Exploring the Virial Theorem: Understanding the Derivative of G Over a Period T

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Homework Help Overview

The discussion revolves around the application of the Virial theorem, specifically focusing on the scalar virial G defined as the dot product of momentum and position vectors. The original poster questions why the mean of the derivative of G over a period T equals zero.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the periodic nature of the function G and its time-derivative. Questions arise regarding the calculation of the mean of a function and the implications of integrating a derivative over a period.

Discussion Status

Some participants have provided insights into the mathematical formulation of the mean of the derivative and have noted that G at the endpoints of the period are equal, suggesting a potential reason for the mean being zero. However, the discussion remains open with various interpretations being explored.

Contextual Notes

There is an emphasis on understanding the periodicity of G and the implications of its derivative, with participants engaging in clarifying the mathematical relationships involved.

Karol
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Homework Statement


In the Virial theorem The scalar virial G is defined by the equation:
$$G=\vec{p}\cdot \vec{r}$$
Where ##\vec{p}## is the momentum vector and ##\vec{r}## the location vector.
When i take the mean of the derivative ##\bar{\dot{G}}## over a whole period T it equals 0. why?

Homework Equations


$$\vec{p}\cdot \vec{r}=(mv)\cdot \cos \theta \cdot r$$

The Attempt at a Solution


I understand this scalar product is zeroed during one period, but why?
 
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You said the function G is periodic. You want the mean of its time-derivative.

How do you calculate the mean of a function?

What is the integral of the derivative?

ehild
 
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Likes   Reactions: 1 person
I think i understand.
$$\bar{\dot{G}}=\frac{1}{T}\int_{0}^{T}\frac{dG}{dt}dt=\frac{1}{T}(G(T)-G(0))$$
Because the end point and the start point are identical G(T)=G(0)
 
Karol said:
I think i understand.
$$\bar{\dot{G}}=\frac{1}{T}\int_{0}^{T}\frac{dG}{dt}dt=\frac{1}{T}(G(T)-G(0))$$
Because the end point and the start point are identical G(T)=G(0)

Correct :smile:

ehild
 

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