Is light time dilated?

[Ziggurat]

Take a clock, or almost anything that oscillates reliably and can therefore function to some degree in a clocklike manner. Accelerate it and observe that as it goes nearer the speed of light it oscillates slower and slower, and see how if it were to go at the speed of light it would apparently stop. We can agree on that much.

Then tell me why I had to use the word 'almost' just then, because the most accurate clocks in existence rely on something for which this isn't true - the photon. The photon, which when emitted in a particular manner from a particular kind of caesium gives us the very definition of a second, as it oscillates 9,192,631,770 times in that period.

Simultaneity has reared its ugly head once again. That clock that you just chucked at close to c? Its ticks aren't happening in the same place in the frame it's being measured in. But the time period for oscillations of a photon ARE measured in the same place. And as we all know, measuring the time interval between events which are not happening in the same place becomes very different in relativity than it is in Newtonian mechanics. So while you can use the same word for both, you're really not measuring the same thing at all.

 

[sol invictus]

Simultaneity has reared its ugly head once again. That clock that you just chucked at close to c? Its ticks aren't happening in the same place in the frame it's being measured in. But the time period for oscillations of a photon ARE measured in the same place.

I don't follow you.

A clock flies along at speed v. All along its route stand researchers with synchronized watches. Every time the clock's nanosecond hand ticks, the researcher standing next to it records the time.

A circularly polarized EM wave propagates down that same path at speed c. All along its route stand researchers with synchronized watches. Every time the wave is vertically polarized, the researcher standing next to it records the time.

The difference is that as v->c, those researchers will get farther and farther apart (in their own frame) - the clock will freeze. But the wave has a perfectly finite wavelength even though it moves at speed c.

 

[epepke]

That author is hopefully dumbing things down for his audience.

He does not mention that a photon does not have a "point of view". It is not an observer. It cannot observe anything such as the passing of time. You can add an hypothetical observer (as in Einstiens thought experiment of travelling at the speed of light beside a light ray)
From the point of view of the observer, their clock keeps on ticking away.

Yes, you keep on saying this, and it's getting very annoying.

There are particles that decay. They can be thought of as clocks (or are you going to argue with that word because they do not sit on the mantlepiece?) Some are made in the upper atmosphere and, without taking into account time dilation, most of them should have decayed before reaching the surface. However, lots of them arrive. If you take into account time dilation, it works out just right.

Photons do not have this property. If you have a problem with the word "experience," then kindly state a word that will fill in the blank of "Photons don't ________ time" that will make you all happy so that people can get back to science instead of argumentative semantics.

 

[ynot]

Yes but in Braille !

I thought of another way to explain that a clock traveling at the speed of light still ticks. So if a photon somehow had a clock attached to it (or had a "point of view" that could measure time) then time would pass for it.
This does not need SR - just one of its postulates:
"Light in vacuum propagates with the speed c (a fixed constant) in terms of any system of inertial coordinates, regardless of the state of motion of the light source".

Consider an observer with a clock that consists of a photon bouncing between 2 mirrors. The photon will always travel at the speed c in the observer's reference frame according to the postulate. Thus the clock will always tick, even if the observer is traveling at the speed c.
Now throw away the observer (or blind them!). What happens to the clock?
I think that the clock continues to tick.

All things experience "proper time" in their own frame but that's not what I'm talking about. I'm saying that the "proper time" of the frame of light should be time dilated relative to the "proper time" all other frames to the extent that time stops. A clock travelling at the speed of light would experience it's own "proper time" but it would stop relative to all other frames. If a clock could ride a photon away from you and and then back it wouldn't have ticked relative to your "proper time" in your frame. I don't understand when people say that light doesn't experience time as it should experience it's own "proper time". It should be the observer of light in another frame that experiences light as being fully time dilated. So why isn't the speed of light instantaneous relative to (or observed from) all other frames? I don't understand why time dilation doesn't dilate the very speed that creates it relative to other frames.

 

[Reality Check]

All things experience "proper time" in their own frame but that's not what I'm talking about. I'm saying that the "proper time" of the frame of light should be time dilated relative to the "proper time" all other frames to the extent that time stops. A clock travelling at the speed of light would experience it's own "proper time" but it would stop relative to all other frames. If a clock could ride a photon away from you and and then back it wouldn't have ticked relative to your "proper time" in your frame. I don't understand when people say that light doesn't experience time as it should experience it's own "proper time". It should be the observer of light in another frame that experiences light as being fully time dilated. So why isn't the speed of light instantaneous relative to (or observed from) all other frames? I don't understand why time dilation doesn't dilate the very speed that creates it relative to other frames.

That is correct - a clock travelling at the speed of light will stop according to the other frames.

It is the way the universe is. Light in vacuum propagates with the speed c (a fixed constant) in terms of any system of inertial coordinates, regardless of the state of motion of the light source. The speed c is measured to be non-zero and that is the reason that it is never zero ("speed of light instantaneous relative to (or observed from) all other frames").

