$E_{tot}$ is conserved but not invariant. What does that mean? 1. It's the same at any time in every reference frame. 2. It's the same at a given time in every reference frame. 3. It's the same at any time in a given reference frame. 4. Something else Note: * Correct answer: C

$m$ is invariant but not conserved. What does that mean? 1. It's the same at any time in every reference frame. 2. It's the same at a given time in every reference frame. 3. It's the same at any time in a given reference frame. 4. Something else Note: * Correct answer: B

Charge is invariant and conserved. What does that mean? 1. It's the same at any time in every reference frame. 2. It's the same at a given time in every reference frame. 3. It's the same at any time in a given reference frame. 4. Something else Note: * Correct answer: A

Do you see a problem do you see with $\mathbf{F} = \dfrac{d\mathbf{p}}{dt}$ with regard to relativity? We still define $\mathbf{p} \equiv \gamma m\mathbf{v}$. 1. There's no problem at all 2. Yup there's a problem, and I know what it is. 3. There's probably a problem, but I don't know what it is.

Can we define a 4-force via the 4-momentum? $$\dfrac{dp^{\mu}}{d\tau} = K^{\mu}$$ Is $K^{\mu}$, so defined, a 4-vector? 1. Yes, and I can say why. 2. No, and I can say why. 3. None of the above. Note: * Correct Answer: A

To match the behavior of non-relativistic classical mechanics, we might tentatively assign which of the following values to $\mathbf{K} = K^{1,2,3}$: 1. $\mathbf{K} = \mathbf{F}$ 2. $\mathbf{K} = \mathbf{F}/\gamma$ 3. $\mathbf{K} = \gamma\mathbf{F}$ 4. Something else Note: * Correct Answer: C

A charge $q$ is moving with velocity $\mathbf{u}$ in a uniform magnetic field $\mathbf{B}$. $$\mathbf{F} = q\mathbf{u}\times\mathbf{B} = m\mathbf{a}$$ If we switch to a different Galilean frame (a low speed Lorentz transform), is the acceleration $\mathbf{a}$ different? 1. Yes 2. No Note: * Correct answer: B

A charge $q$ is moving with velocity $\mathbf{u}$ in a uniform magnetic field $\mathbf{B}$. $$\mathbf{F} = q\mathbf{u}\times\mathbf{B} = m\mathbf{a}$$ If we switch to a different Galilean frame (a low speed Lorentz transform), is the particle velocity $\mathbf{u}$ different? 1. Yes 2. No Note: * Correct answer: A

A charge $q$ is moving with velocity $\mathbf{u}$ in a uniform magnetic field $\mathbf{B}$. $$\mathbf{F} = q\mathbf{u}\times\mathbf{B} = m\mathbf{a}$$ If we switch to a different Galilean frame (a low speed Lorentz transform), is the magnetic field $\mathbf{B}$ different? 1. Yes 2. No Note: * Correct answer: A

A charge $q$ is moving with velocity $\mathbf{u}$ in a uniform magnetic field $\mathbf{B}$. $$\mathbf{F} = q\mathbf{u}\times\mathbf{B} = m\mathbf{a}$$ Suppose we switch to frame with $\mathbf{v} = \mathbf{u}$, so that in the primed frame, $\mathbf{u}’ = 0$ (the particle is instantaneously at rest). Does the particle feel a force from an E-field in this frame? 1. Yes 2. No 3. depends on details Note: * Correct answer: A