The Empire Strikes Back length contraction.

[seto6]

Homework Statement

The starships of the Solar Federation are marked with the symbol of the Federation, a circle, whereas starships of the Denebian Empire are marked with the Empire's symbol, an ellipse whose major axis is n times its minor axis (a=nb in the figure ).

Homework Equations

L=(1-(v/c)^2)l

The Attempt at a Solution

x = rest length of major axis = nb
L = observed length of major axis when ship is traveling at speed v and must be equal to b since you want the ellipse to look like federation circle

b = nb[SQRT(1 - v^2/c^2)] then
to this (1/n)^2 = 1 - v^2/c^2
solve for V...v^2 = (c^2)[1 - (1/n)^2] = (c^2)[ (n^2 - 1)/n^2]
then arrive at v = (c/n)*SQRT[(n^2 - 1)].

i also did this.
L=(1-(v/c)^2)l

b=nb[SQRT(1 - (v/c)^2)]

then i got ((1/n)^2)^1/2)= 1-v/c ---> 1-1/n=v/c

solving for v i got v= c(1-c/n)

i'm not sure which one is correct and don't under stand y different answer.

#1v = (c/n)*SQRT[(n^2 - 1)].

#2 v= c(1-c/n)

im lost

 

[seto6]

i think i made a mistake some where not sure where tho

 

[collinsmark]

Homework Equations

L=(1-(v/c)^2)l

It's kinda difficult to read your notation in pure type. This kind of problem is where LaTeX comes in real handy.

Do you mean,

$$L' = L \sqrt{1 - \frac{v^2}{c^2}} \ ?$$

Here is a link to Physic's Forum LaTex guide.
https://www.physicsforums.com/misc/howtolatex.pdf"

The Attempt at a Solution

x = rest length of major axis = nb
L = observed length of major axis when ship is traveling at speed v and must be equal to b since you want the ellipse to look like federation circle

b = nb[SQRT(1 - v^2/c^2)] then
to this (1/n)^2 = 1 - v^2/c^2
solve for V...v^2 = (c^2)[1 - (1/n)^2] = (c^2)[ (n^2 - 1)/n^2]
then arrive at v = (c/n)*SQRT[(n^2 - 1)].

Re-writing in LaTeX gives,
==========================================================

$$b = nb \sqrt{1 - v^2/c^2}$$

$$\frac{1}{n^2} = 1 - \frac{v^2}{c^2}$$

$$v^2 = c^2 \left( 1 - \frac{1}{n^2} \right) = c^2 \left( \frac{n^2 - 1}{n^2} \right)$$

$$v = (c/n) \sqrt{n^2 - 1}$$

============================================================

Okay, so far I follow you.

i also did this.
L=(1-(v/c)^2)l

b=nb[SQRT(1 - (v/c)^2)]

then i got ((1/n)^2)^1/2)= 1-v/c ---> 1-1/n=v/c

Rewriting,

============================================================

$$b = nb \sqrt{1 - v^2/c^2}$$

$$\sqrt{ \left( \frac{1}{n^2} \right)^2} = \left( 1 - \frac{v}{c} \right) \ \longrightarrow \ \left( 1 - \frac{1}{n} \right) = \frac{v}{c}$$

==============================================================

Ouch! I see the problem,

You had,

$$\sqrt{ \left( \frac{1}{n^2} \right)^2} = \left( 1 - \frac{ \color{red}{v}}{ \color{red}{c}} \right)$$

But it should be:

$$\sqrt{ \left( \frac{1}{n^2} \right)^2} = \left( 1 - \frac{v^{\color{red}{2}}}{c^{\color{red}{2}}} \right)$$

solving for v i got v= c(1-c/n)

i'm not sure which one is correct and don't under stand y different answer.

#1v = (c/n)*SQRT[(n^2 - 1)].

The above one looks good to me.

#2 v= c(1-c/n)

And, followed by not so good.