## Tuesday, August 24, 2010

### PH253: HW1 hints

Tuesday, we'll set up most of the homework problems, time permitting. Thursday, we'll set up the remaining problems. Until then, below the fold are some hints on how to get started. I would agree that this HW is conceptually difficult, and requires a good knowledge of calculus in some spots. Later HW sets will be less conceptually challenging, but the math will not let up, sadly.

Also, keep in mind I may have asked some of the same problems last semester, if you dig around the HW directory ... you may also try ph102.blogspot.com. Questions I've asked before, somewhere, sometime, have a * next to them below.

Lastly, the PH102 notes I posted previously will be enough to get you through Tuesday's and Thursday's lectures. Though some of the derivations are missing, the basic results are there. For Thursday's lecture, check out the beginning of the chapter called "Magnetism" for a derivation of B from E.

1*) This is a classic paradox, worthy of its own Wikipedia page. The observer in the barn sees the doors close simultaneously, but the pole contracted. The observer with the pole sees the barn length contracted, but the doors don’t close simultaneously ...

2) a: basic mechanics
b: in one direction, you have to add velocities, and in the other subtract
c: add velocities as vectors. If the pilot wants to go straight, he or she must make the plane’s velocity point into the wind a bit, the net forward progress made is determined by adding the wind and plane’s velocities as vectors. The plane's airspeed is the hypotenuse ...

3*) The length is shorter by &gamma ; the time is longer by &gamma

4*) Small typo in the original problem set, now fixed. Use the Lorentz transformations to relate u and u’ and t and t’, take derivatives of each relationship and divide.

5*) Force in relativity is still dp/dt, but note that p = &gamma mv. Thus,

F = \gamma m \frac{dv}{dt} + mv\frac{d\gamma}{dt}

Find d&gamma/dt in terms of v and dv/dt. Set F=qE and solve for dv/dt=a

6) I should have specified that the particle is moving perpendicular to the magnetic field to make things more clear. From PH106, you know F=qvB, and you know that a particle moving perpendicular to a magnetic field obeys uniform circular motion. From #5, you know how to write F using relativity. In uniform circular motion, the speed |v| is constant, so d&gamma/dt is constant since it only depends on v2 (dv/dt is not constant though). Thus, &gamma(m)(dv/dt) = qvB. You want &omega=v/r.

7) The energy of photon is hf = hc/&lambda, we’ll get to this. Use that, watch the units (nm=1e-9), and that's about it.

8*) The length along x is contracted, but the length along y is not.

9*) The x velocity in the new frame ux’ is given by the velocity addition formula. The y velocity we’ll derive in class, but note that

u_y^{\prime} = \frac{dy^\prime}{dt^{\prime}} = \frac{dy^\prime/dt}{dt^\prime/dt}

The numerator is the same as uy=dy/dt, since there is no length contraction along y, and the denominator you can find from the Lorentz transformation between t and t’. The speed is found from the components ux’ and uy’ as usual.

10) We’ll do this in class on Thursday. Think about the dimensions of the capacitor and how they may or may not be contracted. Charge is invariant, it is the same in all reference frames. How does the field between the plates depend on the charge *per unit area* on the plates, and what is that for each observer? One field remains the same, the other is a factor &gamma larger for the moving observer.