensurate with a speed of c from the receding source. However a quantity of Energy E emitted cannot accurately be said to have arrived at the destination until the same quantity of energy E has been absorbed at the destination at time T2 resulting in (by this definition) an effective speed of light less than c.
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The train analogy doesn't really work here.
Depends on how you look at it. The loco arrives after the first carriage, but it has travelled at the same speed. Isn't that a simple answer to the original question?
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Problem with the "stretched wave" model is that light from astronomically distant sources arrives pretty much as individual photons, not as a continuous wave. We measure their energy and find it has shifted. The Pound-Rebka experiment measured the gravitational energy shift of single photons by comparing it with Doppler shift.
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Is red shifted light travelling at a speed less than c?
In a vacuum, no. Still c. So I can reflect a beam of light with say a receding mirror and the light comes back to me at c, but lower energy (red-shifted).
Light emitted from a light source moving away from an observer at a speed v would intuitively be expected to be travelling at a speed c – v
This isn't even true of waves in a medium like sound or water waves. You're correct that in fact it is still measured at c.
One can visualise it as follows:
A Quanta of light (Photon) is released from moving light source. The next quanta (Photon) is released at a distance d from the first. Thus a relatively stationary observer will observe a greater distance between each quanta than an observer in the same inertial frame of reference as the moving light source; this is manifested as an increase in wavelength or decrease in frequency.
This makes it sound like the frequency of light is the rate at which the quanta arrive. Not so. Each photon has a frame dependent frequency, and a light source emitting photons at 10x the rate of another is just brighter, not shifted to a different frequency.
If the wave from a stationary light source has a length L then the wave from a moving light source has a length L + n.
If we consider that the full energy of the photon only arrives at the crest of the wave then the amount of energy arriving per second from the moving light source is less than that from the stationary light source. It takes longer for a FULL quanta of light to reach a point A where the light source is moving in a direction away from A than light from a relatively stationary source.
This is a way of looking at it, yes. It works when you do the moving mirror thing I mentioned, but the bit about partial-quanta of light makes no sense. It wouldn't be quanta if it could be emitted and detected over a space of time.
Although energy from each quanta of light will arrive in a continuous stream as its waveform unfolds it cannot accurately be said to have arrived until the whole packet of energy has been absorbed at the destination point. As an analogy a locomotive leaves station A and collects one mile of carriages in front of it on its way to station B. The first carriage being pushed by the locomotive may arrive at a station B at 09:00 but the locomotive doesn’t arrive until 09:03.
Photons are detected as a single event, not some spread out stream. They behave like particles when absorbed. The train analogy doesn't really work here.
The speed of each quanta should be more accurately calculated as distance/time where time is the interval between the FULL quanta being discharged at source and the FULL quanta being fully absorbed at the destination. As its wavelength increases there can be a considerable interval between the arrival of the front of the wave and the back of the wave.
Umm... no.
In conclusion red shifted light from a receding light source can be measured in terms of the quantity of energy transmitted and received per second as travelling at a speed less than c.
Yes except for the speed less than c bit. Yes, it is slowed by lack of a vacuum between source and detection, but there is no slowing because of spread-out emission and detection.
Is red shifted light travelling at a speed less than c?
In a vacuum, no. Still c. So I can reflect a beam of light with say a receding mirror and the light comes back to me at c, but lower energy (red-shifted).
Light emitted from a light source moving away from an observer at a speed v would intuitively be expected to be travelling at a speed c – v
This isn't even true of waves in a medium like sound or water waves. You're correct that in fact it is still measured at c.
One can visualise it as follows:
A Quanta of light (Photon) is released from moving light source. The next quanta (Photon) is released at a distance d from the first. Thus a relatively stationary observer will observe a greater distance between each quanta than an observer in the same inertial frame of reference as the moving light source; this is manifested as an increase in wavelength or decrease in frequency.
This makes it sound like the frequency of light is the rate at which the quanta arrive. Not so. Each photon has a frame dependent frequency, and a light source emitting photons at 10x the rate of another is just brighter, not shifted to a different frequency.
If the wave from a stationary light source has a length L then the wave from a moving light source has a length L + n.
If we consider that the full energy of the photon only arrives at the crest of the wave then the amount of energy arriving per second from the moving light source is less than that from the stationary light source. It takes longer for a FULL quanta of light to reach a point A where the light source is moving in a direction away from A than light from a relatively stationary source.
