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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?

Latest comment: 7 days ago20 comments7 people in discussion

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

August 26

Pharmacology

Latest comment: 6 days ago3 comments3 people in discussion

A friend once mentioned a book similar to an Encyclopedia, describing background events behind the development of many well known medicines . Please inform if a similar book can be found and how to "custom search" at any of the sites of WIKI for such a book . Thnx Dr chifti (talk) 05:05, 26 August 2025 (UTC)Reply

You might find such a work used as one of the many references for the article History of medicine, athough what you describe would be a Tertiary source (like Wikipedia itself) rather than a Secondary source which Wikipedia prefers for article sources.

Searching Wikipedia for the term "Encyclopedia of pharmacology" led me to the article Pharmacology Research & Perspectives whch uses as its reference 4 The Sage Encyclopedia of Pharmacology and Society – see that article for its bibilographical details. {The poster formerly known as 87.81.230.195} 90.210.150.115 (talk) 08:04, 26 August 2025 (UTC)Reply

Searching Archive.org for history of medicines turns up many candidates, including Our Modern Medicines (F Bandelin, 1986) which seems to match your description. -- Verbarson  ^talk_edits 15:04, 26 August 2025 (UTC)Reply

August 27

Lorentz factor derivarion

Latest comment: 1 day ago11 comments7 people in discussion

In special relativity, a common way to derive the Lorentz factor ${\displaystyle \gamma }$  is via the pythagorean theorem, without using the complicated Lorentz transformations (which I myself have yet to study).

By the equation

${\displaystyle (c\,{\rm {T_{s}}})^{2}=(v\,{\rm {T_{s}}})^{2}+(c\,{\rm {T_{v}}})^{2}}$ ,

such that

${\displaystyle c}$  speed of light in a vaccum

${\displaystyle v}$  speed ​​of a moving observer relative to a stationary observer

${\displaystyle {\rm {T_{s}}}}$  stationary observer time

${\displaystyle {\rm {T_{v}}}}$  moving observer time

hence

${\displaystyle \gamma ={\frac {\rm {T_{s}}}{\rm {T_{v}}}}={\frac {1}{\sqrt {1-{\big (}{\frac {v}{c}}{\big )}^{2}}}}}$ 

My questions are as follows:

1. Is this derivation actually legitimate, or does it contain any non-proven hidden assumptions?
2. I have not yet seen any mathematical reasoning showing these transformations must be linear.
   In my humble opinion, most sources completely ignore this point, or treat it superficially at most.
   I would like to know if there is any such reasoning.
3. Why is the factor not written ${\displaystyle \gamma (v)}$ ?

יהודה שמחה ולדמן (talk) 15:29, 27 August 2025 (UTC)Reply

It is not clear where the first equation comes from. Ruslik_Zero 20:25, 27 August 2025 (UTC)Reply

Watch this short video here. It is very common. יהודה שמחה ולדמן (talk) 04:11, 28 August 2025 (UTC)Reply

That the equation is common doesn't mean it's clear where it comes from. It's not a priori clear that in a triangle with edges ${\displaystyle cT_{s}}$ , ${\displaystyle vT_{s}}$  and ${\displaystyle cT_{v}}$  the latter two edges must be perpendicular. When I learned about special relativity, it started with the Lorentz transformations. Those aren't terribly hard; this was my first week at university. It could be done in the last year of secondary school. PiusImpavidus (talk) 09:14, 28 August 2025 (UTC)Reply

It's worth highlighting that the 4-space Lorenz transformations take place in is non-Euclidian. So it doesn't use the Euclidian metric, normally calculated with Pythagoras's theorem. Instead it uses a version with the squares of the three length components added but the time component subtracted.

If you eliminate two of the length components this looks like a difference of squares. Rearrange these you get a sum and it looks like Pythagoras's theorem, but because it's been rearranged it isn't really the same thing. The minus sign in the the second expression more correctly expresses the metric. --217.23.224.20 (talk) 10:38, 28 August 2025 (UTC)Reply

It is not clear who introduced the convention of using the Greek letter ${\displaystyle \gamma }$  for the Lorentz factor (not Lorentz himself, nor Einstein, who used the letter ${\displaystyle \beta }$  in "Zur Elektrodynamik bewegter Körper"*), but since it tends to be all over the place in derivations in special relativity, the one-letter notation is obviously more convenient than writing each time ${\displaystyle \gamma (v)}$ .  ​‑‑Lambiam 10:41, 28 August 2025 (UTC)Reply

---

 • Perhaps Bethe in "Quantenmechanik der Ein- und Zwei-Elektronenprobleme" (1933). This is behind a paywall; I cannot check if he actually used this notation.  ​‑‑Lambiam 11:08, 28 August 2025 (UTC)Reply

