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Author Topic: If a photon is travelling at the speed of light, does time not exist for it?  (Read 23742 times)

Offline Bill S

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Quote from: JP
There is a third option.  The photon has no reference frame (at least not one we know how to describe in terms of it's motion).  That's the one that's true since the above options are not allowed in relativity.

What about the possibility that the photon has a frame of reference in which it is stationary relative to itself and time?

Is that more of an assumption than saying that because we cannot describe the F of R of a photon, it cannot have one?
 

Offline JP

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Quote from: JP
There is a third option.  The photon has no reference frame (at least not one we know how to describe in terms of it's motion).  That's the one that's true since the above options are not allowed in relativity.

What about the possibility that the photon has a frame of reference in which it is stationary relative to itself and time?

Is that more of an assumption than saying that because we cannot describe the F of R of a photon, it cannot have one?

If you can't scientifically describe the F of R of a photon, does it matter if it has one?  It's just a matter of opinion unless there's an actual model for it.
 

Offline yor_on

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I
Matthew, I'm not entirely sure what happens to a photon in matter.  Is the slow down in a superconductor sufficiently different to, say, glass?  The way I usually understand a photon moving in matter is that it is propagating at the speed of light in between the bits of matter, but interactions with the matter essentially cause it to take longer to get through.  I don't think this should change its decay properties, or the fact that it moves at the speed of light in between those interactions.

I am trying to split two phenomena - you rightly describe how a photon moves through material via interactions; this is the case within glass.  The photon moves at light speed between interactions but overall at a slower speed- I think that's the best explanation I have read, not sure if it isnt a bit hand-wavy though.

But within a superconductor, electron pairs (cooper pairs) form a condensate.  It is thought that this condensate can act like a bosonic field and emulate the higg's field.  But in a superconductor (as opposed to everywhere else) this higgs-like field will provide mass (very very small) for the photon.  I would imagine a photon with mass would be forced to travel slower than a photon without mass.  A subluminal photon (and I still don't think it really works) would be free from any argument that time is meaningless for it as it is moving in a normal frame of reference.

I will dig out references for the cooper pair condensate field - I hope.

What would happen to its spin if it gained a mass? It's spin 1 normally having two degrees of freedom instead of three due to its 'masslessness' But gaining a mass should change that, shouldn't it?
 

Offline imatfaal

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To be honest Yoron - I am struggling to find a good write-up on it.  Lots of papers mention it in passing, and several refer to a section in Ryder but no decent explanation.  Not sure what exactly you mean by tying up the degrees of freedom and masslessness
 

Offline yor_on

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The spin of a photon is defined as having two degrees of freedom, that in its turn is a headache. "Let us go back to the photon statistics formula derived by Bose. There is a factor "2" sitting on the numerator of this formula. The usual explanation is that it is because photons are massless particles. Then why not 1 or 3 ? Bose argued that the photon can have two degenerate states. This eventually led to the concept of photon spin parallel or anti-parallel to the momentum.

The question of why the photon spin should be only along the direction of momentum has a stormy history. Eugene Wigner (1939) showed that the internal space-time symmetry of massless particles is isomorphic to the symmetry of two-dimensional Euclidean space consisting of one rotation and two translational degrees of freedom. It is not difficult to associate the rotational degree with the photon spin either parallel or anti-parallel to the momentum, but what physics is associated with the translational degrees of freedom. These translational degrees were later identified as gauge transformations. This does not solve the whole problem because there is one gauge degree of freedom while there are two translational degrees of freedom. How do they collapse into the one gauge degree of freedom? This problem was not completely solved until 1990."

"If electrons had total spin s=1 then there would be three possible spin projections sz={−1, 0, or +1} ... and you would find orbitals with three electrons in them. You can’t occupy a given orbital more times than are allowed by the spin multiplicity because of the (Pauli) exclusion principle."

So a photon that actually is of spin 1 still doesn't have three degrees of freedom. But if we give it a mass?
==

Spin is weird.
« Last Edit: 02/02/2011 21:44:04 by yor_on »
 

Offline Bill S

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Quote from: JP
If you can't scientifically describe the F of R of a photon, does it matter if it has one?  It's just a matter of opinion unless there's an actual model for it.

I'm not really nit-picking here, honest,  :) just trying to get things straight in my own mind.