The speed that the other frames measure is measured in those frames. Thus time dilation has no effect on the speed that is measured in those frames.

 

[BTMO]

Yes. It also has infinite mass, due to its speed.

Which is why, when you turn a torch on, there is a black ring around the spot. The infinite mass of the photons ejected from the torch creates a local black hole...

 

[Reality Check]

Yes, you keep on saying this, and it's getting very annoying.

There are particles that decay. They can be thought of as clocks (or are you going to argue with that word because they do not sit on the mantlepiece?) Some are made in the upper atmosphere and, without taking into account time dilation, most of them should have decayed before reaching the surface. However, lots of them arrive. If you take into account time dilation, it works out just right.

That is correct.

Photons do not have this property. If you have a problem with the word "experience," then kindly state a word that will fill in the blank of "Photons don't ________ time" that will make you all happy so that people can get back to science instead of argumentative semantics.

I have no problem with "Photons never experience time" so long as people do not think that this means that time does not pass for a photon.
I do have a problem with "Photons never experience time and so from their point of view time does not pass" when science says that this is wrong.

 

[ynot]

That is correct.


I have no problem with "Photons never experience time" so long as people do not think that this means that time does not pass for a photon.
I do have a problem with "Photons never experience time and so from their point of view time does not pass" when science says that this is wrong.

Me too. Photons should experience their own "proper time".

 

[ynot]

If light always has a speed of c relative to everything else does it mean that light occupies an infinite number of simultaneous frames?

 

[Reality Check]

If light always has a speed of c relative to everything else does it mean that light occupies an infinite number of simultaneous frames?

Everything occupies an infinite number of simultaneous frames. There are an infinite number of possible coordinate systems.

ETA
But that seems too simple to be what you mean. Can you define "simultaneous frames" in more detail?

 

[ynot]

Everything occupies an infinite number of simultaneous frames. There are an infinite number of possible coordinate systems.

ETA
But that seems too simple to be what you mean. Can you define "simultaneous frames" in more detail?

As I understand it an inertial frame is essentially defined by things that don't move relative to each other and a thing can only be in one particular frame at a particular time. Light however is always simultaneously in an infinite number of frames that are moving at c relative to all other frames.

 

[sol invictus]

As I understand it an inertial frame is essentially defined by things that don't move relative to each other and a thing can only be in one particular frame at a particular time.

No. An inertial frame is nothing more or less than a coordinate system which is moving at constant velocity with respect to any other inertial frame. It's not defined by any particular thing, and there is no requirement that anything be at rest.

A very close analogy is the origin of coordinates in space. You might want to choose it to coincide with the object of most interest, but of course you don't need to.

Light however is always simultaneously in an infinite number of frames that are moving at c relative to all other frames.

It makes no sense to speak of something being "in" a frame. Everything is in every frame all the time. What is true is that it is impossible to choose an inertial frame in which a photon is at rest (i.e. remains at constant spatial position). As I explained above, however, one can choose null coordinates, in which a photon is at constant null position.

 

[Reality Check]

As I understand it an inertial frame is essentially defined by things that don't move relative to each other and a thing can only be in one particular frame at a particular time. Light however is always simultaneously in an infinite number of frames that are moving at c relative to all other frames.

Why do you think that light is in more then 1 frame?
How do you define " simultaneously in an infinite number of frames"?

I would say light falls under the "a thing can only be in one particular frame at a particular time" category.

 

[Ziggurat]

I don't follow you.

A clock flies along at speed v. All along its route stand researchers with synchronized watches. Every time the clock's nanosecond hand ticks, the researcher standing next to it records the time.

A circularly polarized EM wave propagates down that same path at speed c. All along its route stand researchers with synchronized watches. Every time the wave is vertically polarized, the researcher standing next to it records the time.

A circularly polarized photon is not vertically polarized at a certain time, it's vertically polarized in certain places, spaced apart by one wavelength (edit: well, half a wavelength, since up and down are both vertical). But it takes on all polarization directions (in different places) at a given time, and there is always some place at which it's vertically polarized. I'm assuming, of course, that extends over multiple wavelengths, otherwise we can't talk very meaningfully about its frequency at all. The places it's vertically polarized propagate at c, and they pass by fixed points in space at the frequency of the photon. When you measure the photon's frequency, you're measuring the oscillations of the field at a fixed spot in space.

 

[sol invictus]

A circularly polarized photon is not vertically polarized at a certain time, it's vertically polarized in certain places, spaced apart by one wavelength. But it takes on all polarization directions (in different places) at a given time, and there is always some place at which it's vertically polarized. I'm assuming, of course, that it's got multiple wavelengths, otherwise we can't talk very meaningfully about its frequency at all. The places it's vertically polarized propagate at c, and they pass by fixed points in space at the frequency of the photon. When you measure the photon's frequency, you're measuring the oscillations of the field at a fixed spot in space.

Still not following you.

