How does light know the destination?

Question

According to Fermat's principle, light travels the fastest path from dot A to dot B. I wondered how light knows which path is the fastest, and found out that light actually goes all path, but non-optimal paths are destructed. However, when the photon is at dot A, how does it know that it will eventually arrive at dot B?

Answer 1

Light doesn't necessarily take the fastest path from point A to point B. For example, if you place a block of optically dense material between points A and B, such that the straight line from A to B is normal to two faces of the block, then a ray of light can travel in the straight line path from A to B. (It doesn't bend when it enters or exits the block, because the angle of incidence is zero.) But this is not necessarily the fastest path. The fastest path might be for light to go around the block. Alas, it can't do that, unless you place a mirror to allow the light to take the bent path around the block. And even if the mirror were there, the light wouldn't go that way if it were initially aimed along the straight line path.

Light takes a path of stationary action that is consistent with its initial conditions. So does matter.

A path that has globally stationary action must have locally stationary action at each point along the path. Thus, the particle doesn't need to know its ultimate destination. Instead, at each instant of time between $t$ and $t + dt$, the particle merely has to decide which direction to take an infinitesimal step in. You can imagine that it remembers where it was at time $t - dt$. The particle then picks the direction that results in the action between $t - dt$ and $t + dt$ being stationary.

Thus, when light is travelling through vacuum, it "knows" that it mustn't bend at any point, because if it were to bend, the action wouldn't be locally stationary. By not bending at any instant along the path, the light manages to execute a globally straight path through vacuum.

Answer 2

The propagation of light waves satisfies Maxwell's equations and in the high frequency limit, that is when the permittivity and permeability changes slowly over the free-space wavelength $c/\omega$, the light rays satisfy Fermat's principle. Here rays are defined as the envelope of the gradients of the equal phase surfaces $\mathcal S$, in other words, these are the orthogonal trajectories to the equal phase surfaces.

Specifically, the unit tangents to the rays, $\mathbf t $, i.e., the unit tangent vectors of the orthogonal trajectories satisfy the eikonal equation, see this post, $$\nu \mathbf t = \nabla \mathcal S \tag{3}$$ where the refractive index at location $\mathbf r$ is defined by $\nu = \nu(\mathbf r)=\sqrt{\mu(\mathbf r)\epsilon(\mathbf r)}$, an dthe gradient is taken with respect to $\mathbf r$, $\nabla \mathcal S = \frac{d\mathcal S}{d\mathbf r}$. From Eq.(3) follows that the various formulations of geometrical optics, such as, Snell's law, Fermat's principle, Malus-Dupin law, etc., are essentially equivalent; for details see Born & Wolf.

At interfaces where $\mu$ or $\nu$ are discontinuous, this includes ideal reflectors, the standard boundary conditions apply and can be shown that Fermat's principle still holds in the short wavelength limit asymptotically as $\omega \to \infty$.

There is nothing teleological in Eq. (3) or in the other formulations (Snell, Malus-Dupin) although they are mathematically equivalent to Fermat's principle.

Answer 3

Actually, the Fermat's principle is a way of describing the behavior of light, not the intention or knowledge of light. It simply follows the laws of physics, which can be expressed mathematically using the - principle of least action.

The principle of least action states that the path of a physical system between two states is the one that minimizes the action, which is a quantity that depends on the system's energy and momentum.

In the case of light, the action is proportional to the optical path length, which is the product of the distance traveled and the index of refraction of the medium. Therefore, minimizing the action is equivalent to minimizing the optical path length, which is Fermat's principle. So, light doesn't know rather it follows the principle of least action.

Answer 4

I will first give a precise formulation of Fermat's stationary time, and then I will give a derivation:

With that derivation stated I will address the question: "Does light have to know the destination?"

We have:
As light is propagating the true tajectory of the light has the property that if you sweep out variation: the true trajectory is the one for which the derivative of the transit time with respect to the variation is zero.

It is crucial to be aware that there are also cases such that the true trajectory is a maximum of the transit time.

Take the following case:
An ellipse, reflective on the inside, with a point of emission at one focus of that ellipse, and an a light detector at the other focus of the ellipse. As we know: for every direction of emission of that light the reflection against the ellipse will proceed to the other focus.

Next we alter the above case in two opposite ways:
While keeping the emission point and detection point at the same location:
-increase the size of the major axis of the ellipse
-decrease the size of the major axis of the ellipse

With the major axis of the ellipse increased:
That makes the reflecting surface less concave, and then the light from the emission point that reaches the detector has travelled along a trajectory of least time.

With the major axis of the ellipse decreased:
That makes the reflecting surface more concave, and then the light from the emission point that reaches the detector has travelled along a trajectory of most time.

