Describing the Life of a Photon

[enceladus_]

I heard Neil deGrasse Tyson talk about the life of a photon that was emitted from the Sun and hit oneself. He said that its entire life was an instant. I hope I'm not misquoting or misunderstanding what he said.

With that in mind, does that instant take into account the thousand of years (relative to us) it spends trapped in the mantle and core of the Sun, trying to get out?

Also, how would one describe the life of a photon that is emitted from the Sun, or any Star for that matter, and travels indefinitely throughout the Universe, never coming into contact with anything? Is this an indefinite instant?

 

[ghwellsjr]

I heard Neil deGrasse Tyson talk about the life of a photon that was emitted from the Sun and hit oneself. He said that its entire life was an instant. I hope I'm not misquoting or misunderstanding what he said.

With that in mind, does that instant take into account the thousand of years (relative to us) it spends trapped in the mantle and core of the Sun, trying to get out?

Also, how would one describe the life of a photon that is emitted from the Sun, or any Star for that matter, and travels indefinitely throughout the Universe, never coming into contact with anything? Is this an indefinite instant?

A photon only exists from its point of emission to its point of contact with something so you don't have to worry about all the other photons that are trapped in the sun with regard to one that escapes.

I don't know what Neil said but he should have made it clear that time does not apply to photons instead of implying that time for a photon is zero.

 

[PeterDonis]

I heard Neil deGrasse Tyson talk about the life of a photon that was emitted from the Sun and hit oneself. He said that its entire life was an instant. I hope I'm not misquoting or misunderstanding what he said.

I don't think you're misquoting or misunderstanding; but I do think Tyson is stating things in a way that is likely to cause confusion. (He's not the only one, btw: we often get threads here on topics like this after one of Brian Greene's specials airs on PBS.)

The correct way to say what Tyson was saying is that a photon's worldline has zero length. In the simplest case, where gravity is negligible so we can use the formulas of special relativity, the length of a worldline is given by the Minkowski metric:

\tau^2 = c^2 t^2 - x^2

which is basically a spacetime version of the Pythagorean theorem: t is the travel time of the photon in some inertial frame, and x is the distance the photon covers in that time in the same frame. Zero length for a photon's worldline means \tau^2 = 0, which just means that ct = x: the photon moves at the speed of light.

Really what this is saying is that objects that move at the speed of light, like photons, are fundamentally different, physically, from objects like us, that move slower than light, and for which \tau^2 > 0. In that case, of objects moving slower than light, \tau is just the time elapsed on a clock that moves with the object. By analogy, then, pop science discussions often say that "no time elapses for a photon" because \tau = 0 for a photon.

However, the analogy is flawed because \tau has another meaning for objects that move slower than light (we call the worldlines these objects move on "timelike", by contrast with "null" worldlines that photons move on): each event on a timelike worldline can be labeled by a unique value of \tau (which we normally refer to as the "proper time" of the event according to the observer following the worldline). Interpreting \tau = 0 for a photon as meaning the photon "only takes an instant" to go from, say, the Sun to the Earth, would imply that the photon's worldline is not a line but a single point--a single event. Obviously it's not, because we could put a photon detector anywhere between the Sun and the Earth and detect the photon, and each possible location for the detector must be a distinct point on the photon's worldline. So \tau for the photon simply doesn't mean "time" the way it does for a timelike object.

With that in mind, does that instant take into account the thousand of years (relative to us) it spends trapped in the mantle and core of the Sun, trying to get out?

No; that's a whole separate discussion. The "instant" only refers to the photon traveling from the surface of the Sun to Earth, through vacuum.

Also, how would one describe the life of a photon that is emitted from the Sun, or any Star for that matter, and travels indefinitely throughout the Universe, never coming into contact with anything? Is this an indefinite instant?

Not really. The condition \tau = 0 for the photon's worldline still holds, but that doesn't make it an "instant". See above.

 

[A.T.]

By analogy, then, pop science discussions often say that "no time elapses for a photon" because \tau = 0 for a photon.

Interpreting \tau = 0 for a photon as meaning the photon "only takes an instant" to go from, say, the Sun to the Earth, would imply that the photon's worldline is not a line but a single point--a single event.

I see no problem with "no time elapses for a photon", because it indicates that proper time is meant. I agree that "only takes an instant" is misleading, because it leads to confusion with coordinate time.

 

[PeterDonis]

I see no problem with "no time elapses for a photon", because it indicates that proper time is meant. I agree that "only takes an instant" is misleading, because it leads to confusion with coordinate time.

