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

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Homework Help Overview

The discussion revolves around the differences between the notations \( \dot{X}^{\mu} \) and \( X \) in the context of a problem from Zwiebach's text, specifically QC 6.4. Participants are exploring the implications of these notations within the framework of tensor and matrix representations in physics, particularly in relation to derivatives of a 4-vector.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants discuss the nature of \( \dot{X}^{\mu} \) as a derivative, questioning whether it represents a tau or sigma derivative. There is also confusion regarding the notation used in the function \( L(\dot{X}^{\mu}, X^{\mu '}) \) and its implications about the dependence on components of the vectors.

Discussion Status

The discussion is active, with participants raising questions about notation and its interpretation. Some have offered insights into the nature of the derivatives and the implications of the notation used in the equations. There is recognition of potential ambiguity in Zwiebach's notation, but no consensus has been reached on how to resolve these ambiguities.

Contextual Notes

Participants note that certain pages of the text are not available for review, which may limit their understanding of the context. The discussion also highlights the complexity of interpreting notation in the context of multiple components of a vector.

ehrenfest
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What do you know, pages 92 up to 115 are not part of the free preview. [tex]\dot{X}^{\mu}[/tex] is probably the tau or sigma derivative of the X^{\mu}, that's my guess.
 
It is the tau derivative. And the X^mu' is the sigma derivative.
 
Equations (6.49) and (6.50) on page 101 are straight forward partial derivatives of equation (6.46).
[itex]X[/itex] is a 4-vector, as are [itex]\dot{X}[/itex] and [itex]X'[/itex] which are the [itex]\tau[/itex] and [itex]\sigma[/itex] derivatives of [itex]X[/itex] respectively. There are 4 [itex]X^{\mu}[/itex] and each one is one of the 4 components of [itex]X[/itex]. [itex]\dot{X}^{\mu}[/itex] and [itex]X^{\mu\prime}[/itex] are defined in equation (6.40) on page 100.
 
What is confusing me is that [tex]L(\dot{X}^{\mu}, X^{\mu '})[/tex] in 6.46 implies that L is a function of only a single component of [tex]\dot{X}[/tex] and [tex]X^{'}[/tex], 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 [tex]L(\dot{X}, X^{'})[/tex] ?
 
ehrenfest said:
What is confusing me is that [tex]L(\dot{X}^{\mu}, X^{\mu '})[/tex] in 6.46 implies that L is a function of only a single component of [tex]\dot{X}[/tex] and [tex]X^{'}[/tex], 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 [tex]L(\dot{X}, X^{'})[/tex] ?

What do you mean by "simgle component"? The notation implies that L depends on all four [tex]\dot{X}^{\mu}[/tex] and all four [tex]X^{\mu '}[/tex].
 
ehrenfest said:
What is confusing me is that [tex]L(\dot{X}^{\mu}, X^{\mu '})[/tex] in 6.46 implies that L is a function of only a single component of [tex]\dot{X}[/tex] and [tex]X^{'}[/tex], 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 [tex]L(\dot{X}, X^{'})[/tex] ?
I agree with you. Zwiebach's notation here is a little 'funny'. He has for equation (6.46)

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

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 [itex]\mathcal{L}[/itex] is a function of only one of the [itex]X^{\mu}[/itex], 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

[tex]\mathcal{L}(\dot{X},X')[/tex]

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.
 
Last edited:
OK. But when he takes the partial derivative with respect to [tex]\dot{X}^{\mu}[/tex], this is really a partial with respect to the component not [tex]\dot{X}[/tex], right?
 
ehrenfest said:
OK. But when he takes the partial derivative with respect to [tex]\dot{X}^{\mu}[/tex], this is really a partial with respect to the component not [tex]\dot{X}[/tex], right?
Yes, in fact there are 4 such equations, one for each value of [itex]\mu[/itex]
 
  • #10
I finished the quick calculation. Thanks.
 

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