This is a way of looking at it, yes. It works when you do the moving mirror thing I mentioned, but the bit about partial-quanta of light makes no sense. It wouldn't be quanta if it could be emitted and detected over a space of time.
Although energy from each quanta of light will arrive in a continuous stream as its waveform unfolds it cannot accurately be said to have arrived until the whole packet of energy has been absorbed at the destination point. As an analogy a locomotive leaves station A and collects one mile of carriages in front of it on its way to station B. The first carriage being pushed by the locomotive may arrive at a station B at 09:00 but the locomotive doesn’t arrive until 09:03.
Photons are detected as a single event, not some spread out stream. They behave like particles when absorbed. The train analogy doesn't really work here.
The speed of each quanta should be more accurately calculated as distance/time where time is the interval between the FULL quanta being discharged at source and the FULL quanta being fully absorbed at the destination. As its wavelength increases there can be a considerable interval between the arrival of the front of the wave and the back of the wave.
Umm... no.
In conclusion red shifted light from a receding light source can be measured in terms of the quantity of energy transmitted and received per second as travelling at a speed less than c.
Yes except for the speed less than c bit. Yes, it is slowed by lack of a vacuum between source and detection, but there is no slowing because of spread-out emission and detection.
I have now rewritten and hopefully clarified my original post which appears to have been misinterpreted. The point I am trying to make here is that the speed calculation should be based not on the time (T) it takes for a short pulse of light to begin arriving at the destination but on the time (T+t) it takes for the total quantity of energy emitted in that short pulse of light to be absorbed at the destination.
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Same applies. Photon absorption time is independent of red shift.
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Same applies. Photon absorption time is independent of red shift.
I may not be explaining myself very well. I am not focusing on absorption time of individual photons but rather the overall rate that energy arrives at the destination (joules per second) compared with the rate energy was emitted at the source.
Are we agreed that Red shifted light arrives at the destination with less energy than was emitted by the receding source?
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Photon absorption time is independent of red shift.
No problem with that, but, quickly dinning my nit-picker’s hat, I look for something that might be still unanswered.
If a redshifted photon has reduced energy when it arrives, where is the missing energy that point?
Has it been lost (absorbed?) along the way, or will it arrive later, as OP seems to suggest (T+t)?
What role did the increasing distance between source and target play?
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Sorry, James; we seem to have crossed over, there.
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Problem with the "stretched wave" model is that light from astronomically distant sources arrives pretty much as individual photons, not as a continuous wave. We measure their energy and find it has shifted. The Pound-Rebka experiment measured the gravitational energy shift of single photons by comparing it with Doppler shift.
In my understanding light arriving from any distance can be characterised as a wave or as quantum particles. In any event though it doesn't alter my point which I have now updated to be hopefully clearer!
On the basis that red shifted light is less energetic than the light that was originally emitted from the receding light source it will take longer for a given quantity of energy E to arrive at the destination than it took to be emitted.
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Photon absorption time is independent of red shift.
No problem with that, but, quickly dinning my nit-picker’s hat, I look for something that might be still unanswered.
If a redshifted photon has reduced energy when it arrives, where is the missing energy that point?
Has it been lost (absorbed?) along the way, or will it arrive later, as OP seems to suggest (T+t)?
What role did the increasing distance between source and target play?
yes it will arrive later (T+t). Hence by defining speed in terms of the total elapsed time involved in transmitting a quantity of energy E and absorbing the same quantity of energy E at the destination we arrive at a speed < c.
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Just a thought, could be very wide of the mark.
All light = energy, but not all energy = light. Therefore, the only factor in this scenario that is relevant to the speed of light is T; t relates to energy that is not light.
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Just a thought, could be very wide of the mark.
All light = energy, but not all energy = light. Therefore, the only factor in this scenario that is relevant to the speed of light is T; t relates to energy that is not light.
The key factor is the rate of Light Energy being emitted from the receding light source which will be greater than the rate of Light Energy arriving at the destination. This we know from the fact that red shifted light is less energetic than the non red shifted light that was emitted.
Hence by defining speed in terms of the total elapsed time involved in transmitting a quantity of energy E and absorbing the same quantity of energy E at the destination we arrive at a speed < c.
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Photon absorption time is independent of red shift.
No problem with that, but, quickly dinning my nit-picker’s hat, I look for something that might be still unanswered.
If a redshifted photon has reduced energy when it arrives, where is the missing energy that point?
