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

In summary, the equation (5.30) implies that the second term on the first term on the RHS is always nonzero.
  • #1
ehrenfest
2,020
1

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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  • #2
By the chain rule,

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

You need some [itex]\tau'[/itex]s on the right.
 
  • #3
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?
 
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  • #4
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]
 
  • #5
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?
 
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  • #6
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.
 
  • #7
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.
 

1. What is the significance of showing the derivative of p with respect to Tau being 0 for an arbitrary parameter?

The derivative of p with respect to Tau being 0 for an arbitrary parameter is significant because it means that the parameter Tau does not affect the value of p. This can be useful in analyzing the relationship between p and other variables or parameters.

2. How is the derivative of p with respect to Tau calculated?

The derivative of p with respect to Tau can be calculated using the rules of differentiation, specifically the chain rule. This involves taking the derivative of p with respect to all variables or parameters that it is dependent on, and multiplying them together.

3. Can the derivative of p with respect to Tau ever be non-zero for an arbitrary parameter?

No, the derivative of p with respect to Tau will always be 0 for an arbitrary parameter. This is because an arbitrary parameter is a constant and does not change as Tau varies, so its derivative will always be 0.

4. How does showing the derivative of p with respect to Tau being 0 relate to Quality Control (QC) processes?

In QC processes, it is important to understand the relationship between various parameters and their impact on the final product or outcome. Showing that the derivative of p with respect to Tau is 0 for an arbitrary parameter can help identify which parameters have no effect on p and can therefore be controlled or adjusted without affecting the final outcome.

5. Can the derivative of p with respect to Tau being 0 be used to determine the optimal value of Tau?

No, the fact that the derivative of p with respect to Tau is 0 for an arbitrary parameter does not provide information about the optimal value of Tau. It only indicates that Tau does not impact the value of p, but further analysis and experimentation is needed to determine the best value for Tau.

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