Can anyone give me a simple explanation for the virial theorem and an example. I understand it's something like <T>=(1/2)U where T=total kinetic energy and U=total potential energy. Now I saw a solution on physicsGRE.net for 0177 problem 3 where somebody says U=3/2T. I may understand 3 as being something like the number of degrees of freedom for translational motion but I don't get how the 2 comes down instead of up.
Thanks a lot
help virial theorem

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 Joined: Wed Mar 26, 2008 9:14 am
Re: help virial theorem
why do you need virial thm to answer this question.. simply equate the centrifugal with the gravitational force (non inertial frame)? in physicsgre.net there are many redundant NEC's

 Posts: 163
 Joined: Sat Jul 18, 2009 7:24 pm
Re: help virial theorem
I will explain briefly, not providing a full proof. Consider a collection of particles whose position vectors r_a and momenta p_a are both bounded (i.e. they remain finite for all values of the time). The virial theorem is in a more general case an expression about the average kinetic energy of the system of the forementioned particles. Specifically it is:
<T> = (1/2) * <Σ F_a*r_a> (1)
where
T = total kinetic energy
Σ = summation over a = the number of particles of the system
F_a = the force on the ath particle
r_a = the position of the ath particle
where the right hand side was named by Clausius "virial". Notice also the factor 1/2 on the right hand side (RHS). If the forces on each particle are derivable from some potential U_a, then (1) transforms to the following form:
<T> = (1/2) <Σ r_a * grad(U_a)> (2), since in this case F_a =  grad(U_a).
Of particular interest is the case of 2 particles that interact via some central force> F analogous to (r^n) , where n = some exponent.
In this case the potential energy of interaction takes the form U = k*r^(n+1), where k = proportionality constant. Thus, for this particular case it is (r = relative position of the 2 particles, U = potential energy of interaction), in spherical coordinates (see the RHS side of (2)):
r*grad(U) = r*dU/dr = k(n+1)*r^(n+1) = (n+1)*U (3), since U = k*r^(n+1).
Combining (3) and (2) (for the case of 2 particles always) we find that
<T> = (n+1)/2 * <U> (4).
If the mutual interaction is gravitational then n = 2 (see the force dependence on r above, how it was defined) and from (4) it comes out that
<T> = (1/2) * <U> .
This is a useful relation for calculations concerning the planetary motion.
In the alternative answer of GR0177 #3, the argumentation is correct but the virial theorem does not seem to be employed correctly. In general, think carefully before reading the solutions given on this site. I have met several errors in the quoted answers. Think before you "digest" the quoted answer.
Physics_auth
<T> = (1/2) * <Σ F_a*r_a> (1)
where
T = total kinetic energy
Σ = summation over a = the number of particles of the system
F_a = the force on the ath particle
r_a = the position of the ath particle
where the right hand side was named by Clausius "virial". Notice also the factor 1/2 on the right hand side (RHS). If the forces on each particle are derivable from some potential U_a, then (1) transforms to the following form:
<T> = (1/2) <Σ r_a * grad(U_a)> (2), since in this case F_a =  grad(U_a).
Of particular interest is the case of 2 particles that interact via some central force> F analogous to (r^n) , where n = some exponent.
In this case the potential energy of interaction takes the form U = k*r^(n+1), where k = proportionality constant. Thus, for this particular case it is (r = relative position of the 2 particles, U = potential energy of interaction), in spherical coordinates (see the RHS side of (2)):
r*grad(U) = r*dU/dr = k(n+1)*r^(n+1) = (n+1)*U (3), since U = k*r^(n+1).
Combining (3) and (2) (for the case of 2 particles always) we find that
<T> = (n+1)/2 * <U> (4).
If the mutual interaction is gravitational then n = 2 (see the force dependence on r above, how it was defined) and from (4) it comes out that
<T> = (1/2) * <U> .
This is a useful relation for calculations concerning the planetary motion.
In the alternative answer of GR0177 #3, the argumentation is correct but the virial theorem does not seem to be employed correctly. In general, think carefully before reading the solutions given on this site. I have met several errors in the quoted answers. Think before you "digest" the quoted answer.
Physics_auth

 Posts: 116
 Joined: Sat May 09, 2009 10:14 am
Re: help virial theorem
Thank you physics_auth. I was scared that all I know suddenly became wrong...