The IB Physics Compendium 2005: Mathematical Physics

The IB Physics Compendium 2005: Mathematical physics

12.* MATHEMATICAL PHYSICS

12.1.* Mathematical physics

This section is intended for those who want to explore the applications in physics of the calculus commonly learned in high school. Calculus = the mathematics of derivatives and integrals.

12.2.* The one-hour calculus course

1. Derivative = gradient function

The gradient shows how steeply inclined a graph is. For a straight line, the gradient has one specific value. For y = x its 1, for y = 2x it is 2, for y = ½x it is ½; for a horizontal line the gradient is 0.

Fig. calc1: Graphs of y = 0, y = ½x, y = x, y = 2x,

For a nonlinear curve, the gradient is changing. Below is the graph of y = x2 :

Fig. calc2 : y = x2 with tangents at x = 0, x =1 and x =2

The gradient in any point shows how the curve is inclined right there; it can be found by making a very large graph millimeter graph paper and drawing a tangent at the point in question; that is a line showing roughly how the curve is going very near the point. The gradient of this tangent is the gradient of the parabola at this point. For the y = x2 curve above, one can find that the gradient is 0 when x = 0, it is 2 when x = 1 and it is 4 when x = 2.

Fig. calc3: The points (0,0) , (1,2) , (2,4) making the graph of y = 2x

We notice that the gradients make the graph of a new function, y = 2x. This - another function which shows the gradient in the first function - is the "derivative".

 A function takes a value gives a new (or the same) value; y = x2 changes 1 to 1, 2 to 4, 3 to 9,....

 The derivative takes a function and gives another function: it changes y = x2 to y = 2x

2. Some derivative rules

We found that y = 2x is the derivative of y = x2. Instead of finding the gradient function = derivative graphically it can be proved that certain rules give the derivative of any function we have. (This will be left to later mathematics lessons).

When we take the derivative of a function it can be symbolised with "D" or "(d/dx)". So for example:

Dx2 = 2x or (d/dx)x2 = 2x

Rule A:Dxn = nx(n-1)

 Dx2 = 2x is an example with n = 2 where Dx2 = 2x(2-1) = 2x1 = 2x

 Dx3 = 3x(3-1) = 3x2

 Dx4 = 4x(4-1) = 4x3

Note that

 Dx = Dx1 = 1x(1-1) = x0 = 1

Rule B: Constants are not involved in the derivation

 D5x2 = 5Dx2 = 5*2x = 10x

 D7x4 = 7Dx4 = 7*4x3 = 28x3

Rule C: If we have several terms, we take the derivative of one at a time

 D(x3 + x2) = Dx3 + Dx2 = 3x2 + 2x

 D(2x5 - 7x) = 2Dx5 - 7Dx = 2*5x4 - 7*x0 = 10x4 - 7

3. Derivatives in physics

The derivative tells what the gradient of something else is. This is common in physics; for example the velocity is the gradient of the displacement when both are graphed with the time on the horisontal axis. From Mechanics we know that

s = ut +½at2 and v = u + at

Here a is a constant (it is uniformly accelerated motion) and u is a constant (we can reach higher final velocities by accelerating for a longer time, but that does not change what the initial velocity was). We can take the derivative of the s-function with t instead of x as the variable:

 D(ut + ½at2) = Dut + D(½at2) = uDt + ½aDt2 = u*t0 + ½a*2t = u + at as expexted

4. Integral = area function or antiderivative

Some physical quantities are the gradient of something else, like the velocity is the gradient of the displacement. Others are the area under a graph of the other, for example the displacement is the area under the velocity graph. Since from a velocity graph we get back the same quantity (displacement) as we used when taking the derivative we may accept that the integral is the "opposite" of the derivative.

Fig. calc 4: Graphs of y = 2x with areas from x = 0 to x = 0, 1 and 2 shadowed

If we make a function to show how much area we have under the graph we can use the triangles above.

 from x = 0 to x = 0 no area is yet found

 from x = 0 to x = 1 we have the area (1*2)/2 = 1

 from x = 0 to x = 2 we have the area (2*4)/2 = 4

If we now make a graph of the accumulated area under the y = 2x graph we get:

Fig. calc5: Graph of (0,0) , (1,1), (2,4) or y = x2

The function we get in this way is, as expected, y = x2. We can write this as

2xdx = x2

This looks a bit confusing, but just note that the integral symbol consists of two parts, a  and a dx which tells that x is the variable (may be useful if we have several letters involved). Whatever is in between the  and the dx is the function to integrate.

