1Intro to math modelingDimensional analysis and scalingDimensional analysis is a tool based on the observations that:1. physical quantities have dimensions (usually, mass, length, and time, denoted by M,L, and Trespectively);2. physical laws must remain unaltered when the fundamental units for measuringdimensions are changed.This requires that the law of physics be dimensionally homogeneous, that is, they can be stated as(Buckingham pi theorem)f(1, T2,...) = 0,where f is some function and r1, T2, .. are dimensionless products and quotients of thevariables and constants in theproblem.Dimensional analysis alone does not give the exact form of an equation, but it can leadto a significant reduction of the number of variables and thereby provide some non-trivialinformation.Illustration:the projectile problem.Suppose that a projectile of mass m is thrownvertically into the air with an initial velocity V.What is the maximal elevation h that theprojectile will reach? In a first approximation (if Vis not to big,air friction is negligible,...), it isreasonable to assume that the projectile is only subject to the action of gravity, withacceleration g. Thus, the variables are m, V, h, g, and any dimensionless product must beoftheformT = m'Vbh'gd,where the exponent a, b, c, d are to be determined. Counting dimensions of the right handsidewehave[mVbh g*] = M(LT-1)L(LT-2)d = MLb++dT-(b+2d)From this we see thatis dimensionless if and only if(1)a=0,b+c+d=0..b+2d=0.The first equation gives a = 0, implying that π is independent of m. We are left with two(linearly independent) equations for three unknown, b, c, d. Elementary linear algebra tells
Intro to math modeling 1 Dimensional analysis and scaling Dimensional analysis is a tool based on the observations that: 1. physical quantities have dimensions (usually, mass, length, and time, denoted by M, L, and T respectively); 2. physical laws must remain unaltered when the fundamental units for measuring dimensions are changed. This requires that the law of physics be dimensionally homogeneous, that is, they can be stated as (Buckingham pi theorem) f(π1, π2,.) = 0, where f is some function and π1, π2, . are dimensionless products and quotients of the variables and constants in the problem. Dimensional analysis alone does not give the exact form of an equation, but it can lead to a significant reduction of the number of variables and thereby provide some non-trivial information. Illustration: the projectile problem. Suppose that a projectile of mass m is thrown vertically into the air with an initial velocity V. What is the maximal elevation h that the projectile will reach? In a first approximation (if V is not to big, air friction is negligible, .), it is reasonable to assume that the projectile is only subject to the action of gravity, with acceleration g. Thus, the variables are m, V, h, g, and any dimensionless product π must be of the form π = ma Vb hc gd, where the exponent a, b, c, d are to be determined. Counting dimensions of the right hand side we have [ma Vb hc gd] = Ma (LT−1 ) bLc (LT−2 ) d = Ma Lb+c+dT−(b+2d) . From this we see that π is dimensionless if and only if a = 0, b + c + d = 0, b + 2d = 0. (1) The first equation gives a = 0, implying that π is independent of m. We are left with two (linearly independent) equations for three unknown, b, c, d. Elementary linear algebra tells
2Dimensional analysis and scalingus that we can choose one of those arbitrarily.For convenience we take c=1 which from (1)yields b = -2, d = 1. It follows that(2)π= hg/ V2,and any physical law involving m, V, h, gmust beof the formf(hg/ V2) = 0,(3)for some function f.Notice that this relation does not involve the mass m. Suppose that f issuch that the equation f(x) = 0 has a unique root x = C (note that Cis dimensionless; it is anumerical constant).Then (3) can besolved as(4)hg/V=C → h=CV/g.The value of C cannot be determined by dimensional analysis (we will show below thatC= ), but (4) already gives some non-trivial information. For instance, it shows that twoprojectiles (possiblywith different masses)thrownvertically into theairwith velocities Viand V2 will reach altitudes hy and h2 with ratio given byh /h2 = Vr/ V2.