Calculus! For beginners, with a ‘no theorems’ approach!

2.
$$\Big(\frac{1}{1+x}\Big)' = \Big((1+x)^{-1}\Big)' = -1(x+1)^{-2} = \frac {2}{-1(x+1)} = -\frac{2}{x+1}$$
... is correct up to and including the second identity. Then, for unknown reasons you turn the - 2 exponent into a coefficient!

Correct answer:

$$(x+1).-1(x+1)^{-2}=-(x+1)^{-1}=\frac{1}{1-x}$$
3.
$$-\sin (2x).4(\cos (2x)+3)^3 = -4\sin (2x)(\cos (2x)+3)^3$$

Correct apart from one minor mistake:

$$(\cos (2x)+3)'=(\cos(2x))'+0=-2\sin(2x)$$

The cos(2x) also needed chain ruling!

4. is correct.

5. is also correct.

6. is also correct and nicely reworked.

<hr>

Seems to me the chain gremlins have been dealt a lethal blow and apart from one or two still bleeding out, we won't hear from them again!

Remember also that for simple functions:

$$f(g)=\cos g, \sin g, e^g, \ln g, g^n$$

If (with a and b constants):

$$g(x)=ax\:\text{...OR...}\:g(x)=ax+b$$

Then:

$$\frac{dg}{dx}=a$$

So:

$$f'(x)=a\frac{df(g)}{dg}$$

e.g.:

$$f(x)=\sin(ax+b)$$
$$f'(x)=a\cos(ax+b)$$

These are the simplest but also most frequently encountered cases of Chain Rule application!



[Edited on 24-5-2016 by blogfast25]

Quote: Originally posted by aga

(blush) Still not comfortable with dg/dx etc [...]

Not entirely sure what you mean there. The whole Chain Rule business can be proved from the limit theorem.

For a function:

$$f\big(g(x)\big)$$
Then:
$$[f\big(g(x)\big)]'=\lim_{\Delta x \to 0}\frac{f\big(g(x+\Delta x)\big)-f\big(g(x)\big)}{\Delta x}$$
How this leads to:
$$[f\big(g(x)\big)]'=\frac{df(g)}{dg(x)}\times \frac{dg(x)}{dx}$$
... can be found in this formal proof:

http://kruel.co/math/chainrule.pdf

Enjoy!

Quote: Originally posted by aga

OK. You've cited Limit Theorem about 3 times now. Best come clean and unload. What's Limit Theorem all about ?

It would have been more correct for me to refer to the 'limit definition of derivatives' (it's not really a theorem ):

$$f'(x)=\frac{df}{dx}=\lim_{\Delta x \to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$$

All the rules of derivation (sum/difference, product, quotient, chain,...) can be derived from that simple principle.

Here are the actual computations of these limits (i.e. first derivatives) of a few simple functions:

https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/defderd...

(most of them require rather boring and lengthy algebra but aren't really that hard)

But in real life we simply use tabled derivatives of simple functions and the derivation rules for more complicated ones, instead of using the harder limit taking route.

[Edited on 25-5-2016 by blogfast25]

Second order DEs (continued) (II):

Quote: Originally posted by aga

Gulp !

Ct'ued from above:

Music notes!

Ever wondered why a plucked guitar string sounds the way it does? How precisely it vibrates? It’s a surprisingly cool calculus problem. It’s also a big derivation, so we’ll do it in bite-size chunks.

Imagine a flexible string with linear density ρ (kg/m) clamped between x=0 and x=L, with a tension T:



The string is deformed at time t=0 as shown by the red line, then released.

Mathematically the motion the string will assume is governed by the wave equation, partial differential equation (PDE):

$$\frac{\partial^2y}{\partial t^2}=\frac{T}{\rho}\frac{\partial^2y}{\partial x^2}$$

Where y is the displacement of a mass element of the string, as a function of x and t:
$$y(x,t)$$
Also we call:
$$\frac{T}{\rho}=c^2$$
(A full derivation of the PDE can be found here. It’s quite simple)

So:
$$\frac{\partial^2y}{\partial t^2}=c^2\frac{\partial^2y}{\partial x^2}$$
We also need some initial conditions:
$$y(x,0)=f(x)\:\text{and }\frac{\partial y(x,0)}{\partial t}=0$$
The left one describes the string’s position at t=0 (don’t worry about f(x) just yet) and the right one the string’s initial velocity (in the y-direction), that is zero.

