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

AI Thread Summary
The discussion centers on the scalar virial G, defined as the dot product of momentum and location vectors. It is established that the mean of the derivative of G over a period T equals zero because G is periodic, leading to G(T) equaling G(0). The calculation of the mean of a function involves integrating its derivative over the period and dividing by T. This results in the conclusion that the change in G over one complete cycle is zero, confirming the periodic nature of the function. Understanding this relationship is crucial for applying the Virial theorem effectively.
Karol
Messages
1,380
Reaction score
22

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?
 
Physics news on Phys.org
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
 
  • Like
Likes 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
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top