# Why does in the definition of 'optical path', only spatial phase change is taken into consideration & not that of time?

In Concepts of Physics by H.C.Verma, I got the definition of optical path. However, one thing that I couldn't understand is why only the spatial phase change is considered in its definition.

As he writes:

Consider a light wave travelling in a medium of refractive index $\mu$. Its equation may be written as $$E = E_0\sin \omega\left(t - \frac{x}{v}\right ) = E_0\sin \omega\left(t - \frac{\mu x}{c}\right )$$. If the light wave travels a distance $\Delta x$, the phase changes by $$\delta_1= \mu\frac{\omega}{c} \Delta x\tag{I}$$. Instead, if the light wave travels in vacuum, the equation will be $$E = E_0\sin \omega\left(t - \frac{x}{v}\right )$$ If the light travels through a distance $\mu \Delta x$, the phase changes by $$\delta_2= \omega \frac{(\mu \Delta x)}{c}= \mu \frac{\omega}{c} \Delta x \tag{II}$$. From $\text{I}\, \& \, \text{II}$ , we see that a wave travelling through a distance $\Delta x$ in a medium of refractive index $\mu$ suffers the same phase change as when it travels a distance $\mu \Delta x$ in vacuum. ... The quantity $\mu \Delta x$ is the optical path of the light.

Apparently, it was an easy reading. But then I wondered why the phase change was $\mu \dfrac{\omega}{c} \Delta x$. I noticed the words If the light wave travels a distance $\Delta x$; now doesn't travelling needs time? Say, it needs time of $\Delta t$ to come at $x + \Delta x$; hence the phase-change should be $\omega \left((t +\Delta t) - \mu\dfrac{x + \Delta x}{c} \right) -\omega\left(t - \dfrac{\mu x}{c}\right ) = \underset{\text{phase-change}}{\boxed{\omega\left(\Delta t -\mu\frac{\Delta x}{c}\right)}}$ . This should be the phase-change which is not at all equal to $\delta_1 \, \text{or}\, \delta_2 =\mu \dfrac{\omega}{c} \Delta x$.

So, my questions are:

1) Why is time not taken into account, after all travelling needs time?

2) Even if time is ignored, phase-change is not equal to $\mu \dfrac{\omega}{c} \Delta x$ but $-\mu \dfrac{\omega}{c} \Delta x$. Where did the '$-$' sign go then?

The wave is not an object travelling through space and time. At any fixed time the field has values at each location.

If you want to find the phase difference between two times and two places you have a fine formula. But if you are taking an imaginary snap shot and looking how the field is different at different places then the $\Delta t$ will be zero. That's just what they are asking about and looking at.

And the reason this snapshot is good enough is because you made a big assumption. You assumed at every place in space the field changes harmonically in time with the same angular frequency in time, $\omega.$ It is this assumption that tells you everything if you only know the phase.

So if you assume the same amplitude everywhere and the same temporal frequency everywhere then the only thing left to know is the phase so now you are trying to figure out how the phase changes between here and there. Because then you know everything.

As for the minus sign the goal is to define the optical path which is like a distance (it is the distance you'd have to go in a vacuum to get the same phase difference). So they just want a positive distance, that's why they look at the magnitude instead of a signed difference.

• +1; The author shouldn't have used the word travel as it is rather ambiguous.
– user36790
Aug 29, 2015 at 17:25