I Photon states should not evolve?

[PeterDonis]

Saying that the concept of "time" does not apply to them' also often leads to incorrect inferences

How so?

I was simply trying to indicate that it's never necessary to have photons' energy, momentum etc change.

It is if you want to analyze actual experiments, since in actual experiments the energy relative to the source is often different from the energy relative to the detector.

Since an object can always be chosen such that the absorbed photon is in the same state as when it was emitted

No, this is not correct. For example, if you are here on Earth, and you are looking at photons coming from a distant galaxy, you don't get to choose the state of motion of the source or the detector.

Also, your use of the word "state" is incorrect here. The energy of the photon is an inner product of the photon's 4-momentum and the object's 4-velocity; that's not just a matter of the photon's state. (And the photon's state is not just its 4-momentum; it also includes polarization, which doesn't affect the photon's energy relative to an object, but does affect other measurements.)

A particle's state 'really' evolving during flight, like a neutrino, is generally regarded as proof mass is associated with it.

Do you have a reference for this?

 

[Tio Barnabe]

After reading again my first post and all responses that followed from it, I think we could realize that time doesn't pass for light in Einstein's theory. But photons are of Quantum Mechanics and in the latter theory, time evolves. So my OP doesn't make sense. Is this conclusion ok?

Never the less, I learned some interesting things here:

1 - The definition of a reference frame
And that

2 - There's no reference frame where time doesn't evolve, which follows from 1 just above.

 

[jerromyjon]

what do you mean "slows light"

Google "slowed light" and pick one of the 37 million hits.
The speed of light NOT in a vacuum is not "c". I'm still not sure if it slowed between atoms or if it is just delayed by interactions or if it is even relevant to anything in physics.

 

[Herman Trivilino]

I was wondering about the possibility of decay of photons, but read that this could not occur because the photon's time stood still. Since this argument isn't valid at all, - it leaves my question: is decay of photons possible?

The argument is being stated in an imprecise and erroneous way, but the conclusion is valid.

Look at neutrino oscillations. Just a few decades ago it was discovered that the neutrino can undergo a process called oscillation where it changes flavor. If the neutrino were massless, as it had been thought to be up until that time, this would not be possible; so it constitutes evidence that the neutrino has mass. You can apply the same argument, if a neutrino is massless it travels at speed $c$ and therefore, loosely speaking, can't experience time and therefore cannot change flavor.

If a single photon can decay then it cannot travel at speed $c$ and it must have mass.

 

[nitsuj]

Google "slowed light" and pick one of the 37 million hits.
The speed of light NOT in a vacuum is not "c". I'm still not sure if it slowed between atoms or if it is just delayed by interactions or if it is even relevant to anything in physics.

in aggregate light seems like it is going slower...its absorbed by atoms and emitted, is that even the "same" photon?. But goes c in between...light, or photons always go c. Vacuum or not.

 

[PeterDonis]

After reading again my first post and all responses that followed from it, I think we could realize that time doesn't pass for light in Einstein's theory.

No, that's not correct. In classical relativity, the concept of "time passing" does not apply to light.

But photons are of Quantum Mechanics and in the latter theory, time evolves.

What does "time evolves" mean? And how does it make QM different from classical relativity?

There's no reference frame where time doesn't evolve

If this is the case, then "time evolves" in classical relativity too. So I don't understand what point you're trying to make.

A key issue here appears to me to be that you are trying to give precise meanings to ordinary language terms like "time evolves". That is not going to be a good strategy. It's far better to look at the math.

 

[phinds]

.light, or photons always go c. Vacuum or not.

That is not correct. "c" is taken as the universal speed limit and it light were found to have mass, "c" would remain the same but would no longer be the speed of light. The point, then, is that light does NOT travel at "c" in various media, so your statement "vacuum or not" is incorrect.

 

[PeterDonis]

You can apply the same argument, if a neutrino is massless it travels at speed $c$ and therefore, loosely speaking, can't experience time and therefore cannot change flavor.

No, that's not correct. The argument from "massless" to "can't change flavor" does not rely on any version of "can't experience time"--or, to state this in a more technically correct fashion, it doesn't rely on the fact that massless particles travel on worldlines of zero arc length. A discussion of the actual argument really belongs in a separate thread in the QM forum, but briefly, the key point is that if neutrinos have more than one flavor eigenstate and can switch between them (i.e., change flavors), they must also have more than one mass eigenstate, with different masses. One of those mass eigenstates could still have zero mass (i.e., be massless), but the others could not.

