#ph301

Work and power

See "/tagged/work/" and "/tagged/power/"

Using the equations of motion under constant acceleration (SUVAT), we know that

v2 = v02 + 2a · (r - r0)

where v is the final velocity, v0 is the initial velocity, a is the acceleration, r is the final position and r0 is the initial position. Note that v2 represents the scalar (dot) product v · v, hence why v2 is a scalar quantity in the equation above.

This can be rearranged to make a a(r - r0) the subject

a · (r - r0) = 1/2v2 - 1/2v02

and multiplied by mass m on both sides

ma · (r - r0) = 1/2mv2 - 1/2mv02

Using Newton's second law of motion we know that, under constant acceleration, a force F = ma. In addition we know the kinetic energy K = 1/2mv2. Therefore the above equation becomes

F · (r - r0) = ΔK

Work

If we define the mechanical work done W as

W = F · (r - r0)

i.e., the work done by a force F to produce a change in displacement Δr is a result of the scalar (dot) product between the two. Then it can be inferred that

W =  ΔK

In addition, the first law of thermodynamics states that

W = Q - ΔU

where Q is the total energy input to the system and ΔU is the internal potential energy. Under constant acceleration, Newton's first law of motion implies that Q = 0. Therefore,

W = - ΔU

Power

If we define the power as the rate that work is done within a system,

P = dW/dt

where dt represents the infinitesimal change in time. If we consider an infinitesimal change in W

dW = F · dr.

Given that dv = dr/dt, we find that

dW = F · v  dt,

and so

Motion of a point charge in a magnetic field

A particle of charge q moving with velocity v experiences a force F as it travels through a magnetic field B which is expressible as

F = q ( v × B )

where v × B represents the cross (vector) product of v and B. Diagrammatically, this can be shown as

  and the force F can be seen to be acting perpendicular to both v and B unless v and B are parallel. This forms a centripetal force Fc and produces a resultant circular motion.

Therefore, F = Fc. Above, ω represents the angular velocity. We know that

Fc = mv²/r

where m is the mass of the object the force is exerting upon and r is the radius of the circle formed by motion due to Fc.

If we consider a special case (as above) where the velocity and magnetic field are at right angles to each other, the magnitude of F, |F| = F, can be expressed as

F = qvB sin 90° = qvB

and the magnitude of Fc, |Fc| = Fc, is

Fc = mv²/r

We can equate Fc = F

qvB = mv²/r

and rearrange for r

r = mv/qB

This is the orbital radius. The time taken to complete one revolution, the time period τ, is given by

τ = 2πr/v

Using our expression for r we can hence express τ as

Physics

Compton Scattering Cont'd from "The de Broglie hypothesis"

The de Broglie hypothesis Cont'd from "Energy of an electron"

With the breakthrough of quantum physics, de Broglie (pronounced, "de broy") worked to relate the quantum-like and wave-like properties of particles. He typified this by expressing the energy E of a photon in terms of its wavelength λ, which soon became known as the de Broglie wavelength.

E = hν = hc/λ

where h is Planck's constant, ν is the wave frequency and c is the speed of light. He next proposed the momentum p of a particle can be given by

p = h/λ

Furthermore, he suggested this would apply to other quanta, including electrons. If relativistic effects did not apply he would be completely correct, however we know a particle with mass has a frequency ν ≠ c/λ since its velocity v ≪ c. This was confirmed by Davisson and Germer who, by inspecting the diffraction pattern of electron waves reflected off a plate of nickel crystal, quantitatively proved that

 d·sin θ = n·λ

The Bohr atomic model Cont'd from "Deriving Wien’s displacement law from Planck’s radiation formula"

Bohr reconciled the findings from the Rutherford scattering experiments and Planck's interpretations of emission line spectra by proposing a new atomic model. He proposed that:

Negative charges (electrons) made stable, circular orbits around the nucleus under the influence of the electrostatic force.

The angular momentum of orbiting electrons is quantised.

He mathematically expressed postulate 1 by first considering the electrostatic force Fe between the positive nucleus and the electron,

Fe= e²/4πε0r²

where e is the fundamental charge, ε0 is the permitivitty of free space and r is the seperation distance (or orbital radius). He then considered the centripetal motion of the electron and its centripetal force Fc, wherein

 Fc = mev²/r

where me is the rest mass of the electron and v is the velocity of the electron. By equating Fe = Fc we get the expression

① F = mev²/r = e²/4πε0r²

He proposed that postulate 2 could be expressed as the angular momentum L of a particle and the quantised angular momentum of orbit, wherein

② L = mvr = nh/2π

where h is Planck's constant and n is the quantisation number (n = 1, 2, 3,...).

We can solve ① and ② simultaneously to find the quantised orbit radius and velocity of an electron.

