Lecture 27 Wednesday - November 9, 2005 Written or last updated: November 9, 2005 P441 – Analytical Mechanics - I Virial Theorem The attached are the notes from the lecture of Prof. Matt Shepherd. 1 The Virial Theorem in Central-Force Motion M. Shepherd November 9, 2005 The Virial Theorem If we have a system of particles such that the positions of the particles, rα, and the momenta of the particles, pα, are bounded, then define X S ≡ pαrα. (1) α That is, S is the sum over all particles of the product of the position and momenta. Note that the condition that rα and pα be bounded is true for many physical systems like a galaxy that is held together by gravitational attraction or a gas that has been confined to a container. Differentiating S with respect to time gives dS X = (pαr˙α +p ˙αrα) . (2) dt α dS We are interested in computing the average value of dt over a large interval of time, τ. So we could write dS 1 Z τ dS = dt. (3) dt τ 0 dt For periodic motion it is clear that if we pick τ to be a multiple of the period, then S(τ) = S(0) and the right hand side of the equation is zero. However, if the position and momentum of particles in the system is bounded, as we initially assumed, then S itself will be bounded. In that case all we need to do is make τ very long to force hS˙i → 0 – we don’t need to demonstrate periodicity. Therefore in the long time limit of such as system we have * + * + dS X X = pαr˙α + p˙αrα = 0. (4) dt α α The terms pαr˙α are equal to twice the kinetic energy of the particles, 2Tα. The termsp ˙αr can be written as Fαr, or the product of the force on the particle times the position of the particle. Making these substitutions we have the virial theorem: * + 1 X hT i = − Fαrα . (5) 2 α 1 The the theorem gets its name from the right hand side of the equation which Clausius called the virial1 of the system. In words, the theorem is a method of relating the average kinetic energy of a particle in a system to the force on the particle. In the case that the forces can be described by a potential, i.e. they are conservative forces, we can rewrite the viral theorem as * + 1 X hT i = rα∇Uα . (6) 2 α Power-law forces, F ∝ rn, can be derived from potentials of the type, U = krn+1. For such a potential r∇U = k(n + 1)rn+1 = (n + 1)U, (7) and the virial theorem can be written as n + 1 hT i = hUi. (8) 2 1 Therefore, for any “inverse-square force”, like gravity, we have n = −2, and hT i = − 2 hUi. Verification for F (r) = −kr Let’s verify that for a central-force of the type F (r) = −kr the value of hT i and hUi are as we would expect from the virial theorem. To do this we would like to compute the average of either T or U for one period of the motion, τ. Since we have the force, U is a natural choice. So let’s attempt to evaluate 1 Z τ hUi = U dt. (9) τ 0 kr2 Note that the corresponding potential for F = −kr is U = 2 . In order to compute the integral above it is convenient to transform the variable of integration from r to t. Starting with 1 1 l2 E = µr˙2 + + U(r), (10) 2 2 µr2 we can solve forr ˙ and get the expression dr dt = r . (11) 2 h l2 i µ E − U − 2µr2 We also need to change the limits of integration. If we integrate over a period of the motion, we can start at some time such that r = rmin. Since the subsequent motion from rmin to rmax is symmetric in time, integrating from 0 to τ is equivalent to integrating from rmin to rmax and doubling the result. We can therefore recast hUi as 2 Z rmax kr2 dr hUi = r (12) τ rmin 2 2 h kr2 l2 i µ E − 2 − 2r2µ k rµ Z rmax r3 dr = r . (13) h 4 2 i τ 2 rmin 2 kr l Er − 2 − 2µ 1The root of this word comes from Latin vires for “force.” 2 To compute the integral make a change of variables from r to x ≡ r2: 2 k rµ Z rmax x dx hUi = r . (14) 2τ 2 2 h 2 i rmin k 2 l − 2 x + Ex − 2µ This is now an integral of the form Z x dx √ . (15) ax2 + bx + c In the case that a < 0, which is in our case means that k > 0 or the potential is attractive, we can use the expression Z x dx 1 p b 2ax + b √ = ax2 + bx + c + √ sin−1 √ (16) ax2 + bx + c a 2a −a b2 − 4ac and evaluate rmax r s 2 r 2 k µ 2 k 4 2 l E 2 −1 E − kr hUi = − − r + Er − − sin q . (17) 2τ 2 k 2 2µ k k 2 kl2 E − µ rmin We can simplify this expression by noting that rmin and rmax are the turning points of the motion. This means that they are roots of the equation k l2 E − r2 − = 0. (18) 2 2µr2 Therefore, the first term in the bracketed expression above vanishes at the limits of integration, and we are left with rmax E rµ E − kr2 hUi = − sin−1 . (19) q 2 2τ k 2 kl E − µ rmin Now all we need to do is develop an expression for τ. Starting with Equation 11 we can integrate both sides over one period of the motion to get τ. Remember that because of symmetry this is equivalent to integrating r from rmin to rmax and doubling the answer. We have Z rmax dr τ = 2 r (20) h 2 i rmin 2 k 2 l µ E − 2 r − 2µr2 2 rµ Z rmax dx = q (21) 2 r2 k 2 l2 min − 2 x + Ex − 2µ rmax rµ E − kr2 = − sin−1 . (22) q 2 k 2 kl E − µ rmin This result is determined by changing variables from r2 → x and evaluating the integral in the same fashion as before. 3 Plugging this result for τ into Equation 19 provides a dramatic simplification and we are left with E hUi = . (23) 2 k 2 Since hUi + hT i = E, we can write, for the potential U = 2 r , hT i = hUi. (24) This is exactly what we expect by using the general expression for the virial theorem in central-force motion, Equation 8, with n = 1. Applications of the Virial Theorem The virial theorem in the context of a central-force can be extremely useful. It has applications, for example, in statistical mechanics where it can be used to describe molecules of an interacting gas. Another application arises in cosmology where one would like to analyze the motion of galaxies. In the latter case one is analyzing a complicated system of particles that interacting through the gravitational force. Very naively applying the virial theorem to some mass m a distance R away from the center of the galaxy we expect: 1 1 GMm mv2 = , (25) 2 2 R where here M is the mass of the galaxy inside of the radius R. This gives v2R M = . (26) G We can make astronomical observations of all these quantities. M can be estimated by the amount of luminous matter observed in the galaxy. Likewise the velocity profile as a function of radius can be measured through doppler shift experiments. When the data are analyzed, we find they disagree dramatically with the virial theorem. This is a strong piece of evidence for so-called dark matter! 4.