Given that you can't scientifically describe the F of R of a photon, would it also be true to say that you cannot scientifically establish that it does not have one?  I accept that even if this is the case, the whole thing comes down to a matter of opinion, and that some opinions have more weight behind them than others.  In the midst of a sea of strongly held views, though, I often find myself wondering what is established fact and what is opinion.



 
 

Offline JP

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It depends on your definitions, Bill.  Frame of reference in physics tends to refer to Einstein's inertial frames of special relativity.  In that case, you can say conclusively that the photon has no frame of reference.  (The same is true for local coordinate frames in general relativity.)

If you want to use a different definition of frame of reference in order to allow the photon to have one, you'll have to define it first, and justify it's usefulness as a physical model.
 

Offline Geezer

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JP,

What happens in the case of matter? I realize that matter can't travel at the speed of light, but is it possible to extrapolate the effects on time as it approaches c to arrive at a conclusion about what might happen at c, or does the math simply go haywire long before you get there?
 

Offline JP

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The math goes haywire.  The problem is that simply taking the limit as v->c doesn't account for the fact that the mass goes to zero.  If you do the appropriate limit taking, you get something like 0/0, I recall...

That, and all of special relativity deals with observers who always observe light as moving at c.  The math just doesn't hold if you throw out that assumption.
 

Offline Geezer

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The math goes haywire.  The problem is that simply taking the limit as v->c doesn't account for the fact that the mass goes to zero.  If you do the appropriate limit taking, you get something like 0/0, I recall...

That, and all of special relativity deals with observers who always observe light as moving at c.  The math just doesn't hold if you throw out that assumption.

I had a nasty feeling that might be the case  ;D  Thanks!
 

Offline Bill S

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Thanks, JP, but I'm still struggling a bit here.

We seem to have moved from saying that we can't scientifically describe the F of R of a photon, to saying conclusively that the photon has no frame of reference.

let's stick with "inertial frame", so you know I'm not trying to come up with my own definition of F of R.  :)

The reasoning seems to go something like this: A photon must always be observed as travelling at c, so it cannot share an inertial frame with anything else. 
Arguing that  photon must be stationary relative to itself is meaningless; therefore it is meaningless to suggest that a photon occupies an inertial frame.

Am I getting closer?
 

Offline JP

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I think you're understanding me, Bill.  The reason I was saying things like "science can't describe the F of R of a photon" rather than categorically ruling them all out is just because some of the suggestions here have been to introduce new models specifically to define a F of R of a photon. 

But if you stick, like you say, to inertial frames defined by relativity, then there is none for the photon, simply because they don't exist within special relativity.  You correctly point out that it would be wrong to simply add on an inertial frame in which the photon is at rest.  (I'd add on to your description that a photon can't be at rest with respect to itself because the frames described by special relativity all require that light moves at light speed in those frames.) 
 

Offline Locke

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Matter is a concentrated energy. If you like, radiation is a free energy - a diffused type of matter. If all matter is made from photon energy, then it contributes to the kinetic energy of the system, and the inertial energy of the system. It does not mean that matter will move at lightspeed. This is in direct contradiction of relativity theory.

Are you saying that matter and photons are actually two totally different things? I thought you maintained that matter was made from photons, or did I get that wrong?

Yes you are wrong, again.

Matter can be made out of photons, it does not necesserily mean that attributes of the photon are carried on. Intrinsic properties in the form of information is almost certainly passed on in order to conserve quantities like charge and spin. It does not mean that the manifestation (matter) obeys the rules which radiation does. Matter and energy are still quite different, even though they are interchangeable via E=Mc2.
If matter is made out of photons and matter has mass, that would give photons mass as well...
And photons, by my understanding, don't have mass.
 

Offline imatfaal

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If matter is made out of photons and matter has mass, that would give photons mass as well...
And photons, by my understanding, don't have mass.

It's a disputed point Locke - but the answer some might give is:  that mass and energy are equivalent (everyone should agree that E=mc^2) and thus the energy of a photon pair can become the mass of a particular (pair of) particles (this is also agreeable); others go on to say that ALL matter can be considered to be made of photons (this is where the argument lies).

Bear in mind you need really energetic photons - the lightest particles you are gonna create will prob be a positron electron pair  and even these will require light well into the gamma rays
 

Offline yor_on

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Some questions and thoughts :)

First of all. Isn't 'c' a valid frame for photons in any frame seen? Accelerating or inertial (uniformly moving)? And if I argue that light is 'timeless'? Well, that depends on what you define 'time' as, I would say. Is it 'time' it takes for light to wander 13.7 billion light-years from the 'visible' edge of our universe? If you to that add what 'expansion' did to the space it 'wandered' in it has been awfully stretched (like 47 Billion light years one-way), as a wave that is, as a photon it can't be stretched at all, as far as I know?