And some boundary conditions:
$$y(0,t)=0\:\text{and }y(L,t)=0$$
These represent the fact that the string is clamped at 0 and L and thus y=0 at all times in these points.

Ok, so far so good (and so little!) For solving this type of equations, as for DEs in general, no exact recipe exists. Instead a mixture of known techniques, useful theorems and a good dollop of guesswork are used. Here we’ll try separation of variables for a starter: that means separating all the xs from all the ts.

To do so, we make an important assumption:
$$y(x,t)=\phi(x)h(t)$$
Where φ(x) and h(t) are separate functions, resp. in x and t. Our sought after function y(x,t) is then presumed to be the product of the two, φ(x)h(t).

Firstly, we insert the original boundary conditions into the proposed solution, so we get:
$$y(0,t)=\phi(0)h(t)=0\:\text{and }y(L,t)=\phi(L)h(t)=0$$
And assuming:
$$h(t) \neq 0$$
We get the boundary conditions for φ(x):
$$\phi(0)=0\:\text{and }\phi(L)=0$$
Next we plug the proposed solution y=(x,t)= φ(x)h(t) into the original PDE.
$$\frac{\partial^2}{\partial t^2}\big(\phi(x)h(t)\big)=c^2\frac{\partial^2}{\partial x^2}\big(\phi(x)h(t)\big)$$

This look a lot scarier than it is because of course φ(x) is independent of t and h(t) is independent of x, so both can be multiplied out of the resp. derivatives. That also means we can dispense with the partial derivatives and we obtain:
$$\phi(x)\frac{d^2h}{dt^2}=c^2h(t)\frac{d^2\phi}{dx^2}$$
Minimal reworking yields separation of variables:
$$\frac{1}{c^2h}\frac{d^2h}{dt^2}=\frac{1}{\phi}\frac{d^2\phi}{dx^2}$$
Nice, huh? Hmmm... more like a ‘nice mess’, right now. And still no idea if this is even going to lead to a comprehensive solution of the PDE.

Stay tuned!

[Edited on 9-6-2016 by blogfast25]

Quote: Originally posted by aga

I'm starting to understand what they mean when they say a formula has a certain 'beauty'. These hieroglyphs are incredible ! They look more like Art than Maths.

Music notes! (Ct'nued)

Quote: Originally posted by aga

Scary ? Terrifying more like ! There's gonna be chain rules everywhere, probably whips and gimp suits too.

Quote: Originally posted by aga

Erm, are you seriously suggesting that i already have enough knowledge (thanks to you) to be able to compute equations at these levels of complexity ?

I oppose the Culture of Low Expectations^(TM).

Seriously, that depends on ones level of dedication, of course. At the very least you should be able to follow the derivations presented here, without too many problems but remember that reading math isn't like reading some flimsy novella. More like reading Shakespeare 'in Ye Olde English': it takes concentration and some study.

There won't be any exercises on PDEs, just one more example of a simple PDE governing Diffusion. And maybe a chemical example, if I can find a suitable one...

Music notes! (Ct'nued)

Music notes! (Ct'nued)

The Amplitude Spectrum:

We now need to derive values for A_n in:
$$y(x,t)=\displaystyle\sum_{n=1}^{\infty}A_n\cos\Big(\frac{n\pi ct}{L}\Big)\sin\Big(\frac{n\pi x}{L}\Big)$$
We do this by using the initial condition (see first page of the full derivation):
$$y(x,0)=f(x)$$
Since as t=0, all the time dependent cosine terms drop out, which leaves us with:
$$y(x,0)=f(x)=\displaystyle\sum_{n=1}^{\infty}A_n\sin\Big(\frac{n\pi x}{L}\Big)$$
In material not covered in this thread it’s proved that:
$$A_n=\frac{2}{L}\int_0^{L}f(x)\sin\Big(\frac{n\pi x}{L}\Big)dx$$
Here’s f(x):