Summary:

 derivative: D(functiontoderive) or (d/dx)functiontoderive

 integral: (functiontointegrate)dx

Like the derivative, the integral takes one function and returns another function.

5. Some integration rules

Rule A:xndx= x(n+1)/(n+1)

xdx = x1dx = x(1+1)/(1+1) = ½x2

Rule B: Constants are not involved in integration

2xdx = 2x1dx = 2*½x2 = x2 as expected

5x4dx = 5x4dx = 5*x(4+1)/(4+1) = x5

Rule C: If several terms are integrated, we can take one at a time

(3x2 - 7x3)dx = 3x2dx - 7x3dx = 3x2dx - 7x3dx =

3*x(2+1)/(2+1) - 7*x(3+1)/(3+1) = x3 - 7x4/4

6. Integrals in physics

We expect the integral of the velocity as a function of time (v = u +at) to be the displacement as a function of time (s = ut + ½at2)

(u + at)dt = udt + atdt = (u*1)dt + atdt = ut0dt + at1dt

=u*t(0+1)/(0+1) + a*t(1+1)/(1+1) = ut + ½at2 as expected

We may also use integrals in other cases where an area under a graph is used. The work W = Fs when the force F is constant, but if it is not, then W = the area under the graph of F as a function of time. For example, the work done in stretching a spring against the force F = kx requires the work given by the area of the triangle under the F-graph; streching it from s = 0 to s = x gives W = kx*x/2 = ½kx2, which is then the elastic potential energy stored in the stretched spring. This could also have been found with an integral:

W = kxdx = kxdx = kx1dx = k*x(1+1)/(1+1) = kx2/2 = ½kx2 = Ep,elas

In Mechanics, we had for objects far out in space that the force of gravity is

F = Gm1m2/r2

where r is the variable, the distance from the center of the planet, and the others are constants. We also had that

Ep = - Gm1m2/r

We can sort of understand this if we think of Ep as the amount of work done when an object is falling or rising in the graviational field; so it should have something to do with the integral of the force function. Here Gm1m2 is a constant, call it k. Now

(k/r2)dr = kr-2dr = k*r(-2+1)/(-2+1) = -kr-1 = -k/x

which may explain the minus sign and the change from r2 to r. Note that we use r = infinity as the zero level since that is where the force function is zero; at r = 0 we would have F = infinite. (When we integrated y = 2x we could start from x = 0 as a "zero level" since y = 2x = 0 when x = 0).

7. Some other rules

Integration constants

If we take the derivative of a function y = k where k is a constant, then

Dk = D(k*1) = D(k*x0) = kDx0 = k*0*x0-1 = 0 independent of x

Therefore any constant will disappear in derivation, for example

Dx2 = 2x, D(x2 + 1) = 2x, D(x2 + 2) = 2x, D(x2 -127) = 2x etc.

On the other hand, if we integrate 2xdx = x2, the result could as well have been x2 +1 or x2 +2 or x2 - 127 or anything similar. We could then write x2 + C where C is an unknown (positive or negative) integration constant. In physics we can mostly set C = 0, for example as in:(u + at)dt = ... = ut + ½at2 but we should really have ... = s0 + ut + ½at2 where s0 is the displacement at the time t = 0. But this can usually be set to zero by defining s = 0 when t = 0; that is we start measuring the time when an object passes a chosen starting line.

The integral of y = 1/x

We would get x-1dx = x(-1+1)/(-1+1) which involves division by zero. It can be shown in mathematics that x-1dx = ln x (the natural logarithm of x). This is useful in thermodynamics where we find the work done in an isothermal process as the area under a graph in a diagram with pressure P as a function of volume V:

PV = nRT gives P = nRT/V so the work is found using W = (nRT/V)dV = nRTV-1dV = nRTlnV

when n, R and T are constants.

