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Derivation of the Lorentz Transformations 
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#1
Dec2606, 10:48 AM

P: 392

I've been going through "Relativity", a translation of a book by Albert Einstein about the Special and General theories of Relativity. It is stressed that the book should be understandable to anyone with a high school education. "A clear explanation that anyone can understand," it says on the cover. Well, I'm feeling rather stupid at the moment, understanding quite little of the book.
At the moment though, I'm not interested in most of the book, only the "Simple Derivation of the Lorentz Transformations" in the appendix. I'm finding it far from simple and I'm hoping someone could help me with it. First of all, I'll not assume anyone to actually have this book, so here is a link to an online version (of poorer quality) of the derivation: http://www.bartleby.com/173/a1.html My first problem starts at equation (3) (and (4)). It's not all that obvious to me why we're adding that constant. The rest of my problems start after equation (7). I (mathematically) understand everything up to (7), but then I get lost. I have no idea, for example, where the equation between (7) and (7a) is coming from. Given that equation, I can understand (7a), but then going on to (7b), I'm lost again. Why are the two snapshots identical? Then there's (8). I'm willing to believe that by inserting a and b, you get those equations, but how does one acquire the value of b? The last mystery (I haven't gone on to combining everything yet) is (8a). I'm clueless as to what they did to get this equation. I'm sure they did something with the equations of (8), but what exactly... I don't see it. I hope I explained things sufficiently clear and I hope someone will be able to help me. 


#2
Dec2606, 12:06 PM

P: 231

I have read that derivation also, and I find it useless. It's all arithmetic and no physics. The only derivation I find useful is in Einsteins 1905 paper " On the Electrodynamics of Moving Bodies", available in the Dover book The Principle of Relativity. It's not so easy to follow but it uses math no higher than college level.



#3
Dec2706, 05:49 AM

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(Now there is something the book leaves out and that is that the transformation should be linear.) All you know is that if you enter x=ct into the transformation, you should get x'=ct'. By linearity xct and x'ct' should be proportional. The equation [itex]x'ct'=\lambda (xct)[/itex] is linear and has exactly that property that whenever x'ct'=0 we have xct=0. In know that's merely a repetition of the book. Maybe someone else can explain it more clearly. The left endpoint is at x=0 ofcourse, the other at 1/a. So the length is 1/a as seen from K. Now if we reverse the roles, we look at a unit rod in K as seen from K'. We should get the same length 1/a by symmetry, or rather, by the principle of relativity. So we look at the endpoints at a moment in time in K' (t'=0). For this we need to invert the lorentz equations. We want x,t in terms of x' and t'. Since in this case t'=0, we get: x'=axbct 0=actbx Now we eliminate t to find the relation between x and x' (multiply the top one by a and the bottom one by b and add them). We get: ax'=(a^2b^2)x or x'=a(1b^2/a^2)x So again one endpoint is at x'=0 ofcourse, the other (x=1) is at a(1v^2/c^2). (Where bc/a=v is used). Since this should be equal to 1/a by the principle of relativity we find [itex]a^2=1/(1v^2/c^2)[/itex] (Introduce [itex]\gamma=1/\sqrt{1v^2/c^2}[/itex] and use [itex]\gamma^2(1v^2/c^2)=1[/itex] in your calculation so it won't be very messy.) 


#4
Dec2706, 01:43 PM

P: 392

Derivation of the Lorentz Transformations
Thanks for your explanations, Galileo, they've helped me quite a bit, filling in some of the details I was missing. I couldn't gather, for example, that I was to invert the equations for that equation between (7) and (7a). Thanks for that. I still have some questions, however.
About (8a). I'll take your word that calculating those values will get me the same results (I haven't the patience to try), but I'm really looking to know how they got from (8) to (8a) instead of how to check (8a) by means of (8). I also have a new question. Equation (6) is used quite a lot in the reasoning that comes after that equation, but it is constructed under the condition that x'=0. Does it still hold true if x'/=0? It just seems rather odd that in to get (8), (6) is used. In (8) you get an equation for x', while x' is permanently 0 in (6). I hope my messy explanation is clear enough to communicate my problem. At any rate, what am I seeing wrong here? Any lightshedding on my problems would be greatly appreciated. 


#5
Dec2706, 02:55 PM

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I don't know how much math experience you have, but it follows quite directly: If you fix x', say x'=L (so it's like the coordinate of the endpoint of a fixed rod of length L in K'), then from 5: L=axbct or x=(L/a)+(bc/a)t which is the equation for uniform motion with speed bc/a, so evidently bc/a=v always. Incidentally, notice that the endpoint is at [itex]L/a=L/\gamma[/itex], which means the length of the rod is shorter by that factor gamma as seen from K, which is ofcourse the phenomenon of Lorentz contraction. 