 

[Tio Barnabe]

What does "time evolves" mean? And how does it make QM different from classical relativity?

I mean, the proper time variation for light in Relativity is zero. But in Quantum Mechanics there's only one time, namely the coordinate time and it can take on different values. Is it not so?

If this is the case, then "time evolves" in classical relativity too

Sure, that's what I said.

 

[PeterDonis]

I mean, the proper time variation for light in Relativity is zero.

No, the arc length along a null worldline in relativity is zero. But "proper time" is an incorrect term to use to describe that arc length.

in Quantum Mechanics there's only one time, namely the coordinate time

There is coordinate time in relativity too. And the coordinate time changes along the worldline of a light ray. So no, this is not a difference between relativity and QM.

Also, to properly do a comparison here, you should be looking, not at non-relativistic QM, but at quantum field theory. Which is built on relativity.

 

[Carrock]

Saying that the concept of "time" does not apply to them' also often leads to incorrect inferences

How so?

Hard to find a reference, but I've seen claims that photons experience change in some way, but time does not pass for them. It's usually harder to understand when time is implicitly avoided but a vaguely similar concept is used.

I was simply trying to indicate that it's never necessary to have photons' energy, momentum etc change.

It is if you want to analyze actual experiments, since in actual experiments the energy relative to the source is often different from the energy relative to the detector.

Often isn't necessarily. There's no problem e.g. in moving a clock downwards to compensate for gravitational redshift from a lower clock to compare them.

Since an object can always be chosen such that the absorbed photon is in the same state as when it was emitted

No, this is not correct. For example, if you are here on Earth, and you are looking at photons coming from a distant galaxy, you don't get to choose the state of motion of the source or the detector.

Without doing the maths, I'd think a suitable particle at CERN could be created as a detector with an appropriate state of motion.

Also, your use of the word "state" is incorrect here. The energy of the photon is an inner product of the photon's 4-momentum and the object's 4-velocity; that's not just a matter of the photon's state. (And the photon's state is not just its 4-momentum; it also includes polarization, which doesn't affect the photon's energy relative to an object, but does affect other measurements.)

I'm rather rusty on 4-momentum and 4-velocity as you may have guessed...
However I don't see my imprecision/error negating anything else in this post. From your quote, an object with a suitable 4-velocity is all that's required.

A particle's state 'really' evolving during flight, like a neutrino, is generally regarded as proof mass is associated with it.

Do you have a reference for this?

From https://www.nobelprize.org/nobel_prizes/physics/laureates/2015/advanced-physicsprize2015.pdf

The discovery that neutrinos can convert from one flavour to another and therefore have non-
zero masses is a major milestone for elementary particle physics.

Do you have a reference refuting

A particle's state 'really' evolving during flight, like a neutrino, is generally regarded as proof mass is associated with it.

More precisely, refuting "...an object can always be chosen such that the absorbed photon (and absorber I suppose) is in the same state as when it was emitted."

 

[Herman Trivilino]

One of those mass eigenstates could still have zero mass (i.e., be massless), but the others could not.

Ahhh... Yes. I think I see. So the solar neutrino problem of 25 years ago was about missing electron neutrinos. For them to be missing because of a flavor change they would have to have mass, be traveling at speeds less than $c$, and not have a zero arc length (or has been described loosely as not being able to experience time). Do I at least have that part right?

 

[PeterDonis]

Hard to find a reference, but I've seen claims that photons experience change in some way, but time does not pass for them.

Um, have you read this thread? We've already discussed such claims and why they're not valid.

From https://www.nobelprize.org/nobel_prizes/physics/laureates/2015/advanced-physicsprize2015.pdf

I didn't ask for a reference to show that neutrinos oscillate. I asked for a reference supporting your claim that it is the neutrinos "really evolving" that allows them to oscillate. I already explained in post #58, in response to @Mister T , why that claim is not correct (because the inference from neutrino oscillations to them having to have mass does not depend on the worldlines of massless particles having zero arc length). The reference you give is consistent with what I said in post #58.

refuting "...an object can always be chosen such that the absorbed photon (and absorber I suppose) is in the same state as when it was emitted."

