Consider a $S'$ frame moving with a speed $v$ in 1D with respect to a stationary frame $S$. Using your everyday intuition, write down the relationship between a position measurement $x$ and $x'$. *Be ready to explain why this makes sense to you.* Note: This is just to get them thinking about the Galilean transformation again as you will do this with gamma.
The Galilean transformation between $S'$ and $S$ is: $$x = x' + vt$$ The Lorentz transformation will introduce a $\gamma$, where do you think it goes? And why? Note: This is just to get them thinking. The time will dilate and the length with contract, so where does the gamma go? It ultimately multiples both.
I'm in frame $S$, and you are in is in Frame $S'$, which moves with speed $V$ in the $+x$ direction. An object moves in the $S'$ frame in the $+x$ direction with speed $v'_x$. Do I measure its $x$ component of velocity to be $v_x = v'_x$? 1. Yes 2. No 3. ??? Note: * Correct Answer: B * can't be. because of time dilation and length contraction
I'm in frame $S$, and you are in is in Frame $S'$, which moves with speed $V$ in the $+x$ direction. An object moves in the $S'$ frame in the $+y$ direction with speed $v'_y$. Do I measure its $y$ component of velocity to be $v_y = v'_y$? 1. Yes 2. No 3. ??? Note: * Correct Answer: B * again. can't be. because of time dilation and length contraction
With Einstein's velocity addition rule, $$u = \dfrac{u' + v}{1+\frac{u'v}{c^2}}$$ what happens when $v$ is very small compared to $c$? 1. $u\rightarrow 0$ 2. $u\rightarrow c$ 3. $u\rightarrow \infty$ 4. $u \approx u' + v$ 5. Something else Note: * Correct Answer: D * denominator goes to 1 because second term is near zero - get back classical addition
With Einstein's velocity addition rule, $$u = \dfrac{u' + v}{1+\frac{u'v}{c^2}}$$ what happens when $u'$ is $c$? 1. $u\rightarrow 0$ 2. $u\rightarrow c$ 3. $u\rightarrow \infty$ 4. $u \approx u' + v$ 5. Something else Note: * Correct Answer: B * plug it in and you will get c; if something moves with c it does so in every frame - constant speed of light
With Einstein's velocity addition rule, $$u = \dfrac{u' + v}{1+\frac{u'v}{c^2}}$$ what happens when $v$ is $c$? 1. $u\rightarrow 0$ 2. $u\rightarrow c$ 3. $u\rightarrow \infty$ 4. $u \approx u' + v$ 5. Something else Note: * Correct Answer: B * if frame moves at c, all things move at c