Does light always take the shortest path?
And is it possible to change the probability of a photon travelling to a point by only disturbing the paths that are far away from the shortest path?
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Related: https://physics.stackexchange.com/q/167438/2451 and links therein. – Qmechanic Dec 26 '20 at 18:02
2 Answers
@Azzinoth My comment is too long and contains a sketch, so I'm posting it this way.
Your sketch shows only one point Z. For all other points on a radius from Q the same sketch must be drawn. I have done this in a sketch.
The large yellow circle on the left is a light source with a certain intensity represented by the diameter of the source. The yellow circles on the right represent the intensity of light, which in reality is of course a continuum. Except the area around the obstacle.
On obstacles the light gets deflected in a way that fringes of light occur. Not in the best way drawn, but perhaps you see the different intensities represented by (slightly) different diameters of the last three dots on the top. And in between there is no continuum but there are not exposured areas. Because portions of light on obstacles get deflected this way.
The shortest path model works excellently under the assumption of deflection of light at obstacles.

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- I only consider photons that hit my detector at Z, everything else is discarded. I don't need to consider all possible target points to disprove that light always takes the shortest path. Of course the placement of my obstacles is tailored for point Z and will not work for all other points. That does not change any of my conclusions.
- The shortest path model does not work. The shortest path from Q to Z is through M, right? So please explain how a photon traveling along that path is influenced by obstacles on the red paths in your model.
– Azzinoth Dec 26 '20 at 18:54 -
- You can create the same result without diffraction at objects if you replace the window with a mirror if you like that more. Light travels from the source Q to a point on the mirror and is then reflected to Z. The direct path from Q to Z is blocked, such that the light has to use the mirror. I can now scratch out some parts of the mirror such that the amplitude at Z changes, even tough these parts are not on the shortest path (which is the path that has incident angle=reflected angle).
– Azzinoth Dec 26 '20 at 18:54 -
If you want to discuss this further I suggest we move to this chat. https://chat.stackexchange.com/rooms/117709/does-light-always-take-the-shortest-path Long discussions in comments are afaik not allowed. – Azzinoth Dec 26 '20 at 19:01
Consider a light source Q shining through a window at a detector Z. The direct path from Q to Z goes through the windows center M.
According to wave optics to get the amplitude at Z we have to add the contributions (complex amplitudes) of all the "possible paths" through the window from Q to Z. If you graphically respresent each contribution as a vector and add them by putting them head to tail, you will see that the result forms a so called cornu spiral.The straight middle section corrensponds to the path that goes trough M and it's close vicinity.
The paths that are far away from M correnspond to the spiral ends of the cornu spiral. This means all the paths far away from M interfere destructively and only the path through M contributes to the amplitude at Z. This is the classical result that light takes the shortest path. However we can place obstacles at specific points inside the window in such a way that we remove the destructive interference and make the remaining paths far away from M interfere constructively with the main path trough M. This is done by placing the obstacles in paths that correspond to the parts of the cornu spiral that go against the direction of the straight part in the middle of the cornu spiral. This will make it brighter at Z even though we didn't do anything to the main path through M.
I can also do the opposite and place obstacles in the window such that the remaining paths interfere destructively with the main path, such that it is dark at Z even tough there is an unobstructed direct path from from Q to Z through M.
If I now place an obstactle at M to block the main path, it will actually increase the amplitude at Z because the remaining paths interfere constructively and we removed the destructively interfering main path. This means that counter-intuitively we can actually make it brighter at Z by placing obstacles which block the light.
Feynman does also explain how to calculate this cornu spiral in his lectures section 26-6.
If light would always take the shortest path, then I would expect that the amplitude at Z remains constant as long as I don't make changes to the main path. Remember that we didn't use mirrors to reflect light to Z, we placed obstacles that absorb light. But we can still increase the brightness at Z by placing obstacles in the window far away from M. This still works with individual photons.
Does that mean an individual photon takes all paths at once? Or does it choose one of these paths according to the probabilities? In my view quantum objects are not point particles that have a well-defined path in the first place, so a photon does not always travel along the shortest path, because it does not travel along any well defined path at all.
p.s. I created this initially as an answer to this comment.

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