Author Topic: Special Relativity (Read 2368 times) Tweet Share

Andiio · « **on:** May 06, 2011, 08:12:26 pm »

Would anyone by chance have any notes/worksheets/questions/materials on the detailed study Einstein's Special Relativity?

Got a SAC coming up soon! ><

Thanks!

Shark 774 · « **Reply #1 on:** May 06, 2011, 10:31:19 pm »

I'd appreciate the same thing if anyone has some stuff!

Vincezor · « **Reply #2 on:** May 06, 2011, 10:35:22 pm »

How is your detailed study SAC structured? Although I'm not doing relativity, my further electronics SAC was very similar to that of the exam (we had 15 MCQ questions, instead of 12 though)

If it is like that, then why not just try some questions from the detailed study question from trial/vcaa exams?

Lasercookie · « **Reply #3 on:** May 07, 2011, 10:31:49 pm »

If your SAC is a more theory based than exam-style questions, I've found these resources useful.

With online resources:
Einstein Light (by the UNSW) is a really good website
http://www.phys.unsw.edu.au/einsteinlight/

Another good resource:
http://www.rafimoor.com/english/SRE.htm

Translated version of Einstein's original paper, I found it fairly easy to follow, though didn't understand all of the maths.
http://www.fourmilab.ch/etexts/einstein/specrel/specrel.pdf

And these lecture notes from Macquarie University (does go a bit beyond the VCE course and uses calculus)
http://www.physics.mq.edu.au/~jcresser/Phys378/LectureNotes/VectorsTensorsSR.pdf

The MIT OpenCourseWare site has this course:
http://ocw.mit.edu/courses/physics/8-20-introduction-to-special-relativity-january-iap-2005/index.htm
Obviously it will go beyond VCE, but it does have problem sets and two exams.

Shark 774 · « **Reply #4 on:** May 08, 2011, 01:06:03 am »

Just a quick thought experiment, if anyone can be bothered answering: If I travel away from Earth at 0.7c, relative to the Earth and my friend travels in the opposite direction at 0.7c, relative to the Earth, what speed to I measure him to be travelling relative to myself???

xZero · « **Reply #5 on:** May 08, 2011, 02:10:23 am »

http://en.wikipedia.org/wiki/Velocity-addition_formula#Special_theory_of_relativity

This should answer your question, in summary its not 0.7c+0.7c=1.4c

Shark 774 · « **Reply #6 on:** May 08, 2011, 04:41:38 pm »

Cheers, already found it out though haha.

Thu Thu Train · « **Reply #7 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.

Andiio · « **Reply #8 on:** May 09, 2011, 06:02:21 pm »

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.

Just wondering, where is this question from?

forumguy · « **Reply #9 on:** May 09, 2011, 06:27:30 pm »

Quote from: Andiio on May 09, 2011, 06:02:21 pm

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.

Just wondering, where is this question from?

^ what he said

/0 · « **Reply #10 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

TrueTears · « **Reply #11 on:** May 10, 2011, 01:07:48 am »

LOLOLOLOLOLOL

taiga · « **Reply #12 on:** May 10, 2011, 01:31:06 am »

well that's the funniest shit I've seen all day

pi · « **Reply #13 on:** May 10, 2011, 04:37:25 pm »

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

Nice and simple

Thu Thu Train · « **Reply #14 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.

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Author Topic: Special Relativity (Read 2368 times) Tweet Share

Andiio

Special Relativity

Shark 774

Re: Special Relativity

Vincezor

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Lasercookie

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Shark 774

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xZero

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Thu Thu Train

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Andiio

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