Obviously, f(x) is not a continuous function, as it’s gradient changes abruptly changes at L/a but we can construct two linear function that act resp. on 0 < x < L/a and L/a < x < L:
$$f_1(x)=\frac{ay_0}{L}x\:\text{ and }f_2(x)=\frac{ay_0}{(a-1)L}(L-x)$$
A_n then becomes:
$$A_n=\frac{2}{L}\Big[\int_0^{L/a}f_1(x)\sin\Big(\frac{n\pi x}{L}\Big)dx+\int_{L/a}^{L}f_2(x)\sin\Big(\frac{n\pi x}{L}\Big)dx\Big]$$

$$A_n=\frac{2ay_0}{L^2}\Big[\int_0^{L/a}x\sin\Big(\frac{n\pi x}{L}\Big)dx+\frac{1}{a-1}\int_{L/a}^{L}(L-x)\sin\Big(\frac{n\pi x}{L}\Big)dx\Big]$$
$$A_n=\frac{2ay_0}{L^2}\Big[\int_0^{L/a}x\sin\Big(\frac{n\pi x}{L}\Big)dx+\frac{L}{a-1}\int_{L/a}^{L}\sin\Big(\frac{n\pi x}{L}\Big)dx\Big]-\frac{1}{a-1}\int_{L/a}^{L}x\sin\Big(\frac{n\pi x}{L}\Big)dx\Big]$$
This isn’t a particularly difficult integral but it is tedious and I’ll only present the final result:
$$A_n=\frac{4ay_0}{n^2\pi^2}\sin\Big(\frac{n\pi }{a}\Big)$$
The A_n should be seen as amplitudes, associated with each frequency and they vary inversely with n²:
$$f_1,2f_1,3f_2,4f_1,...$$
$$A_1,\frac{A_1}{4},\frac{A_1}{9},\frac{A_1}{16},...$$
So the amplitudes diminish very quickly with increasing values of n.

We can also see that the A_n increase with y₀ but are independent of L.

The spectrum:
$$A_n\cos\Big(\frac{n\pi ct}{L}\Big)$$
... determines the timbre (as do other factors like instrument shape) of the note.

[Edited on 12-6-2016 by blogfast25]

Quote: Originally posted by blogfast25

This isn’t a particularly difficult integral but it is tedious and I’ll only present the final result: $$A_n=\frac{4ay_0}{n^2\pi^2}\sin\Big(\frac{n\pi }{a}\Big)$$

Quote: Originally posted by aga

Looks good to me : an integration problem that Can be solved and has an answer that can be used to check if i did it right. Deffo gonna attempt it. Might need more paper for this one ...

Music notes! (Ct'nued)

Standing waves in the string:

So far our vibrating string time-dependent equation is:
$$y(x,t)=\displaystyle\sum_{n=1}^{\infty}A_n\cos\Big(\frac{n\pi ct}{L}\Big)\sin\Big(\frac{n\pi x}{L}\Big)$$
Where the amplitude coefficients A_n have been determined above.

Now look at an individual term:
$$\cos\Big(\frac{n\pi ct}{L}\Big)\sin\Big(\frac{n\pi x}{L}\Big)$$
Using the well-known trigonometric identity:
$$\cos \alpha \sin \beta=\frac12 \Big(\sin(\alpha+\beta)+\sin(\alpha-\beta)\Big)$$
... y(x,t) can then be reworked to:
$$y(x,t)=\frac12\displaystyle\sum_{n=1}^{\infty} A_n\Big(\sin\frac{n\pi(x+ct)}{L}+\sin\frac{n\pi(x-ct)}{L}\Big)$$
In:
$$\sin\frac{n\pi(x+ct)}{L}+\sin\frac{n\pi(x-ct)}{L}$$
... the left hand term represents a wave travelling from 0 to L and the right hand term an identical wave travelling from L to 0. The superposition (sum) of both is a standing wave. In that interpretation the vibrating string is a superposition of a spectrum of standing waves with harmonic wavelengths and frequencies and diminishing amplitudes:

Standing waves (amplitudes not to scale) in a string:



(Source of pic)

<hr>

Next up: modelling the process of diffusion.