Has it been lost (absorbed?) along the way, or will it arrive later, as OP seems to suggest (T+t)?
What role did the increasing distance between source and target play?
There is no "missing energy". There is just the energy of the light as measured in the frame of the source, and the energy as measured in the frame of the receiver. Energy conservation only works when working in a single frame of reference and can't be applied between frames.
Example, I am in the back of a truck moving at 10 m/sec away from you, and toss a 0.1kg ball back towards you at 15 m/sec as measured relative to myself. The ball has a KE of 0.1kg(15m/sec)^2/2 = 11.25 joules as measured from my frame of reference. But the same ball is only moving at 5 m/sec relative to you and has 1.25 joules of KE as measured from your frame. The ball hasn't "lost" 10 joules of energy on it's way to you, we just measure a different KE for the ball relative to ourselves.
With light there is no difference in velocity in the light as measured by source and receiver, But, just like with the ball, there is a difference in the energy measured by the two frames. With light, this is exhibited by a diffrence in measured frequency.
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Just a thought, could be very wide of the mark.
All light = energy, but not all energy = light. Therefore, the only factor in this scenario that is relevant to the speed of light is T; t relates to energy that is not light.
The key factor is the rate of Light Energy being emitted from the receding light source which will be greater than the rate of Light Energy arriving at the destination. This we know from the fact that red shifted light is less energetic than the non red shifted light that was emitted.
Hence by defining speed in terms of the total elapsed time involved in transmitting a quantity of energy E and absorbing the same quantity of energy E at the destination we arrive at a speed < c.
No. because you get different predicted values for red shift if you assume a constant value of c for light speed than you do if you assume a changing light speed. And every real life experiment ever done gives values of red shift that agree with the speed of light in a vacuum always being c.
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Photon absorption time is independent of red shift.
No problem with that, but, quickly dinning my nit-picker’s hat, I look for something that might be still unanswered.
If a redshifted photon has reduced energy when it arrives, where is the missing energy that point?
Has it been lost (absorbed?) along the way, or will it arrive later, as OP seems to suggest (T+t)?
What role did the increasing distance between source and target play?
There is no "missing energy". There is just the energy of the light as measured in the frame of the source, and the energy as measured in the frame of the receiver. Energy conservation only works when working in a single frame of reference and can't be applied between frames.
Example, I am in the back of a truck moving at 10 m/sec away from you, and toss a 0.1kg ball back towards you at 15 m/sec as measured relative to myself. The ball has a KE of 0.1kg(15m/sec)^2/2 = 11.25 joules as measured from my frame of reference. But the same ball is only moving at 5 m/sec relative to you and has 1.25 joules of KE as measured from your frame. The ball hasn't "lost" 10 joules of energy on it's way to you, we just measure a different KE for the ball relative to ourselves.
With light there is no difference in velocity in the light as measured by source and receiver, But, just like with the ball, there is a difference in the energy measured by the two frames. With light, this is exhibited by a diffrence in measured frequency.
As you state with light there is no difference in velocity in the light as measured by source and receiver if you measure velocity traditionally in terms of the time it takes for the light to begin arriving at the destination.
However if you measure velocity in terms of the time it takes for a given quantity of energy emitted to arrive at the destination then as the light arriving is less energetic than the light emitted there will be a time lag which leads to a measured velocity < c .
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However if you measure velocity in terms of the time it takes for a given quantity of energy emitted to arrive at the destination
But we don't.
The dimensions of velocity are LT-1
Time per unit energy is M-1L-2T3
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However if you measure velocity in terms of the time it takes for a given quantity of energy emitted to arrive at the destination
But we don't.
The dimensions of velocity are LT-1
Time per unit energy is M-1L-2T3
Well yes you are stating the obvious here and I acknowledge this in my post!
The whole essence of my point is that this would be a more meaningful approach to determining the speed of red shifted light.
Imagine a 100 metre race where a competitor leaves the start line with an average chest to back measurement of 40cm which then increases to 40 metres by the time the front of his chest trips the photoelectric cell at a measured time of 11 elapsed seconds. If his back crosses the line 5 seconds later than his chest it raises considerable doubt as to whether he has run the race in 11 seconds or 16 seconds.
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Except that a photon arrives all at once, regardless of its energy.