(5)Thus under our assumptions an object thrown twotimesfaster goes fourtimeshigher.Notethat both the (unknown) constant C and the acceleration of gravity, g, have disappearedfrom this relation (which, by the way, is dimensionally homogeneous as it should be). Infact (5) might be used to test our assumptions (to realize that air friction cannot be neglectedin most situations: see below.).Exercise 1. In the same setting as before, how long the projectile will fly before hittingthe ground again? Show that the only dimensionless product is = gt/V, where t is theduration of the flight.Deducethat(6)t= CV/g,where Cis another numerical constant (C=1,as shown below).Consider nowthefollowingrefinementof theaboveproblemwhereairfriction is takinginto account. For simplicity, we shall assume that air resistance leads to a drag force, Fp,given byFp = -kv.Herevis the instantaneous velocity of the projectile, and k is a constant with dimension(recall that F= ma)[K] = MT-1
2 Dimensional analysis and scaling us that we can choose one of those arbitrarily. For convenience we take c = 1 which from (1) yields b = −2, d = 1. It follows that π = hg/V2, (2) and any physical law involving m, V, h, g must be of the form f(hg/V2) = 0, (3) for some function f . Notice that this relation does not involve the mass m. Suppose that f is such that the equation f(x) = 0 has a unique root x = C (note that C is dimensionless; it is a numerical constant). Then (3) can be solved as hg/V2 = C ⇒ h = CV2/g. (4) The value of C cannot be determined by dimensional analysis (we will show below that C = 1 2 ), but (4) already gives some non-trivial information. For instance, it shows that two projectiles (possibly with different masses) thrown vertically into the air with velocities V1 and V2 will reach altitudes h1 and h2 with ratio given by h1/h2 = V2 1 /V2 2 . (5) Thus under our assumptions an object thrown two times faster goes four times higher. Note that both the (unknown) constant C and the acceleration of gravity, g, have disappeared from this relation (which, by the way, is dimensionally homogeneous as it should be). In fact (5) might be used to test our assumptions (to realize that air friction cannot be neglected in most situations: see below.). Exercise 1. In the same setting as before, how long the projectile will fly before hitting the ground again? Show that the only dimensionless product is π = gt/V, where t is the duration of the flight. Deduce that t = CV¯ /g, (6) where C¯ is another numerical constant (C¯ = 1, as shown below). Consider now the following refinement of the above problem where air friction is taking into account. For simplicity, we shall assume that air resistance leads to a drag force, FD, given by FD = −kv. Here v is the instantaneous velocity of the projectile, and k is a constant with dimension (recall that F = ma) [k] = MT−1
3Intro to math modelingWhat is the maximal altitude h reached by the projectile in this case? Using m, V, h, g.together with the additional parameterk (vin Fp =-kv is the instantaneous velocity,henceit isnota parameterto include)wemusthaveT=mVh'gdke.Dimension countgives[maVbh g"ke] = M(LT-1)bLc(LT-2)dLe = Ma+eLb+c+dT-(b+2d+e)Itfollowsthatisdimensionless ifandonlyif(7)a+e=0,b+c+d=0,b+2d+e=0In contrast with (1),we have now three (linearly independent) equations for five unknowns,and we can choose, say, c and e arbitrarily. Since (c, e) = c(1, O) + e(0, 1), all dimensionlessproducts can be obtained from the two cases (c, e) = (1, 0) and (c, e) = (0, 1). These give(8)T1 = hg/ V2,2 = kV/(gm).The dimensionless product Ti is the same as the one we obtained by neglecting friction(see (2): Clearly, including more variables into a problem can only increase the number ofdimensionless products. Thus, any law involving m, V, h, g, R, must be of the form(9)f(hg/ V2, kV/(gm) = 0.for some function f. Solving this equation in h gives(10)h= Vg-n(kV/(gm),for some function n. Notice that this equation is consistent with (4) if limx-0 n(x) = C; infact, we will show later thatn(x) = 1/x- ln(1 + x)/x2,implying limx-0 n(x) = 1 /2.It is interesting to interpret /2 since it gives a hint about when it is appropriate to neglectfriction.T2 can be written as the ratio T2= V/Vr,where V is the initial velocity of theprojectile, and Vr =gm/k is proportional to the terminal velocity of a projectile releasedwith zero velocity at high altitude and falling down to earth (free fall problem; see the nextexercise). Neglecting air friction is appropriate if the instantaneous velocity of the projectile,V, satisfies v< Vr, which is true (why?) if V< Vr, ie. T2 ~ 0