Trigonometric functions

Most useful are :

Dsin x = cos x Dcos x = - sin x

cos x dx = sin x sin x dx = - cos x

We can also differentiate (= take the derivative of) or integrate a function more than once. If we have a displacement function (with the time as a variable) and differentiate it once we get a velocity function. If we differentiate that again, we get an acceleration function. If we try trigonometric functions, then Dsinx = cosx, and then Dcosx = -sin x. We got the original function back, only with an extra minus sign! Now think of a mass oscillating on a spring. Then F = ma and F = kx or -ks (with directions) for the force on it. Then

ma = -ks gives a = -(k/m)s

that is, the acceleration function is the displacement function, give or take a negative constant. This may help explain why sine-functions are important for oscillating things, like the water molecules in an ocean wave. More about this in the Waves topic.

12.3* Further calculus in physics (and why E = mc2)

Using the "dx": Gradient (slope) = derivative

For constant velocity v = s / t and for a changing velocity, the average velocity is s/t. If t gets smaller and smaller, we reach the instantaneous velocity which is the gradient of the displacement-as-function-of-time curve or: the time derivative of displacement. When s and t become infinetely small we write v = ds/dt or velocity is the "rate of change" in the displacement. We will when necessary calculate with ds and dt in the same way as with any other variable. Note that "ds" then is not "d times s" but one variable.

Area under curve = integral

For constant velocity, s = vt which is the area of a rectangle in a velocity-as-function-of-time curve. If the velocity is not constant but the acceleration is then the v-curve is a straight line, and the area under it can be found as a trapeze (here a triangle on top of a rectangle). From this the usual formulas v = u +at, s = ut + ½at2 etc. were found. If the acceleration is not constant either, the area under the curve is found by splitting the t-axis into short intervals with the length t and using the v-value in the beginning or end of the interval to find an approximate area for it. Summing up all the thin vertical slices roughly gives the area under the curve as s =  vt. If the t becomes infinitely small and the slices infinitely many, this gives s =  vdt.

Changing variables in integration

Here we will assume that you learned the usual rules of deriving and integrating functions in maths, like that if y = x2 then dy/dx = 2x or that x2dx = x3/3. In physics it is sometimes convenient to change one variable in integration to another. Let us take (using something else than x as a variable!)

a2da = a3/3 (plus an integration constant, if you like)

We can find the same result in a different way (which here is unnecessary, but shows the steps to take).

 Let b = a2 and therefore a = b = b½.

 One would think that we simply get bdb, but this is ½b2 and since b = a2 it would give ½b2 = ½a4 which is not the correct result a3/3

To get the correct result we must find out what to replace da with like this: Start with b = a2 We take the derivative of both sides (with respect to something, we don't care what):

 db = 2ada where 2a is multiplied with the "inner" derivative of a; then

(e.g. if we derived for time then we would have db/dt = 2ada/dt, but multiply the whole with dt and we have the db = 2ada left)

 da = db/2a and since a = b we get da = db/2b so

a2da = b(db/2b) = (b/2)db = ½(b)db and since (x)dx = (2x3/2)/3 we get

 ½(2b3/2)/3 = b3/2/3 but since a = b½ this = a3/3 as it should

Here the whole substitution of b = a2 was unnecessary, but in some cases a substitution would give a function easier to integrate.

Example 1: Simple derivative

Since v = ds/dt and s =  vdt we can for a constant acceleration take v = u +at and get s = (u+at)dt which with conventional integrating rules gives the familiar s = ut + ½at2.

Example 2: Inner derivative

We have seen the formula P = Fv and used it for situations of constant velocity where F is e.g. the force pulling a train, and other forces (friction) keep the velocity constant. But if no other forces act (a rocket force F accelerates a spaceship) the power = the rate of work done which goes to increasing the kinetic energy.

E = ½mv2 => dE = d(½mv2) = ½(2mvdv) = mdv so dE = mvdv and dividing with dt:

dE/dt = mvdv/dt = mva = mav and with F = ma we get dE/dt = P = Fv.