[Edited on 13-6-2016 by blogfast25]

Modelling diffusion

Diffusion is an important mass transport phenomenon and a part of many chemical, biological and physical processes. There are also some really interesting and colourful experiments that illustrate it (I’ll suggest one further down).



On the left hand side concentration of a species (e.g. a dye) is high. Diffusion causes concentration to level out over time. The rate of diffusion is proportional to the concentration gradient of the solute, acc. to Fick’s first law, here in one spatial dimension only:
$$J=-D\frac{\partial u}{\partial x}$$
Where J is the diffusion flux (in mol m^-2 s^-1, D the diffusion coefficient (aka the diffusivity) (in m² s^-1 and u the concentration of the species (in mol m^-3). The negative sign is needed because the partial derivative is negative (for a positive J).

From it, Fick’s second law, aka the diffusion equation, can be derived:
$$\frac{\partial u}{\partial t}=D\frac{\partial^2 u}{\partial x^2}\:\text{for: } 0\leq x \leq L$$
In shorthand:
$$u_t=Du_{xx}$$
We’ll apply this to a long, narrow tube, filled with a dye solution and closed off at both ends:



As initial condition we set:
$$u(x,0)=f(x)$$
Where f(x) is the initial distribution of dye at t=0.

Now for some boundary conditions. As the tube is closed off at both ends, no mass transfer can take place through these boundaries. With Fick’s first law this means:
$$0=D\frac{\partial u}{\partial x} \implies \frac{\partial u}{\partial x}=0\:\text{at } x=0, x=L$$
... at all times. In shorthand:
$$u_x(0,t)=0\:\text{and: }u_x(L,t)=0$$
Now for our Ansatz, we assume that:
$$u(x,t)=X(x)T(t)$$
Where X(x) depends only on x and T(t) only on t. Inserting into the diffusion equation we get:
$$X(x)T'(t)=DT(t)X''(x)$$
Divide both sides by X(x)T(t), rearrange slightly and dispense with the (x) and (t) notation and we get:
$$\frac{1}{D}\frac{T'}{T}=\frac{X''}{X}$$
Introducing the separation constant λ:
$$\frac{1}{D}\frac{T'}{T}=\frac{X''}{X}=-\lambda$$
Splitting then gives us two ordinary DEs:
$$X''+\lambda X=0\:\text{and: }T'+\lambda DT=0$$
<hr>

Next: solving the separated DEs.

[Edited on 15-6-2016 by blogfast25]

Modelling diffusion (Ct'nued)

Quote: Originally posted by Richard3050

Chemistry! For beginners, with a ‘no elements’ approach!

Modelling diffusion (Ct'nued)

To approximately evaluate the diffusion equation graphically, let’s try two different initial distributions of the dye in the tube.

1. Dye is initially concentrated in the centre of the tube:

We have a constant concentration u₀ over a narrow length a, centred about L/2, like this:



Now we calculate the coefficients A_n:

