Quantum mechanics relation between p, λ, E, f in a wave

[Samuelriesterer]

Problem statement, equations, and work done:

In quantum mechanics, there is a relation between momentum and wavelength and between energy and frequency. These are:

$p=\hbar k = \frac{h}{\lambda}$
$E = hf = \hbar \omega$

A wave with an amplitude of 10cm is traveling on a string in the +x direction. The distance between wave crests (tops of oscillations) is 0.5 meters and the string oscillates up and down with a period of 0.10 seconds.

[1] Calculate the wavelength, write the equation for the wave and calculate the wave speed.

$T=0.1 s$
$f = 10 Hz$
$\lambda = 0.5 m$
$A = 10 cm$
$k = \frac{2\pi}{\lambda} = 12.566$
$v = \frac{\lambda}{T} = 5 m/s$
$\omega = 2\pi f = 62.8 rad/sec$
$y(x,t) = A sin(kx - \omega t) = 10 sin(12.566x – 62.8t)$

[2] If the tension in the string is 0.01N, determine the mass per unit length of the string.

$F_t = \mu v^2 → \mu = \frac{F_T}{v^2} = \frac{0.01 N}{(5 m/s)^2} = 0.0004 kg/m$

[3] Use the quantum relations above to substitute momentum and energy for k and ω in the wave equation,; make sure to put the constant in the right places.

$p = \hbar k = \frac{h}{\lambda}$
$E = h f = \hbar \omega$
$k = \frac{p}{\hbar} = \frac{h}{\lambda \hbar}$
$\omega = \frac{E}{\hbar} = \frac{h}{f \hbar}$
$Asin(kx-\omega t) = A sin (\frac{px}{\hbar} - \frac{Et}{\hbar}) = Asin (\frac{hx}{\lambda \hbar} - \frac{ht}{f \hbar})$

[4] Show that kx-ωt is invariant under the Lorentz transformation. That is, with the E and p substitutions, show that, for an observer moving in the direction of wave travel, transforming x,t,E and p produces the same expression as in the original frame. It is helpful to think of vectors and dot products here.

OK this is where I am stuck.

[5] Waves carry energy from place to place. To calculate the energy density and power of a wave, start with Power = F•v. Using the note posted to the Canvas Syllabus on the speed of a wave on a string, show that the power is equal to:

$P = F_y \cdot v_y = -F_T tan(\theta) \cdot \frac{\partial y}{\partial t}$

Next, use:
$tan(\theta) = \frac{\partial y}{\partial t}$

to get:
$P = -F_T \frac{\partial y}{\partial x} \frac{\partial y}{\partial t}$

Given y = Asin(kx-ωt), calculate the two partial derivatives and write the full expression for the power. Eliminate FT using v2 = FT/μ and substitute for one of the velocity terms the correct combination of wave parameters ω and k.

Finally for the power, calculate the average power during one cycle by calculating the average of cosine squared over one period.

$f_{average} = <f> = \frac{1}{b-a} \int_a^b f(x) dx$
$<cos^2 (\theta)> = \frac{1}{2\pi} \int 0^{2\pi} cos^2(\theta) d\theta$

The graph attached may be of help in determining that average value.

[6] Lastly, to get the energy density, use the power as the energy delivered per unit time, so E(Δt) = P Δt and the energy density will be the energy per unit length of string , so Energy density u = E(Δt)/Δx. Carry out these steps to get an expression for the energy density of a wave on a string.

 

[haruspex]

Use the quantum relations above to substitute momentum and energy for k and ω

I think the form wanted is this one:
$A sin (\frac{px}{\hbar} - \frac{Et}{\hbar})$

transforming x,t,E and p

What are the Lorentz transformations for those variables?

 

[Samuelriesterer]

I think I see something but we haven’t got that far in the Lorentz transformations. So would it be:

$x’ =\gamma (x-vt)$
$t’= \gamma (t - \frac{v}{c^2} x)$
$p=\gamma mv$
$E=\gamma mc^2$

 

[haruspex]

I think I see something but we haven’t got that far in the Lorentz transformations. So would it be:

$x’ =\gamma (x-vt)$
$t’= \gamma (t - \frac{v}{c^2} x)$
$p=\gamma mv$
$E=\gamma mc^2$

I guess so. (Never studied either quantum theory or relativity myself.)

 

[HalfLostAndSearching]

To calculate the partial derivatives, we can use the chain rule:

$\frac{\partial y}{\partial x} = Ak \cos(kx-\omega t)$
$\frac{\partial y}{\partial t} = -A\omega \cos(kx-\omega t)$

Substituting these into the expression for power, we get:

$P = -F_T Ak \cos(kx-\omega t) (-A\omega \cos(kx-\omega t))$
$P = F_T Ak^2 \omega \cos^2(kx-\omega t)$

Using the relationship v^2 = F_T/μ, we can substitute for F_T and for one of the velocities using ω = v k:

$P = \frac{\mu v^2}{k} Ak^2 v k \cos^2(kx-\omega t)$
$P = \mu v^3 A k \cos^2(kx-\omega t)$

To get the average power, we can use the relationship given in the problem:

$<cos^2(kx-\omega t)> = \frac{1}{2\pi} \int_0^{2\pi} \cos^2(kx-\omega t) d(kx-\omega t)$
$<cos^2(kx-\omega t)> = \frac{1}{2\pi} \int_0^{2\pi} \frac{1}{2} (1 + \cos(2(kx-\omega t))) d(kx-\omega t)$
$<cos^2(kx-\omega t)> = \frac{1}{2}$

Substituting this into the expression for average power, we get:

$<P> = \frac{\mu v^3 A k}{2}$

To get the energy density, we can use the formula given in the problem:

$u = \frac{E(\Delta t)}{\Delta x}$

Substituting in the expression for energy, we get:

$u = \frac{P \Delta t}{\Delta x}$
$u = \frac{\mu v^3 A k}{2} \frac{\Delta t}{\Delta x}$

Since Δx = λ and Δt = T, we can substitute in the expressions for wavelength and period:

##u = \frac{\mu v^3 A k}{2