Chapter Biography of J. C. Maxwell Derivation of the Maxwell Speed Distribution Function


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1 Chapter Biography of J. C. Maxwell 10.2 Derivation of the Maxwell Speed Distribution Function
2 The distribution of molecular speeds was first worked out by Maxwell before the development of statistical methods. Later, Boltzmann derived the speed distribution function using statistical methods. We present here the method used by Maxwell rather than first developing the statistics of Boltzmann. Maxwell s method essentially employs the concepts we developed previously for kinetic theory. We start by assuming the number of molecules or particles with velocity components in the x direction, between v x and v x + dv x, that is dn vx, is some fraction of the total number of molecules in the system, N. This fraction should depend on v x and dv x in some way. That is fraction = dnv x /N = f(v x )dv x (1) Or dnv x = N f(v x )dv x (2) where f(v x ) is some function of v x that is now to be determined. Since, as was assumed previously in kinetic theory, all directions are equally probable or equivalent, we may write the same functional form for each of the Cartesian components of the velocity vector v, i.e.: f(v x ) = f(v y ) = f(v z ), (3) and hence, dnvy = N f(v y )dv y (4) and dnvz = N f(v z )dv z (5) Maxwell then assumed that the fraction dnv x /N is the same value for any subgroup of the N particles. This is valid if the volume of the subgroup is not so small that the stochastic variations from the volume element to another volume element would be significant. Now we ask: what is the number of particles that have velocity component between v x and v x + dv x while simultaneously having a velocity component between v y and v y + dv y? Let this be dnv x v y, which is a subgroup of dnv x. Using the subgroup assumption we made above, dnv x v y is a fraction of dnv x equal to: from equation (1) above. Hence, dnv x /N = f(v x )dv x,
3 But from (3) and (4), Then, using (7), equation (6) may be written as f(v x )dv x = f(v y )dv y (7) dnv x v y = dnv x f(v x )dv x = dnv x f(v y )dv y (8) Geometrically, dnv x v y is the number of molecules in velocity space that are located in a columnar volume formed by the intersection of two slabs that are perpendicular to the v x and v y axes respectively. See the diagram below. Each slab has thickness dv x or dv y. We may use the same arguments to define dnv x v y v z, which is a subgroup or subset of dnv x v y. Doing this, we obtain dnv x v y v z = N f(v x ) f(v y ) f(v z )dv x dv y dv z (10) In velocity space, this is a cube formed by a slab of thickness dv z that is perpendicular to the v z axis and intersecting the column we had previously. The volume of this cube is dv x dv y dv z, which is depicted in the diagram below. The density of points (number of particles) in this cube is Now substitute for dnv x v y v z from (10) and we get = dnv x v y v z / dv x dv y dv z (11) = N f(v x ) f(v y ) f(v z )dv x dv y dv z / dv x dv y dv z Or, = N f(v x ) f(v y ) f(v z ) (12)
4 Since the velocity distribution is isotropic, then v 2 = v x 2 + v y 2 + v z 2 is the radius of a sphere in velocity space, centered at the origin. Now consider moving along such a radius from some arbitrary value of v to v + dv. The density then varies as:
5 Here f (v i ) is df(v i )/dv and the index i is to be permutated among x, y and z, while the indices j and k are the other two coordinates. When this is done, (13) becomes: Now consider the special case where the changes in dv x, dv y and dv z are such that we move from one volume element to another one that lies in the same spherical shell as the first volume element. That is, v is constant. But the density,, is the same for any volume element at the same radial distance, v, from the origin in a shell of thickness dv. Hence, d = 0 and we have: Since the two volume elements are in the same shell in velocity space: v 2 = v x 2 + v y 2 + v z 2 (18) The total differential of (18) is v x dv x + v y dv y +v z dv z = 0 (19) Now, since dv is in the same shell in velocity space so that d = 0, then dv x, dv y and dv z are not independent. That is, they can not be assigned arbitrary values but must have values so that equation (19) is satisfied. Equation (19) is called the conditional equation. We can find an equation in which dv x, dv y and dv z are independent by using a method invented by Lagrange. This method is called: The Method of Undetermined Multipliers. (See Boas, Chapter 4.9. Also see Carter, Chapter 13.2). We proceed by multiplying (19) by a constant,, and then add this to equation (17). We then obtain:
6 We may constrain the value of once, so we choose it to have a value so that Then equation (20) becomes: But any two of the variables dv x, dv y or dv z can be considered independent, so assume that dv y and dv z are independent. In order for equation (22) to be valid, the coefficients of dv y and dv z must be equal to zero, that is, Now we can determine the form of the function f(v i ) in either equation (21), (23) or (24). From (21), or Integrating (26) we get:
7 where ln is the constant of integration. Hence, If we let 2 = /2, then ln f(v x ) =  ½( v x 2 ) + ln (27) f(v x ) = exp( v x 2 /2) (28) f(v x ) = exp( 2 v x 2 ) (29) Since f(v y ) and f(v z ) obey the same form of the differential equation as f(v x ), it follows that: and f(v y ) = exp( 2 v y 2 ) (30) f(v z ) = exp( 2 v z 2 ) (31) The form of the function f(v i ) is therefore determined, although we now need to determine the meaning of the constants and. We put this task off until later. Now insert the expression for f(v x ), f(v y ) and f(v z ) into equation (10) and get: dnv x v y v z = N 3 exp[ 2 (v x 2 + v y 2 + v z 2 )]dv x dv y dv z (32) Let dnv x v y v z = d 3 N, and since v 2 = v x 2 + v y 2 + v z 2, we may write (32) as: d 3 N = N 3 exp[ 2 (v 2 ]dv x dv y dv z (33) From (11), the number of points per unit volume of velocity space is = d 3 N / dv x dv y dv z or = N 3 exp[ 2 v 2 ] (34) This is the Maxwell velocity distribution function. Hence, the density function for v is in accord with our assumption that the distribution is isotropic. The density function as given by (34) is shown plotted versus v in the adjacent diagram. The density is a maximum, o, at the origin and then falls off exponentially with v 2. Even though is a maximum when v = 0, dn v = 0. This is because the volume element has radius v
8 = zero and therefore, the volume element has zero volume. Hence, dn v = dv = 0. That is, essentially no molecules have v = 0. Now we may calculate the number of molecules that have speeds between v and v + dv. Recall that the density of points in velocity space is uniform in any thin shell of radius v. The volume of such a shell is and the number of points in such a shell is Now substitute into (36) the expression for as given in (34): d = 4 v 2 dv (35) dn v = d = 4 v 2 dv (36) dn v = 4 N 3 v 2 exp[ 2 v 2 ] dv (37) The ratio dn v /dv is the Maxwell distribution of speeds. Unlike the Maxwell velocity density function, (37) is not the number of points per unit volume in vspace, but rather the number of per unit range of speed dv. The ratio dn v /dv is plotted as shown in the following figure: Note the difference between this function and the density function. The v 2 factor causes the function to be zero at the origin before rising to a maximum. The parameter v max is the most probable speed. The area dn v is the number of molecules with speeds between v and v+dv. The total area under the curve is N. The number of molecules that have a speed less than some value v o is the area under the curve to the left of v o, that is, up to the red line shown above. Although the density of points (molecules) in velocity space is a maximum at the origin, the spherical shell centered at radius v max, and of width dv, encloses the maximum number of molecules. That is, more molecules have speed v max than any other speed.