$$A_n=\frac{2}{L}\int_0^Lf(x)\cos\Big(\frac{n\pi x}{L}\Big)dx$$
$$A_n=\frac{2}{L}\int_{\frac{L-a}{2}}^{\frac{L+a}{2}}u_0\cos\Big(\frac{n\pi x}{L}\Big)dx$$
$$A_n=\frac{2u_0}{n\pi}\big[\sin\Big(\frac{n\pi x}{L}\Big)\Big]_{\frac{L-a}{2}}^{\frac{L+a}{2}}$$
$$A_n=\frac{2u_0}{n\pi}\Big[\sin\Big(\frac{n\pi}{2}+\frac{n\pi a}{2L}\Big)-\sin\Big(\frac{n\pi}{2}-\frac{n\pi a}{2L}\Big)\Big]$$
Using the well-known trigonometric identity:
$$\sin(\alpha + \beta)-\sin(\alpha - \beta)=2\cos\alpha \sin \beta$$
$$A_n=\frac{4u_0}{n\pi}\cos \frac{n\pi}{2} \sin \frac{n\pi a}{2L}$$
Now we apply the small angle approximation for sine functions:
$$2L \gg n\pi a \implies \sin\frac{n\pi a}{2L} \approx \frac{n\pi a}{2L}$$
$$A_n\approx \frac{2u_0a}{L}\cos \frac{n\pi}{2}$$
Which gives:
$$A_0=\frac{2u_0a}{L}$$
$$A_1=0$$
$$A_2=-\frac{2u_0a}{L}$$
With:
$$u(x,t)=A_0+\displaystyle \sum_{n=1}^{\infty}A_ne^{-\Big(\frac{n\pi}{L}\Big)^2Dt}\cos\Big(\frac{n\pi x}{L}\Big)$$
Applied to the first three terms only (n=0, n=1, n=2):
$$u(x,t) \approx \frac{2u_0a}{L}-\frac{2u_0a}{L}e^{-4\Big(\frac{\pi}{L}\Big)^2Dt}\cos\Big(\frac{2\pi x}{L}\Big)$$
$$u(x,t) \approx \frac{2u_0a}{L}\Big[1-e^{-4\Big(\frac{\pi}{L}\Big)^2Dt}\cos\Big(\frac{2\pi x}{L}\Big)\Big]$$
Before we do some graphical plotting, a word about time scales. To allow dimensionless plotting we’ll use dimensionless time t_r:
$$t_r=\frac{D}{L^2}t$$
$$\implies t=\frac{L^2}{D}t_r$$
The following resource:
http://webserver.dmt.upm.es/~isidoro/dat1/Mass%20diffusivity...

... lists some values for D for common chemicals in water. They are typically in the order of 10^-9 m² s^-1!

Assuming a tube length of 10 cm (0.1 m), then t= 1 x 10⁷ x t_r = 16 weeks x t_r! Diffusion in liquids is a decidedly slow process.

Here’s the result using the first three terms approximation. All parameters are relative values.



Vertical axis: relative concentration, horizontal axis: relative time t_r.

2. Dye is initially concentrated in the left hand side of the tube:

If a narrow region 0 < x < a has an initial concentration of u₀, then calculation of A_n yields:

$$A_n \approx \frac{2u_0a}{L}$$
We’ve used the following 4 term approximation for u(x,t):
$$u(x,t) \approx \frac{2u_0a}{L}\Big[1+e^{-\frac{\pi^2}{L^2}Dt}\cos\Big(\frac{\pi x}{L}\Big)+e^{-\frac{4\pi^2}{L^2}Dt}\cos\Big(\frac{2\pi x}{L}\Big)+e^{-\frac{9\pi^2}{L^2}Dt}\cos\Big(\frac{3\pi x}{L}\Big)\Big]$$



Vertical axis: relative concentration, horizontal axis: relative time t_r.

3. Suggested diffusion experiment:

Place a small amount of potassium permanganate crystals at the bottom of a test tube. Add about 0.5 ml of water and swirl gently to dissolve some of the crystals. Now carefully (so as not to disturb the permanganate layer) pipette some water into the test tube, filling it to about half way. Stopper and place in an undisturbed place, preferably of constant temperature (convection currents will promote advective diffusion!)

Be patient and watch the permanganate diffuse through the water.


[Edited on 18-6-2016 by blogfast25]

Particle in a 2D circular box with zero potential energy

1. Introduction:

We previously derived the wave functions of a particle in 1D box, here. The wave functions for 2D rectangles and 3D cuboids can be found easily by means of Separation of Variables of the Schrödinger equation.

The solutions lead to some interesting plots, like these:

ψ_2,2 for a particle in a rectangular box (with zero potential energy, length twice the width):



The probability density ψ²_2,2 of the same wave function:



But things tend to get a whole lot more interesting when they’re circular (or cylindrical or spherical – see hydrogenic atoms!).