As a 'light-quanta' it shouldn't be able to lose any energy without a 'interaction' either, as far I know? As such all energy it has should be the same from 'source' to 'sink'. And all of those descriptions are valid as I understand it. 'scientifically so.' But the 'trick' is to choose your definition to what the radiation does, isn't it? Why I differ a light quanta and a photon is in the indeterminacy of a photon seen as a wave-packet, where it start and ends as I understands it. But as a light quanta, again as I understands it, we expect it to be of the same defined 'energy' from source to sink. And what we call a photons red or blue shift will then only be a relation to the 'frame of reference' observing it.

Now, this is a questionable approach, but never the less the one taken as I understands it, and from 'Science' no less. A better approach is to admit that if you observe a photon, or light-quanta you will annihilate it. To observe or interact with a wave is not the same though :) as we can see when looking in a mirror, although there you can actually use both descriptions, photons or waves, to describe it. As with waves, the red and blue-shift makes more sense in form of frequency and energy getting 'stretched out'.

When it comes to a expansion though, stretching a 'light-quanta' or photon? Even when it comes to wave packet this sounds very far fetched, how exactly can a mere 'distance' down-shift a photon? It won't stretch at all as far as I can see, it's with gravity and speed we use red or blue shift and then as only relative the observer, if we we define them of invariant light quanta. To make a light-quanta 'stretch' you better start with giving it a 'size', otherwise it's a exercise in futility. and if you don't define it a 'size' then you have distance doing what we only attributed to gravity and speed before, 'down-shifting' light-quanta.
==

The real problem here is when someone suggests that photons just are like bullets coming from our universal machine guns (Suns). Then a red-shift should be fewer 'bullets' per 'time unit' and a blue-shift more bullets per 'time unit'. That one seems to work until we consider one light-quanta/photon moving. Because if we do so, it is my understanding that this light quanta will express the same red and blue shift, and just as easy as any amount of photons then would do by being 'together'. Now, if my understanding of this is wrong :) Then I'll go with the 'machine gun description'. But if I'm right a photon somehow manage to express its red and blue shift some other way. And if it does I suggest it to be a relation, that is, two different 'frames of reference' interacting.

(The Pound–Rebka experiment did just that, if I understood it right, testing one photon 'jumping' between atoms.)

" When an atom transits from an excited state to a base state, it emits a photon with a specific frequency and energy. When the same atom in its base state encounters a photon with that same frequency and energy, it will absorb that photon and transit to the excited state. If the photon's frequency and energy is different by even a little, the atom cannot absorb it (this is the basis of quantum theory).

When the photon travels through a gravitational field, its frequency and therefore its energy will change due to the gravitational redshift. As a result the receiving atom can no longer absorb it. But if the emitting atom moves with just the right speed relative to the receiving atom the resulting doppler shift  will cancel out the gravitational shift and the receiving atom will be able to absorb the photon. The "right" relative speed of the atoms is therefore a measure of the gravitational shift. The frequency of the photon "falling" towards the bottom of the tower is blueshifted. Pound and Rebka countered the gravitational blueshift by moving the emittor away from the receiver, thus generating a relativistic Doppler redshift"
==

Anyway, now I will use my own trick. If we keep on persisting that photons 'propagates' then we know that they don't 'die', by that alone. And then I will use that for defining them as 'timeless', that is as observing them from my frame of reference, or you from yours JP :) Or anyones 'frame', willing to dispute it. And to dispute, you will need to prove that a photon 'decay', if you now missed that. My definition is valid for all frames I know of too, as long as you don't let a light-quanta/photon 'interact'.

And you can start with defining if it interacts in a redshift/Hubble/expansion. Then you need to define that from two scientifically equally valid truths at least the light-quanta/photon and the wave. And as far as I know they all two/three are experimentally proven.

(Then we come to what I find the nicest solution, that light don't 'move' at all :)
That takes care of it. Now we only have the photons/waves relations defining how it will behave, and even though 'c' will be as valid a concept as ever it now becomes a axiom, or a constant if you like, just as the Feigenbaum constant.)


Heh.
So, yes JP, I will call them 'timeless' :)
« Last Edit: 08/02/2011 05:54:06 by yor_on »
 

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