# Special Relativity with no short cuts

I am currently a 3rd year undergraduate electronic engineering student. I have completed a course in dynamics, calculus I, calculus II and calculus III. I have decided to self study a basic introduction to special relativity as it was not part of my physics syllabus and is not included in any of my other modules. The source I am using is the textbook "Fundamentals Of Physics(9th Edition)" by Halliday and Resnick.

I understand the concept of proper time and length, but I am interested in solving problems using the Lorentz transformations rather than using the length contraction and time dilation "short cuts."

For example, the following textbook sample problem:

I understand how the the answer was obtained, however I am wondering if it is possible to tackle the problem as follows; using just the Lorentz transformation equations and the equation of motion; x = vt and defining 2 events, i.e. [event 1] = point B passes Sally & [event 2] = point A passes Sally. I have tried this but I keep getting the wrong answer unless I use the length contraction formula. Is there a way to work out this problem in this manner based on the given information?

All answers on this will be greatly appreciated.

First set up space and time origins for each frame. Say the spatial origin of Sally's frame is right in point A, and her time origin is right when Sam's point B passes by her. For Sam, say his space origin is in point B and his time origin is also when his point B passes Sally at her point A.

Now write down event coordinates for both frames, keeping just the relevant axis (x) along the direction of motion.

In Sally's frame, if she observes Sam's ship to have length L and pass by at velocity $v >0$, then

Event 1 = $(x_1 = 0, \;t_1 = 0)$

Event 2 = $(x_2 = 0, \;t_2 = t_1 + \Delta t = \frac{L}{v})$

In Sam's frame, his ship is known to have rest length $L_0$ and he observes Sally pass by at velocity $-v < 0$, so the same events must read

Event 1 = $(x'_1 = 0, t'_1 = 0)$

Event 2 = $(x'_2 = - L_0, t'_2 = t'_1 + L_0/v)$

The Lorentz transformation from Sally's frame to Sam's frame must take $(x_1, t_1)$ into $(x'_1, t'_1)$ and $(x_2, t_2)$ into $(x'_2, t'_2)$. The way we chose coordinate origins renders the first set trivial, $(0,0) \rightarrow (0,0)$, but the 2nd set yields $L$ in terms of $L_0$: $$x'_2 \equiv - L_0 = \gamma(x_2 - v t_2) = \gamma \left(0 - v\frac{L}{v}\right) = -\gamma L\\ t'_2 \equiv \frac{L_0}{v} = \gamma\left(t_2 - \frac{v}{c^2}x_2\right) = \gamma\left(\frac{L}{v} - \frac{v}{c^2}0\right) = \gamma \frac{L}{v}$$ which give of course the length contracted length seen by Sally as $$L = \frac{L_0}{\gamma}$$ Then the ship speed observed by Sally follows from $L \equiv L_0/\gamma = v\Delta t$ etc.

• And what if I wanted to Lorentz transform from Sam's frame to Sally's frame? – Salim Sep 3 '16 at 5:11
• Apply the inverse transform, $$x = \gamma(x' + vt')\\t = \gamma\left(t' + \frac{v}{c^2}x'\right)$$ – udrv Sep 3 '16 at 7:16
• That's how I attempted the problem, but I get stuck because I can't explicitly solve for v, at one point it becomes impossible to isolate v – Salim Sep 3 '16 at 7:36
• No, it works the same way. The difference is that the transformation for $x$ gives another trivial result, $x_2 \equiv 0 = \gamma(x'_2 + vt'_2) = \gamma(-L_0 + v L_0/v) =0$. But the one for $ct$ gives $t_2 \equiv L/v = \gamma(t'_2 + vx'_2/c^2) = \gamma[L_0/v + (v/c^2)(-L_0)] = \gamma(1- v^2/c^2) L_0/v = L_0/(v\gamma)$. This retrieves again length contraction and the relation for speed, $L = L_0/\gamma = v\Delta t$. – udrv Sep 3 '16 at 15:27
• Welcome, and good luck :) – udrv Sep 3 '16 at 18:57