[jeffreyH (OP)]

If a photon is at point p1 at time t1 and later is at point p2 at time t2 how far has it actually travelled? "With respect to whom?", you ask. Is that the wrong question?

[alancalverd]

Nothing wrong with the question at all. We measure distance by bouncing photons off things (radar, lidar, interferometry) or exchanging encoded photons (DME) on the presumption that c is constant.

[Bill S]

Is that the wrong question?

Obviously, Alan's response is spot on.  As I see it, the only way to raise any question about this is to introduce the idea that the photon might "experience" something different, and we all know where that goes.  

Did you have something else in mind?

[PmbPhy]

If a photon is at point p1 at time t1 and later is at point p2 at time t2 how far has it actually travelled? "With respect to whom?", you ask. Is that the wrong question?

That depends on whether the spacetime is curved or not.

[jeffreyH (OP)]

Is that the wrong question?

Obviously, Alan's response is spot on.  As I see it, the only way to raise any question about this is to introduce the idea that the photon might "experience" something different, and we all know where that goes.  

Did you have something else in mind?

It is not about a photon experiencing anything. It is about the distance covered by the photon. There is no fixed background against which to measure the distance a photon has travelled so how can we possibly measure the actual distance. Things are all relative. It is akin to asking what stationary means. Again the response can only be "With respect to what?"

[Janus]

Is that the wrong question?

Obviously, Alan's response is spot on.  As I see it, the only way to raise any question about this is to introduce the idea that the photon might "experience" something different, and we all know where that goes.  

Did you have something else in mind?

It is not about a photon experiencing anything. It is about the distance covered by the photon. There is no fixed background against which to measure the distance a photon has travelled so how can we possibly measure the actual distance. Things are all relative. It is akin to asking what stationary means. Again the response can only be "With respect to what?"

Generally, the answer to with "respect to what?" is " with respect to the frame from which it is being measured" (the frame in which t1 and t2 is being measured and which p1 and p2 are at rest with respect to ). 

[alancalverd]

Pete's point is also true. The photon has travelled p2 - p1 on the geodesic of whatever surface constitutes the local spacetime, since that is the definition of geodesic.

[guest45734]

As I see it, the only way to raise any question about this is to introduce the idea that the photon might "experience" something different, and we all know where that goes.   

I am guessing . Are you referring to time stopping for a photon and distances shrinking to zero from the photons view point. Am I correct in thinking the distance between p1 and p2 does not exist for a photon

Edit found this link https://phys.org/news/2014-05-does-light-experience-time.html

[jeffreyH (OP)]

As I see it, the only way to raise any question about this is to introduce the idea that the photon might "experience" something different, and we all know where that goes.   

I and am guessing . Are you referring to time stopping for a photon and distances shrinking to zero from the photons view point. Am I correct in thinking the distance between p1 and p2 does not exist for a photon

My original question had nothing to do with a photon experiencing anything. A photon always travels a finite distance in a finite time. At the particle level you can talk about particles experiencing a force but not the passage of time. Even then the word experience is interpreted differently to everyday use. The particle doesn't contemplate the force that affects it.

[Bill S]

My original question had nothing to do with a photon experiencing anything.

Good. That never goes anywhere.

A photon always travels a finite distance in a finite time.

In the RF of a finite observer in 3+1D spacetime; which is all we can reasonably deal with. (?)

[Bill S]

Pete's point is also true. The photon has travelled p2 - p1 on the geodesic of whatever surface constitutes the local spacetime, since that is the definition of geodesic.

Would I be right in thinking that the definition of a geodesic also maintains that it is the “shortest” route, in curved spacetime, between two points?

The shortest distance between two points on a sphere would vary if the diameter of the sphere changed. Is my thinking correct?

[PmbPhy]

My original question had nothing to do with a photon experiencing anything. A photon always travels a finite distance in a finite time. At the particle level you can talk about particles experiencing a force but not the passage of time.