I didn't say this statement was wrong. I just said it wasn't useful, and I explained why. Did you read my post?

 

[PeterDonis]

Do I at least have that part right?

Not quite. Neutrino oscillations imply that some neutrinos must have mass. They do not require that all neutrinos have mass. I explained why briefly in post #58. For a more detailed explanation, the PDF that @Carrock linked to gives a good discussion in the section "Theory of neutrino oscillations", starting at the bottom of p. 13.

 

[PeterDonis]

Often isn't necessarily.

You claimed that it is never necessary to have the photon's energy/momentum change. One counterexample, which I gave (and it's not the case of gravitational redshift, which you mention), is sufficient to refute such a strong claim.

Before making even a weaker claim, I strongly suggest that you spend some time reviewing the kinds of actual experiments in which photon energies are relevant, so you can see how often actual experimenters have to use models in which photon energies and momenta are allowed to change in order to explain their results.

Without doing the maths, I'd think a suitable particle at CERN could be created as a detector with an appropriate state of motion.

See what I just said above.

From your quote, an object with a suitable 4-velocity is all that's required.

Sure, but you can't always guarantee that such an object exists, or make one if it doesn't. To do so, you would have to already know the photon's 4-momentum, and in many, if not most, cases of interest, you don't. That's why you have to make measurements on it--to find out what its 4-momentum is.

Furthermore, even if you could construct such an object, you might not care what its measurements are telling you.

 

[Tio Barnabe]

No, the arc length along a null worldline in relativity is zero. But "proper time" is an incorrect term to use to describe that arc length.

Why? It's even written down as $d \tau^2 = \ ... \$. How can it not be the proper time?

to properly do a comparison here, you should be looking, not at non-relativistic QM, but at quantum field theory. Which is built on relativity.

Is there any difference as to how time is treated in Quantum Field Theory and non-relativistic Quantum Mechanics?

 

[PeterDonis]

It's even written down as dτ2=

So what? $d\tau^2$ is a symbol, not physics. I could just as easily write the same timelike interval as $ds^2$ (in fact, many textbooks do exactly that); would that make it a proper length?

Is there any difference as to how time is treated in Quantum Field Theory and non-relativistic Quantum Mechanics?

Most definitely.

 

[Tio Barnabe]

So what? $\tau^2$ is a symbol, not physics. I could just as easily write the same timelike interval as $ds^2$ (in fact, many textbooks do exactly that); would that make it a proper length?

It's just a symbol, yes. But it's just a symbol with a well defined meaning. As long as we define $d \tau^2$ as the proper time squared, so it is the proper time squared.

 

[PeterDonis]

it's just a symbol with a well defined meaning.

Yes, but not the meaning you think it has. See below.

As long as we define $d \tau^2$ as the proper time squared, so it is the proper time squared.

"Proper time squared" is not the definition of $d\tau^2$. The definition of $d\tau^2$ is "the square of the infinitesimal spacetime interval". There is nothing in the definition that tells you the physical meaning of that spacetime interval. For that you have to look at its actual value. "Proper time" is only the physical meaning of the interval if the interval is timelike. That is true regardless of what symbol you use.

 

[Tio Barnabe]

Hmmm, good to know. I thought $d \tau^2$ could always be regarded as a proper time, because in the reference frame of the "particle" there's no spatial displacement. So $d \tau^2 = dt^2$ in the particle reference frame.

"Proper time" is only the physical meaning of the interval if the interval is timelike

because if it is space-like the "proper time" would be negative?

 

[PeterDonis]

I thought $d \tau^2$ could always be regarded as a proper time, because in the reference frame of the "particle" there's no spatial displacement.

This will be true if the particle has nonzero rest mass and $d\tau^2$ is an interval along its worldline. But the definition of $d\tau^2$ makes no such assumptions. What you're describing is just one particular special case.

because if it is space-like the "proper time" would be negative?

The square root of $d\tau^2$, assuming you were using a timelike signature convention, would not be negative for a spacelike interval, it would be imaginary.

