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Statement: "In optics, you can take the example of a concave mirror: the optical path chosen by the light to join two fixed points A and B is a maximum."

The statement gives the impression that the light ray travels the path with maximum optical length (maximum time) between any two fixed points when reflected off the surface of a concave mirror. (Usually, a light ray takes the path of least optical length, such as the case when reflected off a flat mirror or a refraction between mediums with positive refractive index). However, if you use a ray simulator you can configure an infinite number of point pairs so that the optical path is not maximised, as I have found out. So the statement can't be correct. right?

A generative AI answer for the query above (namely Bing AI):

The statement is false. The optical path chosen by the light to join two fixed points A and B is not always a maximum when reflected off a concave mirror. It depends on the position and shape of the mirror, as well as the position of the points A and B. There may be other ray paths that have longer or shorter optical paths than the one chosen by the light. The correct statement is that the optical path chosen by the light to join two fixed points A and B is stationary with respect to variations of the path when reflected off a concave mirror. This means that a small change in the path does not change the optical path length significantly. This is a consequence of Fermat’s principle of stationary time, which states that light travels between two points in such a way that the time taken is stationary with respect to variations of the path.

Original link to the statement: https://physics.stackexchange.com/a/144362/366787

Someone (earlier I also erroneously thought so) can say "The statement is made regarding symmetrical points off the axis and for such points, the ray should contact the pole of the mirror in order to traverse both points and such that, it is the maximum distance that the ray can traverse assuming only one reflection off the mirror." But such an assumption is wrong because we can always introduce kinks to the path of the ray like zigzag or bending to increase path length hence it is still an inflexion point. Legendary Richard P. Feynman had this say on a similar scenario in one of his lectures at Caltech :

"Actually, we must make the statement of the principle of least time a little more accurately. It was not stated correctly above. It is incorrectly called the principle of least time and we have gone along with the incorrect description for convenience, but we must now see what the correct statement is. Suppose we had a (flat) mirror. What makes the light think it has to go to the mirror? The path of least time is clearly AB. So some people might say, “Sometimes it is a maximum time.” It is not a maximum time, because certainly a curved path would take a still longer time! The correct statement is the following: a ray going in a certain particular path has the property that if we make a small change (say a one percent shift) in the ray in any manner whatever, say in the location at which it comes to the mirror, or the shape of the curve, or anything, there will be no first-order change in the time; there will be only a second-order change in the time. In other words, the principle is that light takes a path such that there are many other paths nearby which take almost exactly the same time."

Further explanation: The question is not about whether or not such rays have the ability to form usable images for practical applications, but about the stationery action of a light ray reflecting off a concave mirror. It is about whether the stationary action of such a ray is minimum, maximum or in-between (saddle point). For a plane mirror, for example, any ray travelling from point A to B through point C (Point C must be on the surface of the mirror) takes always the geometrically shortest path no matter where you geometrically configure those two points. (The point C automatically selected by the ray will be on the shortest path or rather only the ray that reaches B will be the one that reflected off point C on the shortest geometrical path). So what about a concave mirror? My guess is the stationary action of a ray being minimised, maximised or saddle point depends on the configuration of the mirror and the three points. So the statement "In optics, you can take the example of a concave mirror: the optical path chosen by the light to join two fixed points A and B is a maximum." is poorly conceived.

Duke William
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    Note that using AI for physics questions is terrible idea. – joseph h May 11 '23 at 04:45
  • I actually agree with it in this case, because it at least gives a starting point to the question @Duke William – tryst with freedom May 11 '23 at 04:47
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    @joseph h I agree that it is a terrible idea, but not so terrible regarding that they give instant answers, whereas in a forum you have to wait for an unknown time for some benevolent member kindly take interest in answering. No disrespect intended. – Duke William May 11 '23 at 04:54
  • You'll have to excuse us all for not immediately answering. The moment I saw this post I wanted to go straight home, since answering from my office won't do because of the possibility of work-related stuff. That's the least all "benevolent members" can do. Now, sarcastic comments aside on your and my part, there are on average five to ten members that read each post at this time of day/night, so having an expectation that you'll get a quick response and that members don't take an interest, comes off as disrespectful and slightly entitled. No disrespect either. Thanks. – joseph h May 11 '23 at 05:29
  • @joseph h Sometimes it is hard to express tone through a text medium. "Benevolent" in my previous comments was not intended as sarcasm. It is actually used as a metaphor to drive the point in using Generative AI for answering queries. As you are aware, not all persons looking for answers in physics or any field are students or academic communities of relevant fields. Some or most can be from completely different fields or just laymen like me looking for an answer, so they can move on to the next phase of whatever they are doing. As such using AI is very convenient, but also very risky. – Duke William May 11 '23 at 05:56
  • @joseph h I understand it takes reasonable time for a post to get answered, after all, we have other things to do in life or some questions may not even warrant an answer. Thank you for taking an interest in my question anyway (that is if your comment is not entirely meant for humour.) I did not intend to imply discontent about the delay in getting answers. Just that in some situations, generative AI is very convenient. But users have to understand the tendency of AI to misinform/ misguide users. So use common sense and through cross checking. – Duke William May 11 '23 at 06:11
  • You're absolutely welcome, and yes, my comment was intended as humor and as an attempt to explain the amount of member activity we have this time of day and that patience will give you the result you want. Cheers. – joseph h May 11 '23 at 06:19
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    Does this answer your question? When is the principle of stationary action not the principle of least action? – Miyase May 11 '23 at 06:21
  • @Miyase Thanks, but if you read my question entirely it is actually itself regards an answer to the very post you linked. I have even linked the page itself in my OP. – Duke William May 11 '23 at 06:49

