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 of any point shows how the curve is inclined right there; it can be found by making a very large graph using 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 and 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 as the derivative of y = x². Instead of finding the gradient function = derivative graphically it can be proven 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 symbolized 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 us what the gradient of something else is going to be. This is common in physics; for example the velocity is the gradient of the displacement when both are graphed with the time on the horizontal 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 expected

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. 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

ò2xdx = x²

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

Summary:

·        Derivative: D(function to derive) or (d/dx)function to derive

·        Integral: ò(function to integrate)dx

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

5. Some integration rules

Rule A: òxⁿdx= x⁽ⁿ⁺¹⁾/(n+1)

· òxdx = òx¹dx = x⁽¹⁺¹⁾/(1+1) = ½x²

Rule B: Constants are not involved in integration

· ò2xdx = 2òx¹dx = 2*½x² = x² as expected

· ò5x⁴dx = 5òx⁴dx = 5*x⁽⁴⁺¹⁾/(4+1) = x⁵

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

· ò(3x² - 7x³)dx = ò3x²dx - ò7x³dx = 3òx²dx - 7òx³dx =

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)

· ò(u + at)dt = òudt + òatdt = ò(u*1)dt + aòtdt = uòt⁰dt + aòt¹dt

=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; stretching 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 = òkxdx = kòxdx = kòx¹dx = k*x⁽¹⁺¹⁾/(1+1) = kx²/2 = ½kx² = E_p,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

E_p = - Gm₁m₂/r

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

· ò(k/r²)dr = kòr^-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*x^0-1 = 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 ò2xdx = 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: ò(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 òx^-1dx = x^(-1+1)/(-1+1) which involves division by zero. It can be shown in mathematics that òx^-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 = ò(nRT/V)dV = nRTòV^-1dV = nRTlnV

when n, R and T are constants.

Trigonometric functions

Most useful are :

Dsin x = cos x Dcos x = - sin x

òcos x dx = sin x òsin 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 systems, 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 Ds/Dt. If Dt 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 Ds and Dt become infinitely 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 Dt 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 = S vDt. If the Dt becomes infinitely small and the slices infinitely many, this gives s = ò vdt.

Changing variables in integration

Here we will assume that you learned the usual rules of deriving and integrating functions in math, like that if y = x² then dy/dx = 2x or that òx²dx = 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!)

òa²da = 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 = Öb = b^½.

·        One would think that we simply get òbdb, 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 = Öb we get da = db/2Öb so

· òa²da = òb(db/2Öb) = ò(Öb/2)db = ½ò(Öb)db and since ò(Öx)dx = (2x^3/2)/3 we get

·        ½(2b^3/2)/3 = b^3/2/3 but since a = b^½ this = a³/3 as it should

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

Example 1: Simple derivative

Since v = ds/dt and s = ò vdt we can for a constant acceleration take v = u +at and get s = ò(u+at)dt so using 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 horizontal surface by a force F acting in a direction at an angle q up from the horizon one will find that:

• the normal force is now mg - Fsinq
• the resultant horizontal force is Fcosq - F_fr = Fcosq - m(mg-Fsinq)
• so F_tot = F(cosq + msinq)-mg
• which has a maximum when cosq + msinq has a maximum
• the derivative is sinq - mcosq
• let sinq - mcosq = 0 to find the maximum when
• m= tanq or q = arctan m

Example 5 : deriving E_k = 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 = FDt = change in momentum = Dp = 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 =òFds = ò(dp/dt)ds òd(mv)ds/dt which gives W = ò 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 = E_k = ò d(mv)v is, we can find out what ò(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 = òc²dm = c²òdm which if we integrate 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,

E_k = 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 in some nuclear reactions it does ...

12.4 Constants and formulas available in IB examinations

Page 1: Fundamental constants

Gravity acceleration g 9.81 ms^-2

Gravitational constant G 6.67 x 10^-11 Nm²kg^-2

Avogadro's constant N_A 6.02 x 10²³ mol^-1

Gas constant R 8.31 JK^-1 mol^-1

Boltzmann's constant k 1.38 x 10^-23 JK^-1

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

Coulomb constant k 8.99 x 10⁹ Nm²C^-2

Permittivity of free space e₀ 8.85 x 10^-12 C²N^-1m^-2

Permeability of free space m₀ 4p x 10^-7 TmA^-1

Speed of light in vacuum c 3.00 x 10⁸ ms^-1

Planck's constant h 6.63 x 10^-34 Js

Charge on electron e 1.60 x 10^{-19 C}

Electron rest mass m_e 9.11 x 10^{-31 kg = 0.000549u = 0.511 MeV/c2}

Proton rest mass m_p 1.673 x 10^{-27 kg = 1.007276 u = 938 MeV/c2}

Neutron rest mass m_n 1.675 x 10^{-27 kg = 1.008665 u = 940 MeV/c2}

Unified atomic mass unit u 1.661 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 = m = 10^-6, nano = n = 10^-9, pico = p = 10^-12, femto = f = 10^-15

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

1 astronomical unit(AU) = 1.50 x 10¹¹ m 1 radian (rad) = 180^o/p

1 kilowatt-hour (kWh) = 3.60 x 10⁶ J 1 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: A_H = Acosq