However, this is still focusing on superficial features instead of the fundamental definition. The fundamental definition just says that timelike, null, and spacelike intervals are physically different. How that difference is modeled in the math depends on your choice of signature convention--note that I specified a timelike signature convention above, which means that timelike intervals have positive $d\tau^2$ and spacelike intervals have negative $d\tau^2$. Conversely, the spacelike signature convention means that timelike intervals have negative $d\tau^2$ (the symbol $ds^2$ is normally used in this case--but as I said before, it's often used in the timelike signature case as well) and spacelike intervals have positive $d\tau^2$. Either way, null intervals have zero $d\tau^2$.

 

[Tio Barnabe]

This will be true if the particle has nonzero rest mass and $\tau^2$ is an interval along its worldline.

This is going against what is said in Weinberg's book. I remember a part of the book that says for light the LHS of the equation $d \tau^2 = \ ... \$ is equal to zero.

 

[PeterDonis]

This is going against what is said in Weinberg's book.

No, it isn't. I said "if the particle has nonzero rest mass". For something with zero rest mass, like light, the interval along the worldline is zero, yes, but the term "proper time" is not a proper description of it, for that very reason--because the worldline is null, not timelike.

 

[Tio Barnabe]

No, it isn't. I said "if the particle has nonzero rest mass". For something with zero rest mass, like light, the interval along the worldline is zero, yes, but the term "proper time" is not a proper description of it, for that very reason--because the worldline is null, not timelike.

So, also, is the statement that times stop at the speed of light incorrect?

 

[PeterDonis]

is the statement that times stop at the speed of light incorrect?

I would say it's "not even wrong", because it assumes that the concept of "proper time" or "elapsed time" is even meaningful for light, or anything that moves at the speed of light.

 

[Dale]

Hmmm, good to know. I thought $d \tau^2$ could always be regarded as a proper time, because in the reference frame of the "particle" there's no spatial displacement. So $d \tau^2 = dt^2$ in the particle reference frame.?

If $d\tau^2=0$ then there is no particle reference frame.

 

[Dale]

Since an object can always be chosen such that the absorbed photon is in the same state as when it was emitted, and emission and absorption are the only times a photon's state can be observed, there is no requirement for a photon's state to evolve over time.

This is wrong. The wave function representing the photons state evolves over time according to the schrodinger equation. At least in standard QFT it is required.

 

[PeterDonis]

The wave function representing the photons state evolves over time according to the schrodinger equation. At least in standard QFT it is required.

Just to clarify: the Schrodinger equation is non-relativistic, and you can't really apply it to photons/light. (You can apply it to atoms or particles interacting with an electromagnetic field, but the field itself will be modeled as an external potential.) A better way of expressing what I think you're trying to express here would be that, in QFT, you can't assume that, if the photon field (electromagnetic field) is in a given state at a particular event, it will be in the same state at any other event that is null separated from that event.

 

[PeterDonis]

This is wrong.

To clarify a bit further: I was interpreting @Carrock's use of the term "photon" classically, i.e., it should really be "very short pulse of EM radiation", which we can model as a "particle" with a well-defined 4-momentum vector which is null. Given a known null 4-momentum vector, and some chosen value for energy, we can always find a local inertial frame at any event on the null worldline of the photon in which the inner product of that 4-momentum vector and the timelike basis vector of the frame yields the chosen value. If we imagine a series of such local inertial frames all along the null worldline, we could say that, in these frames, the photon's energy is "unchanged" as it travels.

The problem, as I pointed out in response, is that in order to know which frames these are, you have to already know the photon's 4-momentum, and in most cases of practical interest, you don't. Also, even if you do, the frame given by the above prescription might not be the one you actually care about.

 

[Carrock]

Hard to find a reference, but I've seen claims that photons experience change in some way, but time does not pass for them.

Um, have you read this thread? We've already discussed such claims and why they're not valid.

Um, have you read the post you quoted from? ( https://www.physicsforums.com/threads/photon-states-should-not-evolve.925675/page-4#post-5847397 )

Carrock: Saying that 'the concept of "time" does not apply to them' also often leads to incorrect inferences.
PeterDonis: How So?
Carrock: Hard to find a reference, but I've seen claims that photons experience change in some way, but time does not pass for them.

So you request an example of an incorrect inference; when I provide it you infer I haven't read this thread.

I didn't ask for a reference to show that neutrinos oscillate. I asked for a reference supporting your claim that it is the neutrinos "really evolving" that allows them to oscillate.

