## Wednesday, January 20, 2010

### PH253: Homework 1 hints

Mostly this is what I ran through quickly at the end of Tuesday's class, but here you go. A bit more extensive help on the tougher problems. The weekly homework hints will not always be this extensive (hey, it is like the first week), but they will typically come just a couple of days before the due date to encourage you to mull them over on your own first. Technical note, this may look weird in an RSS feed due to my use of an equation typesetting plug-in in a few places.

(1) First, remind yourself how this problem works out without relativity, it is a standard PH106 problem. You should get
r=mv/eB
Now, what has to be different? In relativity, using F=ma is not really OK any more. You *can* still use F=dp/dt and this is where you want to start. Given that relativistic momentum is p = &gamma mv, the ordinary force is then, for a body of constant mass,
\frac{dp}{dt} = \gamma m \frac{dv}{dt} + mv \frac{d\gamma}{dt}
If the electron is on a circular path, its speed is constant, even though its velocity is not (it changes direction ...). If speed is constant, then so is &gamma, and our expression above has only one term. Further, the constraint on circular motion you remember from mechanics is purely geometric, and it is still OK in relativity:
\frac{dv}{dt} = \frac{v^2}{r}
This is centripetal acceleration.'' Plug this into the expression for force above, the total force, and equate that to the magnetic force.

(2) A classic relativity "paradox." The internet is your friend.

(3) What is the distance according to the astronaut?

(4) We did this in lecture on Tuesday ... and again the internet is useful.

(5) Start similarly to the first one. The electric force can be equated to the time rate of change of momentum:
qE = \frac{dp}{dt} = \gamma m \frac{dv}{dt} + mv \frac{d\gamma}{dt}
This time, the second term is not zero, so you'll have to grind through it and collect the terms you like. Recall &gamma depends on time through the velocity ...

(6) One direction is contracted, one isn't. Find the resulting length via the distance formula.

(7) You'll need the general formulas for velocity addition (both along the direction of relative motion and at right angles) we went over at the beginning of Tuesday's class. Transform the x and y components of velocity to the new frame, then you just need the speed and angle. The expressions will not be aesthetically pleasing; so it is. If you did everything correctly, the angle should look like this:
\tan{\theta^{\prime}} = \left(\frac{u_y}{u_x-v}\right)\sqrt{1-v^2/c^2}

(8) Shamelessly manipulate differentials. Note the following:
a_x^{\prime} = \frac{dv_x^{\prime}}{dt^{\prime}} = \frac{dv_x^{\prime}/dt}{dt^{\prime}/dt}
Really, it is OK, even if you can't quite recall how to show it formally. Once you have this, you can find the derivatives in the numerator and denominator by exploiting the velocity addition formula and Lorentz transformation for time, respectively. Much algebra ensues.

(9) If force is constant, then so is dp/dt. So set it equal to a constant and integrate ... momentum increases linearly with time. Use the relativistic formula for momentum, solve for velocity, and integrate that with respect to time to get position. Remember that the integration will yield an arbitrary constant (or you need appropriate limits) you'll need to find ... apply the boundary (initial) conditions you are given.