Intro to math modeling 3 What is the maximal altitude h reached by the projectile in this case? Using m, V, h, g, together with the additional parameter k (v in FD = −kv is the instantaneous velocity, hence it is not a parameter to include) we must have π = ma Vb hc gdke . Dimension count gives [ma Vb hc gdke ] = Ma (LT−1 ) bLc (LT−2 ) dLe = Ma+e Lb+c+dT−(b+2d+e) . It follows that π is dimensionless if and only if a + e = 0, b + c + d = 0, b + 2d + e = 0. (7) In contrast with (1), we have now three (linearly independent) equations for five unknowns, and we can choose, say, c and e arbitrarily. Since (c,e) = c(1, 0) + e(0, 1), all dimensionless products can be obtained from the two cases (c,e) = (1, 0) and (c,e) = (0, 1). These give π1 = hg/V2, π2 = kV/(gm). (8) The dimensionless product π1 is the same as the one we obtained by neglecting friction (see (2)): Clearly, including more variables into a problem can only increase the number of dimensionless products. Thus, any law involving m, V, h, g, R, must be of the form f(hg/V2, kV/(gm)) = 0. (9) for some function f . Solving this equation in h gives h = V2g−1η(kV/(gm)), (10) for some function η. Notice that this equation is consistent with (4) if limx→0 η(x) = C; in fact, we will show later that η(x) = 1/x − ln(1 + x)/x2 , implying limx→0 η(x) = 1/2. It is interesting to interpret π2 since it gives a hint about when it is appropriate to neglect friction. π2 can be written as the ratio π2 = V/VT, where V is the initial velocity of the projectile, and VT = gm/k is proportional to the terminal velocity of a projectile released with zero velocity at high altitude and falling down to earth (free fall problem; see the next exercise). Neglecting air friction is appropriate if the instantaneous velocity of the projectile, v, satisfies v VT, which is true (why?) if V VT, i.e. π2 ≈ 0
4Dimensional analysis and scalingExercise 2.What is the terminal velocity,Vr,of a projectilereleased with zero velocityat high altitude and falling down to earth (free fall problem)? Vr is a function of the onlyrelevant parameters in this situation: g, k, and m. Show that 2 with Vr substituted for V isthe only dimensionless product based on g,k,m, and Vr and, hence, any law for the freefallproblem should be of the form f(kVr/(gm) for some function f. Deduce that Vr = Cgm/kfor somenumerical constant C (belowwewill getC=1).Exercise 3. Find an expression for the period of a perfect pendulum whose amplitude ofoscillation is o.Howdoes this formula change if there is friction?Putting a differential equation in dimensionless form (scaling).Many laws come in theform of differential equations.As for functional relations, it is useful to nondimensionalizedifferential equations since it usually reduces thenumber of parameters.Consider our first example of the projectile when air friction can be neglected. Themotionoftheprojectile isgoverned bydx些(0) = V.n(11)x(0) = 0,mat = -mg,dtHere x(t) is the distance of the projectile from the ground at time t, and (11) is F = ma forF = -mg (notice that m can be simplified out of (11), indicating right away that the mass ofthe projectile is irrelevant, as we alreadyconcluded).The solution of this equation is(12)x(t) = Vt- gt /2Themaximal value for x(t) is attained when dx/dt = 0 (why?),i.e.at a time t, satisfying(13)V-gt=0→t=V/g.Thus h = x(t) = V/(2g), consistent with (4), and the time of fight is given by (why?)t=2t, =2V/g, consistent with (6).Let us now put (11) into a dimensionless form. We use the following two step procedurevalid in other situations:STEP A.List all parameters and variables together with their dimensions. Here thedependent variable is x, with dimension [x] = L, the independent variable is t, withdimension [t] = T, and the parameters are V and g, with dimensions [V = LT-1 and[g] = LT-2.STEP B. For each variable, here x or t, form a combination of parameters, here v and g.withthe samedimensionasxand t,sayliandti,andintroducey=x/landT=t/t as new
