# Homework Help: Another question on special relativity (sorry lol)

1. Aug 22, 2013

### PsychonautQQ

1. The problem statement, all variables and given/known data

On my ever hopeful quest to come to a personal understanding of special relativity, I have stumbled upon a question while following my textbooks explanation of deriving an equation for length contraction.

My textbook word for word:
S' is moving at v relative to S
Consider a rod at rest in frame S' with one end at x'2 and the other end at x'1. The length of the rod in this frame is its proper length Lp = x'2-x'1. Some care must be taken to find the length of the rod in frame S. In that frame, the rod is moving to the right with speed v, the speed of frame S'. The length of the rod in frame S is defined as L = x2 - x1, where x2 is the position of one end at some time t2, and x1 is the position of the other end at the same time t1 = t2 as measured in frame S. to calculate x2 - x1 at some time t, we use these equations
x'2=γ(x2-vt2)
x'1=γ(x1-vt1)
x'2-x'1 = γ(x2-x1)

Okay... So I understand all that.. what I DON"T understand is why using the inverse equations doesn't lead to the same result.. you are relating all the same variables to each other, shouldn't the answer be the same? But of course when you do the inverse equation of
x2=γ(x'2+vt'2)
x1=γ(x'1+vt'1) you end up with
(x'2-x'1)γ=x'2-x'1

Are t'2 and t'1 not equal to each other? Why would the same variables give different answers? Does it have to do with "perspective"/"What frame your looking from" or something like that? And if so shouldn't that information be inside of the equation somewhere? I hope somebody understands the dilemma i'm seeing here... I just feel like the equations aren't being consistent.. Does it have something to do with maybe t'2 doesn't equal t'1 so they don't cancel in this situation? If so why don't they equal?

2. Aug 23, 2013

### voko

The "inverse equations" tell you that if you look at a rod in frame S from frame S', then the rod will appear contracted - just like a rod in S' appears contracted when looking from S.

This is to be expected because motion is relative: if you can say S' is moving with velocity v relative to S, you can also say S is moving with velocity -v relative to S'.

3. Aug 23, 2013

### vanhees71

I can't follow in detail what you wrote in #1, because it's awfully hard to read. Please use LaTeX for complex formulae.

To derive the length constraction you can of course use the Lorentz transformation. For a Lorentz boost in $x$ direction it reads, setting $c=1$
$$t'=\gamma (t-v x), \quad x'=\gamma(x-v t)$$
with
$$\gamma=\frac{1}{\sqrt{1-v^2}}.$$
$$t=\gamma(t'+v x'), \quad x=\gamma(x'+v t')$$

Now let a rod with one end in the origin $x'=0$ and the other end at $x'=L'$ be at rest in the primed reference frame.

Now for an observer in the umprimed frame this rod is moving with velocity $v$ along the $x$ axis. He is measuring the length of this moving rod by reading off at $t=0$ the coordinate of the end not in the origin. This "event" of reading off happens at another time $t'$ in the rest frame
$$t'=-v L'.$$
Plugging this into the formula for $x$, using the fact that this end is always at $x'=L'$, because the rod is at rest in the primed frame, gives
$$L=\gamma(L+v t')=\gamma (L'-v^2 L')=L' \gamma (1-v^2)=L' \sqrt{1-v^2}=\frac{L'}{\gamma}.$$
The observer in the frame, where the rod moves with velocity $v$ reads off a length that is shorter by a factor $1/\gamma=\sqrt{1-v^2}$.

4. Aug 23, 2013

### PsychonautQQ

So are these equations unique in the fact that just because they have the same variables doesn't mean you'll get the same answer? Like one of the equations is seeing from one reference frame, while the other reference frame is seeing from the other point of view. Is this information in the equations somewhere or do you just have to realize their limitations so to speak?

5. Aug 23, 2013

### voko

If you ask the same question, the equations give you the same answer. If you are getting different answers, then you are probably not asking the same question.

6. Aug 23, 2013

### Staff: Mentor

Using the inverse transformation does lead to the same result.

$$x=γ(x'+vt')$$
$$t=γ(t'+\frac{vx'}{c^2})$$
So,$$t_2=γ(t_2'+\frac{vx_2'}{c^2})$$
$$t_1=γ(t_1'+\frac{vx_1'}{c^2})$$
$$t_2-t_1=γ(t_2'-t_1'+\frac{v(x_2'-x_1')}{c^2})$$
So, if t1=t2,
$$(t_2'-t_1')=-\frac{v(x_2'-x_1')}{c^2}$$
From the first inverse equation, we have:
$$x_2-x_1=γ((x_2'-x_1')+v(t_2'-t_1'))$$
If we combine the above two equations, we obtain:
$$x_2-x_1=γ(x_2'-x_1')(1-\left(\frac{v}{c}\right)^2)=\frac{(x_2'-x_1')}{γ}$$

You need to come to grips with the idea that if t2=t1, then t2'≠t1', and vice versa.