Part A: Units, Dimensions and Dimensional Analysis

Transcript

0:00Introduction

Hello everyone! Here is one small puzzle before we start. You know this formula from kinematics: the distance covered in the nth second, s n, equals u, plus a by two, times two n minus one. Now check its dimensions. The left side is a distance, so it is a length. But on the right, u is a velocity! Length equals velocity? Wait, has the textbook broken the most basic rule of physics? Today, in this mixed problem class, we will catch the culprit, and pick up the tricks that make JEE dimension questions quick. So, let's begin.

0:38Topper's trick: spot the building blocks

Okay, first problem from the bank. Show that the square root of B squared, over mu nought rho, is a speed. Here B is a magnetic field, mu nought is the Greek letter mu with a small zero, and rho, the letter that looks like a p, is a density, okay? The long way is to break B and mu nought into mass, length, time and current. Lots of algebra, right? Toppers just ask: where have I seen B squared over mu nought before? Yes, energy density! The energy density of a magnetic field is B squared over two mu nought. So B squared over mu nought is just twice an energy density, and its dimensions are M, L to the minus one, T to the minus two. The square brackets on screen just mean, the dimensions of. The two is a pure number, so it changes nothing. Clear? Now divide by the density rho, M, L to the minus three. Mass cancels. For length, minus one, minus minus three, is plus two. So we get L squared, T to the minus two. Take the square root, and you get L, T to the minus one: a speed! In fact, this is called the Alfvén speed, in a plasma. No heavy algebra. Nice and clean. Same trick again. In v equals the square root of gamma P over rho, v is a speed, P is a pressure, and gamma is the Greek letter that looks like a y. Why must gamma have no dimensions? Because the square root of P over rho is already a speed, as we saw in Chapter three. So gamma has to be a pure number. In fact, it is the ratio of two specific heats. That's it!

2:17Dimensionless? Make the blocks cancel

Right, on to the next problem. Find integers p, q and r for which e squared, times epsilon nought to the p, h to the q, and c to the r, has no dimensions. Here e is the electron's charge, epsilon nought is the Greek letter that looks like a curly e, h is Planck's constant, and c is the speed of light, okay? Again, don't expand everything. Spot the block. Take e squared over epsilon nought. This pair comes together in Coulomb's law, and it works out to energy times length. The cross on screen just means, times. And h c, Planck's constant times the speed of light? That is also energy times length. See that? Now think. We have e squared over epsilon nought, which is energy times length. What should we divide it by, to kill all the dimensions? Exactly, by h c! So, e squared over epsilon nought h c is dimensionless: energy times length, over energy times length. Now read off the powers. Epsilon nought, h and c all sit in the denominator, once each. So p, q and r are all minus one. Remember this block, e squared over epsilon nought. It comes back in example two. Got it?

3:35JEE trap: the hidden one second

Alright, now back to our puzzle. s n equals u, plus a by two, times two n minus one. Left side, a length. Right side, u, a velocity. So the formula is dimensionally wrong, and we should throw it out, right? No, no, careful! This is a classic JEE trap. Look, the formula has a hidden quantity. Written fully, s n equals u tau, plus a by two, times two n minus one, times tau squared. Tau is the Greek letter that looks like a small t, and it is the length of each time interval. In this formula, each interval is one second, so tau equals one second. Write tau as just one, and the second vanishes from view. That hides the dimensions. Now check again. The u tau term: velocity times time, a length. The a tau squared term: acceleration times time squared, also a length. So with tau shown, every term is a length. The formula was right all along! So, before you call a standard formula wrong, look for a hidden tau like this. With me?

4:46Example 1: dimensions of radiation pressure I/c

Let's do example one, an easy one. Radiation of intensity capital I falls on a surface that absorbs it, and presses on it with a pressure p equal to I over c. Check the dimensions. Pause the video, and try it first. Done? Okay, let's check. The intensity of a wave, from our table in Chapter two, is M, T to the minus three. And c, the speed of light, is L, T to the minus one. Now divide. Mass stays. Length: zero minus one, so L to the minus one. And time? Minus three, and then minus minus one. T to the minus four, right? Hold on, careful! Subtracting minus one means adding one. So minus three plus one: the power of time is minus two. We get M, L to the minus one, T to the minus two. And that is exactly the dimension of pressure. The formula passes. Absolutely correct.

5:49Example 2: Larmor's formula by the method of dimensions

Now a harder one, JEE level. The power radiated by an accelerated charge depends on the charge q, through q squared over epsilon nought, on its acceleration a, and on the speed of light c. Find the relation. Pause and try. Hint: remember the block from earlier. Done? Okay, let's check. Write capital P equals k, times the whole block, q squared over epsilon nought, to the power x, then a to the y, and c to the z. Here k is a pure number. And q squared over epsilon nought, we know, is energy times length: M, L cubed, T to the minus two. Now the left side: power. Quick, what are its dimensions? M, L squared, T to the minus two? No, no, careful! That is energy. Power is energy per second, so one more T goes in the denominator. So the left side is M, L squared, T to the minus three. On the right: M, L cubed, T to the minus two, to the power x, then acceleration, L, T to the minus two, to the power y, and the speed c, L, T to the minus one, to the power z. Now, which base dimension appears in only one quantity on the right? Mass! Only the charge block has it. So x equals one, straight away. Put x equal to one. Length: three, plus y, plus z, equals two. Time: minus two, minus two y, minus z, equals minus three. Tidy them up. The first gives y plus z equals minus one. The second gives two y plus z equals one. Subtract the first from the second, and y equals two. Then z is minus one minus two, that is, minus three. Put the powers back. P is proportional to q squared a squared, over epsilon nought c cubed. The curly sign on screen means, is proportional to. This is Larmor's formula. See? Once you knew the block, a hard problem became a few short lines. Got it?

8:02Quick revision: building blocks, powers, the hidden 1 s

So, quick revision. One: spot the known blocks, like B squared over mu nought, e squared over epsilon nought, and h c, and don't expand everything. Two: for unknown powers, equate the powers of mass, length and time. Three: if a standard formula looks wrong, look for a hidden tau equal to one second. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!