4 Dimensional analysis and scaling Exercise 2. What is the terminal velocity, VT, of a projectile released with zero velocity at high altitude and falling down to earth (free fall problem)? VT is a function of the only relevant parameters in this situation: g, k, and m. Show that π2 with VT substituted for V is the only dimensionless product based on g, k, m, and VT and, hence, any law for the free fall problem should be of the form f(kVT/(gm)) for some function f . Deduce that VT = Cgm ˜ /k for some numerical constant C˜ (below we will get C˜ = 1). Exercise 3. Find an expression for the period of a perfect pendulum whose amplitude of oscillation is φ. How does this formula change if there is friction? Putting a differential equation in dimensionless form (scaling). Many laws come in the form of differential equations. As for functional relations, it is useful to nondimensionalize differential equations since it usually reduces the number of parameters. Consider our first example of the projectile when air friction can be neglected. The motion of the projectile is governed by m d2x dt2 = −mg, x(0) = 0, dx dt (0) = V. (11) Here x(t) is the distance of the projectile from the ground at time t, and (11) is F = ma for F = −mg (notice that m can be simplified out of (11), indicating right away that the mass of the projectile is irrelevant, as we already concluded). The solution of this equation is x(t) = Vt − gt2/2. (12) The maximal value for x(t) is attained when dx/dt = 0 (why?), i.e. at a time t? satisfying V − gt? = 0 ⇒ t? = V/g. (13) Thus h = x(t?) = V/(2g2), consistent with (4), and the time of flight is given by (why?) t = 2t? = 2V/g, consistent with (6). Let us now put (11) into a dimensionless form. We use the following two step procedure valid in other situations: STEP A. List all parameters and variables together with their dimensions. Here the dependent variable is x, with dimension [x] = L, the independent variable is t, with dimension [t] = T, and the parameters are V and g, with dimensions [V] = LT−1 and [g] = LT−2. STEP B. For each variable, here x or t, form a combination of parameters, here v and g, with the same dimension as x and t, say `1 and t1, and introduce y = x/`1 and τ = t/t1 as new
5Introtomathmodelingvariables. In terms of y and , the dynamical equation, here (1l), will be in dimensionlessform (that is, it will only involve some -sometimes none- of the dimensionless products oftheparameters)Here it is easy to see that the only choice for li and ti is(14)li= V/g,ti = V/g.liand ty are intrinsic length and time scales and by introducing the new variablesy= x/l1 = xg/V2,(15)T = t/ti = tg/ V,we decide to measurelength and time in units of l and ty. Substituting (15) into (11)requiresthechain rule.Weuseddg d(dr)dtdt) drVdr'anddy.dx_ g d(yV2 /g)dyg? d(yv /g)dydtdt2V2VddTd-2Thus (11) becomesdy%(0) =1.-1y(0) = 0,(16)d-2drThis system of equations is dimensionless (in fact, it is even parameterfree)Now, let's do the same exercise if air friction is not neglected. The differential equationis in this case (this is F= ma again, with F=-mg- Fp)dx些() = V..dx(17)x(0) = 0,kat -mg,mdt=dtThe parameters are now m, g, v and V. Clearly, li = V2/g and ti = V/g still are intrinsiclength and time scales for the problem.Thus, the change of variable in (15) remainsappropriate; itcan be checked (check it) that it turns (17) intodydy_(0) =1,(18)1.y(0) = 0,㎡2drdr2dwhere 2=kV/(mg) is theonlydimensionless parameterleft in the system.In fact, we shallshowbelow that (18)is appropriate to study (using regular perturbations techniques)themotion of a projectile in the limit as 0< 2< 1,that is, if the initial velocity is small enoughcompared to Vr so that air friction is a small effect.Exercise 4.Deduce from (17)that theterminal velocity in the feee fall problem, Vrsatisfies 0 =-kVr- mg.Hence Vr =mg/k, consistent with what we found before if C-1