#6
Dec2706, 03:06 PM

P: 392

Thanks again, Galileo, for your explanations. I'm not sure what you mean about fixing x', but the first, simple explanation came through.
Going through the derivation again, I've stumbled on yet another problem I'm hoping someone could help me with. To obtain the equations on (5), equations (3) and (4) are added and subtracted. What puzzles me is why one should want to do that. Why add and subtract (3) and (4)? 


#7
Dec2806, 01:22 AM

P: 997

special relativity and its experimental foundations yuan zhong zhang world scientific 1996 in order to see how a guessed shape of the transformation equations and some physical conditions imposed to it lead to expressions for the 16 coefficients. 


#8
Dec2806, 07:17 AM

P: 392

I haven't yet been able to find that particular piece, but from your description of it, I gather that the derivation isn't entirely based on pure logic and mathematics, am I correct?
In the mean time, I've continued on through the derivation and stumbled yet again upon problems. I hope someone will be able to clear things up for me. First of all, I was expecting to see the equations of (8) and then similar ones for y' and z'. It would seem the most logical approach, or am I mistaken in this? Then there's (11a). There's a large piece of text between (11) and (11a), but what I gather is that (11) needs to be generalized. (11a), however, is just (11) with sigma=1. I'd sooner think that specializing the equation, simplifying it to an extended version of (8a), so that it only holds for the xaxis. Perhaps I'm just reading it completely wrong (as you may or may not have noticed, English isn't my native language), or perhaps just misunderstanding. In general, the entire course of the derivation from (10) onwards seems rather vague. I can follow things through to (11), but the point of it all seems to escape me. I anyone could clarify things a little, I'd much appreciate it. 


#9
Dec2806, 07:50 AM

P: 154

I found http://www.mth.uct.ac.za/omei/gr/chap1/node4.html from http://www.mth.uct.ac.za/omei/gr/ useful when I was first learning SR. It uses a physical example of light being emitted and reflected off of something back towards the emitter as a starting point.
I own the book you mention, I found it okay to read but the translation is quite poor in places. I've lent it to someone recently and they said that they couldn't get into it at all. I can't actually remember whether or not there is a derivation of the Lorentz transformation in the book or whether it is just stated that "these are the transformations you must use to get the right answers" (EDIT: just noticed you mentioned the appendix there). I was also bitterly disappointed with the section on General Relativity, it barely scratches the surface. 


#10
Dec2806, 01:16 PM

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[tex] x'ct'=\lambda (xct)[/tex] [tex] x'+ct'=\mu (x+ct) [/tex] for x' and t'. The fastest way is clearly to add and subtract the 2. Then a and b are introduced for notational convenience. You could keep on working with lambda and mu, but that's a bit clumsy because of the lengthy expressions you might get. 


#11
Dec2806, 01:31 PM

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To obtain the transformation for a system K' going, say, in the positive ydirection just switch x and y. It says the relation (11a) always holds for arbitrary Lorentz transformations. 


#12
Dec2806, 02:56 PM

P: 392

Thanks again, Galileo. Looking through things again, I think I've come to understand that (11a) is simply an equation that holds true, not one that is used for calculations. Am I correct?
I understand that if the system would be moving along one of the other axes, you could switch the variables, but what if it's not moving along an axis, what if it's moving in a random direction? Is that what you meant when you talked about applying rotation transformations? What are rotation transformations? 


#13
Dec2806, 03:15 PM

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Check here for more: http://en.wikipedia.org/wiki/Coordinate_rotation If you want the Lorentz transformation for arbitrary orientations and velocity, then simply apply rotations to K and K' such that their orientations are the same and the xaxes point along the direction of motion, then apply the 'standard' Lorentz transformation, then rotate them back to the oiginal orientations. The principle is simple, but the math is tedious. I haven't seen anyone use it in practice. 


#14
Dec2806, 03:31 PM

P: 392

Ah, I get it, thanks. One question arises from your answer, however. If no one uses this method in practice, how is it done usually?



#15
Dec3006, 04:58 AM

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#16
Dec3006, 06:23 AM

P: 392

Ah, I understand from your description, thank you.



#17
Jun611, 10:33 PM

P: 1

Thnx, Galileo, I had exactly the same question!



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