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If I understand properly your question, there is a very simple example: the elliptical mirror. It is obvious that the foci $F_1$ and $F_2$ are conjugate since the optical path is constant: neither minimum nor maximum. This is a well-known property of the ellipse: the distance $F_1MF_2 = 2a$ is constant if $M$ is on the ellipsoid.

If $M$ is inside the ellipsoid, $F_1MF_2 < 2a$ and If $M$ is outside the ellipsoid, $F_1MF_2 > 2a$

Now, consider a curved mirror tangent to the ellipsoid (any shape, but tangent) at a point $H$: the ray $F_1HF_2$ verifies Descartes' law for this new mirror: it is a path actually followed by light.

If this tangent mirror is entirely inside the ellipsoid, for a point $M$ close to $H$ on the mirror, the distance $F_1MF_2$ is always smaller than $F_1HF_2$: we are therefore dealing with a maximum.

If this mirror is entirely outside the ellipsoid, for a point $M$ close to $H$ on the mirror, the distance $F_1MF_2$ is always greater than $F_1HF_2$: we are therefore dealing with a minimum.

If this mirror is on one side outside the ellipsoid and on the other inside, we are dealing with a saddle point.

EDIT : I specified that the tangent mirror in $H$ is not necessarily a plane mirror.

Hope it can help and sorry for my poor english.

Vincent Fraticelli
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  • In other words, you are saying for an elliptical mirror maximum and saddle points for optical length is possible right? This means that It is a valid example where stationery action becomes maximum or saddle point action rather than the usual least action we experience in classical mechanics, although it may be due to convention. That's why Fermats original wording, the principle of least time was later weakened for wide use as "time that is "stationary" with respect to slight variations of the path" – Duke William May 11 '23 at 12:25
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    Yes, that is what I think. When we use the principle of least action (to obtain Lagrange's equations for example), we never calculate the terms of order 2: we simply say that the variation is zero at first order. – Vincent Fraticelli May 11 '23 at 12:45
  • I do not understand this "If this tangent mirror is entirely inside the ellipsoid". How can a straight line (plane) be entirely inside a convex domain such as an ellipsoid? – hyportnex May 11 '23 at 14:31
  • The tangent mirror is not a plane mirror : It can have any shape. Simply tangent in H to the ellipsoid. – Vincent Fraticelli May 11 '23 at 15:55
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    @DukeWilliam A visualization of the concave-mirror-fermat-time case is on my website. The initial view is an ellipse shaped reflector, with point of emission and point of reception at $F_1$ and $F_2$. I refer to that as 'the critical case'. There is a slider to modify the shape of the ellipse, keeping the point of emission and point of reception fixed. Less concave than the critical case: Fermat time is a minimum; more concave than the critical case: Fermat time is a maximum. Scroll to: picture 6 Reflection and Fermat time – Cleonis May 11 '23 at 17:40
  • @Cloenis Thanks, it is a nice website. And yes, assuming the actual ray path locates the reflection point on the pole, it shows that depending on certain criteria for curvature, optical length can be either minimum or maximum for two fixed symmetrical points. I observed at 1.41 curvature (critical point) that it has the same constant length for any path connecting the three points. I do not know what to call that instance. Minimum or maximum or fixed optical length? It can't be a saddle point as there is no inflection. – Duke William May 11 '23 at 21:37
  • @Cloenis And there will be an infinite number of individual rays for that particular scenario that travel between points A and B reflected off everywhere along the mirror surface. – Duke William May 11 '23 at 21:37
  • @DukeWilliam On the question of how to refer to the neither minimum nor maximum case. That critical case is a cross-over point; it crosses from: 'the stationary time is a minimum of the Fermat time' to 'the stationary time is a maximum of the Fermat time'. The fact that a cross-over point exists demonstrates that Fermat's stationary time is not about minimum/maximum. Minimum or maximum is irrelevant. The actual criterion: the true path is the path such that the derivative of the Fermat time is zero. – Cleonis May 12 '23 at 03:32
  • @DukeWilliam Let me address why some people offer the suggestion that Fermat's time can be thought of as either a minimum, or a saddle point. This comes about when the variation space is expanded to allow the light to meander in an infinite number of ways. Then light propagating in a straight line is still possible, just not mandatory. Allowing meandering paths expands the variation space to infinitely many dimensions. The simplest instance where a saddle point can occur is when there are two degrees of freedom. By extension: infinite number of degrees of freedom allows saddle point. – Cleonis May 12 '23 at 03:57
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    @DukeWilliam One more point to address: why allow meandering? Here is, I think, the context of that. Given a light emitter A and a detector B: we can conceive of a hypothetical variation space such that the light can meander in any way as it travels from A to B. Within that hypothetical variation space the path along a straight line is uniquely the path of shortest time. In order to apply Fermat's stationary time to propagation from A to B in free space one allows meandering paths. If you wouldn't allow the hypothetical meandering then you wouldn't have a variation range to choose from. – Cleonis May 12 '23 at 04:18
  • @Cleonis Upon further contemplation upon the curved mirrors, it became obvious they don't actually have maximums as stationary paths, but rather saddle points because we can always come up with a longer path zigzagging or bending the ray. So actually I cannot think of a instance where maximum action comes into play in Optics. I have heard negative refraction as another possible phenomenon that action is maximum but how can that be? A flat mirror reflection can be considered also saddle point as a direct ray gives the shortest path. – Duke William May 16 '23 at 12:17
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The answer for a concave, rotational symmetric mirror:

