On one of my many short-lived attempts to study for the International Physics Olympiad, I picked up a copy of a well-known textbook by Halliday and Resnick. My high school didn’t offer a course in electricity and magnetism, so I figured I would have to teach myself about the topic if I wanted to solve problems competitively.

I only managed to solve about 3 problems from the book before a friend called me up asking if I wanted to play Frisbee. After that, I don’t think I ever looked at the text again, although for some reason I did bring it to college – maybe to make sure I had enough things gathering dust on my dorm room shelf. That’s not to say it’s a bad book! In fact, I learned quite a bit from the three problems I solved. Here’s one of them:

Problem: There are three fixed positive charges equally spaced on a circle (see the picture). At the center of the circle is a negative charge . Prove that the negative charge at the center of the circle does not move.

My first instinct was to introduce a coordinate system, calculate all the Coulomb forces on the negative charge, and show that these forces all summed to 0. But there is a much more beautiful and powerful way to solve the problem! All you need to know is that the force one charge exerts on another is directly proportional to the product of their charges, and inversely proportional to the square of the distance between them – and you don’t need to do any calculations.

Let’s suppose the negative charge does move. Then, it has to choose a certain direction in which to start moving (see picture below). So far, so good. Now, move the central charge back to the center and rotate the entire system 120 degrees clockwise about the center. The initial direction the negative charge will move towards is now rotated 120 degrees. But wait – the entire system is exactly the same as before – all the charges are in the same place (imagine someone called your name and you looked away while your sneaky friend rotated the whole system 120 degrees – when you looked back, you would think that nothing happened)! This means, in fact, that the negative charge should move in the direction it started moving before we rotated the system! This is a contradiction, since the negative charge cannot move in two different directions at the same time. If you rotate the system by 120 degrees again, you can see that the negative charge actually has to move in three different directions all at once! You can see why this is a problem. The negative charge gets confused and cannot move.

Notice how much simpler the above analysis, a symmetry argument, was compared to calculating three Coulomb forces individually and using trigonometry to show the forces cancelled (assuming you even know how to compute Coulomb forces)! The symmetry argument also works with any number of equally-spaced positive charges on the circle, where (why doesn’t it work for ). Can you imagine calculating forces and showing they all summed component-wise to 0?

Symmetry arguments are extremely powerful in physics, and many physicists will use them without even thinking about it. If you want some more practice, try Problem 1. If you’re feeling daring, try Problem 2 below (though you might need more than symmetry arguments to solve it)!

Problem 1: An infinite sheet of charge is located on the x-y plane. The charge on the sheet is uniformly distributed. A test charge is placed at some point outside the plane. Prove that the force felt by the test charge does not depend on its x- or y-coordinate. (I haven’t yet told you how to calculate the Coulomb force due to an extended body, but you don’t need that. Either you can break the sheet into tiny pieces which you treat as point charges, or better yet, just ask yourself how the charge configuration appears to change when you move a little in the x- or y-direction).

Problem 2: Prove that the force felt by the test charge does not depend on its z-coordinate. (This is harder, but you can do it! And you can do it without performing any explicit calculations. Break the sheet into concentric rings and think about the contribution from each ring. You may find a discussion of Olbers’ paradox helpful).

To mathematicians, the word elegant has a very specific meaning. A mathematically elegant proof is one that is short, simple, and reveals some profound new insight. A mathematically elegant problem is one that has an elegant solution. It was really the elegant problems that got me excited about math and physics while I was in high school.

For two other arguments that struck me so poignantly that I remember them to this day, continue reading. We’re outside the realm of symmetry arguments, now, and Problem 4 is completely unrelated to everything else in this post, but these problems are very elegant nonetheless!

Problem 3: Prove that the rate at which a body falls is independent of its mass. (Neglect air resistance)

Solution 3 [due to Galileo]: Suppose that heavier objects fall faster than lighter objects. Take two balls – one a 10 kg ball, the other a 5 kg ball, and drop them off the Leaning Tower of Pisa. The 10 kg ball should land first. Now, tie the two balls together with a piece of rope. One line of reasoning will tell you that the 5 kg ball will slow down the 10 kg ball since the latter is attached to a slower falling object, so the time it takes for the combined system to reach the ground will be somewhere between the times it took the individual balls to reach the ground. Another line of reasoning will tell you that the combined system is heavier (15 kg + weight of rope) and thus will hit the ground faster than the standalone 10 kg ball did. We’ve reached a contradiction – it can’t be the case that the combined system reaches the ground both before and after the 10 kg ball would. Thus heavier objects cannot fall faster than lighter objects. I leave to you the case where lighter objects fall faster and the case where the acceleration due to gravity is not a monotonic function of mass.

Problem 4: Prove that there exist infinitely many prime numbers.

Don’t look at the following hint unless you’re really stuck!

Hint: There are many solutions to Problem 4, but the most elegant is due to Euclid. Let’s assume there are finitely many primes . Can you think of a way to create a new prime number from this list?

I hope these problems will inspire you as they inspired me when I was younger. Good luck!

P. S. I realized I forgot to introduce myself! My name’s Mo. I’m a rising senior at Caltech. This summer I’ve been working directly under Spiros in the Preskill group. More on me to come, but for now I’ll just say that like Shaun, I’m very surprised they’re giving a hooligan like myself the keys to the blog.