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