In: Physics
II. Dimensions Involving Derivatives When we do dimensional analysis, we do something analogous to stoichiometry, but with multiplying instead of adding.
Part A) Consider the diffusion constant that appears in Fick’s first law: J = −D dn dx. In this expression, J represents a flow of particles (the number of particles per unit area per second), n represents a concentration of particles (the number of particles per unit volume), and x represents a distance. We can assume that they have the following dimensionalities: [J] = 1 L 2T , [n] = 1 L 3 , [x] = L.
1. What is the dimensionality of dx? Explain. (Hint: Remember from calculus that dx is just an infinitessimally small portion of x.)
2. Based on part (A), what is the general rule for the dimensionality of any differential? That is, given any physical quantity, Z, how does the dimensionality of dZ compare to that of Z?
3. Given any two physical quantities, Y and Z, what is the
dimensionality of the derivative dY dZ in terms of [Y ] and
[Z]?
Part B) Einstein discovered a relation that expresses how D depends on the parameters of the system: the size of the particle diffusing (R), the viscosity of the fluid it is diffusing in (η), and the thermal energy parameter (kBT). Assume that we can express D as a product of these three quantities to some power, like this: D = (kBT) a (η) b (R) c .
1. Rewrite the equation above in terms of M, L, and T, by substituting in your answer for the dimensionality of D from question 4 above, and the dimensionalities of the other constants, which are: [kBT] = ML2 T
2 , [η] = M LT , [R] = L 2. By “balancing” M, L, and T on each side of the expression above, determine three equations in terms of a, b, and c that will guarantee that D will have the correct dimensionality.
3. Solve these equations and write an expression for how D depends on the three parameters. (The correct equation has a factor of 1/6π that cannot be found from dimensional analysis.)
Let us assume Diffusion coefficient D = ( kB T)ab Rc .....................(1)
where is dimensionless constant, kBT is thermal energy, kB is Boltzman constant, T is temperature,
is coefficient of viscosity and R is size of diffusing particle .
Diffusion coefficient D is given as
J = -D ( dn/dx ) ..............(2)
where J is the rate of particle flux , i.e., number of particles per unit area per unit time , (dn/dx) is the spatial gradient of density, i.e number of particles per unit volume per unit distance.
Let us get the dimension of D from eqn.(2)
[D] = [J] [dx] / [ dn ] = [ L-2 T-1 ] [ L ] / [ L-3 ] = [ L2 T-1 ] ...................(3)
Using dimensional formula of diffusion coefficient D as given in eqn.(3), and applying respective dimensional formula for thermal energy, coefficient of viscosity and size in eqn.(1), we get the dependence power a, b, c as given below
dimension of thermal energy kBT = [ ML2 T-2 ]
dimension of viscosity = [ M L-1 T-1 ]
dimension of size R = [ L ]
Dimensional analysis of eqn.(1) :- [ L2 T-1 ] = [ ML2 T-2 ]a [ M L-1 T-1 ]b [L]c
equating power of [ M ] on both side, we get , a + b = 0 ....................(4)
equating power of [ L ] on both side, we get, 2a - b + c = 2 ...................(5)
equating power of [ T ] on both side, we get, 2a + b = 1 ...................(6)
By solving equations (4), (5) and (6), we get a =1 , b = -1 and c=-1
Hence from eqn.(1), we get D = ( kBT ) / ( R )
If we assume the constant = (1/6) , then D is given as