Example 3 : Newton's II law in relativity

Instead of F = ma we will for linear acceleration have

F = dp/dt = (d/dt)mv = (d/dt)(m0v/(1 - v2/c2)½ , that is we need to derive the ratio of two functions, and need to use the derivation rule

D(f/g) = (f'g - g'f)/g2

where now f = m0v and g = (1 - v2/c2)½

so f' = m0*dv/dt = m0a remembering the inner derivative dv/dt = a
and g' = ½(1 - v2/c2)½-1 *( (- 2/c2)v*dv/dt) where two layers of inner derivatives are needed; we get
g' = ½(1 - v2/c2)-½(- 2va/c2) then giving the expression f'g - g'f as
m0a(1 - v2/c2)½ - (1 - v2/c2)-½(- va/c2)m0v since ½*2 = 1
the downstairs part is g2 = (1 - v2/c2)

We will now multiply both the upstairs and downstairs part with (1 - v2/c2)½ as a result of which the downstairs becomes (1 - v2/c2)3/2 and in the second term of the upstairs (1 - v2/c2)½(1 - v2/c2)-½ = 1 so that the whole upstairs now is

m0a(1 - v2/c2)1 - ( - va/c2)m0v which discarding the parentheses gives
m0a - m0av2/c2 + m0av2/c2 = m0a so the whole result is
m0a / (1 - v2/c2)3/2 or using γ = 1 / (1 - v2/c2)1/2 finally

F = γ3m0a

Example 4: The ideal angle for pulling a sleigh

When a an object is pulled on a horisontal surface by a force F acting in a direction at an angle  up from the horison one will find that:

the normal force is now mg - Fsin
the resultant horisontal force is Fcos - Ffr = Fcos - (mg-Fsin)
so Ftot = F(cos + sin)-mg
which has a maximum when cos + sin has a maximum
the derivative is sin - cos
let sin - cos = 0 to find the maximum when
= tan or  = arctan 

Example 5 : deriving Ek = mc2 - m0c2 in relativity

(Note that the word "derive" can mean both "take the derivative of" and "show why a formula is true"!). In relativity we must change Newton's II law F = ma since m is not constant. We can use the momentum instead : F = ma = m(v-u) = mv - mu and then impulse = Ft = change in momentum = p = mv - mu which for an infinitely short time dt becomes:

Fdt = dp and dividing with dt then: F = dp/dt (force is the rate of change in momentum).

We also know from relativity that mass increases as m = m0/(1- v2/c2)½ . If a resultant force F acts on an object accelerating it from rest we will assume that all the work done goes into increasing its kinetic energy. Before, the work W = Fs. Here we use

 W =Fds = (dp/dt)ds d(mv)ds/dt which gives W =  d(mv)v since ds/dt = v

Then

 d(mv) = mdv + vdm (the usual rule for deriving the product of two functions, note that both m and v are variables here)

 multiplying both sides with v gives d(mv)v = mvdv + v2dm

So instead of finding out what W = Ek =  d(mv)v is, we can find out what (mvdv + v2dm) is. Now we first use the formula for mass increase:

 m = m0/(1- v2/c2)½ square both sides so

 m2 = m02/(1- v2/c2) ; mult. with parenthesis and div. with m2 :

 m02 / m2 = (1- v2/c2) divide with m02

 1 / m2 = (1- v2/c2) / m02; this expression for 1/m2 will be useful later

Then we differentiate (take the derivative of) both sides, where m0 and c are constants:

 start with m02 / m2 = (1- v2/c2)

 on the left we get: - 2m02dm/m3 since the derivative of 1/x2 is -2/x3 (remember the inner dm!)

 on the right the 1 is lost and we get: -2vdv/c2 (remember the inner dv!)

 so - 2vdv/c2 = - 2m02dm/m3 ; cancel -2 and multiply with mc2 :

mvdv = m02c2dm/m2 = m02c2dm(1/m2)

From the equation before the differentiation we take our expression for (1/m2):

 mvdv = m02c2dm(1/m2) now gives

 mvdv = m02c2dm((1- v2/c2) / m02) which gives

mvdv = dmc2(1- v2/c2) = dm(c2 - v2)

Now we can return to the mvdv + v2dm we had in the beginning and get

 mvdv + v2dm = dm(c2 - v2) + v2dm = (c2 - v2 + v2)dm = c2dm

And this finally means that the integral simply is:

 W = c2dm = c2dm which if we intgrate from the rest mass m0 to the mass m after acceleration gives

 W = c2(m - m0) and since we assumed that the work goes to kinetic energy,

Ek = mc2 - m0c2

That is, terms of the form mass*c2 express some kind of energy that the particle has by virtue of having a mass. Even at rest there is some energy E0 = m0c2 which can be released (the commonly known E = mc2 really refers to this) if, in any reaction, the mass would decrease. Which it in some nuclear reactions does ...

