Waves. Reply

Optics Series Lecture, Lecture – XII and – XIII.

“Traveling waves, Differential wave equations, Particle and wave velocities.” These lectures were delivered on 17th and 20th February 2017, in two lecture sessions of 1 and 1/2 hours each. The web version has been named “Waves.” and the lectures were delivered to Physics honors students.

In one of our earlier optics session lecture I had hinted at having waves defined by their pulse shape called as wave profile — or alternatively wave shape or wave form, and transcribing them into forms that represent actual wave motion. The later are then called as traveling or progressive waves. The former, the so called wave shape or wave profile are then time-snapshots of the full fledged time varying waves that we just called traveling waves. Remember that stationary or standing waves are not wave profiles or any snapshots of a single traveling wave, they are rather the superposition of an advanced and a retarded wave — that is one traveling wave moving forward and another exactly shaped traveling wave moving in the reverse direction. We studied advanced and retarded waves, here. We have also already dealt with traveling waves in much detail, eg, here and here. This lecture will justify what we have been espousing all along. More…

Waves, particles and Einstein ! 1

Waves are something that have no mass and move at the maximum speed, mass m = 0. speed c = 1. So whats their momentum? p = m.v = 0? Right?

No. For pure waves; momentum does not come from mass. It comes only from motion.

(pure wave; they do not have mass)

For matter waves, on the other hand, momentum comes in two ways, mass as well as motion.

(impure, now they have mass)

Albert Einstein recognized this fact and derived his relation; $latex E = \sqrt {(pc)^2+(mc^2)^2}$

This relation is called as Einstein’s relativistic equation, also Einstein’s mass-energy relation. But more appropriately mass-energy-momentum relation.

Let us consider E as the hypotenuse, p and m; as base or perpendicular as is your choice.

triangle_copyThen $latex E = \sqrt {(pc)^2+(mc^2)^2}$ is Pythagoras Theorem; when p is momentum and m is mass.

For pure waves such as photon … the quanta of light, m = 0.

Hence the Pythagorean Triangle is now one, where the mass side is arbitrary small. Thus E = p. More…