Bethe uses ${\displaystyle \varepsilon =E/E_{0}}$  (his eq. 9.16; ${\displaystyle E_{0}}$  is the rest energy), which is the Lorentz factor; I don't think he expresses it in terms of v and c anywhere. (Accessed via Wikipedia Library, which once again proved more powerful than my university account...). --Wrongfilter (talk) 11:34, 28 August 2025 (UTC)Reply

To point 3: Writing ${\displaystyle \gamma (v)}$  declares only that ${\displaystyle \gamma }$  is a function of a variable ${\displaystyle v}$  thus ${\displaystyle \gamma =\gamma (v)}$  which is true but conceals the nature of the function. Lorentz factor ${\displaystyle \gamma }$  is a dimensionless ratio that is a non-linear function of a velocity ${\displaystyle v}$  that for any real mass is constrained to be less than ${\displaystyle c}$ . For a massless particle such as a photon, calculating a unityan infinite Lorentz factor adds no useful information. To point 2: Given the Lorentz equation confirmed in Special relativity

${\displaystyle \gamma ={\frac {1}{\sqrt {1-{\big (}{\frac {v}{c}}{\big )}^{2}}}}}$ 

the OP chooses by algebraic reasoning to derive an arbitrary relation

${\displaystyle (c\,{\rm {T_{s}}})^{2}=(v\,{\rm {T_{s}}})^{2}+(c\,{\rm {T_{v}}})^{2}}$ 

but this result is flawed. It predicates two different observers each of whom would have to be moving at c, as only a photon can, if expressions ${\displaystyle (c\,{\rm {T_{s}}})}$  and ${\displaystyle (c\,{\rm {T_{v}}})}$  are to be real distances moved in a real time interval. It is hard for me to conceive of even one photon observing another photon and somehow reporting an observed distance. I don't believe that Pythagoras who relied more on the geometrical axioms of Euclid ca. 300 BC than on Einstein publishing in 1905 would appreciate or endorse being cited here. I see no other good source for the latter "pseudo-Pythagorean" equation and I conclude that it is introduced here as algebraic sleght-of-hand. The trick is to quote a supposed definition ${\displaystyle \gamma ={\frac {\rm {T_{s}}}{\rm {T_{v}}}}}$  into which the trickster plugs ${\displaystyle {\rm {T_{s}}}}$  and ${\displaystyle {\rm {T_{v}}}}$  that have been synthesized to give the standard result. To point 1: The ploy is clever but hardly legitimate. 2A02:FE1:4088:5E00:DC53:CFFC:3F4A:F5BE (talk) 15:33, 28 August 2025 (UTC)Reply

This is not a "Pseudo-Pythagorean" equation.

Perhaps this derivation here will be even clearer.

I do apologise for not being able to show a geometric sketch in advance. יהודה שמחה ולדמן (talk) 18:27, 28 August 2025 (UTC)Reply

Isn't this the same derivation as given in Time dilation § Simple inference?

The presentation in the YouTube video is (IMO) in one respect somewhat confusing. There are two observers, one inside a moving train car, one standing outside next to the railroad tracks in the grass. There are two time values for the different times clocked by these observers, ${\displaystyle T_{stationary}}$  and ${\displaystyle T_{moving}.}$  Now in the narrative of the video ${\displaystyle T_{stationary}}$  stands for the time clocked by the observer in the moving car, while ${\displaystyle T_{moving}}$  is the time clocked by the observer patiently standing outside till this ordeal is over. I beg your pardon. (To be fair, the narrative gives a reasonable explanation for this choice, but note that the notation in the question has this swapped: ${\displaystyle {\rm {T_{s}}}=T_{moving}}$  and ${\displaystyle {\rm {T_{v}}}=T_{stationary}.}$ )

The simple derivation given is based on vector decomposition of a velocity into a component in a given direction (in this case that of the velocity of the moving car) and an orthogonal component, and that this remains valid for a moving system observed from the outside. I don't know if this can be called an assumption – if so, it is hardly hidden — but the validity of this step may not be that obvious to the confused student all of whose implicit assumptions based on the Newtonian model of absolute time and space have lost their certainty and who may wonder what happened to length contraction.

The presenter himself states that this is not his favourite derivation, and that there is a better, but more difficult explanation.  ​‑‑Lambiam 08:47, 31 August 2025 (UTC)Reply

September 1

Getting rid of ants

Latest comment: 1 hour ago3 comments3 people in discussion

I am looking for a natural remedy to stop ants from entering my home. Is there any solution that I can make or create to get rid of them? Caralynn8 (talk) 20:28, 1 September 2025 (UTC)Reply

Maybe invest in an anteater? ←Baseball Bugs ^{What's up, Doc?} carrots→ 20:56, 1 September 2025 (UTC)Reply

Old Farmer's Almanac's website addresses this here. Modocc (talk) 21:09, 1 September 2025 (UTC)Reply