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Consider two observers, travelling away from each other, after meeting (at which time they sync their clocks). $O$ sends a photon towards $O'$ at times $t$, received by $O'$ at time $t'=kt$, where we have defined Bondi's k factor.

In the following reasoning I know I have made a mistake: So by Milne's radar definition of simultaneity, if the photon is instantly sent back from $O'$ and received by $O$ at $t_2$, then: $$t'=\frac12(t+t_2)=kt\Rightarrow t_2=2t'-t=(2-\frac{1}{k})t'$$ But $t_2$ is meant to equal $kt'$

Could someone explain to me where I have I gone wrong here?

EDIT: Basically I am trying to show that $t_2=kt'$ working from only the two assumptions that:

(i) they both observe the speed of light as c

(ii) Only relative motion is observable

enter image description here

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Using the definition of simultaneity on Page 13 here: maths.ox.ac.uk/system/files/coursematerial/2012/2393/8/… –  Freeman Mar 27 '13 at 13:07
    
Thanks for that. I see that line is correct after all. –  Mew Mar 27 '13 at 13:09
    
So it doesn't make any sense to write the equality between that and $t'$? How would you go about showing the result that $t_2=kt'$? –  Freeman Mar 27 '13 at 13:17

2 Answers 2

Your formula $$ t' = \frac{t+t_2}{2} $$ implicitly uses the time measured by the observer with the vertical world line – which is the same thing as the vertical $y$-coordinate. However, the tilted/moving observer measures a shorter time $t'$ by the time dilation factor $\sqrt{1-v^2/c^2}$.

I know that my method of calculation isn't following the pedagogical goal of Bondi to convert everything to the Bondi's $k$-factor, the Doppler factor $$ k = \sqrt{\frac{1+v/c}{1-v/c} }$$ but I haven't learned relativity in Bondi's way and the main goal is to fix the discrepancy which should be allowed to be done in any way. So let me say that the time dilation occurs by the factor $\sqrt{1-v^2/c^2}\lt 1$. So let's revert the relationships. For the $k$ defined above, we have $$ \frac vc = \frac{k^2-1}{k^2+1},\quad \sqrt{1-\frac{v^2}{c^2}} = \frac{2k}{k^2+1} $$ This is the time dilation factor you have omitted so the right equation replacing $$ t' = \frac{t+t_2}{2} $$ is $$ t' = \frac{k}{k^2+1} (t+t_2) $$ This is right. Substitute $t_2=kt'$ and you get $$ t' = \frac{kt}{k^2+1} + \frac{k^2}{k^2+1}t' $$ i.e., after multiplication by $k^2+1$, $$ t' (k^2+1-k^2) = kt,\quad t'=kt$$ which you wanted to get.

Inverted calculation

For Freeman, let me also revert the logic and order of the calculation. By the definition of Bondi's $k$, we may assume $t'=kt$. We want to calculate the time dilation factor $1/\gamma \lt 1$ – I want to avoid new symbols. The time $t'=kt$ may also be expressed as $$ t' = \frac{1}{\gamma} \frac{t+t_2}{2} $$ i.e. $$ kt = \frac{1}{\gamma}\frac{t+k^2 t}{2} $$ where I used $t_2=kt'=k^2 t$. From the last equation, it is easy to calculate $$\frac{1}{\gamma} = \frac{2k}{k^2+1},$$ just like expected.

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Thanks for your answer, this is great! But the result of this problem was to derive the time dilation factor -as a consequence of $t_2=kt'$, working from two assumptions that (i) they both observe the speed of light as c, and (ii) Only relative motion is observable. I should have written that I now see. –  Freeman Mar 27 '13 at 13:24
    
I see. It's my guess that there are about 3 relationships and each 2 of them imply the third, so you may just revert the logic and direction of the calculation. Apologies if it ain't the case. I used the known form of the dilation factor $2k/(k^2+1)$ and some other data (2) to derive some other data (3). So you probably may assume (2) and (3) to derive that the time dilation factor is $2k/(k^2+1)$. –  Luboš Motl Mar 27 '13 at 13:27
    
Yes exactly, so your answer is very helpful, I will try and do that. Thank you :) –  Freeman Mar 27 '13 at 13:28
    
Incidentally, I do think that the Bondi approach to special relativity is at least equally natural as the normal one. But habits influence one's thinking... –  Luboš Motl Mar 27 '13 at 13:30
    
Yes exactly, I've only been taught this one and this is the only step that isn't quite clicking for me. Everything else is just algebra, I know it's something to do with only relative motion being observable... hmm... I think I just figured it out, I think it has to be the case by considering similar triangles. (or that might be bullshit.. just remembering something my tutor said..) –  Freeman Mar 27 '13 at 13:33
up vote 2 down vote accepted

enter image description here

Is it due to this similar triangle relation, derived from the fact light rays will always be seen at the same speed, so we draw them at $\pi/4$

(larger link http://i.stack.imgur.com/ZaVOd.jpg)

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