Intro to math modeling 5 variables. In terms of y and τ , the dynamical equation, here (11), will be in dimensionless form (that is, it will only involve some –sometimes none– of the dimensionless products of the parameters). Here it is easy to see that the only choice for `1 and t1 is `1 = V2/g, t1 = V/g. (14) `1 and t1 are intrinsic length and time scales and by introducing the new variables y = x/`1 = xg/V2, τ = t/t1 = tg/V, (15) we decide to measure length and time in units of `1 and t1. Substituting (15) into (11) requires the chain rule. We use d dt = dτ dt d dτ = g V d dτ , and dx dt = g V d(yV2/g) dτ = v dy dτ , d2 y dt2 = g2 V2 d2(yV2/g) dτ 2 = g d2 y dτ 2 . Thus (11) becomes d2 y dτ 2 = −1, y(0) = 0, dy dτ (0) = 1. (16) This system of equations is dimensionless (in fact, it is even parameter free). Now, let’s do the same exercise if air friction is not neglected. The differential equation is in this case (this is F = ma again, with F = −mg − FD) m d2x dt2 = −k dx dt − mg, x(0) = 0, dx dt (0) = V. (17) The parameters are now m, g, v and V. Clearly, `1 = V2/g and t1 = V/g still are intrinsic length and time scales for the problem. Thus, the change of variable in (15) remains appropriate; it can be checked (check it!) that it turns (17) into d2 y dτ 2 = −π2 dy dτ − 1, y(0) = 0, dy dτ (0) = 1, (18) where π2 = kV/(mg) is the only dimensionless parameter left in the system. In fact, we shall show below that (18) is appropriate to study (using regular perturbations techniques) the motion of a projectile in the limit as 0 < π2 1, that is, if the initial velocity is small enough compared to VT so that air friction is a small effect. Exercise 4. Deduce from (17) that the terminal velocity in the feee fall problem, VT, satisfies 0 = −kVT − mg. Hence VT = mg/k, consistent with what we found before if C˜ = 1
6Dimensional analysisandscalingNotice that because of the presence of the additional parameterk,the intrinsic lengthand time scales l and ti are not unique. In fact, infinitely many different intrinsic lengthand time scales can be defined because there exists a dimensionless parameter, e, based onm, k, g, and V (for instance, e f(e) is a length for an arbitrary function f). Another pair ofintrinsic lengthand time scales isl2 = Vm/k,(19)T2 = m/ k,whichleadtothechangeofvariable(20)s= t/t2= tk/m.Z= x/l2 = xk/(Vm),It can be checked that in terms of z and s, (17) becomesdz_dz 1浆(0) =1.S(21)z(0) = 0,ds2= s-元2"dsThis equation is appropriate for regular perturbation techniques when T2 > 1, since then0 < 1/π2 1; this corresponds to a situation where gravity is small compared to friction(for instance, if the projectile is a ping-pong ball).Exercise 5.The equation for a pendulum of length l which initially forms an angle with the vertical axis and has zero angular velocity is (accounting for friction)nedekedede.-kedr(0) = 0. (22)mlae=ke%-mgsin(0), 0(0)=0,-Put this equation in dimensionless form.There are manyways of doing this.Find theequation appropriate in the low friction limit. Howdo you interpret the dimensionlessparameter entering this equation?Find also theequation appropriatein thelowamplitudelimit.Again, what is the dimensionless parameter entering this equation?
6 Dimensional analysis and scaling Notice that because of the presence of the additional parameter k, the intrinsic length and time scales `1 and t1 are not unique. In fact, infinitely many different intrinsic length and time scales can be defined because there exists a dimensionless parameter, ε, based on m, k, g, and V (for instance, `1 f(ε) is a length for an arbitrary function f). Another pair of intrinsic length and time scales is `2 = Vm/k, τ2 = m/k, (19) which lead to the change of variable z = x/`2 = xk/(Vm), s = t/t2 = tk/m. (20) It can be checked that in terms of z and s, (17) becomes d2z ds2 = −dz ds − 1 π2 , z(0) = 0, dz ds(0) = 1. (21) This equation is appropriate for regular perturbation techniques when π2 1, since then 0 < 1/π2 1; this corresponds to a situation where gravity is small compared to friction (for instance, if the projectile is a ping-pong ball). Exercise 5. The equation for a pendulum of length ` which initially forms an angle φ with the vertical axis and has zero angular velocity is (accounting for friction) m` d2θ dt2 = −k` dθ dt − mg sin(θ), θ(0) = φ, dθ dt (0) = 0. (22) Put this equation in dimensionless form. There are many ways of doing this. Find the equation appropriate in the low friction limit. How do you interpret the dimensionless parameter entering this equation? Find also the equation appropriate in the low amplitude limit. Again, what is the dimensionless parameter entering this equation?