12.4 Constants and formulas available in IB examinations

Page 1: Fundamental constants

Gravity acceleration g9.81 ms-2

Gravitational constantG6.67 x 10-11 Nm2kg-2

Avogadro's constantNA6.02 x 1023 mol-1

Gas constantR8.31 JK-1 mol-1

Boltzmann's constantk1.38 x 10-23 JK-1

Stefan-Boltzmann constant5.67 x 10-8 Wm-2K-4

Coulomb constantk8.99 x 109 Nm2C-2

Permittivity of free space08.85 x 10-12 C2N-1m-2

Permeability of free space04 x 10-7 TmA-1

Speed of light in vacuumc3.00 x 108 ms-1

Planck's constanth6.63 x 10-34 Js

Charge on electrone1.60 x 10-19 C

Electron rest massme9.11 x 10-31 kg = 0.000549u = 0.511 MeV/c2

Proton rest massmp1.673 x 10-27 kg = 1.007276 u = 938 MeV/c2

Neutron rest massmn1.675 x 10-27 kg = 1.008665 u = 940 MeV/c2

Unified atomic mass unitu1.661 x 10-27 kg = 931.5 MeV/c2

Page 2 : SI prefixes and unit conversions

tera = T = 1012, giga = G = 109, mega = M = 106, kilo = k =103, hecto = h = 102, deca = da = 101, deci = d = 10-1, centi = c = 10-2, milli = m = 10-3, micro =  = 10-6, nano = n = 10-9, pico = p = 10-12, femto = f = 10-15

1 light year (ly) = 9.46 x 1015 m1 parsec = 3.26 ly

1 astronomical unit(AU) = 1.50 x 1011 m1 radian (rad) = 180o/

1 kilowatt-hour (kWh) = 3.60 x 106 J1 atm = 1.01 x 105 Nm-2 =101 kPa = 760 mmHg

Page 3 : Electrical circuit symbols

Symbols are given for: cell, battery, lamp, ac supply, switch, ammeter, voltmeter, galvanometer, resistor, potentiometer, transformer, heating element

Page 4 : Measurement and mechanics

Horizontal,vertical components vector A: AH = AcosAV = Asin

Uncertainties:If y = a  b then y = a + b

If y = ab/c then y/y = a/a + b/b + c/c

v = s/ta = v/tg = F/m g =Gm/r2

Page 5 : Mechanics

v = u + ats = ((u+v)/2)t s = ut + ½at2 v2 = u2 + 2as

s : displacement, t : time, u : initial speed (?), v : final speed (?), a : acceleration

F = map =mvF = p/tEk =p2/2m

Impulse = Ft = pW = (Fss =) Fs cosEk = ½mv2Ep= mgh

P (= E/t or W/t) = work/time = FvF = (-) kx Eelas=½kx2

a = v2/r = 42r/T2 = FrsinFfr = kN Ffr < or = sN

F = Gm1m2/r2Ep = -Gm1m2/r V=-Gm/rT2/R3=const.

Page 6Thermal physics and Waves

Q = mcTQ = mLp = F/ApV = nRT

Q = U + W W = pVefficiency = (QH - QC)/ QH

 = = (TH - TC)/ TH(Carnot efficiency)QC/QH = TC/TH(Carnot cycle)

f = 1/Tv = ffbeat = f1 - f2

Moving source: f' = f ( 1 / (1  vs/v))Moving observer: f' = f (1  vo/v)