The geometric path is the shortest path between A and B and a mirror point C on the surface, where the length is measured by multiples of the wavelength. Point C can be determined by the reflection principle: The path triangle has to be perpendicular to the tangent plane in C and both rays against the tangent plane in C have to be equal. Variation of C in a small circle around the point of reflection gives a quadratic variation of the distance of C.

This proof follows the same principles as for plane mirrors. The shortest path is the shortest time of course by the constancy of the speed of light c in vacuum for all wavelengths.

I have to admit, that the reflection pinciple for the shortest straight path between two points and a point of any concave manifold is not quite transparent to me by elementary geometry. Needs an afternoon of minimum detection in differential geometry.

  • Thanks, but the question is about whether actual rays reflected off a concave mirror follow the maximum time path hence the path with maximum geometrical length always. Regarding plane mirrors light rays "reflected off the surface" travel the least distant path (true when you consider only the rays reflected off, but not rays directly reach). But what about concave mirrors? Is it maximum action, least action or saddle point action or it differs according to the choice of mirror and points? I think we consider stationary action minimum when ray takes the path with the least time. – Duke William May 11 '23 at 07:08
  • Its the invariant length by wavenumber principle in general, that results in the minimum principle, if there exists one. Saddle principle would be relevant, if the mirror is convex-concave with curvature zwero in one point or everwhere, but such things dont make images, except for a laser ray of diameter zero. E.g an elliptic concave mirror maps tiny circular spot sources to ellipses, On a screen perpendicular to the ray, a s should add. As everybody knows, only rotation-paraboloid mirrors are devices, that are able to map points to points and are in use widely for telephoto lenses. –  May 11 '23 at 08:46
  • @ Roland F Thanks, can you give a definitive answer to the question? If a light ray connecting any two fixed points and the surface of a concave mirror follow the shortest geometrical path or maximal geometrical path or path in between? I am guessing that it depends on the choice of points and the mirror. (I understand that such rays might not be able to form coherent images.) – Duke William May 11 '23 at 09:19
  • The extreme of a concave mirror is an ellpsoid. Any two points connected over a point on the surface form a possible path of light with the conditions said.Take your points on the two focal points of any ellipse inside, then any path in this plane is a light path. The same is true for the rotation-paraboloid. Any point has an image over all straight paths. On the hyperboloid, the image is on the back somwehre inside the other sheet of the cone section, if 'reflexion' is defined accordingly. –  May 11 '23 at 10:41
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It is possibly helpful to stress that [within the set of (virtual) paths between 2 points A and B] there is never$^1$ a path of maximal time/optical length in the Fermat's principle. It is always possible to construct an even longer (virtual) path for the variational principle.

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$^1$ If space is 1-dimensional $d=1$, one can consider back-and-forth paths. If one rules out back-and-forth paths, then assume that $d>1$, so that paths are not spacefilling.

Qmechanic
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  • If A and B located symmetricaly in other words if A an B are mirror points of the main axis, then shouldn't the optical path (where point c is located at pole of the mirror), be geometrical maximum for a circular arc mirror? – Duke William May 11 '23 at 12:07
  • The main point is that the variational principle is over the set of virtual paths; not the set of actual paths. – Qmechanic May 11 '23 at 12:14
  • I think my original post is explicitly about actual paths as it also involves optical paths and action. Virtual path is usually helpful when the image forming is considered. In my question, point C is on the surface of the mirror whereas for virtual paths point C is formed behind the mirror. – Duke William May 11 '23 at 12:36
  • Well, the referenced post is about the variational principle and hence about virtual paths. – Qmechanic May 11 '23 at 12:41