# Testing Contest - Win "Conquering the Physics GRE" book!

Tags:
1. Mar 13, 2018

### Greg Bernhardt

Undergrad physics students, are you taking the physics GRE in April or next fall? We have a contest for you! Big thanks to Cambridge University Press for donating a copy of the newly published 3rd edition of Conquering the Physics GRE!

How to win: Post your favorite physics problem you've run into during your studies and show how to solve it step by step. The community will vote through "likes". Member with the most "likes" wins. Contest ends 3/23/18.

International students are welcome to participate as the publisher has kindly offered to ship from their warehouse worldwide!

2. Mar 18, 2018

### lekh2003

At this rate, Greg's original post is the only one with any likes or a post at all. Where are our PF members?

3. Mar 18, 2018

### Staff: Mentor

Spring break!

4. Mar 18, 2018

### opus

Would love to participate, but I havent taken any Physics courses yet!
But now that I know this is a thing, Ill horde all my future physics problems for cases such as this

5. Mar 18, 2018

### Staff: Mentor

I'm not interested in the book, but one of my favorites is the demonstration that every straight tunnel through a ball of constant density has the same oscillation period for frictionless motion, which also matches the period of an orbit directly above the surface.
I didn't post steps, so feel free to use that if you want.

6. Mar 19, 2018

### lekh2003

This is very sad. May I post a mathematics problem? It was a very neat derivation of PI, which I particularly enjoyed. I can post here a scanned copy of my working (was very long).

Last edited: Mar 19, 2018
7. Mar 21, 2018

I'm interested! Is this still open?

8. Mar 21, 2018

### Greg Bernhardt

Indeed!

9. Mar 22, 2018

### PumpkinCougar95

Not interested in the book, but here is a problem I find interesting.

Consider 2 point charges Q1 and Q2, separated by a distance d. (Q1> 0, Q2<0)If an electric field line originates from Q1 and makes an angle with the line joining the two charges near Q1(say $\alpha$), then what angle would the same electric field line make near Q2?

image.

#### Attached Files:

• ###### a42cfd163c0c841047907b69f5f9d7b874e3e4ae.jpg
File size:
51.8 KB
Views:
86
10. Mar 23, 2018

### long758

I'm interested in the book! While there are a multitude of interesting problems I could choose, I'll present a fun problem I came across and solved last week:

Problem: Write down the conserved quantities for a classical mechanical system described by the Lagrangian $L = \frac{\dot{x}^2+\dot{y}^2}{2} - xy$.
Solution: I can readily observe that this Lagrangian does not explicitly depend on time, so by Noether's theorem, the total energy $E = \frac{\dot{x}^2+\dot{y}^2}{2} + xy$ must be conserved.​
However, this Lagrangian does depend on $x$ and $y$, breaking translational symmetry. Thus, by Noether's theorem, there is no simple momentum conservation. Furthermore, the system is not radially symmetric, because the potential energy $V=xy=r^2\cos\theta\sin\theta$ depends on the angle when written in polar coordinates. Using Noether's theorem again, we can assert that angular momentum about the origin is not conserved.​
Should we conclude at this point that there are no other conserved quantities in this system?​
No! We have more work to do. Lets look at the potential energy term more carefully. The lines of constant potential $V=xy$ are hyperbolas with asymptotes along the x and y axes. The potential is positive in the first and third quadrants and negative in the second and fourth quadrants. ​
Here is where the slight problem-solving leap comes in, but I'll do my best to make this next step as natural as possible. Looking at the graphs of hyperbolas described by the equation $xy=const.$, I notice that those hyperbolas have reflection symmetry along the lines $y=x$ and $y=-x$. Thus, the potential must also have this symmetry. This can also be seen by visualizing or plotting the potential directly.​
Why wasn't this symmetry obvious from the beginning? Well, the $x-y$ coordinate system does not make this symmetry manifest. To get the reflection symmetries out in the open, lets try to find a better coordinate system. Well, the lines $y=x$ and $y=-x$ are the naturally suggested axes, so we define variables $u = \frac{x-y}{\sqrt{2}}$ and $v=\frac{x+y}{\sqrt{2}}$ along these axes.​
Writing the Lagrangian in terms of our new variables, we get $L= \frac{\dot{u}^2+\dot{v}^2}{2} - \frac{v^2-u^2}{2}$. Low and behold, the Lagrangian now has no cross terms between the two different variables, which means that the dynamics in these two directions will be entirely independent. In other words, the Lagrangian was separable.​
Looking at the Lagrangian, we now have a harmonic oscillator in $v$ and an inverted quadratic potential for $u$. Isn't that so cool? Who would have guessed that a harmonic oscillator was hiding in this system? Anyway, the energies of the two independent systems will be independently conserved: $E_1 = \frac{\dot{u}^2}{2} - \frac{u^2}{2}$ and $E_2=\frac{\dot{v}^2}{2} + \frac{v^2}{2}$.​
We found the second conserved quantity. One can rewrite these conserved quantities in terms of $x$ and $y$ again. After taking convenient linear combinations, one nice way to represent the two conserved quantities is: $E = \frac{\dot{x}^2+\dot{y}^2}{2}+xy$ and $C = \frac{x^2+y^2}{2}+\dot{x}\dot{y}$.​
The problem is solved. However, there are a few further comments to make.​
First, the inverted harmonic oscillator can be converted to a harmonic oscillator by making the substitutions $u=i u'$ and $t = -i\tau$. We know the solution to the harmonic oscillator, so $u'(\tau)$ is sinusoidal. Making the reverse substitution, we get that the trajectory $u$ will be written in terms of $cosh$ and $sinh$ functions. Thus, we get the trajectories for free, without having to solve a different differential equation.​
Second, I would note that the two reflection symmetries do not guarantee that the coordinate transformation that we performed would always separate such Lagrangians. A simple counterexample is to instead use the potential $V=x^2y^2$, which leaves a cross term $\frac{u^2v^2}{2}$ after the transformation. Thus, the success of our attempt was rather fortuitous. ​
Third, I have a question of my own (whose answer I don't know). It seems to me that finding the second conserved quantity in this problem required an insight that could not easily be generalized and reduced to a routine. Is there a systematic way of finding all the conserved quantities in a classical system? What about a quantum system (I imagine the answer is similar)? What about in quantum field theories? ​
On a related note, I am pretty sure that the system considered does not have a third conserved quantity. But how can I be sure. Is there a necessary property of maximally superintegrable systems (which this system would be if it had three conserved quantities) which is not true of this system?​
Finally, I'll point out that this problem suggests many other possible problems to be solved. In particular, I find one to be especially interesting:​
Suppose we added an uncoupled harmonic oscillator to the Lagrangian: $L'=\frac{\dot{x}^2+\dot{y}^2}{2} - xy +\frac{\dot{z}^2}{2}- \frac{z^2}{2}$. Do you think the same logic would work as before? Would we just get three conserved energies corresponding to the three directions $u, v,$ and $z$? ​
The answer is that there is a fourth conserved quantity hiding in this system, and your job is to find it. Good luck!​