So let’s try and determine the wave functions for a circular 2D box of radius R, with zero potential (V) inside the box and infinite potential outside the box. As with the case of the 1D box, this means the wave function is zero outside the box:

Particle in a 2D circular box with zero potential energy (Ct'ued)

2. The problem described in polar coordinates:

In Cartesian coordinates (x,y), a 2D, one particle quantum system is described by the Schrödinger equation (here time-independent):
$$- \frac{\hbar}{2m} \bigg( \frac{\partial^2}{\partial x^2} + \frac{ \partial^2}{\partial y^2} \bigg) \psi(x,y) = [E-V(x,y)] \psi(x,y)$$
In our case we have:
$$V(x,y)=0$$
$$\implies - \frac{\hbar}{2m} \bigg( \frac{\partial^2}{\partial x^2} + \frac{ \partial^2}{\partial y^2} \bigg) \psi(x,y) = E\psi(x,y)$$
But this problem, because of the circular geometry, is easier to solve in polar coordinates:



$$x=r\cos \theta\:\text{and }y=r\sin \theta\:\text{and }r^2=x^2+y^2$$
So we need to convert the Cartesian coordinates based equation to a polar coordinates based equation. I’ll also switch to u instead of ψ, because of easier notation:
$$\psi(x,y) \to u(r,\theta)$$
This transformation of coordinates is tedious and finicky but is also well known (it can be found on plenty web pages), so I can just present the polar coordinates Schrödinger equation, here:
$$-\frac{\hbar^2}{2m}\Big(\frac{\partial^2u}{\partial r^2}+\frac{1}{r}\frac{\partial u}{\partial r}+\frac{1}{r^2}\frac{\partial^2u}{\partial \theta^2}\Big)=Eu$$
Slight reworking:
$$\frac{\partial^2u}{\partial r^2}+\frac{1}{r}\frac{\partial u}{\partial r}+\frac{1}{r^2}\frac{\partial^2u}{\partial \theta^2}+\frac{2mE}{\hbar^2}u=0$$
For shortness of notation we set:
$$k^2=\frac{2mE}{\hbar^2}$$
$$\frac{\partial^2u}{\partial r^2}+\frac{1}{r}\frac{\partial u}{\partial r}+\frac{1}{r^2}\frac{\partial^2u}{\partial \theta^2}+k^2u=0$$
Due to:
$$u=0\:\text{for: }V=+\infty\:\text{,that is for: }r \geq R$$
... we have a boundary condition:
$$u(R,\theta)=0$$
<hr>
Next: separating the variables r and theta.


[Edited on 24-6-2016 by blogfast25]

Quote: Originally posted by aga

Polar co-ordinates. Now we're talking business ...

You haven't lived if you haven't used polar coordinates!

Grab a piece of chalk, the blackboard is yours!

In 'normal' Cartesian co-ordinates, every point is defined as having an X,Y value relative to the Zero point where the X and Y axes cross.

Polar co-ordinates can define exactly the same spatial position using Distance (d) from a zero-point on the x-axis, and an Angle relative to that x-axis.



If you think about it, 'where the thing is' in Cartesian (normal X,Y) co-ordinates generates an Angle from the zero-point anyway, so you could use trigonometry to work out what the Polar co-ordinates would be and vice versa.

For some problems, Polar co-ordinates simplify the maths enormously.

For example, in the one and only case where i used them, i defined a shape on a computer screen in Polar co-ordinates, each point desribed by a Distance from zero and an Angle.

Simply by adding 1 to the Angle of every point, the entire shape rotated.

Changing the Zero point alters the position of the entire shape.

Adding/subtracting 1 from the Distance of each point alters the size of the entire share, retaining the overall structure.

In a Cartesian system this would be Much more difficult (take more time to solve) for each point in the plotted image.

Yes, i'm Pedagogically disabled.

You can take the chalk back now, Please.

[Edited on 25-6-2016 by aga]

Particle in a 2D circular box with zero potential energy (Ct'ued)

Quote: Originally posted by blogfast25

$$\frac{Y''}{Y}=-m^2$$ The value of the separation constant m<sup>2</sup> is yet to be determined.

