Author Topic: Special Relativity (Read 2378 times) Tweet Share

/0 · « **Reply #15 on:** May 11, 2011, 08:01:26 pm »

Quote from: Rohitpi on May 10, 2011, 04:37:25 pm

Quote from: /0 on May 10, 2011, 01:00:51 am
Nice and simple

lulz
Nice = I got it right (maybe? oh well comeatmebrah says it's right and I'm not going to check it again)
Simple = I completed it within my lifetime

If only my assignments had more nice simple problems

Thu Thu Train · « **Reply #16 on:** May 11, 2011, 08:12:49 pm »

Quote from: /0 on May 11, 2011, 08:01:26 pm

Quote from: Rohitpi on May 10, 2011, 04:37:25 pm
Quote from: /0 on May 10, 2011, 01:00:51 am
Nice and simple

lulz
Nice = I got it right (maybe? oh well comeatmebrah says it's right and I'm not going to check it again)
Simple = I completed it within my lifetime

If only my assignments had more nice simple problems

It looks similar to the answer I got (I used k-calculus though which makes it easier to work with)

Asx4Life · « **Reply #17 on:** May 11, 2011, 11:45:29 pm »

Quote from: comeatmebrah on May 10, 2011, 05:02:12 pm

Quote from: /0 on May 10, 2011, 01:00:51 am
Quote from: comeatmebrah on May 08, 2011, 05:34:04 pm
I have a question:

A star measures a second star to be moving away at speed v = 0.85c, measured
in light-meters. The second star measures a third to be receding in the same
direction at 0.85c. Similarly, the third measures a fourth, and so on, up to some
large number n of stars. What is the velocity of the n^th star relative to the first?
Give an exact answer and an approximation useful for large n.

Suppose the nth star moves at velocity $v$ relative to the first planet, and the nth star moves at velocity $v'$ relative to the (n-1)th star. Then we can apply $(n-2)$ lorentz boosts to transform the 4-velocity of the $n$ th star from the frame of the first star into the frame of the $(n-1)th$ star. (for $n \geq 2$ )

$\left[\begin{matrix} \gamma(v')c \\ \gamma(v')v' \end{matrix}\right]=\left[\begin{matrix}\gamma(v') & -\frac{v'}{c}\gamma(v') \\ -\frac{v'}{c} \gamma(v') & \gamma(v') \end{matrix}\right]^{n-2}\left[\begin{matrix} \gamma(v)c \\ \gamma(v)v \end{matrix}\right]$

$\Rightarrow \frac{1}{\sqrt{1-\frac{v'^2}{c^2}}}\left[\begin{matrix} c \\ v' \end{matrix}\right]=\left(\frac{1}{\sqrt{1-\frac{v'^2}{c^2}}}\right)^{n-2}\left[\begin{matrix}1 & -\frac{v'}{c} \\ -\frac{v'}{c} & 1 \end{matrix}\right]^{n-2}\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\left[\begin{matrix}c \\ v \end{matrix}\right]$

$\Rightarrow \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\left[\begin{matrix} c \\ v \end{matrix}\right]=\left(\sqrt{1-\frac{v'^2}{c^2}}\right)^{n-3}\left(\frac{c}{c^2-v'^2}\right)^{n-2}\left[\begin{matrix} c & v' \\ v' & c \end{matrix}\right]^{n-2}\left[\begin{matrix} c \\ v'\end{matrix}\right]$

In our case, $v'=0.85c$ , so

$\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\left[\begin{matrix} c \\ v \end{matrix}\right]=c(0.2775)^\frac{1-n}{2}\left[\begin{matrix} 1 & 0.85 \\ 0.85 & 1 \end{matrix}\right]^{n-2}\left[\begin{matrix} 1 \\ 0.85\end{matrix}\right]$

Let's try to find an explicit expression for $v$ ... Because because!