To find the radius of orbit we rearrange ② for v and substitute into ①.

v = nh/2πmr

v² = e²/4πε0rme =  n²h²/4π²m²r²

e²/ε0 =  n²h²/π²mr

let r = rn

rn = ε0 · n²h²/π²me²

This implies that

rn ∝ n²

To find the velocity of the electron, rearrange ② for r and substitute into ①.

r = nh/2πmv

and so,

vn =  1/ε0 · e²/2nh

This implies that

vn ∝ 1/n

We can find the Bohr radius a0 (the radius of an atom) at the point where n = 1 (ground state).

a0 = ε0h²/π²me²

and so we can express rn as

Deriving Wien's displacement law from Planck's radiation formula Cont'd from "Planck's quantum world"

As mentioned in a previous post, Wien's displacement law can be derived by finding the maximum of Planck's radiation formula, which can be treated as any other function to find minima or maxima. We begin by finding dI/dλ and setting it equal to zero.

By resolving the constants into a single constant, the solution becomes much simpler. We can allow α = 2πhc² and β = hc/kT, where h is Planck's constant, c is the speed of light, k is the Boltzmann constant and T is the absolute temperature of the radiating body. The maximum then becomes

d/dλ(α/λ5[exp(β/λ) - 1]) =  α · d/dλ(λ-5[exp(βλ-1) - 1]-1) = 0

where λ is the wavelength of the emitted wave. In an astronomical setting we can assume that exp(βλ-1) ≫ 1, therefore the expression for spectral emittance becomes

I(λ) = α · λ-5 · exp(-βλ-1)

for which we can take logs.

ln[I(λ)] = ln(α) + ln(λ-5) - βλ-1 = ln(α) - 5 ln(λ) - βλ-1

and find d/dλ{ln[I(λ)]}, for which I apologise about the notation.

d/dλ{ln[I(λ)]} = - 5/λ + β/λ²

Set this d/dλ{ln[I(λ)]} = 0 and rearrange

0 = - 5/λm + β/λm²

5/λm = β/λm²

λm = β/5

If we substitute back in β, we get

λm/T = hc/5k = 2.9×10-3 m·K

Redefining SUVAT Cont'd from "Instantaneous velocity and acceleration" 

Some of the equations we're deriving may seem familiar to you but with a new 'twist'. We'll start by redefining some of the quantities from my post about instantaneous velocity and acceleration and making some important assumptions.

x2 = x, x1 = x0

t2 = t, t1 = 0

v2 = v, v1 = v0

a = constant, so a = Δv/Δt = (v - v0)/(t - 0)

and by rearranging a, we get

① v = v0 + at 

(we know this as the first SUVAT equation)

We can define average velocity vav  in two ways: in terms of v and v0 and in terms of Δx and Δt. Hence, we can equate these

vav = ½(v + v0 ) = Δx/Δt

Taking the ½(v + v0 ), we can substitute equation ① for v

½(v + v0 ) = ½(v0 + at + v0 ) = ½(2v0 + at) = v0 + ½at

and equate as before

v0 + ½at = Δx/Δt

v0 + ½at = (x - x0)/t

by rearranging, we can get an expression for x in terms of a, t, v0 and x0

② x = x0 + v0t + ½at2

(we know this as the second SUVAT equation with an new term of x0)

We'll rearrange equation ② 

x - x0 = v0t + ½at2

and our expression for a for t

t = (v - v0)/a

and substitute t into our rearranged expression of ② 

x - x0 = v0(v - v0)/a + ½a(v - v0)²/a²

x - x0 = 2v0(v - v0)/2a + (v - v0)²/2a

2a(x - x0 ) = 2v0(v - v0) + (v - v0)²

2a(x - x0 ) = 2v0v - 2v0² + v² - 2v0v + v0²

and so, by canceling 2v0v and one v0², we get 

③ 2a(x - x0 ) = v² - v0²

(we know this as the third SUVAT equation with an new term of x0)

An alternative way of reaching equation ① is be defining the velocity in terms of the integral of instantaneous acceleration with limits t, 0, v and v0.

{a = dv/dt } · dt

{a dt = dv } · ∫t=0t=t

∫0t a dt' = ∫v0vdv' 

[at ]0t = [v ]v

a(t - 0) = v - v0

① v = v0 + at

and by following the same technique for v, we can reach a new equation.

{v = dx/dt } · dt

{v dt = dx } · ∫t=0t=t

∫0t v dt' = ∫x0xdx' 

∫0t (v0 + at) dt = ∫0t v0 dt + ∫0t at dt

[v0t + ½at²]0t = [x ]x0x

v0t + ½at² = x - x0

so, by rearranging

④ x = v0t + ½at² + x0

Instantaneous velocity and acceleration

Velocity

Consider two points at time t with displacement x as follows:

t1, x1 t2, x2

We can define the total displacement Δx between the two as the difference between the x values

Δx = x2 - x1

and the total time Δt spent travelling from x1 to x2

Δt = t2 - t1

since we know total time and total displacement we can hence find the average velocity vav

vav = Δx/Δt

To find the instantaneous velocity v, we need to allow the difference between t values to go infinitesimally small so that t1 → t2 ⇒ t1 = t2 (this is only an implication to illustrate the idea of limits, not an exact equality!) and so Δt → 0. Since Δx is the total distance travelled in time Δt we don't have to consider the limit of this, this can be represented as

v = limΔt → 0  Δx/Δt = dx/dt

Acceleration

Defining average aav acceleration in the same way, we know that it is rate of change of velocity, and so

aav = (v2 - v1)/(t2 - t1) = Δv/Δt

Taking the limit of Δt → 0 for aav we can find the instantaneous acceleration a