QC 5.2: Show Derivative of p w/ Respect to Tau' is 0 for Arbitrary Parameter

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Homework Help Overview

The discussion revolves around a problem from Zwiebach's QC 5.2, which involves the relativistic momentum \( p \) and its derivative with respect to a parameter \( \tau' \) that is a function of another parameter \( \tau \). The goal is to show that the derivative \( \frac{dp_{\mu}}{d\tau'} = 0 \) holds for an arbitrary parameter \( \tau'(\tau) \) and to explore the conditions under which \( \tau' \) is a valid parameter when \( \tau \) is a good one.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the chain rule in the context of derivatives, questioning how to relate \( \frac{dp_{\mu}}{d\tau} \) and \( \frac{dp_{\mu}}{d\tau'} \). There is uncertainty about the implications of setting these derivatives to zero and what restrictions this places on \( \frac{d\tau'}{d\tau} \). Some participants suggest visualizing the problem using a figure from the text to clarify the relationship between the parameters.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some have offered insights into the relationship between the parameters and the implications of their derivatives, while others are questioning the connections between the parts of the question. There is no explicit consensus yet, but productive lines of reasoning are being explored.

Contextual Notes

Participants note that the validity of \( \tau' \) as a parameter depends on the positivity of \( \frac{d\tau'}{d\tau} \) and its implications for the behavior of spacetime coordinates along the worldline. There is also mention of specific equations and sections in the text that are relevant to the discussion.

ehrenfest
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Homework Statement


Zwiebach QC 5.2
Tau is the parametrization of a worldline. p is the relativistic momentum
Show that [tex]\frac{ dp_{\mu}}{d \tau'}} = 0[/tex]implies that [tex]\frac{ dp_{\mu}}{d \tau'}} = 0[/tex] holds for an arbitrary paramter [tex]\tau'(\tau))[/tex]
What needs to be true about the derivative of tau' with respect to tau for tay' to be a good parameter when tau is a good one?

Homework Equations


The Attempt at a Solution


By the chain rule dp/dtau *dtau'/dtau = dp/dtau * dtau/dtau' but we only have tau' as a function of tau and I am not sure whether you can just flip the differentials in that derivative?
 
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By the chain rule,

[tex]\frac{dp_\mu}{d\tau}=?[/tex]

You need some [itex]\tau'[/itex]s on the right.
 
George Jones said:
By the chain rule,

[tex]\frac{dp_\mu}{d\tau}=?[/tex]

You need some [itex]\tau'[/itex]s on the right.

[tex]\frac{dp_\mu}{d\tau}= \frac{dp_\mu}{d\tau'}\frac{d\tau'}{d\tau}[/tex]

But how does that help? We only know that this equals zero. Either term on the right could equal 0. How does this put any restrictions on dtau'/dtau?
 
Last edited:
ehrenfest said:
How does this put any restrictions on dtau'/dtau?

Imagine Figure 5.2, but with both tau and tau' on it. Also, look at the second paragraph of section 5.2. Can

[tex]\frac{d\tau'}{d\tau} = 0?[/tex]
 
George Jones said:
Imagine Figure 5.2, but with both tau and tau' on it. Also, look at the second paragraph of section 5.2. Can

[tex]\frac{d\tau'}{d\tau} = 0?[/tex]

No, because then dx/dtau' (where x is a spacetime coordinate on the worldline) is not monotonically increasing. So it must be positive. I see.

So, the second question in this QC should really come before the first because you need the fact that dtau'/dtau is positive in order to prove that dx/dtau' must be zero, right?
 
Last edited:
The first part of the question asks you to show that any [itex]\tau'[/itex] would satisfy equation (5.30) and the second part of the question asks you to remember that only certain [itex]\tau'[/itex] are completely satisfactory. They are almost unrelated questions.
 
I do not see why they are unrelated. To show that 5.30 implies this, we have that

[tex](dp_{\mu}/dtau') (dtau'/dtau) = dp_{\mu}/dtau = 0[/tex]

It seems like the second term on the first term on the RHS is only identically zero when the second term on the RHS is always nonzero. This seems to answer both parts of the question but maybe there is something I am missing.
 

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