$\left[\begin{matrix} 1 & 0.85 \\ 0.85 & 1 \end{matrix}\right]^{n-2}=\left(I+\left[\begin{matrix} 0 & 0.85 \\ 0.85 & 0 \end{matrix}\right]\right)^{n-2}=I+\sum_{k=1}^{n-2} {n-2 \choose k}\left[\begin{matrix} 0 & 0.85 \\ 0.85 & 0 \end{matrix}\right]^k$

Then

$I+\sum_{k=1}^{n-2} {n-2 \choose k}\left[\begin{matrix} 0 & 0.85 \\ 0.85 & 0 \end{matrix}\right]^{k}=I+\sum_{k=1}^{\lfloor\frac{n-1}{2}\rfloor}{n-2 \choose 2k-1}\left[\begin{matrix} 0 & (0.85)^{2k-1} \\ (0.85)^{2k-1} & 0 \end{matrix}\right]+\sum_{k=1}^{\lfloor\frac{n-2}{2}\rfloor}{n-2 \choose 2k}\left[\begin{matrix}(0.85)^{2k} & 0 \\ 0 & (0.85)^{2k}\end{matrix}\right]$

In other words,

$\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}v=c(0.2775)^{\frac{1-n}{2}}\left[\sum_{k=1}^{\lfloor \frac{n-1}{2}\rfloor}{n-2 \choose 2k-1}(0.85)^{2k-1}+0.85\left(1+\sum_{k=1}^{\lfloor \frac{n-2}{2}\rfloor}{n-2 \choose 2k}(0.85)^{2k}\right)\right]$

Solving for $v$ we get:

$v=\frac{c(0.2775)^{\frac{1-n}{2}}\left[\sum_{k=1}^{\lfloor \frac{n-1}{2}\rfloor}{n-2 \choose 2k-1}(0.85)^{2k-1}+0.85\left(1+\sum_{k=1}^{\lfloor \frac{n-2}{2}\rfloor}{n-2 \choose 2k}(0.85)^{2k}\right)\right]}{\sqrt{1+\frac{1}{c^2}\left(c(0.2775)^{\frac{1-n}{2}}\left[\sum_{k=1}^{\lfloor \frac{n-1}{2}\rfloor}{n-2 \choose 2k-1}(0.85)^{2k-1}+0.85\left(1+\sum_{k=1}^{\lfloor \frac{n-2}{2}\rfloor}{n-2 \choose 2k}(0.85)^{2k}\right)\right]\right)^2}}$ , $n \geq 2$ .

Nice and simple

CORRECT.

Quote from: /0 on May 10, 2011, 01:00:51 am

Quote from: comeatmebrah on May 08, 2011, 05:34:04 pm
I have a question:

A star measures a second star to be moving away at speed v = 0.85c, measured
in light-meters. The second star measures a third to be receding in the same
direction at 0.85c. Similarly, the third measures a fourth, and so on, up to some
large number n of stars. What is the velocity of the n^th star relative to the first?
Give an exact answer and an approximation useful for large n.

Suppose the nth star moves at velocity $v$ relative to the first planet, and the nth star moves at velocity $v'$ relative to the (n-1)th star. Then we can apply $(n-2)$ lorentz boosts to transform the 4-velocity of the $n$ th star from the frame of the first star into the frame of the $(n-1)th$ star. (for $n \geq 2$ )

$\left[\begin{matrix} \gamma(v')c \\ \gamma(v')v' \end{matrix}\right]=\left[\begin{matrix}\gamma(v') & -\frac{v'}{c}\gamma(v') \\ -\frac{v'}{c} \gamma(v') & \gamma(v') \end{matrix}\right]^{n-2}\left[\begin{matrix} \gamma(v)c \\ \gamma(v)v \end{matrix}\right]$