I never claimed that it is 'the neutrinos "really evolving" that allows them to oscillate.' Still, the time I wasted searching for a good reference has saved you the necessity of looking for yourself.

refuting "...an object can always be chosen such that the absorbed photon (and absorber I suppose) is in the same state as when it was emitted."

I didn't say this statement was wrong. I just said it wasn't useful, and I explained why. Did you read my post?

I read your posts but don't recall which is the relevant one. I'm not going to trawl through this thread yet again and perhaps pick a post you didn't mean. Perhaps you should read my posts to avoid changing their meaning with misleading editing and inaccurate precis.

I see in post #79 most of your objections seem rather forgotten as well as the reason I introduced this rather old concept in the first place. If you'd posted something like this earlier I'd have clarified this concept; as it is, when I spend this amount of time on a post like this, I know it's time to quit.

 

[Dale]

we could say that, in these frames, the photon's energy is "unchanged" as it travels.

Sure, but the state is more than just the energy. In classical EM the fields evolve over time per Maxwell's equations. In QED the wave function evolves over time per Schrodinger's equation. Either way, the state evolves over time.

 

[Dale]

Just to clarify: the Schrodinger equation is non-relativistic

Hmm, I thought that the Schrodinger equation, in a suitably generalized form, still determines the evolution of the wavefunction. Is it called another name?

 

[PeterDonis]

In QED the wave function evolves over time per Schrodinger's equation.

Not really. See below.

Either way, the state evolves over time.

Agreed, although there are caveats when we are talking about QFT; a better way to put it in QFT would be the way I put it before, that if we know the state of the quantum EM field at a particular event, we cannot just assume that the quantum EM field will be in the same state at any other event that is null separated from the first one.

I thought that the Schrodinger equation, in a suitably generalized form, still determines the evolution of the wavefunction.

In QFT, there isn't a "wave function" in the sense that there is in non-relativistic QM. It's a very different framework conceptually. We should probably start a separate thread in the quantum forum if this needs to be discussed further.

 

[PeterDonis]

you request an example of an incorrect inference; when I provide it you infer I haven't read this thread.

I asked for an example of "the concept of time does not apply to photons" leading to an incorrect inference. You gave me an example of "time does not pass for photons" leading to an incorrect inference. But "time does not pass for photons" implies that the concept of time does apply to photons. So your example was not responsive to my question.

I never claimed that it is 'the neutrinos "really evolving" that allows them to oscillate.'

I suppose that's true; here's what you said:

A particle's state 'really' evolving during flight, like a neutrino, is generally regarded as proof mass is associated with it.

Which is still wrong, as I have already explained in other posts.

 

[PeterDonis]

I read your posts but don't recall which is the relevant one.

It was the exact same one you were already responding to in the quote you gave (your quote starting with "refuting..."), which is from your post #61. In other words, in your post #61, you were responding to the relevant post of mine, which gave the explanation you are looking for.

 

[nitsuj]

That is not correct. "c" is taken as the universal speed limit and it light were found to have mass, "c" would remain the same but would no longer be the speed of light. The point, then, is that light does NOT travel at "c" in various media, so your statement "vacuum or not" is incorrect.

You're right, but think you should read why. By the same interaction light may go faster than light...I mean c.

"If light were found to have mass..." has nothing to do with it.

 

[phinds]

You're right, but think you should read why. By the same interaction light may go faster than light...I mean c.

"If light were found to have mass..." has nothing to do with it.

I don't understand any of this. Can you expound on it please?

 

[nitsuj]

I don't understand any of this. Can you expound on it please?

sure, for light going through something where it is not absorbed and emitted it easiest to think of it as a wave as opposed to a photon. the EM of the light wave when it begins to pass through the medium, such as glass induces a wave in the glass, the sum of these waves is what we see leave the glass, most often this amounts to the original light wave as slowing. This induced wave and original wave can also amount to the the phase velocity bring greater than c.

You should read up on it it's pretty interesting...

 

[phinds]

sure, for light going through something where it is not absorbed and emitted it easiest to think of it as a wave as opposed to a photon. the EM of the light wave when it begins to pass through the medium, such as glass induces a wave in the glass, the sum of these waves is what we see leave the glass, most often this amounts to the original light wave as slowing. This induced wave and original wave can also amount to the the phase velocity bring greater than c.

You should read up on it it's pretty interesting...

Ah. Phase velocity. You're right, I should read up on it, since I keep hearing about it here on PF.