dsin = ns = D/dsin1/sin 2 = v1/v2n = c / vn1sin1 = n2sin2

Page 7 : Electricity and magnetism, electromagnetism

F = kq1q2/r2 k = 1/40E = F/qE = kq / r2E = - V/xE = V/d

V = kq/r I = q/tR = V / IP = VI = I2R = V2/RR=R1+R2

1/R = 1/R1 + 1/R2B = 0I/2rB = 0NI/L = 0nIF = ILBsin

F = qvBsinF/L = 0I1I2/2r = BLv=BAcos = - N/t

Irms = I0/2Vrms = V0/2Vp/Vs = Np/Ns

Page 8 : Atomic, nuclear and quantum physics

E = mc2 E = hfp = h/ hf = φ + Ek,maxhf = hf0 + eVs

N = N0e-tT½=ln 2/

Page 9-10 : SL options

Page 11 : Biomedical physics and The history and development of physics

 = 10 log ( I / I0 ) where I0 = 10-12 Wm-2I = I0e-x x½ = ln2/

Mechanical Advantage = load/effort;Velocity Ratio = distance moved by effort/ distance moved by load; Absorbed dose = Absorbed Energy / mass; Exposure = total charge / mass;

Dose equivalent = quality factor x Absorbed dose

1/TE = 1/TB + 1/TR1/ = RH (1/n2 - 1/m2)xp  h/2Et  h/2

Page 12 : Astrophysics and Relativity

L = AT4max = 2.90 x 10-3/T d(parsec) = 1/p(arc-second)b = L/4d2

v = Hd/ v/c = 1/( 1 - v2/c2 )½t = t0L = L0 / 

ux' =(ux - v)/(1 - uxv/c2) m = m0p =m0uE0= m0c2E = mc2

E2 = p2c2 + m02c4f/f = gh/c2RSch = 2GM / c2

Page 13 : Optics

n1sin1 = n2sin2n = 1/sinc1/f = 1/u + 1/vm = hi / ho = v / u

M = i / o = /b = 1.22/bdsin = n

12.5 Exercise problems

1.1.1. Can you think of a physical quantity that could not be used to describe some aspect of living beings?

1.2.1. Change a) 50 cg to kg b) 0.034 kg to ng c) 23400 kN to MN

1.3.1. A lifeboat drifts 14 km east one day and 12 km north the following. How far from the starting point is it then? In what direction (angle from the direction due north from the starting place). [18 km, 49o]

1.4.1. Manipulate the values on the axis until these (x,y)-data points give approximately a straight line. What function do you think describes them best?

(1.01, 0.98), (2.04, 0.26), (2.99, 0.11), (4.87, 0.05), (6.91, 0.02)

1.5.1. A plane drops from the altitude 5500 m to 2700 m. If the uncertainty of the altimeter is 40 m, how much altitude was lost (with uncertainty?) [280080m]

1.5.2. A cube with the side 857 cm has the mass 42030 kg. What is its density, with uncertainty? [684218 kgm-3 or approx. 700200 kgm-3 ]

2.2.1. If you move 45 meters forwards and 20 backwards, what is the a) distance b) displacement ?

2.2.2. If you move in a half-circle with the radius 40 m, what was the a) distance b) displacement ? [a) 130m, b) 80m]

2.2.3. Draw a qualitative graph of a) the velocity b) speed c) displacement d) the distance as a function of time when a billiard ball moves towards a wall, bounces straight back without losing energy and comes to rest in a "hole" some distance from the starting point in the opposite direction to where it first moved.

2.3.1. What was the (average) speed and velocity of the boat in problem 1.3.1, if the "days" are 24 hours ? [0.54 kmh-1, 0.38 kmh-1]

2.4.1. Draw two possible graphs each of a) acceleration b) velocity as a function of time for a ball thrown upwards and falling back down to the ground.

2.4.2. What is the acceleration of a) a car that speeds up from 50 kmh-1 to 80 kmh-1 in 4.5 s b) of one that slows down from 90 kmh-1 to 30 kmh-1 in 7.1 s? [a) 1.9 ms-2 b) -2.3 ms-2]

2.5.1. If an object is uniformly decelarated to rest in 2.34 s while covering a displacement of 5.05 m, what was its initial velocity? [4.3 ms-1]

2.5.2. Draw a graph of the velocity as a function of time for an object which accelerates from rest with 4.0 ms-2, for t = 1, 2, 3 and 4 seconds.