$\Rightarrow \frac{1}{\sqrt{1-\frac{v'^2}{c^2}}}\left[\begin{matrix} c \\ v' \end{matrix}\right]=\left(\frac{1}{\sqrt{1-\frac{v'^2}{c^2}}}\right)^{n-2}\left[\begin{matrix}1 & -\frac{v'}{c} \\ -\frac{v'}{c} & 1 \end{matrix}\right]^{n-2}\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\left[\begin{matrix}c \\ v \end{matrix}\right]$

$\Rightarrow \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\left[\begin{matrix} c \\ v \end{matrix}\right]=\left(\sqrt{1-\frac{v'^2}{c^2}}\right)^{n-3}\left(\frac{c}{c^2-v'^2}\right)^{n-2}\left[\begin{matrix} c & v' \\ v' & c \end{matrix}\right]^{n-2}\left[\begin{matrix} c \\ v'\end{matrix}\right]$

In our case, $v'=0.85c$ , so

$\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\left[\begin{matrix} c \\ v \end{matrix}\right]=c(0.2775)^\frac{1-n}{2}\left[\begin{matrix} 1 & 0.85 \\ 0.85 & 1 \end{matrix}\right]^{n-2}\left[\begin{matrix} 1 \\ 0.85\end{matrix}\right]$

Let's try to find an explicit expression for $v$ ... Because because!

$\left[\begin{matrix} 1 & 0.85 \\ 0.85 & 1 \end{matrix}\right]^{n-2}=\left(I+\left[\begin{matrix} 0 & 0.85 \\ 0.85 & 0 \end{matrix}\right]\right)^{n-2}=I+\sum_{k=1}^{n-2} {n-2 \choose k}\left[\begin{matrix} 0 & 0.85 \\ 0.85 & 0 \end{matrix}\right]^k$

Then

$I+\sum_{k=1}^{n-2} {n-2 \choose k}\left[\begin{matrix} 0 & 0.85 \\ 0.85 & 0 \end{matrix}\right]^{k}=I+\sum_{k=1}^{\lfloor\frac{n-1}{2}\rfloor}{n-2 \choose 2k-1}\left[\begin{matrix} 0 & (0.85)^{2k-1} \\ (0.85)^{2k-1} & 0 \end{matrix}\right]+\sum_{k=1}^{\lfloor\frac{n-2}{2}\rfloor}{n-2 \choose 2k}\left[\begin{matrix}(0.85)^{2k} & 0 \\ 0 & (0.85)^{2k}\end{matrix}\right]$

In other words,

$\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}v=c(0.2775)^{\frac{1-n}{2}}\left[\sum_{k=1}^{\lfloor \frac{n-1}{2}\rfloor}{n-2 \choose 2k-1}(0.85)^{2k-1}+0.85\left(1+\sum_{k=1}^{\lfloor \frac{n-2}{2}\rfloor}{n-2 \choose 2k}(0.85)^{2k}\right)\right]$

Solving for $v$ we get:

$v=\frac{c(0.2775)^{\frac{1-n}{2}}\left[\sum_{k=1}^{\lfloor \frac{n-1}{2}\rfloor}{n-2 \choose 2k-1}(0.85)^{2k-1}+0.85\left(1+\sum_{k=1}^{\lfloor \frac{n-2}{2}\rfloor}{n-2 \choose 2k}(0.85)^{2k}\right)\right]}{\sqrt{1+\frac{1}{c^2}\left(c(0.2775)^{\frac{1-n}{2}}\left[\sum_{k=1}^{\lfloor \frac{n-1}{2}\rfloor}{n-2 \choose 2k-1}(0.85)^{2k-1}+0.85\left(1+\sum_{k=1}^{\lfloor \frac{n-2}{2}\rfloor}{n-2 \choose 2k}(0.85)^{2k}\right)\right]\right)^2}}$ , $n \geq 2$ .

Nice and simple

WTF LOL

schnappy · « **Reply #18 on:** May 12, 2011, 12:03:26 am »

Matrices. Urgh.

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