9 10.3 Evaluation of and Now N = dn v = ρdv = 4 N exp( 2 v 2 )dv (40) This integral cannot be evaluated in the usual way, but may found from tables (Carter, Appendix D) or through Mathematica. The result is N = 4 N
10 Now in Chapter 9, we found that v rms = (3kT/m) 1/2 and from equation (51), that = (1/v rms )(3/2) 1/2. Hence, Now substitute this value for into equation (43): = (m/2kt) 1/2 (52) dnv = N(4/ 1/2 )(m/2kt) 3/2 v 2 exp(mv 2 /2kT)dv (53) This is the Maxwell speed distribution function. Since Boltzmann derived the same relationship using a statistical approach, (56) is generally called the MaxwellBoltzmann speed distribution function Most Probable Speed The most probable speed, v m, corresponds to the radius of a shell in velocity space that contains the largest number of representative points. In other words, more molecules have speed v m that any other speed. An expression for v m may be found by taking the first derivative of the speed distribution function and setting this to zero. It is easier to start with (43) rather than (56), since does not depend on v. So, from (43) we get: We now take the derivative and let v be v m :
11 4Nπ 1/2 3 [2v m exp( 2 v m 2 ) + v m 2 ( 2 )2v m exp( 2 v m 2 )] = 0 Canceling and rearranging the terms leads to: exp( 2 v 2 m ) = v 2 m 2 exp( 2 v 2 m ) So 1 = v 2 m 2 or v m = 1/ (54) But from (52), = (m/2kt) 1/2, so (55) E. Discussion of the Probability Function. Dividing both sides of (53) by dv, we may write dn v /dv = NP(v), where P(v) is the coefficient of N on the left side of (53). P(v) is the probability of finding a particle or molecule with speed between v and v+dv. It must be then that and that, (56) (57), which is the area under the curve shown in the diagram below. Substituting for in (46) we get (58) Also (59) These parameters are shown in the figure below, where v P is the most probable speed, v m. The ordinate is improperly labeled to be in units of 103, implying that the numbers along the vertical
12 axis must be multiplied by this factor. In most physics journals, the numbers along the scale have already been multiplied by the factor on the axis label. So to be correct, the factor should be The area Pdv is the fraction of molecules that have speed between v and v + dv. When this fraction is multiplied by N, one gets the number of molecules in the system that have speeds between v and v + dv. One can also show the following: = 1 : : (60) Also = v rms (61) and v m = v rms (62) The next diagram shows the probability function versus speed for two different temperatures. Notice how the value of the most probable speed increases as the temperature increases and that the probability curve also tends to flatten out. The area under either curve must be 1.00.
13 F. The Density Function Revisited The density function as given by equation (34) may be written in a form that facilitates calculating the number of molecules that have a specific velocity component such as v x. The density function is where 3 = 3/2 3 from (42). But 3 = 1/v m 3, so = N 3 exp[ 2 v 2 ] (34), So (34) becomes 3 = 1/( 3/2 v m 3 ) (63). (64)
14 in Fig 10.1 below: Then But v 2 =v x 2 + v y 2 + v z 2. Substitute this into (66) for v 2 :
15 Geometrically, this is equivalent to moving the little cube in Fig along the v y  axis to form a parallelepiped and then move the parallelepiped along v z axis to make a slab or lamina of thickness v z. G. The Energy Distribution Function The translational kinetic energy of a molecule is w = 1/2mv 2, while the kinetic energy associated with a component of the velocity such as v x is w x = (1/2)mv x 2 (70) Note that the exponential factor in (53), viz., exp(mv 2 /2kT), may be written as exp(w/kt). That is, the exponent is the negative ratio of an energy, w, to the quantity kt, which is also a quantity of energy. That is, the argument of the exponential function is always dimensionless. It can be shown that the Maxwell distribution function is a special case of a more general distribution function derived by Boltzmann, using statistical methods. This general distribution function is not limited to kinetic energy, but energy in general. That is, it contains the factor e (E/kT ).
16 It is useful to have an expression for the number of molecules that have kinetic energies within a certain prescribed range, say between w and w + dw. Since, w = mv 2 /2, then dw = mv dv, or dv = dw/vm. But v = (2w/m) 1/2. Hence dv = dw/[m(2w/m) 1/2 ] = dw/[(2m 2 w/m) 1/2 ] = (2mw) 1/2 dw (71) Then in equation (53), replace v and dv with the above and we obtain the following expression for dn w : (72) A graph of dn w /dw is shown below. The maximum is at w = 1/2kT, that is, the most probable energy is 1/2kT, not 3/2kT. Why? The explanation is subtle
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