A lot of people have a problem with duality because they confuse the mathematical model of the universe with what actually happens. It turns out that we need two different models to predict our observations, but that doesn't mean that a photon or electron "is" a wave or a particle depending on who is looking at it or when.
The principal characters in "The Big Bang Theory" are very well drawn. Theoretical physics is vanity, experimental physics is humility. And cheesecake is delicious - but that's for another thread.
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Except that a photon arrives all at once, regardless of its energy.
A lot of people have a problem with duality because they confuse the mathematical model of the universe with what actually happens. It turns out that we need two different models to predict our observations, but that doesn't mean that a photon or electron "is" a wave or a particle depending on who is looking at it or when.
The principal characters in "The Big Bang Theory" are very well drawn. Theoretical physics is vanity, experimental physics is humility. And cheesecake is delicious - but that's for another thread.
I don't think you are clear on the point I am making.
The rate of light energy arriving at a destination is something that is measured and the measurement is not altered by any consideration of whether light is a wave or a particle.
I don't see the relevance of Big Bang theory to this topic.
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The rate of arrival of photon energy is called illuminance (if it's visible light) signal intensity (radio waves) or dose rate (ionising radiation). Clearly if the photons are of lower energy, than a given photon flux (number of photons hitting the target per second) will deliver less energy per second, but the photon interaction time is not dependent on photon energy, which is why the "stretched wave" model isn't appropriate.
"The Big Bang Theory" is the best TV comedy series ever, nothing to do with the creation of the observable universe (except as a minor sub plot).
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The rate of arrival of photon energy is called illuminance (if it's visible light) signal intensity (radio waves) or dose rate (ionising radiation). Clearly if the photons are of lower energy, than a given photon flux (number of photons hitting the target per second) will deliver less energy per second, but the photon interaction time is not dependent on photon energy, which is why the "stretched wave" model isn't appropriate.
"The Big Bang Theory" is the best TV comedy series ever, nothing to do with the creation of the observable universe (except as a minor sub plot).
Sorry your Big Bang Theory reference went right over my head!!
My proposition rests on the fact that less energy arrives per second at the destination regardless of whether light is treated as a wave or a particle . Thus I can't see the relevance of Photon interaction time or the "stretched wave" model in this context.
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The stretched wave model is what you described. It implies that some of the energy of a photon arrives before the rest, so that the rate of arrival of energy is reduced by the amount of stretch. It simply doesn't happen.
The fact is that received photon energy is reduced by doppler red shift. Neither wave nor particle model is necessary to calculate this - it's a consequence of c being constant, which is an experimental fact.
If I've opened your eyes to The Big Bang Theory, you are in for a treat. There are lots of repeats on E4 in the UK, and there is no better investment in DVDs if you are anywhere else.
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James, Alan is a professional physicist with real qualifications. He has worked in the field FOREVER! What are your qualifications?
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The stretched wave model is what you described. It implies that some of the energy of a photon arrives before the rest, so that the rate of arrival of energy is reduced by the amount of stretch. It simply doesn't happen.
The fact is that received photon energy is reduced by doppler red shift. Neither wave nor particle model is necessary to calculate this - it's a consequence of c being constant, which is an experimental fact.
If I've opened your eyes to The Big Bang Theory, you are in for a treat. There are lots of repeats on E4 in the UK, and there is no better investment in DVDs if you are anywhere else.
Yes I know and fully agree and have already acknowledged that as you correctly state "received photon energy is reduced by doppler red shift. Neither wave nor particle model is necessary to calculate this - it's a consequence of c being constant, which is an experimental fact."
It appears that I still have not made my proposition clear ::)
I will have another go:
The inarguable fact relevant to my proposition is that Red shifted light from a receding light source is less energetic than the light that was emitted.
The speed of light is measured by switching on a light emitter and recording the elapsed time between switching on the emitter and the first photon being received by a detector without taking into account the rates of energy being emitted and received respectively. The speed measured is c.
If the light emitter is receding from the light detector the speed measured is still c. However in this case the light will be less energetic (Red shifted). So although the speed is still measured by this method as being c there is a discrepancy between what is emitted and what is received. In order to account for this discrepancy we can propose an alternative method of measuring the speed whereby we measure the elapsed time for a given quantity of light energy that has been emitted from the source to be received at the detector.
By adopting this alternative method of measuring the speed of a light beam from a receding light in terms of the rates of energy transfer at source and destination respectively we arrive at a speed that is less than c.
