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August 19

Is the speed of light constant, when measured by a remote inertial observer, who non-locally measures a photon traveling a long way in a curved vacuum?

I'm asking, because our article speed of light states confusingly: "In non-inertial frames of reference (gravitationally curved spacetime or accelerated reference frames)...the speed of light can differ from c when measured from a remote frame of reference". This sentence seems to ignore the situation I'm asking about, when the remote observer's frame of reference is inertial, but the spacetime the light travels through is curved. HOTmag (talk) 08:32, 19 August 2025 (UTC)[reply]

The length traveled by a photo should be the path length as measured along its curved trajectory, a geodesic of the manifold that is spacetime. I am not sure how you propose the stationary observer is going to measure this. It is in fact not even clear how to define the path length (in the mathematical model of curved spacetime, a Lorentzian manifold) with respect to a given, fixed frame of reference. Inertial frames of reference are useful in special relativity, when objects not acted upon by a force travel in straight lines. Space curvature means that there are no "straight lines", so the inertial model for establishing a reference frame breaks down. ‑‑Lambiam 14:07, 19 August 2025 (UTC)[reply]

Let's assume we (as inertial observers) see a photon travel near the sun in a curved trajectory. Do you claim we can't use any tool (e.g. a telescope or whatever) for measuring the length of this photon's curved trajectory? HOTmag (talk) 15:59, 19 August 2025 (UTC)[reply]

We can detect only photons that arrive at our location. If a remote photon interacts with something else in such a way as to cause emission of another photon in our direction, we can detect the resulting photon but we're not directly observing the trajectory of the initial one.

Saying "what if as remote observers we see a photon travel near the sun" is like saying "what if as fans watching a soccer match from 10 miles away, we get hit by the ball on its way from the players foot to the goal". A remote observer can't observe a photon's trajectory. -- Avocado (talk) 17:49, 19 August 2025 (UTC)[reply]

So what does the quote (from Wikipedia) in my original post mean, about when c is "measured from a remote frame of reference"? Doesn't the measurement of c made by a remote observer, mean measuring the ratio between, the photon's trajectory measured by that remote observer, and the time it takes the photon to travel this trajectory - when this time is measured by that remote observer? HOTmag (talk) 18:29, 19 August 2025 (UTC)[reply]

I'm not a physicist nor the person who wrote the article. I would assume that we can know the time of the photon's origin based on whatever caused it to be emitted also having other effects (gravitational waves, other photons, etc) that reach us directly. And that we can measure the time of the photon's arrival at another point based on the effects of its arrival (reflected or re-emitted light, for instance) that reach us directly. And that we can thus measure the time elapsed between departure and arrival and deduce its speed. But we can't observe its trajectory, only infer it. -- Avocado (talk) 20:15, 19 August 2025 (UTC)[reply]

Please note that the condition of "local measurement" (as opposed to "non-local" one) is a well known requirement for the speed of light to be constant. I've asked whether the requirement of locallity of measurement is also needed when the observer's frame of referenece is inertial. HOTmag (talk) 06:47, 20 August 2025 (UTC)[reply]

You can imagine that you have a torch in your hand and point it towards a remote black hole. The light from the torch will travel in the direction of the event horizons but will never cross it (from the point of view of an external inertial observer). This effectively means that the speed of light becomes zero in the vicinity of the horizon. However the proper speed of light will remain c of course. Ruslik_Zero 20:33, 19 August 2025 (UTC)[reply]

When a photon is approaching a black hole, both the distance traveled by the photon, and the time it takes the photon to travel that distance, approach infinity (from the inertial observer's viewpoint), so the "effective" velocity becomes meaningless rather than "zero". HOTmag (talk) 06:47, 20 August 2025 (UTC)[reply]

Sorry, but the distance cannot become infinite because it is a known quantity. Indeed, you can measure the distance to the black hole and its mass and then calculate the distance to the horizon from the observer.

