The Principles of Relativity - Help Kindly Requested

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The discussion revolves around a user's difficulty in understanding a mathematical derivation in "The Essential Einstein," specifically regarding the transition between two lines involving partial derivatives. It is confirmed that the second line indeed involves partial derivatives and that the function τ represents time as measured by a moving observer. The user learns to apply the approximation correctly, realizing that applying it to the second argument of the left-hand side yields the desired result. There is a sentiment expressed about Einstein's lack of explanatory context in his equations, which could aid readers in following the derivation. Overall, the conversation emphasizes the importance of clarity in mathematical communication, especially in complex topics like relativity.
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Hello,

I'm new to these forums so I hope this first post is ok.

I'm currently reading & (slowly) working my way through the book "The Essential Einstein - His Greatest works" published by Penguin books but I'm stuck on a derivation on page 11 !

I've attached the derivation as a jpg image to this post (I currently don't know how to work latex as yet) and it's just two lines of mathematics.

My question is, how does he get from the first line to the second ?

It looks to me as thought it's partial differentiation as there are partial derivatives on the second line but when I try doing so it just doesn't work out right. Am I right in thinking it's partial derivatives ? Also he quotes a function, Tau but does not define the function only its parameters so that confuses me slightly as to how he differentiates this and ends up with constants.

Your help is appreciated

Bongos
 

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Yes, these are partial derivatives.
\tau is the time as measured by the moving observer. Generally, it is a function of t,x,y,z.
You know that for any function
\tau(x,t+dt) \simeq \tau(x,t) + \frac{\partial \tau}{\partial t}dt?
Do these steps, subtract \tau(0,0,0,t) and divide by x'.
 
Ich said:
Yes, these are partial derivatives.
\tau is the time as measured by the moving observer. Generally, it is a function of t,x,y,z.
You know that for any function
\tau(x,t+dt) \simeq \tau(x,t) + \frac{\partial \tau}{\partial t}dt?
Do these steps, subtract \tau(0,0,0,t) and divide by x'.

Thanks Ich,

It took a bit of working out. I originally started applying the approximation you gave to the WHOLE of the LHS of the first line, but I ended up with a half of Tau in the result. I then realized I just apply it the the second argument of the LHS and again on the right (after a bit of re-arranging) and the Tau function drops out.

Thanks for that !
 
DrGreg said:
As luck would have it, I answered this very question 4 years ago in post #6 of the thread Understanding Einstein's Math.

Thanks DrGreg.

I remember actually working this out before but second time round I completely forgot ! The PDF is useful as well as I needed a hint to understand the next line in einstein's paper.

I wonder why Einstein never put a few words into explain how he got from one equation to the next, even if he put 'an approximation' it might give people a hint at how he arrived at the second line.
 
Bongos said:
I wonder why Einstein never put a few words into explain how he got from one equation to the next, even if he put 'an approximation' it might give people a hint at how he arrived at the second line.

Well, he could have added a few words of explanation, but I guess he just assumed his readers would follow the step. Experienced mathematicians (and physicists) can become so familiar with applying the chain rule in situations like this, they can do it in their heads and assume the reader can, too. Bear in mind he was (I believe) writing a technical paper aimed at the physicists and mathematicians of the time, not the general public.
 
Bongos said:
I've attached the derivation as a jpg image to this post (I currently don't know how to work latex as yet) and it's just two lines of mathematics.


\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}

Left click on the above equation and a small window pops up showing the latex code that was used to generate it. Enter the code exactly as shown including the red text in square brackets. You can also left click on the equation posted by Ich to see how the symbols he used are generated. That should be enough to get you started.
 

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