I think you're operating under a false assumption. First off particles don't "experience" anything no matter what the particle is. If the particle is an electron we have knowledge of where it started and where it ended including the path it took they we in the lab frame use our clocks to measure "our" time of flight. We then imagine what a clock moving with the electron measures. This can't be done with photons and it is therefore impossible to say anything about its time of flight. Einstein was the first to show why. When you said "the photon is at point 1" and the same for point 2 then you in reality set up a coordinate system without telling us, you sly daug!   Then, assuming it moves in a straight line, you know the time of flight.

Its a subtle point since most people say things like that without knowing its meaning.

[Bill S]

Pete; I think you may be blaming Jeffrey for something for which I am responsible.  I actually introduced the idea of photons experiencing something.  I was expressing the hope that that was not the direction in which Jeffrey was going.  I should have known better.

[Bill S]

As geodesics figure relevantly in this thread, I have a couple of questions that could be on topic.

1. Is the curvature of spacetime only a mathematical factor, used because it describes gravity better than does Newton's law?
2. If the answer to “1” is yes; does the distance between two points on a geodesic actually vary with any change in this mathematical “curvature”?

[PmbPhy]

As geodesics figure relevantly in this thread, I have a couple of questions that could be on topic.

1. Is the curvature of spacetime only a mathematical factor, used because it describes gravity better than does Newton's law?
2. If the answer to “1” is yes; does the distance between two points on a geodesic actually vary with any change in this mathematical “curvature”?

No. It's actually changes in spatial distances and is in part responsible for the measured Shapiro time delay.

[Bill S]

Thanks Pete.  I still trying to clarify my understanding of the actual, physical, curvature of spacetime. 

In relation to what is it curved?  Itself?

Dr Baird says of spacetime: "..., it's not curved in the sense of a bar being bent to one side. It's curved in the scientific sense of spacetime behaving differently from one point to the next.”

[jeffreyH (OP)]

The potential acceleration due to gravity varies with change in radial distance from the source. This relates to the concept of curvature of spacetime.

[PmbPhy]

Thanks Pete.  I still trying to clarify my understanding of the actual, physical, curvature of spacetime. 

In relation to what is it curved?  Itself?

The curvature is intrinsic to the manifold itself. E.g. intelligent ants of a sphere can measure the curvature at any on the manifold itself. Whereas the and can't detect it on a cylinder or cone.

Time cannot be said to bend or curve.

[evan_au]

Time cannot be said to bend or curve.

In my primitive understanding, as a gravitational wave passes, it distorts spacetime.
- So the width of the Earth changes (very slightly - less than the width of a Hydrogen atom)
- And time goes faster & slower (very slightly)
- It is apparently the simultaneous distortion of space and time that makes the effect detectable

So maybe time doesn't bend or curve, but it does compress & expand?
Listen to Brian Cox (some comedy included): https://www.bbc.co.uk/programmes/b09kxt28

[Janus]

Getting back to the original question and how the length of the photon's path can depend on the reference frame, consider the following scenerio:

You have a room 299.792458 m long.  You fire a photon from one end to the other. At the ends of the room are clocks that have been synchronized.  If both clocks read 0 when you fire the photon, then the clock at the receiving end will read 1 us (microsecond) upon the photon reaching it,    The length of its path is 299.792458m over a time interval of 1 us.

However, if you were measuring this from a frame in which the room was traveling at 0.5c in the same direction as the room, then you would measure the following:
The length of the room would be 259.6278845 m  due to length contraction.  The photon, traveling at c relative to you would have to case after and catch the receiving clock which is moving at 0.5 c.  By this frame's clock this will take ~1.732050808 us.  This means that, in this frame, the photon will travel a distance of ~519.255769 m
 
During the ~1,732050808 us, the clocks in the room will be ticking slow due to time dilation and 1.5 us will tick off on each clock.   Due to the relativity of simultaneity, the clock in the room will not be synchronized, but will be offset by 0.5 us, with the receiving clock being 0.5 us behind the transmitting clock.  Thus when the photon leaves the first wall, the clock there reads 0 and the far wall clock reads -0.5 us.    In the time it takes the photon to catch the receiving clock, both clocks advance 1.5 us,  the transmitting clock reads 1.5 us, and the receiving clock will read 1 us ( the same as what it reads in the frame of the room).