1
$\begingroup$

I'm working through SR along with a lecture/class series. Last week, we worked through an elementary proof of length contraction after already deriving that $t = \gamma t_0$ with $\gamma$ the familiar Lorentz factor.


This is the proof in the lecture notes:

Consider an observer B and a mirror. They are at rest with respect to each other, separated by a distance $L_0$. The observer reflects a beam of light onto the mirror to measure its distance.

$$L_0 = \frac{ct_0}{2}$$

Now consider an observer A. Observer B is moving away with velocity $v$ with respect to A. A also measures the length with B's experiment:

$$L = \frac{c}{2} (\Delta t_1 + \Delta t_2)$$

Where the terms in brackets represent the motion of the light to the mirror, and the motion of the light from the mirror back to B.

$$\Delta t_1 = \frac{L}{c} + \frac{\Delta t_1 v}{c} \text{ and } \Delta t_1 = \frac{L}{c - v}$$

Analogously follows $\Delta t_2 = \frac{L}{c+v}$.

We now obtain:

$$\Delta t_1 + \Delta t_2 = \frac{2L/c}{1-v^2/c^2}$$

Using that $\Delta t_1 + \Delta t_2 = t_A = \gamma t_0$ we derive

$$\gamma t_0 = \frac{2L/c}{1-v^2/c^2}$$

$$t_0 = \gamma \frac{2L}{c}$$ $$\frac{ct_0}{2} = \gamma L$$ $$L_0 = \gamma L$$

Which is the familiar equation for length contraction.


Here my issue starts.

I was recreating the derivation when I figured "hey, why not make the substitution for gamma earlier." So, I took

$$\Delta t_1 + \Delta t_2 = t_A = \gamma t_0$$

And substituted it into

$$L = \frac{c}{2} (\Delta t_1 + \Delta t_2)$$

This gives

$$L = \frac{c}{2} \gamma t_0$$ $$L = \frac{ct_0}{2}\gamma$$ $$L = L_0 \gamma$$

The reverse of the equation I derived and should have derived. Thus, there must be an illegal operation somewhere. But where?

$\endgroup$
0
$\begingroup$

$$L≠\frac{c}{2}(\Delta t_{1}+\Delta t_{2})$$ See your third equation and the analogous one involving $\Delta t_{2}$, or indeed your fifth equation.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

The idea here is that you make the measurement in a rest frame $S$, where the length of the rod is measured instantaneously. By this, the only thing that should concern you is that in your frame $S^{'}$, $\Delta t=0$, and not the measurement of time in the moving frame.

From this, you will realise that there is a length contraction.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

The second equation in the lecture notes is nonsense, and unfortunately, you have used it in your calculations. Just compare the second and the fourth equations in the notes, they are obviously incompatible with each other. The second (nonsense) equation implies:

$$\Delta t_1 + \Delta t_2= \frac{2L}{c}\space,$$

and the fourth one is:

$$\Delta t_1 + \Delta t_2 = \frac{2L/c}{1-v^2/c^2}\space .$$

Eliminating $\Delta t_1 + \Delta t_2$ in these equations, we get:

$$v=0\space.$$

You can see that the second equation is not directly applied in the note. That is, the rest of the calculation is done independently of the mentioned equation, whereas you mistakenly supposed that the equation is correct and used it.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.