Momentum and Radiation Pressure

Transcript

0:00Introduction

Hello everyone! Let me start with a small puzzle. Light has zero rest mass, right? And yet, light falling on a surface pushes that surface. So how can something with no rest mass push anything at all? Keep that question in your mind. In this video, we will see that light carries momentum, and when it hits a surface, it presses on it. This push is called radiation pressure. Questions on it do come up in JEE Main, and they are quick marks once you know the formulas. So, let's begin.

0:35Light carries momentum

So, here is the big idea. An electromagnetic wave that carries energy, capital U, also carries momentum, small p, equal to U over c, where c is the speed of light. This momentum points along the direction in which the wave travels. Okay? Now, the wave hits a surface and stops there, or bounces back. Momentum cannot simply vanish, right? So where does it go? It goes into the surface, yes, exactly! Whether the wave is absorbed or reflected, momentum is passed on to the surface; for a reflection, it is even more, as we will see. So the surface feels a force, and a pressure. This is radiation pressure. With me? Let us try a quick one from the course. Suppose a surface absorbs three joules of light energy; the J on screen means joule. How much momentum does it receive? Simple, just divide by c. Three over three times ten to the power eight gives ten to the power minus eight kilogram metre per second; that is the k g, m, s to the power minus one on screen. Very small, right? Light pushes, but very gently. And now our puzzle. Here E is the energy of one photon, and its momentum is E over c, even though its rest mass is zero. Momentum does not need rest mass; the fields themselves carry it. So the puzzle is solved. Clear?

2:05Radiation pressure at normal incidence

Now, suppose light falls straight on a surface and is completely absorbed; straight on is called normal incidence, as in the heading. Intensity, capital I, is the average energy per second on each square metre. Divide it by c, and you get the momentum per second on each square metre, which is the pressure. So for a perfect absorber, the radiation pressure, p with a small rad below, equals I over c. Okay? Now, suppose the surface is a perfect mirror, and the light comes straight back. So tell me, is the push the same, smaller, or bigger? It is bigger, twice as big, that is correct! The momentum is reversed: it arrives as plus p and leaves as minus p, so the change is two p. So for a perfect reflector, p rad equals two I over c. Many students use I over c for a mirror too, so be careful here. Got it? Real surfaces are often in between. If a surface reflects a fraction r of the light and absorbs the rest, p rad equals one plus r, in brackets, times I, over c. Right? Now check this formula yourself. Put r equal to one, and then r equal to zero. What do you get? With r equal to one, we get two I over c, the perfect mirror. With r equal to zero, we get I over c, the absorber, very good! Clear? Next, the force. Pressure is force per area, so the force on an area, capital A, is p rad times A, okay? And for a beam of power, capital P, the force is P over c if it is fully absorbed, and two P over c if it is fully reflected. Careful, capital P is power in watts, and small p is momentum. With me?

3:58At an angle, and in terms of the field

Now, look at the figure. The left and middle pictures are the two cases we just found: I over c for the absorber, and two I over c for the perfect mirror. The figure writes the pressure simply as small p. Okay? Now, the picture on the right. The beam hits a perfect mirror at an angle theta, the Greek letter, measured from the normal, the dashed line. Under it, the pressure is two I cos squared theta, over c; the small two up there means cos theta is squared. So let me ask you, why cos squared theta, and not just cos theta? Because two things happen, and each one brings a cos theta. First, only the part of the momentum along the normal is reversed; that brings one cos theta. Second, the beam spreads over a bigger patch of mirror; that brings a second cos theta. So p rad equals two I cos squared theta, over c, which is less than two I over c whenever theta is not zero. Right? Let us test it quickly. If theta is zero, what does this formula give? Cos zero is one, so we are back to two I over c, the mirror hit straight on, perfect! And remember, theta is measured from the normal, not from the mirror surface. Got it? Last, the pressure in terms of the field. For a perfect absorber at normal incidence, I over c equals u in angle brackets, the average energy density. That equals half epsilon naught E naught squared. Epsilon naught, the curly e with a small zero, is the permittivity of free space, and E naught, the capital E with a small zero, is the peak electric field. Nice and simple. Clear?

5:46Example 1: pressure of sunlight on an absorber and a mirror

Now, example one. Sunlight of intensity one point four kilowatt per square metre falls normally on a surface; the m to the power minus two on screen means per square metre. Find the radiation pressure if the surface is, first, perfectly absorbing, and second, perfectly reflecting. Pause the video and try it first. Done? Okay, let's check. First, one point four kilowatt per square metre is one thousand four hundred watt per square metre. Always change to watts first, okay? Part A, the absorber. p rad equals I over c, so one thousand four hundred, over three times ten to the power eight. That gives four point seven times ten to the power minus six pascal; the P a on screen is the pascal, the unit of pressure. With me? Now part B, the mirror. It is twice the absorber, so two times four point seven, which is nine point four times ten to the power minus six, right? Is that really the answer? No, no, careful, because four point seven was already rounded. Two times one thousand four hundred, over three times ten to the power eight, gives nine point three times ten to the power minus six pascal. Double the exact value, not the rounded one. Got it?

7:08Example 2: mirror at 60 degrees to the normal, cos squared

Now a harder one, example two. This time, a beam of intensity one thousand two hundred watt per square metre falls on a perfect mirror, at sixty degrees to the normal. Find the pressure on the mirror. Pause the video and try it first. Done? Okay, let's check. Many students take cos theta only once, and get four times ten to the power minus six pascal. So is that the answer? No, no, that is the trap, because at an angle we need cos squared theta! In the right picture, theta is measured from the dashed normal; here theta is sixty degrees, so cos sixty degrees is one half. Okay? So cos squared theta is one quarter, that is, zero point two five. Now the numbers: two times one thousand two hundred, times zero point two five, over three times ten to the power eight. With me? The top is six hundred. Six hundred over three times ten to the power eight gives two times ten to the power minus six pascal. See how much that second cos theta matters? It halved the answer.

8:19Quick revision: I/c, 2I/c, (1+r)I/c and the angle rule

So, quick revision. Light carries momentum, U over c, and it pushes any surface that absorbs or reflects it. At normal incidence, the pressure is I over c for an absorber, two I over c for a perfect reflector, and one plus r, times I, over c, when a fraction r is reflected. For a mirror at theta to the normal, it is two I cos squared theta, over c, and force is pressure times area. Now go and try the practice problems on this page. All the best!