This is fine as long as people don't immediately infer "only takes an instant" from "no time elapses". However, it seems like a lot of people do.

 

[Dale]

I see no problem with "no time elapses for a photon", because it indicates that proper time is meant.

I still have a problem with that because proper time is only defined along timelike worldlines. I would agree that the spacetime interval is 0, but I wouldn't call that 0 proper time.

 

[enceladus_]

No; that's a whole separate discussion. The "instant" only refers to the photon traveling from the surface of the Sun to Earth, through vacuum.

I don't see how this makes sense. Isn't the photon created by nuclear fusion in the core, and thus exists?

 

[A.T.]

I would agree that the spacetime interval is 0, but I wouldn't call that 0 proper time.

What's the downside of calling it proper time?

 

[JohnWisp]

I still have a problem with that because proper time is only defined along timelike worldlines. I would agree that the spacetime interval is 0, but I wouldn't call that 0 proper time.

May I ask about the origin of photons from the sun's nuclear furnace?

Where do they come from under the standard model?

 

[Bill_K]

I don't see how this makes sense. Isn't the photon created by nuclear fusion in the core, and thus exists?

The photon that escapes from the sun's surface is not the same one as was produced in the core. Photons do not bounce, they only travel in a straight line at velocity c until they hit something. At each collision, one photon A is absorbed and another photon B is created.

 

[Bill_K]

What's the downside of calling it proper time?

The adjective "proper" means "in the particle's own rest frame."

 

[JohnWisp]

The adjective "proper" means "in the particle's own rest frame."

So, is there math to describe the frame of the photon or not?

If not why not?

 

[DrGreg]

So, is there math to describe the frame of the photon or not?

If not why not?

"Frame" usually means "inertial frame", in which case the answer is no: a photon is not at rest in any inertial frame. See our FAQ: Rest frame of a photon.

 

[JohnWisp]

"Frame" usually means "inertial frame", in which case the answer is no: a photon is not at rest in any inertial frame. See our FAQ: Rest frame of a photon.

Yes, I could not agree with you more.

A photon does not occupy an inertial frame.

Now, I did not imply this and am back to my question.

So, is there math to describe the frame of the photon or not?

If not why not?

The frame would be that of the photon, not an inertial frame.

 

[Nugatory]

Is there math to describe the frame of the photon or not?

If not why not?

The frame would be that of the photon, not an inertial frame.

There is no math to describe the frame of a photon because "the frame of <something>" is just a convenient verbal shortcut for "a frame in which that <something> is at rest", and there is no such thing when the <something> is a photon.

Inertial or not is beside the point; a frame in which something is at rest is not necessarily an inertial frame.

 

[sophiecentaur]

I don't see how this makes sense. Isn't the photon created by nuclear fusion in the core, and thus exists?

This assumes that it has to be the 'same photon' throughout the journey. If you think in terms of the energy propagating from inside the Sun to its surface as a series of interactions between ions (nuclei?) on the way through, you are really talking in terms of a whole series of photons, for each step in the journey. Energy finally arrives at the surface, is emitted, in quanta (photons) which have no individual identity, until the energy arrives at Earth, where a further photon interaction occurs when you 'see' the light. Photons don't actually need to 'exist' at all, on the way - just at each end, with the energy existing in the form of waves in between. The 'little bullet' model is not a good one.

 

[Dale]

What's the downside of calling it proper time?

There are several downsides. First, why should it be proper time and not proper distance? Second, proper time is the time measured by a clock traveling along that worldline and no clock can travel that worldline. Third, a timelike worldline can be parameterized by proper time, but a null worldline cannot, you have to introduce an affine parameter instead. Fourth, for geodesics in flat spacetime the proper time is the coordinate time in the reference frame where the starting and ending events are co-located, there is no such frame for light.

 

[Dale]

May I ask about the origin of photons from the sun's nuclear furnace?

Where do they come from under the standard model?

They come from conservation of energy and momentum in many of the nuclear reactions that take place there.

 

[Dale]

Yes, I could not agree with you more.

A photon does not occupy an inertial frame.

Now, I did not imply this and am back to my question.

So, is there math to describe the frame of the photon or not?

If not why not?

The frame would be that of the photon, not an inertial frame.

There is a subtle distinction between a frame and a coordinate system. You can make a coordinate system where a pulse of light is at rest, but he coordinate basis of such a coordinate system is not orthonormal, so it doesn't form a valid reference frame. Frames must, by definition, be orthonormal, but a pulse of light is null.

