Ziwebach: Xdot^{\mu} vs X. Matrix & Tensor Notation

[ehrenfest]

Homework Statement

http://books.google.com/books?id=Xm...xDw&sig=Eqnkxjj7B9gc4-nig1fgUkk7AqQ#PPA101,M1

I am confused about QC 6.4

What is the difference between Xdot^{\mu} and X. ? Is one just tensor notation and one just matrix notation?

By the way how to you get the dot on top of the X?

Homework Equations

The Attempt at a Solution

 

[dextercioby]

What do you know, pages 92 up to 115 are not part of the free preview. \dot{X}^{\mu} is probably the tau or sigma derivative of the X^{\mu}, that's my guess.

 

[ehrenfest]

It is the tau derivative. And the X^mu' is the sigma derivative.

 

[Jimmy Snyder]

Equations (6.49) and (6.50) on page 101 are straight forward partial derivatives of equation (6.46).
X is a 4-vector, as are \dot{X} and X' which are the \tau and \sigma derivatives of X respectively. There are 4 X^{\mu} and each one is one of the 4 components of X. \dot{X}^{\mu} and X^{\mu\prime} are defined in equation (6.40) on page 100.

 

[ehrenfest]

What is confusing me is that L(\dot{X}^{\mu}, X^{\mu '}) in 6.46 implies that L is a function of only a single component of \dot{X} and X^{'}, but then in the function definition all of the other components appear in the dot products. What is wrong with my thinking? Why is it not L(\dot{X}, X^{'}) ?

 

[nrqed]

What is confusing me is that L(\dot{X}^{\mu}, X^{\mu '}) in 6.46 implies that L is a function of only a single component of \dot{X} and X^{'}, but then in the function definition all of the other components appear in the dot products. What is wrong with my thinking? Why is it not L(\dot{X}, X^{'}) ?

What do you mean by "simgle component"? The notation implies that L depends on all four \dot{X}^{\mu} and all four X^{\mu '}.

 

[Jimmy Snyder]

What is confusing me is that L(\dot{X}^{\mu}, X^{\mu '}) in 6.46 implies that L is a function of only a single component of \dot{X} and X^{'}, but then in the function definition all of the other components appear in the dot products. What is wrong with my thinking? Why is it not L(\dot{X}, X^{'}) ?

I agree with you. Zwiebach's notation here is a little 'funny'. He has for equation (6.46)

\mathcal{L}(\dot{X}^{\mu},X^{\mu\prime}) = -\frac{T_0}{c}\sqrt{(\dot{X}\cdot{X}') - (\dot{X})^2(X')^2}

Now the rhs makes it clear that this is a function of all 4 components, but the notation on the lhs might be considered ambiguous. Certainly he does not mean to imply that \mathcal{L} is a function of only one of the X^{\mu}, because then he would have to tell us which one. Do not let this confuse you, he means all 4. If, like me, you are marking up the margins of your book with notes, then simply cross out the lhs and rewrite it as

\mathcal{L}(\dot{X},X')

and similarly for equation (6.45). If you do not mark up your book, then simply note that the rhs of (6.46) shows you what he had in mind.

 

[ehrenfest]

OK. But when he takes the partial derivative with respect to \dot{X}^{\mu}, this is really a partial with respect to the component not \dot{X}, right?

 

[Jimmy Snyder]

OK. But when he takes the partial derivative with respect to \dot{X}^{\mu}, this is really a partial with respect to the component not \dot{X}, right?

Yes, in fact there are 4 such equations, one for each value of \mu

 

[ehrenfest]

I finished the quick calculation. Thanks.