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The inarguable fact relevant to my proposition is that Red shifted light from a receding light source is less energetic than the light that was emitted.
This is not inarguable and I'm saying it's wrong. If 20 joules of light energy reaches Earth in Earth frame, in say an hour, then 20 joules of energy left that star, in Earth frame. Any other answer violates energy conservation.
Those particular 20 joules did admittedly take less than an hour to be emitted since the initial light reaching Earth had less distance to travel than did the final light from that segment. So it may have taken say 50 minutes to be emitted, resulting in more power leaving the star than the power received on Earth. But you said energy above, not power.
The speed of light is measured by switching on a light emitter and recording the elapsed time between switching on the emitter and the first photon being received by a detector without taking into account the rates of energy being emitted and received respectively. The speed measured is c.
While incomplete but not completely wrong, I am unaware of this impractical method being used to measure light speed. There are far simpler ways to go about it.
You talk about measuring the first photon, but measuring the last one will work just as well since it goes no slower. Your text implies otherwise.
You then go on to define speed based on a rate of energy transfer, which has been shown by others above to be something other than speed.
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It is known that the maximum possible speed of movement of material objects or the propagation of any signals is the speed of light in a vacuum. It is denoted by the letter c and is almost 300 thousand kilometers per second; exact value c = 299 792 458 m / s. The speed of light in vacuum is one of the fundamental physical constants. The impossibility of achieving speeds exceeding c follows from Einstein's special theory of relativity (STR). If it were possible to prove that the transmission of signals with superluminal speed is possible, the theory of relativity would fall. So far this has not happened, despite numerous attempts to refute the ban on the existence of speeds greater than s. However, in recent experimental studies, some very interesting phenomena have been discovered that indicate that under specially created conditions superluminal speeds can be observed and the principles of the theory of relativity are not violated.
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It is known that the maximum possible speed of movement of material objects or the propagation of any signals is the speed of light in a vacuum. It is denoted by the letter c and is almost 300 thousand kilometers per second; exact value c = 299 792 458 m / s. The speed of light in vacuum is one of the fundamental physical constants. The impossibility of achieving speeds exceeding c follows from Einstein's special theory of relativity (STR). If it were possible to prove that the transmission of signals with superluminal speed is possible, the theory of relativity would fall. So far this has not happened, despite numerous attempts to refute the ban on the existence of speeds greater than s. However, in recent experimental studies, some very interesting phenomena have been discovered that indicate that under specially created conditions superluminal speeds can be observed and the principles of the theory of relativity are not violated.
That is interesting can you give a link to the experiment please.
My proposition is based upon the well documented observation that the rate of transfer of light energy differs between source and destination where either is receding from the other. In these cases I am proposing that the speed of light can be alternatively measured to provide a more meaningful result by modifying c with respect to this differential factor.
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The inarguable fact relevant to my proposition is that Red shifted light from a receding light source is less energetic than the light that was emitted.
This is not inarguable and I'm saying it's wrong. If 20 joules of light energy reaches Earth in Earth frame, in say an hour, then 20 joules of energy left that star, in Earth frame. Any other answer violates energy conservation.
Those particular 20 joules did admittedly take less than an hour to be emitted since the initial light reaching Earth had less distance to travel than did the final light from that segment. So it may have taken say 50 minutes to be emitted, resulting in more power leaving the star than the power received on Earth. But you said energy above, not power.
The speed of light is measured by switching on a light emitter and recording the elapsed time between switching on the emitter and the first photon being received by a detector without taking into account the rates of energy being emitted and received respectively. The speed measured is c.
While incomplete but not completely wrong, I am unaware of this impractical method being used to measure light speed. There are far simpler ways to go about it.
You talk about measuring the first photon, but measuring the last one will work just as well since it goes no slower. Your text implies otherwise.
You then go on to define speed based on a rate of energy transfer, which has been shown by others above to be something other than speed.
You state " So it may have taken say 50 minutes to be emitted, resulting in more power leaving the star than the power received on Earth. But you said energy above, not power"
I have repeatedly stated the rate of transfer of energy which is the same thing as power.
You state : "You then go on to define speed based on a rate of energy transfer, which has been shown by others above to be something other than speed.".
My proposition is an alternative approach to measuring light speed more accurately by making adjustments for discrepancies in rates of energy transfer between source and destination. Whilst it is open to interpretation it has not yet been shown by anyone to be conceptually wrong or to be something other than speed.
