1
Physics, Astronomy & Cosmology / Re: Can light travel at a speed less than c?
« on: 10/04/2020 14:18:15 »flounceNo quite the opposite of a flounce, I am stating this seriously as my thinking has indeed been flawed.
This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.
flounceNo quite the opposite of a flounce, I am stating this seriously as my thinking has indeed been flawed.
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.THANK YOU FOR ALL YOUR FEEDBACK ON THIS PROPOSITION. I HAVE NOW CONCLUDED THAT MY THINKING WAS FLAWED.
THANK YOU FOR ALL YOUR FEEDBACK ON THIS PROPOSITION. I HAVE NOW CONCLUDED THAT MY THINKING WAS FLAWED.However if you measure velocity in terms of the time it takes for a given quantity of energy emitted to arrive at the destinationBut we don't.
The dimensions of velocity are LT-1
Time per unit energy is M-1L-2T3
Except that a photon arrives all at once, regardless of its energy.THANK YOU FOR ALL YOUR FEEDBACK ON THIS PROPOSITION. I HAVE NOW CONCLUDED THAT MY THINKING WAS FLAWED.
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.
THANK YOU FOR ALL YOUR FEEDBACK ON THIS PROPOSITION. I HAVE NOW CONCLUDED THAT MY THINKING WAS FLAWED.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.Quote from: AlanPhoton 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?
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.
THANK YOU FOR ALL YOUR FEEDBACK ON THIS PROPOSITION. I HAVE NOW CONCLUDED THAT MY THINKING WAS FLAWED.Quote from: AlanPhoton 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?
Just a thought, could be very wide of the mark.THANK YOU FOR ALL YOUR FEEDBACK ON THIS PROPOSITION. I HAVE NOW CONCLUDED THAT MY THINKING WAS FLAWED.
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.
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.THANK YOU FOR ALL YOUR FEEDBACK ON THIS PROPOSITION. I HAVE NOW CONCLUDED THAT MY THINKING WAS FLAWED.
THANK YOU FOR ALL YOUR FEEDBACK ON THIS PROPOSITION. I HAVE NOW CONCLUDED THAT MY THINKING WAS FLAWED.Quote from: HalcThe 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?
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.
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).QuoteLight 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 – vThis 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.QuoteOne can visualise it as follows: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.
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.QuoteIf the wave from a stationary light source has a length L then the wave from a moving light source has a length L + n.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.
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.QuoteAlthough 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.QuoteThe 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.QuoteIn 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.
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.
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.QuoteThe 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.
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.
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.
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.Sorry your Big Bang Theory reference went right over my head!!
"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).
Except that a photon arrives all at once, regardless of its energy.I don't think you are clear on the point I am making.
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.
Well yes you are stating the obvious here and I acknowledge this in my post!However if you measure velocity in terms of the time it takes for a given quantity of energy emitted to arrive at the destinationBut we don't.
The dimensions of velocity are LT-1
Time per unit energy is M-1L-2T3
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.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.Quote from: AlanPhoton 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?
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.
Just a thought, could be very wide of the mark.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.
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.