Photon states should not evolve?

[vanhees71]

Although often we use the term "reference frame" to mean "coordinate system", this is one case where the difference becomes important. The technical term for a reference frame is a tetrad. It consists of a set of four vector fields covering some section of spacetime. Three of the vector fields are spacelike and one is timelike and they are orthonormal. So even though you can define coordinate systems with null basis vectors, the basis vectors of those coordinates do not form a tetrad or reference frame

Well, we had this debate some time ago, and I still stand to what I said: A reference frame is something realizable by an experimentalist. The most simple example we have in everyday life is a clock at the wall of my office and one corner of the office with three (orthogonal) edges, is a realization of a reference frame. This is a pretty accurate realization of @Dale's tetrades. I don't think that Wikipedia is so bad in this case (it's even pretty good).

Of course, I can use any coordinates that please me most in my problem, e.g., light-cone coordinates. These coordinates, however, are not necessarily realizable as a (local) reference frame, and light-cone coordinates are an example. In other words there are more coordinate systems than real-world (local) reference frames :-).

 

[martinbn]

Yes, I guess that we would need to get more authoritative references than Wikipedia!

Now I think that even the serious sources don't follow the same convention.

 

[vanhees71]

Of course you can work with other bases of the tangent spaces than the tetrades described by @Dale and the Wikipedia article, but as I argued, the usual more narrow definition of tetrades are at least in principle realizable with real-world material in the lab :-).

 

[nitsuj]

Photons only move at c in a vacuum...

is there atmospheric influence at such a scale..if its not a vacuum then there is stuff a photon may interact with. A photon always goes c. Some parts of it faster !

 

[PeterDonis]

A photon always goes c. Some parts of it faster !

What are you referring to here?

 

[jerromyjon]

A photon always goes c. Some parts of it faster !

I would like to believe photons always travel at c but there is currently no way to prove that (that I've heard of). By some parts do you mean the phase velocity? That I have heard can exceed "c" in some sense, but it would require multiple photons to have a phase, wouldn't it?

 

[Dale]

I would like to believe photons always travel at c but there is currently no way to prove that (that I've heard of).

See sections 3.3 and 3.4

http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html

 

[jerromyjon]

See sections 3.3 and 3.4

http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html

That's the speed in a vacuum, I meant in matter, where the refractive index "slows" light.

 

[NoTe]

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? This is not just a question; if decay of photons is possible, we may see the universe quite different from what it really IS! (Because the photons arising from far away objects have decayed in the mean time) Let's say a mean decay time of days or so - this doesnot have any influence on our solar system.

 

[PeterDonis]

I was wondering about the possibility of decay of photons, but read that this could not occur because the photon's time stood still.

Which, as should be evident from this thread, is a pop science misconception.

is decay of photons possible?

What does "decay" mean? A pair of photons of sufficient energy can produce a particle-antiparticle pair. Does that count as "decay"?

If you mean "decay" of a single photon in free space, no, that's not possible.

 

[Carrock]

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? This is not just a question; if decay of photons is possible, we may see the universe quite different from what it really IS! (Because the photons arising from far away objects have decayed in the mean time) Let's say a mean decay time of days or so - this doesnot have any influence on our solar system.

The general view, I think, is that time isn't an applicable concept for a photon, since a reference frame 'moving with' a photon isn't possible.
To go from this to saying that photons are timeless (i.e. without defined time) is OK IMO. Whether or not their 'time' 'stands still' is semantics rather than physics.
No one seems to have directly disagreed with my previous post

In special relativity, you can make the the frequency or momentum etc of a photon any arbitrary finite value you want, by choosing an appropriate reference inertial frame from which to measure it.

Similarly, in GR, any apparent bending, doppler shift etc is a result of curved space etc and using a non-local reference frame to calculate the apparently changing state of the photon.It would be possible, by using suitable different local inertial frames near different points on the photon's path, to calculate the photon's state to be unchanging.

i.e. by using 'correct' inertial frames, photons' state can be calculated to never evolve.

It's perhaps worth mentioning that the changing state of moving neutrinos is regarded as strong evidence they have mass.

except

Yes, I'm specifying that [the energy of a photon must be defined by a comoving inertial frame]. Many things affect the wavelength of photons, time is not one of them.

which is simply an assertion.

 

[PeterDonis]

The general view, I think, is that time isn't an applicable concept for a photon, since a reference frame 'moving with' a photon isn't possible.

Yes.

To go from this to saying that photons are timeless (i.e. without defined time) is OK IMO.

Unfortunately, this doesn't seem to work in practice; as soon as you say photons are "timeless" instead of saying that the concept of "time" does not apply to them, people start drawing incorrect inferences.

 

[PeterDonis]

No one seems to have directly disagreed with my previous post

Your statements in that post are not incorrect, but they're not really useful either. When we're talking about the energy of a photon, what we really care about is its energy relative to some observer or object that it is interacting with. That energy is a Lorentz scalar: you take the inner product of the photon's 4-momentum vector with the 4-velocity of the object or observer, and inner products of vectors are scalars, independent of any choice of frame.

So, for example, if we say a particular photon is Doppler shifted, what we mean is that its energy relative to some observer or detector is different from its energy relative to its source--which means that the inner product of the photon's 4-momentum with the detector's 4-velocity is different from the inner product of the photon's 4-momentum with the source's 4-velocity. That is a statement about a comparison between scalars, and is independent of any choice of frame.

The only other technical point involved in all this is how the photon's 4-momentum gets "propagated" from the source to the detector; but the answer to that is simply that the 4-momentum gets parallel transported along the photon's worldline. That is also independent of any choice of frame.

 

[Carrock]

...Unfortunately, [saying that photons are timeless (i.e. without defined time)] doesn't seem to work in practice; as soon as you say photons are "timeless" instead of saying that the concept of "time" does not apply to them, people start drawing incorrect inferences...

'Saying that the concept of "time" does not apply to them' also often leads to incorrect inferences; I subjectively find the latter more annoying.

Your statements in that post are not incorrect, but they're not really useful either. When we're talking about the energy of a photon, what we really care about is its energy relative to some observer or object that it is interacting with...

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

Without anthropomorphising, photons are governed by the laws of QED, which do include the way a photon's state evolves over time. Whether that implies they "experience" time is more a question of semantics than physics

It's often useful to think the photon is changing during its (unobservable) flight. 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.
A particle's state 'really' evolving during flight, like a neutrino, is generally regarded as proof mass is associated with it.

 

[nitsuj]

That's the speed in a vacuum, I meant in matter, where the refractive index "slows" light.

what do you mean "slows light"

 

[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.

 

[Mister T]

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."

 

[Mister T]

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?