 

[enceladus_]

Many thanks for all the responses. Much appreciated.

 

[pervect]

Yes, I could not agree with you more.

A photon does not occupy an inertial frame.

Now, I did not imply this and am back to my question.

So, is there math to describe the frame of the photon or not?

If not why not?

Short answer: there are coordinates, called null coordinates, in which some photons (usually photons going in a particular direction) have constant coordinates.

But I wouldn't call these coordinates a frame. I'd call them coordinates.

I haven't surveyed the literature in great deal on this topic to make sure nobody ever calls them a frame. But I've always seen them referred to as coordinates, specifically null coordinates.

Here's an example of null coordinates:

Suppose you have some inertial frame [t,x,y,z]
Then you define coordinates [U,V,Y,Z] by the following transformation:

U = t-x
V = t+x
Y=y
Z=z

and you have null coordinates. A photon moving in the +x direction has a constant U coordinate, a photon moving in the -x diretion has a constant V coordinate.

You can get a metric for these coordinates that express distance - if you're familiar with metrics.

dt^2 - dx^2 - dy^2 - dz^2 == du dv - dy^2 - dz^2

These coordinates have most of what you'd expect of the usual non-null coordinate system, except any notion of "simultaneity". You can give an objects position in space in time by specifying (U,V,Y,Z). You can determine the invariant interval (the Lorentz interval) between two objects given both their coordinates.

You don't have a time coordinate, so there's no ordinary notion of simultaneity , at least not any obvious one.

You still have the usual notion of "light cones" which determines cause and effect.

 

[JohnWisp]

There is no math to describe the frame of a photon because "the frame of <something>" is just a convenient verbal shortcut for "a frame in which that <something> is at rest", and there is no such thing when the <something> is a photon.

Inertial or not is beside the point; a frame in which something is at rest is not necessarily an inertial frame.

Can you explain why a math theory like SR does not have have a mathematical explanation of the logic/frame of a photon?

It would seem a mathematical theory like SR can explain all the objects it conjectures, like the photon.

Is this false?

 

[JohnWisp]

Short answer: there are coordinates, called null coordinates, in which some photons (usually photons going in a particular direction) have constant coordinates.

But I wouldn't call these coordinates a frame. I'd call them coordinates.

I haven't surveyed the literature in great deal on this topic to make sure nobody ever calls them a frame. But I've always seen them referred to as coordinates, specifically null coordinates.

Here's an example of null coordinates:

Suppose you have some inertial frame [t,x,y,z]
Then you define coordinates [U,V,Y,Z] by the following transformation:

U = t-x
V = t+x
Y=y
Z=z

and you have null coordinates. A photon moving in the +x direction has a constant U coordinate, a photon moving in the -x diretion has a constant V coordinate.

You can get a metric for these coordinates that express distance - if you're familiar with metrics.

dt^2 - dx^2 - dy^2 - dz^2 == du dv - dy^2 - dz^2

These coordinates have most of what you'd expect of the usual non-null coordinate system, except any notion of "simultaneity". You can give an objects position in space in time by specifying (U,V,Y,Z). You can determine the invariant interval (the Lorentz interval) between two objects given both their coordinates.

You don't have a time coordinate, so there's no ordinary notion of simultaneity , at least not any obvious one.

You still have the usual notion of "light cones" which determines cause and effect.

How is cause and effect maintained in the coordinate system of a photon?

Also, if a coordinate system is accepted, we can then ask questions like when will the photon reach a position in the coordinate system.

So, can you explain this?

 

[Dale]

Can you explain why a math theory like SR does not have have a mathematical explanation of the logic/frame of a photon?

SR is a physics theory, not a math theory. It does have a mathematical explanation of the logic of a pulse of light, and that logic forbids a pulse of light from having a frame. There is no reason that everything needs to have a frame. In fact, one of the founding postulate is that ANY inertial frame is just as good as any other, so you never need to use a frame where anything is at rest, you can always use one where the object of interest is moving.

It would seem a mathematical theory like SR can explain all the objects it conjectures, like the photon.

Is this false?

Again, it is a physical theory, and it is QED that conjectures photons, not SR. However QED uses SR, so that is probably an irrelevant distinction. QED can explain everything about a photon and does not require it to be at rest in any inertial frame in order to do so. You are never required to use an object's rest frame to explain it.

 

[Dale]

How is cause and effect maintained in the coordinate system of a photon?

Coordinate systems don't change anything about cause and effect.