Mon, Dec 11, 6:03 p.m. - When exactly do you use distance=speed*time for relativity problems? Is it only when a question is asked and can be answered with info from the same reference frame? (if that makes sense)

Yes, you have that correct. Distance = speed * time is the way to go if you are given two of these quantities -- as measured by a particular observer -- and you are looking for the third quantity as measured by the same observer.

This is really important: the relativity equations for time, distance, length are only used if you are relating how one observer views distance and time to how a different observer views distances or time.

There have been many cases in the past of students getting really easy problems wrong because they forget distance = speed * time.

Sun, Dec 10, 5:20 p.m. - Can you please explain equation (3.4) in chapter 3, supplemental reading, page 69? How do we use this equation and what does the (') on the other side of the equation denote?

This is the invariant spacetime interval. The (') means a different user.'' If you know how one user measures distance and time between two events, and how a different user measures either distance OR time between the same two events, this equation gives you the ability to find that unknown.

Sun, Oct 8, 4:32 p.m. - How do you do A38 and A39. I can't figure out a method even after reviewing the lecture example, and the strategy. The video examples also don't work on my computer. Thank you in advance.

The method discussed in lecture and on the lecture materials'' for this past Thursday's lecture works perfectly for both of these problems, but you can't skip any steps. For each problem:

1. Draw before'' and after'' sketches. For A38, the Before sketch should show the motionless particle and the photon that is about to strike it. The After sketch should show just a single particle that is moving. For A39, the Before sketch should show a moving particle and a stationary particle, and the After sketch should show a motionless particle and a photon.
2. Write down p and E for each particle in your sketches. Some of this information is given in the statement of the problems. For some of it, you will need to use $E^2 = p^2c^2+m^2c^4$. Count how many unknowns you have. If more than two, then use $E^2 = p^2c^2+m^2c^4$ to write some of the unknowns in terms of other unknowns.
3. Once down to 2 unknowns, then $\Sigma p_{x,before} = \Sigma p_{x,after}$ and $\Sigma E_{before} = \Sigma E_{after}$.
4. You now have two equations and two unknowns. Solve.

Tue, Oct 3, 2:06 p.m. - For question A35, how is it possible for the superball to have that much energy in Joules stored in it? It doesn't seem real for it to have more energy in it than an atomic bomb. How does that work?

Excellent question. Actually, it doesn't have more energy than an atomic bomb. It's just that an atomic bomb doesn't convert all of its mass into kinetic energy. In fact, an A-bomb only converts a very small fraction of its mass into energy (and that is terrifying enough as it is). If an atomic bomb converted ALL of its mass to energy, it would be even more catastrophically awful.

So, an atomic bomb would be like taking your superball and converting 1/1000 or 1/1000000 of the mass into kinetic energy.

Mon, Oct 2, 9:13 a.m. - In the velocity transformation equation is the primed observer always the one that is moving (like a train) and the unprimed observer always the one that is at rest (like an observer on the ground)

No. We use the prime simply to have a different symbol for the different observers. It doesn't really mean anything.

I'll be honest -- I don't use that equation. You will notice in the Supplementary Reading, right after Eq. 4.4 on p. 90, it says that you don't have to use the equation with the primes and such. Just get the result classically and divide by a relativistic correction. I'll show you this approach -- which is the approach that I use -- in class tomorrow. With that approach, you don't have to worry about which observer is the prime and which is the unprimed observer.

Tue, Sep 26, 7:44 p.m. - How scientifically accurate is Interstellar?

It's actually pretty good, up until the last 15 minutes when the main character ends up in a black hole and finds himself behind the paneling in his daughter's bedroom. But up until that point, it does a good job of capturing the essence of what it would be like to be on an interstellar mission where your family on Earth is aging at a different rate than you. I like the fact that the movie shows what it would be like to go on one of these missions with the realization that when you get back, your daughter would be significantly older than you.