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an alternative method of measuring the speed whereby we measure the elapsed time for a given quantity of light energy that has been emitted from the source to be received at the detector.
That is the inverse of illuminance, signal intensity or dose rate. I've spent may happy years measuring all those things. In the last case, we did indeed measure elapsed time for a given amount of energy as being the simplest and most accurate method available for x-ray photons: I built the UK national primary standard on that principle.
It is not a measure of speed.
I haven't repeated the Pound-Rebka experiment but those who have, report that it gives the expected result if and only if c is constant.
You can derive c from Maxwell's equations and note that they do not invoke any notion of the source being stationary with respect to the receiver. In fact they don't mention the receiver at all.
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an alternative method of measuring the speed whereby we measure the elapsed time for a given quantity of light energy that has been emitted from the source to be received at the detector.
That is the inverse of illuminance, signal intensity or dose rate. I've spent may happy years measuring all those things. In the last case, we did indeed measure elapsed time for a given amount of energy as being the simplest and most accurate method available for x-ray photons: I built the UK national primary standard on that principle.
It is not a measure of speed.
I haven't repeated the Pound-Rebka experiment but those who have, report that it gives the expected result if and only if c is constant.
You can derive c from Maxwell's equations and note that they do not invoke any notion of the source being stationary with respect to the receiver. In fact they don't mention the receiver at all.
You state that what you measured is "not a measure of speed". I would say that this is open to interpretation.
My proposition is that in the special case of relative motion between source and destination the speed of a light beam measured to be c should be corrected to account for the discrepancy between the rates of energy transfer between source and destination.
Light received from a receding light source is less energetic than the light that was emitted. It is qualitatively and quantitatively different from the light that was emitted. The rate of light energy received is less than the rate of light energy emitted.
Quantitatively there is less energy per second arriving at the destination from a receding light source than light from a relatively stationary light source.
Energy emitted per second = E
Energy received per second = (E – e)
Eventually the total quantity of energy emitted in T seconds will arrive at the destination but with a portion of that energy (e) delayed by t seconds.
Energy emitted in T seconds = ET
Energy received in T seconds = (E – e) x (T)
Energy received in T + t seconds = ET
Energy from the emitted light will start to arrive at the destination at a point in time T1 commensurate with a measured speed of c from the receding source. However a quantity of Energy E emitted cannot accurately be said to have arrived at the destination until the same quantity of energy E has been absorbed at the destination at time T2 resulting in (by this definition) an effective measured speed of light less than c.
As an analogy we can consider a 100 metre race where a competitor leaves the start line with an average chest to back measurement of 40cm which then increases to 40 metres by the time the front of his chest trips the photoelectric cell at a measured time of 11 elapsed seconds. If his back crosses the line 5 seconds later than his chest it raises considerable doubt as to whether he has run the race in 11 seconds or 16 seconds or some averaged time (say 13.5 seconds) between the 2 or indeed if he is really the same competitor that started the race.
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You cannot use Galilean relativity with photons, which is effectively what you are trying to do. With light, kinetic energy does not have a variable speed component (in vacuum). Therefore the wavelengyh has to vary. As per Maxwell. If you don't understand this then you need to start listening rather than preaching.
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Imagine a 100 metre race where a competitor leaves the start line with an average chest to back measurement of 40cm which then increases to 40 metres by the time the front of his chest trips the photoelectric cell at a measured time of 11 elapsed seconds. If his back crosses the line 5 seconds later than his chest it raises considerable doubt as to whether he has run the race in 11 seconds or 16 seconds.
Using that analogy, what you appear to be doing is starting the runner with just one heel behind the start line and stopping the clock as soon as any of their body reaches the line, except you're doing it the other round with a receding object to get a longer time.
It looks like you're starting the clock when the first part of a light wave is emitted from a receding source and stopping the clock when the last part of that wave reaches you. If you want to treat light as an extended object then you need to stop the clock when the first part of the wave reaches you.
According to that argument you could have a 1km long train with its rear slightly over the start line and claim it traveled 1km in less than a second.
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You state that what you measured is "not a measure of speed". I would say that this is open to interpretation.
Speed is distance divided by time. Energy received per unit time is power, not speed. Scientific words have very precise meanings, and they are all different. If you start mixing up dimensions you will confuse yourself and not be taken seriously by others.