I'll do the experiment like this: I'll shoot one photon at a time down the pipe. Before each shot, I'll set up a vertical polarizer at some location along it, with a photodiode just behind it to see if the photon passed through or was absorbed. I'll assume I have a device that generates right-circularly polarized photons with a given phase, the same every time.

What I'll find is that there are certain locations for the polarizer where the photon always makes it through, and others where it never does, and at the rest it passes with some probability. And of course it takes it a time d/c to arrive at the photodiode a distance d away. That sounds absolutely identical to the clock experiment - but the time dilation isn't infinite.

How can this happen? Simple - it's a massless particle, so it can have finite energy (and frequency) at speed c.

 

[Ziggurat]

Still not following you.

I'll do the experiment like this: I'll shoot one photon at a time down the pipe. Before each shot, I'll set up a vertical polarizer at some location along it, with a photodiode just behind it to see if the photon passed through or was absorbed. I'll assume I have a device that generates right-circularly polarized photons with a given phase, the same every time.

What I'll find is that there are certain locations for the polarizer where the photon always makes it through, and others where it never does, and at the rest it passes with some probability.

No, that is not what you will find. You will find that the photon passes through with a 50% probability regardless of orientation and regardless of position, because right circular polarization is a 50/50 superposition of two perpendicular polarizations (doesn't matter which two, any basis will do) with a specific phase offset between them. The polarizer blocks one of those and lets the other through.

In terms of the polarization direction, this means that at the point in space immediately prior to the polarizer, the polarization direction is constantly rotating at the frequency of the photon. The photon does NOT have one polarization direction everywhere which is rotating in time, it has all polarization directions (like a corkscrew) at all times, and the profile moves at c so that any stationary point sees the polarization direction rotating.

 

[Vorpal]

No, that is not what you will find. You will find that the photon passes through with a 50% probability regardless of orientation and regardless of position, because right circular polarization is a 50/50 superposition of two perpendicular polarizations (doesn't matter which two, any basis will do) with a specific phase offset between them. The polarizer blocks one of those and lets the other through.

That reasoning seems broken. If we apply the same logic to the reverse situation of launching linearly polarized photons into a circular polarizer, we get something that's very obviously wrong (because of the way circular polarizers are constructed). Since any such photon would also be a superposition of opposing circular polarizations of arbitrary orientation, we get the same conclusion of a 50% transmission rate regardless of relative orientation of the circular polarizer. And that's just not right.

Are you sure you're not thinking of the average of polarized but otherwise incoherent light?

 

[sol invictus]

No, that is not what you will find. You will find that the photon passes through with a 50% probability regardless of orientation and regardless of position, because right circular polarization is a 50/50 superposition of two perpendicular polarizations (doesn't matter which two, any basis will do) with a specific phase offset between them. The polarizer blocks one of those and lets the other through.

Yes, I think you're right. But that doesn't affect my point so long as I can construct experiments which are sensitive to how far the photon has traveled. But I'm pretty sure I can - for example, I could interfere it with another photon, or with itself traveling along some other path, and the interference pattern that results will depend on the length it propagated down the pipe.

And that's all I need - once you agree some physically observable characteristic of the photon changes in a regular way as it propagates, you've agreed it's a clock that is not infinitely time dilated. Its phase does change, and phase is observable via interference. So I think I could "read" my photon clock as it moves along, just as I can read a regular clock - but the photon's second hand keeps ticking.

 

[Ziggurat]

That reasoning seems broken. If we apply the same logic to the reverse situation of launching linearly polarized photons into a circular polarizer, we get something that's very obviously wrong (because of the way circular polarizers are constructed). Since any such photon would also be a superposition of opposing circular polarizations of arbitrary orientation, we get the same conclusion of a 50% transmission rate regardless of relative orientation of the circular polarizer. And that's just not right.

Except we can't apply the same logic, because circular polarizers (ie, quarter-wave plates) don't work by absorbing the "wrong" circular polarization. They work by retarding one of the linear photons so they have the right phase difference. This means that unlike linear polarizers, they always have (in principle) 100% transmission rates. And the proper basis set (to keep the operator diagonal) is still the linear polarizations, not the circular polarizations.

Are you sure you're not thinking of the average of polarized but otherwise incoherent light?

Yes, I'm sure.

 

[sol invictus]

That reasoning seems broken. If we apply the same logic to the reverse situation of launching linearly polarized photons into a circular polarizer, we get something that's very obviously wrong (because of the way circular polarizers are constructed). Since any such photon would also be a superposition of opposing circular polarizations of arbitrary orientation, we get the same conclusion of a 50% transmission rate regardless of relative orientation of the circular polarizer. And that's just not right.

Let's see: a typical circular polarizer first projects onto a linear polarization, and then rotates that into circular polarization. Correct?

If so, it can't be thought of as a projector onto circular polarization - it's not a projector at all, actually (whereas a linear polarizer really is a projector onto some linear polarization). It's just a linear polarizer followed by a unitary operation that converts that polarization into circular polarization.

So in other words, the situation isn't reversible like that - it's not symmetric.