Understanding Vector Theory Proofs: Solutions to Common Questions

latentcorpse
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In the notes attached here:
https://www.physicsforums.com/showthread.php?p=3042019#post3042019
(apparently I can't attach the same thing in multiple threads?)
I have quite a few problems with one of the proofs. In the proof of the proposition on p15,

a) he says to note that [itex]\nu(0)=0[/itex]. why is this?

b) he goes from
[itex]\{ \frac{d}{dt} [ \alpha ( x^\mu ( \lambda(t)) - x^\mu(p)) + \beta (x^\mu(\kapa(t))-x^\mu(p)) + x^\mu(p)] \}_{t=0} = [ \alpha ( \frac{d x^\mu ( \lambda (t))}{dt})_{t=0} + \beta ( \frac{dx^\mu ( \kappa ( t))}{dt} )_{t=0}][/itex]
I really don't understand how these two lines are equal at all!
And also how can we change the [itex]\phi[/itex]'s to [itex]x^\mu[/itex]'s in going from eqn 25 to the defn of [itex]Z_p(f)[/itex]?

c) where does eqn 27 come from? isn't [itex]( \frac{\partial}{\partial x^\mu})_p (f) = \frac{\partial f}{\partial x^\mu})_p[/itex]
is it something like if we compose the numerator with [itex]\phi^{-1}[/itex] then we have to cancel that out by composing the [itex]p[/itex] with [itex]\phi[/itex] to give the [itex]\phi(p)[/itex]? I don't really get why this is allowed though?

d)Where does eqn 29 come from?

Thanks a lot for any help. I really need to get my head round all this vector business over the holidays!
 
on Phys.org
For a, [tex]\lambda(0)=\kappa(0)=p[/tex], so once you plug those in you get
[tex]\nu(0)=\phi^{-1}(\alpha(\phi(p)-\phi(p)) + \beta(\phi(p)-\phi(p) +\phi(p)) = \phi^{-1}(\phi(p)) = p[/tex] (not 0 like you say in your post)

For (b), just distribute the derivative linearly. Then [tex]\frac{d x^{\mu}(p)}{dt} = 0[/tex] because p is just a fixed point, so that's just the derivative of a number
 
Office_Shredder said:
For a, [tex]\lambda(0)=\kappa(0)=p[/tex], so once you plug those in you get
[tex]\nu(0)=\phi^{-1}(\alpha(\phi(p)-\phi(p)) + \beta(\phi(p)-\phi(p) +\phi(p)) = \phi^{-1}(\phi(p)) = p[/tex] (not 0 like you say in your post)

For (b), just distribute the derivative linearly. Then [tex]\frac{d x^{\mu}(p)}{dt} = 0[/tex] because p is just a fixed point, so that's just the derivative of a number

hi there. thanks for your answers.

do you have any ideas for c) or d) or also, how we get the formula for [itex]Z-p(f)[/itex] in the first place?

thanks!
 

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