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Light received from a receding light source is less energetic than the light that was emitted. It is qualitatively and quantitatively different from the light that was emitted. The rate of light energy received is less than the rate of light energy emitted.Quantitatively there is less energy per second arriving at the destination from a receding light source than light from a relatively stationary light source. Energy emitted per second = EEnergy received per second = (E – e)
So far, so good,. apart from "qualitative". A photon is a photon.
Eventually the total quantity of energy emitted in T seconds will arrive at the destination but with a portion of that energy (e) delayed by t seconds.
And there's the fatal non sequitur. Consider a single photon. It can only travel at c (Maxwell), but the energy it transfers between bodies depends on their relative velocity (Doppler) and gravitational potential (relativity). All as confirmed by umpteen experiments.
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You state " So it may have taken say 50 minutes to be emitted, resulting in more power leaving the star than the power received on Earth. But you said energy above, not power"
OK then, I agree with that. A receding laser pointed at some target puts out more power than is received by the target, but not more energy.
Your problem is somehow just taking that statement and concluding that speed of light somehow changes at some point in the process. You've not justified this assertion.
My proposition is an alternative approach to measuring light speed more accurately by making adjustments for discrepancies in rates of energy transfer between source and destination.
That's not more accurate at all, since speed of light is not related to rates of energy transfer. The rate of energy transfer observed can only be computed by assuming the speed of light is constant the whole time. Any other speed of the laser photons results in energy transfer rates different from those observed.
Whilst it is open to interpretation it has not yet been shown by anyone to be conceptually wrong or to be something other than speed.
It is not open to interpretation, and we're showing how your assertions are wrong, and you're ignoring what we're saying. You've not countered any corrections except to ignore them and re-assert your speed-as-energy-rate thing.
Eventually the total quantity of energy emitted in T seconds will arrive at the destination but with a portion of that energy (e) delayed by t seconds.
And there's the fatal non sequitur.
I actually agree with this. Consider a simple example of a laser pointer at zero distance, receding at .6c. Everything is expressed in the frame of the observer.
It starts emitting at location zero, and is thus observed immediately at time zero.
It turns off 10 seconds later, and since it is moving at .6c, it has moved 6 light seconds during that time. So the last of it, emitted at time 10, arrives at the observer at time 16 seconds.
That means a 10 second burn of our laser pointer was observed for 16 seconds, and yet not one bit of it ever traveled at any speed other than c. It is the conclusion that some of the light moves at some other speed that is the fatal non-sequitur.
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So if the source emits one photon, what portion if its energy arrives t seconds behind it?
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Is red shifted light travelling at a speed less than c?
In a vacuum, no. Still c. So I can reflect a beam of light with say a receding mirror and the light comes back to me at c, but lower energy (red-shifted).
Light emitted from a light source moving away from an observer at a speed v would intuitively be expected to be travelling at a speed c – v
This isn't even true of waves in a medium like sound or water waves. You're correct that in fact it is still measured at c.
One can visualise it as follows:
A Quanta of light (Photon) is released from moving light source. The next quanta (Photon) is released at a distance d from the first. Thus a relatively stationary observer will observe a greater distance between each quanta than an observer in the same inertial frame of reference as the moving light source; this is manifested as an increase in wavelength or decrease in frequency.
This makes it sound like the frequency of light is the rate at which the quanta arrive. Not so. Each photon has a frame dependent frequency, and a light source emitting photons at 10x the rate of another is just brighter, not shifted to a different frequency.
If the wave from a stationary light source has a length L then the wave from a moving light source has a length L + n.
If we consider that the full energy of the photon only arrives at the crest of the wave then the amount of energy arriving per second from the moving light source is less than that from the stationary light source. It takes longer for a FULL quanta of light to reach a point A where the light source is moving in a direction away from A than light from a relatively stationary source.
This is a way of looking at it, yes. It works when you do the moving mirror thing I mentioned, but the bit about partial-quanta of light makes no sense. It wouldn't be quanta if it could be emitted and detected over a space of time.
Although energy from each quanta of light will arrive in a continuous stream as its waveform unfolds it cannot accurately be said to have arrived until the whole packet of energy has been absorbed at the destination point. As an analogy a locomotive leaves station A and collects one mile of carriages in front of it on its way to station B. The first carriage being pushed by the locomotive may arrive at a station B at 09:00 but the locomotive doesn’t arrive until 09:03.
Photons are detected as a single event, not some spread out stream. They behave like particles when absorbed. The train analogy doesn't really work here.
