# Why does light travel in straight paths?

Why light travels in straight paths? What's the real cause that makes light photons to go in a straight line? and what are the factors that could change the path of light externally? (I'm excluding reflection, refraction and deviation here.)

• Just as a side note, in regions where material properties vary continuously light can follow curved paths. This phenomena is a big contributor to road-bad mirages. – dmckee May 17 '14 at 17:40
• What kind of an answer are you looking for? General relativity basically almost postulates this. Quantum or classical domain (or both)? – SMeznaric May 17 '14 at 19:27

## 4 Answers

Light has momentum.

$p = h/\lambda$

Light must travel in a straight line for momentum to be conserved.

• One of the best answers on this site: love how it cuts through to a core physical principle in two lines, although I think the form $\vec{p}=\hbar\,\vec{k}$ better illustrates your pithy point. – WetSavannaAnimal Oct 24 '15 at 9:31

It's valid to look at light as a wave - it's just like a radio wave - and waves travel by self-interference.

If you think of every point on a wave front as if it were a point source making its own ripples, you would see that the ripples cancel each other out in every possible direction except straight forward.

Actually, that's only true in a flat relativistic space, or in a medium where the speed of light is constant. The easiest example where this isn't so is if the light beam goes into a prism or through the surface of water. At a boundary like that, the speed of light changes, so the beam changes direction.

The Lagrangian of a light ray is given by,

$$\mathcal{L} = \mu(y) \left[ 1+ \left( \frac{\mathrm{d}y(x)}{\mathrm{d}x}\right)^2\right]^{1/2}$$

where $\mu$ is the refractive index, a function of the path. The associated equations of motion are,

$$\frac{\partial \mu}{\partial y} \frac{1}{\sqrt{1+y'^2}} + \left[ \frac{y''y'}{\sqrt{(1+y'^2)^3}} - \frac{\mu y''(x)}{\sqrt{1+y'^2}} - \frac{\partial \mu}{\partial y} \frac{y'^2}{\sqrt{1+y'^2}}\right] = 0$$

If we set the refractive index to a constant, i.e. $\mu=\text{const.}$ then the equations of motion reduce to,

$$\frac{y''y'}{\sqrt{(1+y'^2)^3}} - \frac{\mu y''(x)}{\sqrt{1+y'^2}}=0$$

which is a second-order nonlinear differential equation. It can be shown the only solutions are,

$$y(x)=c_1x+c_2$$

These solutions are linear functions, hence simply straight paths which is the required result, providing the medium which the light travels through is uniform and constant.

I'll give other ways of getting to the same answer.

First of all, the wonderful path integral formulation (which can be found in Feynman's excellent book, "QED"):

In order to consider the probability of a particle from going to a point $A$ to a pont $B$, you consider ALL the possible paths. Each path has a probability contribution and we just have to sum them up (integrate, to be more precise). You might want to watch this video, where each contribution is represented by an arrow. The contribution of each trajectory, will roughly depened on the time taken. But this is the Quantum Mechanical view, the thing is that all the deviations from a straight line tend to cancel out, giving us a straight line in the classical limit.

This interpretation has a deep importance in modern physics since it applies not only to photons, but to any other particle. It would be written mathematicaly as:

$$\langle \text{fina state} \vert e^{-iHt} \vert \text{inital state} \rangle = \int e^{iS[x]} Dx(t)$$

Where $S$ is the classical action, as seen in Hamiltonian mechanics.

Another, quite similar view, is that light travels in a way it minimizes the time (this is not true in the case of mirrors, it might be a maximum, but we are no interested in that). This is called Fermat's principle and can be derived from Huygen's principle (which Mike Dunlavey explained).

So in free space, the way you arrive earliest is by going in a straight light.

## protected by Qmechanic♦Oct 24 '15 at 12:07

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