Ever the Master of the Cliff-Hanger

As an aside, when polar co-ordinates are used in n-dimensions, is there any hard-and-fast rule dictating which axis the zero point lies on ?

I.e. in 3-D space it could be x,y,z or time ... or anything.

[Edited on 25-6-2016 by aga]

Quote: Originally posted by aga

Quote: Originally posted by blogfast25   $$\frac{Y''}{Y}=-m^2$$ The value of the separation constant m<sup>2</sup> is yet to be determined. Ever the Master of the Cliff-Hanger As an aside, when polar co-ordinates are used in n-dimensions, is there any hard-and-fast rule dictating which axis the zero point lies on ? I.e. in 3-D space it could be x,y,z or time ... or anything. [Edited on 25-6-2016 by aga]

The choice or origin and axis is usually yours and made out of mathematical convenience (there's no absolute origin for example).

In Cartesian coordinates we can describe a system of n dimensions, e.g. x₁,x₂, x₃, x₄, etc. Of course you can't represent that graphically.

For spherical coordinates there's an unwritten convention:



But really it's up to the user.

Quote: Originally posted by blogfast25

Quote: Originally posted by aga   Polar co-ordinates. Now we're talking business ... You haven't lived if you haven't used polar coordinates!

Quote: Originally posted by aga

I'll swap for my secret formula for actually making Aluminium Sulphate xHydrate crystals.

Particle in a 2D circular box with zero potential energy (Ct'ued)

Quote: Originally posted by aga

Edit: Hold on hoss. So we're in Polar co-ordinates now and we got a Derivative, then we got a derivative of a derivative. In plain english, what's the first derivative mean ? The rate of change of the angular rotation ?

Particle in a 2D circular box with zero potential energy (Ct'ued)

5. Solving the radial equation:

So far we have:
$$r^2R''+rR'+(k^2 r^2-m^2)R=0$$
With boundary condition:
$$u(R,\theta)=R(R)Y(\theta)=0 \implies R(R)=0$$
We make a simple substitution:
$$\rho=kr$$
Minimal reworking then yields:
$$\frac{d^2R(\rho)}{d\rho^2}+\frac{1}{\rho}\frac{dR(\rho)}{d\rho}+\Big(1-\frac{m^2}{\rho^2}\Big)R(\rho)=0$$
This is a Bessel Differential Equation, a type of DE that also pops up in the solution of the hydrogen SE, the Fourier (heat) equation and other important PDEs.

It has a solution for every m, called a Bessel function: J_m:
$$m \to J_m(\rho)=J_m(kr)$$
Where:
$$R(\rho)=J_m(\rho)=\displaystyle \sum_{n=0}^{+\infty}\frac{(-1)^n}{n!\Gamma(n+m+1)}\Big(\frac{\rho}{2}\Big)^{2n+m}$$
The J_m functions are sophisticated infinite term polynomials and they look like this:

J₀ compared a cosine function, to show the resemblance:


J₃ compared a sine function, to show the resemblance:

Later we’ll make use of these resemblances to define some quasi-solutions, because the actual Bessel functions really don’t lend themselves easily to graphical 2D plotting.

Radial wave function quantisation:

The particle in a 2D circular box is a quantum system and its wave functions and energy levels are quantised. Quantisation is derived from the boundary condition:
$$J_m(kR)=0$$
Basically this means that for the roots (zero-points) z_m,n, where n designates the n^th zero of J_m:
$$k_{m,n}R=z_{m,n}$$
The z values can be found here:



(Source).
$$k_{m,n}=\frac{z_{m,n}}{R}$$
Much higher up we stated:
$$k^2=\frac{2mE}{\hbar^2}$$
We can now use that expression to derive the energy spectrum of the system:
$$E_{m,n}=\frac{\hbar^2}{2mR^2}z_{m,n}^2$$
And the ground state energy:
$$E_{0,1}=\frac{\hbar^2}{2mR^2}(2.4048)^2$$

<hr>
Next we'll replace the Bessel functions with appropriate sines/cosines and the plotting can begin!