Fermat's stationary time is not about minimization or maximization; that is immaterial.

What counts is that which 'least time' and 'most time' have in common: the true trajectory has the property that the derivative of the transit time with respect to variation is zero.

I start with using Huygens' Principle to account for Snell's law of refraction:

Diagram 1

A wavefront of width 'd' take a time interval 't' to transit a change of optical medium.

$v_1$ and $v_2$ are the two velocities in the respective mediums.

To prepare for later use I derive an expression for the derivative of the length of a hypotenuse as one of the sides of a right triangle is changed in length.

Diagram 2 (animated GIF)

$$ \frac{dC}{dA} = \frac{d(\sqrt{A^2 + B^2})}{dA} = \frac{A}{\sqrt{A^2+B^2}} = \frac{A}{C} \tag{1} $$

Without the intermediate steps:

$$ \frac{dC}{dA} = \frac{A}{C} \tag{2} $$

The interesting thing about (2): the derivative of the length of $C$ (wrt change of $A$) is the sine of the changing angle.

Diagram 3

In diagram 3 the letter 'S' stands for ‘Snell's point’. I take as starting point that there is a fixed point from where the light is emitted, point 'T', and that there is a fixed point 'R' where the light is received. (T and R not shown in the image; T and R can be arbitrarily far away.)

Let it be granted that the wavefront is perpendicular to the direction of propagation (if the wavefront is perpendicular then the triangles involve are right triangles): it then follows that the angle $\beta_1$ is equal to the angle $\alpha_1$, and that the angle $\beta_2$ is equal to the angle $\alpha_2$.

The variation of the path of the light consists of moving point S along the refraction line. We want to find the criterion that identifies the location of point S in the variation space such that Snell's law is satisfied.

(3) is another way of stating Snell's law:

$$ \frac{\sin(\alpha_1)}{v_1} = \frac{\sin(\alpha_2)}{v_2} \tag{3} $$

We set up expressions for $\sin(\alpha_1)$ and $\sin(\alpha_2)$ according to (2):

$$ \sin(\alpha_1) = \frac{dC_1}{dA_1} \qquad \sin(\alpha_2) = \frac{dC_2}{dA_2} \tag{4} $$

While the speed of light is different in each of the two optical media, within each medium the speed is constant. That is, in each medium the distance traveled and transit time is in a fixed ratio.

$$ \frac{C_1}{v_1} = T_1 \qquad \frac{C_2}{v_2} = T_2 \tag{5} $$

In (6) the division by velocity is accommodated by substituting the distance C with the time T:

$$ \frac{d(v_1 T_1)}{dA_1} = \sin \alpha_1 \qquad \frac{d(v_2 T_2)}{dA_2} = \sin \alpha_2 \tag{6} $$

$v_1$ and $v_2$ are constants, so we can move them outside of the differentiation, and to the other side of the equals sign

$$ \frac{dT_1}{dA_1} = \frac{\sin \alpha_1}{v_1} \qquad \frac{dT_2}{dA_2} = \frac{\sin \alpha_2}{v_2} \tag{7} $$

Hence: in order to satisfy Snell's law:

$$ \frac{dT_1}{dA_1} = \frac{dT_2}{dA_2} \tag{8} $$

Diagram 3 illustrates that when Snell's point 'S' is moved the sides $A_1$ and $A_2$ change by the same amount. Therefore (8) can also be expressed as derivatives with respect to the position of point 'S'

$$ \frac{dT_1}{dS} = -\frac{dT_2}{dS} \tag{9} $$

Which can be rearranged as follows:

$$ \frac{d(T_1 + T_2)}{dS} = 0 \tag{10} $$

Equation (8) looks as if it is about time, but in fact (8) is expressing the relation between the angles $\beta_1$ and $\beta_2$

The crucial step is taking the derivative of the transit time (with respect to variation) Taking the derivative recovers the sine of the angle.

So now the question:
Is it necessary for propagating light to have advanced knowledge of the destination?

The purpose of the above derivation is to show that Fermat's stationary time can be accounted for in terms of Huygens' Principle.

For Huygens' principle it is not necessary for light to have advanced knowledge of the destination. It follows that in the case of Fermat's stationary time advanced knowledge is not necessary either.

Answer 5

Light has long been observed to travel as if prescient of its destination. It is speculated that a two-way, wave-like connection going both forward and backward in time is necessary for an exchange of light energy from one charged particle (usually electrons) to another. In this view, the path of light from signal to receiver is predetermined rather than random. This explains how light ‘knows’ where it is going and it provides the basis for some modern theories of light. The most prominent of these models is John Cramer’s Transactional Interpretation of Quantum Mechanics TIQM. See https://en.wikipedia.org/wiki/Transactional_interpretation