The speed of each quanta should be more accurately calculated as distance/time where time is the interval between the FULL quanta being discharged at source and the FULL quanta being fully absorbed at the destination. As its wavelength increases there can be a considerable interval between the arrival of the front of the wave and the back of the wave.
Umm... no.
In conclusion red shifted light from a receding light source can be measured in terms of the quantity of energy transmitted and received per second as travelling at a speed less than c.
Yes except for the speed less than c bit. Yes, it is slowed by lack of a vacuum between source and detection, but there is no slowing because of spread-out emission and detection.
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So if the source emits one photon, what portion if its energy arrives t seconds behind it?
A photon arrives when it arrives. One photon does not have a meaningful amount of power, but that seems to be what you're asking here.
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The train analogy doesn't really work here.
Depends on how you look at it. The loco arrives after the first carriage, but it has travelled at the same speed. Isn't that a simple answer to the original question?
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Problem with the "stretched wave" model is that light from astronomically distant sources arrives pretty much as individual photons, not as a continuous wave. We measure their energy and find it has shifted. The Pound-Rebka experiment measured the gravitational energy shift of single photons by comparing it with Doppler shift.
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Just a thought, could be very wide of the mark.
All light = energy, but not all energy = light. Therefore, the only factor in this scenario that is relevant to the speed of light is T; t relates to energy that is not light.
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Photon absorption time is independent of red shift.
No problem with that, but, quickly dinning my nit-picker’s hat, I look for something that might be still unanswered.
If a redshifted photon has reduced energy when it arrives, where is the missing energy that point?
Has it been lost (absorbed?) along the way, or will it arrive later, as OP seems to suggest (T+t)?
What role did the increasing distance between source and target play?
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Photon absorption time is independent of red shift.
No problem with that, but, quickly dinning my nit-picker’s hat, I look for something that might be still unanswered.
If a redshifted photon has reduced energy when it arrives, where is the missing energy that point?
Has it been lost (absorbed?) along the way, or will it arrive later, as OP seems to suggest (T+t)?
What role did the increasing distance between source and target play?
There is no "missing energy". There is just the energy of the light as measured in the frame of the source, and the energy as measured in the frame of the receiver. Energy conservation only works when working in a single frame of reference and can't be applied between frames.
Example, I am in the back of a truck moving at 10 m/sec away from you, and toss a 0.1kg ball back towards you at 15 m/sec as measured relative to myself. The ball has a KE of 0.1kg(15m/sec)^2/2 = 11.25 joules as measured from my frame of reference. But the same ball is only moving at 5 m/sec relative to you and has 1.25 joules of KE as measured from your frame. The ball hasn't "lost" 10 joules of energy on it's way to you, we just measure a different KE for the ball relative to ourselves.
With light there is no difference in velocity in the light as measured by source and receiver, But, just like with the ball, there is a difference in the energy measured by the two frames. With light, this is exhibited by a diffrence in measured frequency.
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Except that a photon arrives all at once, regardless of its energy.
A lot of people have a problem with duality because they confuse the mathematical model of the universe with what actually happens. It turns out that we need two different models to predict our observations, but that doesn't mean that a photon or electron "is" a wave or a particle depending on who is looking at it or when.
The principal characters in "The Big Bang Theory" are very well drawn. Theoretical physics is vanity, experimental physics is humility. And cheesecake is delicious - but that's for another thread.
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However if you measure velocity in terms of the time it takes for a given quantity of energy emitted to arrive at the destination
But we don't.
The dimensions of velocity are LT-1
Time per unit energy is M-1L-2T3
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It is known that the maximum possible speed of movement of material objects or the propagation of any signals is the speed of light in a vacuum. It is denoted by the letter c and is almost 300 thousand kilometers per second; exact value c = 299 792 458 m / s. The speed of light in vacuum is one of the fundamental physical constants. The impossibility of achieving speeds exceeding c follows from Einstein's special theory of relativity (STR). If it were possible to prove that the transmission of signals with superluminal speed is possible, the theory of relativity would fall. So far this has not happened, despite numerous attempts to refute the ban on the existence of speeds greater than s. However, in recent experimental studies, some very interesting phenomena have been discovered that indicate that under specially created conditions superluminal speeds can be observed and the principles of the theory of relativity are not violated.
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That's quite some flounce.
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flounce
No quite the opposite of a flounce, I am stating this seriously as my thinking has indeed been flawed.