[Edited on 29-6-2016 by blogfast25]

Particle in a 2D circular box with zero potential energy (Ct'ued)

Particle in a 2D circular box with zero potential energy (Ct'ued)

m=0, n=1



m=0, n=3



m=0, n=5



m=1, n=1



m=1, n=2



To come: m=2, m=3

[Edited on 30-6-2016 by blogfast25]

Particle in a 2D circular box with zero potential energy (Ct'ued)

m=2, n=1



m=2, n=2



m=3, n=1



m=3, n=2

Quote: Originally posted by wg48

Blogfast25: After titillating me with your color rendered 3D graphs, I am waiting for your orbital derivations. I have even fired up a cobbled together xp pc to run my old copy of mathcad so I might give them a try.

What specific orbital derivations are you looking for? I would certainly be interested in assisting with that!

Recently I tried 3D plotting of the H 2p_z orbital but something went wrong. I suspect the error was due to the conversion from polar to Cartesian coordinates. Unfortunately neither matlab nor Wollram's plotting function allow direct plotting of polar functions in 3D.

For now I leave you with the contour plot of the actual wave function (not squared), which also shows the zero-lines (nodal lines), for the system above (m=3, n=2). All straight lines and circles are nodes where the wave function changes sign:



And here's a fascinating tidbit. The energy quantisation of the particle in a circular well with infinite walls is essentially chaotic. And this explains why a circular flexible membrane, like a drum for musical purposes, has no fundamental tone or well defined harmonics. It's basically noise.


[Edited on 27-7-2016 by blogfast25]

Quote: Originally posted by blogfast25

Quote: Originally posted by wg48   Blogfast25: After titillating me with your color rendered 3D graphs, I am waiting for your orbital derivations. I have even fired up a cobbled together xp pc to run my old copy of mathcad so I might give them a try. What specific orbital derivations are you looking for? I would certainly be interested in assisting with that! Recently I tried 3D plotting of the H 2pz orbital but something went wrong. I suspect the error was due to the conversion from polar to Cartesian coordinates. Unfortunately neither matlab nor Wollram's plotting function allow direct plotting of polar functions in 3D. For now I leave you with the contour plot of the actual wave function (not squared), which also shows the zero-lines (nodal lines), for the system above (m=3, n=2). All straight lines and circles are nodes where the wave function changes sign: And here's a fascinating tidbit. The energy quantisation of the particle in a circular well with infinite walls is essentially chaotic. And this explains why a circular flexible membrane, like a drum for musical purposes, has no fundamental tone or well defined harmonics. It's basically noise. [Edited on 27-7-2016 by blogfast25]

Quote: Originally posted by wg48

I thought drums do have a distinct modes but I suspect you mean something different than that. I will reread your particle in box to see if I can understand what you mean or even better play with it in mathcad.

Look here:

http://www.sciencemadness.org/talk/viewthread.php?tid=65532&...

$$k_{m,n}R=z_{m,n}$$

The roots of the Bessel function (z_m,n) are the quantum numbers of the system.

The PDE for a circular, elastic membrane is, bar the coefficients, the SAME as for the particle in a circular box. In that derivation the z_m,n are basically the frequencies of oscillation: the spectrum is inharmonic. Drum beats are noise.

Now look at the vibrating modes of the membrane:

https://en.wikipedia.org/wiki/Vibrations_of_a_circular_membr...

Ring a bell?

Let me know what I can do to help with orbital plotting: it would be a first here at SM!
<hr>

Some excellent sources of orbital representations:

http://www.st-andrews.ac.uk/physics/quvis/simulations_chem/c...

And:

http://winter.group.shef.ac.uk/orbitron/



[Edited on 27-7-2016 by blogfast25]

Quote: Originally posted by wg48

Ok I think I understand what you mean now. Personally I would not use the term noise ( or chaotic) to describe the spectrum of a drum in this technical context. In practice the modes of a drum are excited preferentially by where you hit the drum and by the resilience of the drum stick.