Actually there is no need to use black holes at all. You can put a mirror on the Earth's surface and direct the laser beam at it from a remote location in space. Then since you know the distance and can measure the time when the reflected signal comes back you can calculate the speed by dividing the first quantity by the second. The result will be that the (apparent) speed of light is less than c. Ruslik_Zero 10:39, 20 August 2025 (UTC)[reply]

I can see some practical issues with measuring the distance to a black hole. And also some theoretical issues. ‑‑Lambiam 16:53, 20 August 2025 (UTC)[reply]

Any black hole is just a mass. You need only to measure the orbital parameters of test particles moving around it. Ruslik_Zero 17:34, 20 August 2025 (UTC)[reply]

This is far from the first time I have been exposed to these facts, but this concept still breaks my brain a little. I think it's on account of how we utilize the notion of an observer from an outside frame of reference as an abstraction. Obviously, in terms actual empirical observation at this point, the photon is completely red-shifted and has no chance of ever escaping. So it can't ever be directly observed. And yet we regard it as being unable to ever being able to be observed to have crossed the event horizon. Can someone help me with the structural distinction here? Because obviously if we had a photon's trajectory bent around the gravity well of a black hole (or any mass), we could observe it only by directly interacting with it by intercepting it somewhere along its path. So what do we mean when we talk about observation in an instance that is not in any scenario actually physically possible? SnowRise ^{let's rap} 06:44, 24 August 2025 (UTC)[reply]

Just a small remark: "red-shifted" (as you say), only when it tries to escape a black hole, but here we are talking about a photon approaching a black hole, so it's blue-shifted. 2A06:C701:745A:B800:B559:3320:A4F4:C460 (talk) 10:22, 24 August 2025 (UTC)[reply]

Regardless of their colour (frequency), photons can only be directly observed when they hit the observer. This was already pointed out above by Avocado. They can only be observed, directly or indirectly, when they are detected by some detector, which means in quantum terminology that they are "measured". Measurement of a photon means a change in a macroscopic system (a photoreceptor cell in the observer's eye, a photographic plate or film, a photodetector, ...) as the result of an interaction with that system. Unless the measuring system is close to where the photon is, the probability of an interaction taking place is vanishingly small. ‑‑Lambiam 12:05, 24 August 2025 (UTC)[reply]

Do you claim, any measurement (e.g. by a telescope or whatever) of the length of a photon's curved trajectory - whether near the sun - or in any phenomenon of gravitational lensing, is a local measurement? HOTmag (talk) 13:11, 24 August 2025 (UTC)[reply]

Does Principle of locality help? {The poster formerly known as 87.81.230.195} 90.210.150.115 (talk) 18:03, 24 August 2025 (UTC)[reply]

I think you've mis-interpeted my inquiry here, Lambiam. As it happens, I'm a bit of an expert in visual cognition, and so very familiar with the physics/biophysics of photoreceptive media. That's not the part I am struggling to fix in my mind here. My epistemological confusion about the terminology is this: since a photon trapped at the event horizon never escapes to interact with such a medium, what do we mean when we talk about "observation" when, for example Ruslik0 says The light from the torch will travel in the direction of the event horizons but will never cross it (from the point of view of an external inertial observer).? Is it a conceptual conceit/misnomer for describing the relation of the frames of reference? If so, can you think of a thought experiment that would explain those interactions in such a way that accounts for the fact that, as a strictly empirical and ontological matter, no observation at a distance can be made? Maybe Ruslik0 just mixed their metaphors and terminology a bit? If not, I'm super confused as to what the act of observation means in that description. SnowRise ^{let's rap} 22:10, 24 August 2025 (UTC)[reply]

You are right, I misunderstood the essence of your post. My reaction was triggered by the statement connecting our inability to observe the photon to its colour, which is I think essentially correct – in the model its wavelength tends to zero as it approaches the event horizon – but irrelevant. Scenario's of a photon traveling to an event horizon can be described that conform to a mathematical model of GR, such as Schwarzschild's exact solution to Einstein's equations. Such descriptions need a frame of reference, preferably one that in the limit, away from the mass, is an inertial frame. I too think the wording of these scenario's is sometimes confused. The scenario may include an observer for which this frame is stationary who can observe phenomena as predicted by the model, which in real life would validate the model. But such observation can only be through information that reaches them from afar, such as transmitted by electromagnetic waves. An astronaut approaching the event horizon might broadcast a livestream witness report that reaches the observer, but a photon can do no such thing. The models do not allow an observer to observe the unfolding of the scenario with regard to the traveling photon, so describing the scenario in terms of observations is confused. ‑‑Lambiam 23:58, 24 August 2025 (UTC)[reply]

What I actually meant is shapiro time delay, which can be interpreted as slowing of light in presence of a gravitational field. Ruslik_Zero 20:33, 25 August 2025 (UTC)[reply]