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

Pages:  1  ..  3    5 aga - 24-5-2016 at 00:18

3.
$$f=[\cos (2x)+3]^4$$
$$g=\cos (2x)+3, f=g^4$$
$$(\cos (2x)+3)'.\Big((\cos (2x)+3)^4\Big)'$$
$$-\sin (2x).4(\cos (2x)+3)^3 = -4\sin (2x)(\cos (2x)+3)^3$$

aga - 24-5-2016 at 00:50

4.
$$f=\sqrt [5]{1-2x}$$
$$g=1-2x, f=\sqrt [5] g, f=g^{\frac15}$$
$$(1-2x)'.\Big((1-2x)^{\frac15}\Big)'$$
$$=-2. \frac 15 (1-2x) ^{-\frac 45}$$
$$=\frac {-2}{5} \Big( \frac {1}{\sqrt [5]{1-2x}}\Big)^4$$

aga - 24-5-2016 at 00:57

5.
$$f=\sin\Big(\ln x+\frac{1}{x^2}\Big)$$
$$g=\ln x+\frac{1}{x^2} = \ln x +x^{-2}$$
$$\sin\Big(\ln x+\frac{1}{x^2}\Big)'.(\ln x +x^{-2})'$$
$$=\cos\Big(\ln x+\frac{1}{x^2}\Big). (\frac 1x -2x^{-3})$$

aga - 24-5-2016 at 01:19

6.
$$f=\sqrt [3] {2x^4-x^2-3}$$
$$g=2x^4-x^2-3$$
$$(\sqrt [3] {2x^4-x^2-3})'.(2x^4-x^2-3)'$$
$$= (({2x^4-x^2-3})^\frac 13)'.(8x^3-2x)$$
$$= \frac 13 (2x^4-x^2-3)^{-\frac 23}.(8x^3-2x)$$
$$= \frac {8x^3-2x}{3} \Big( \frac {1}{\sqrt [3] {(2x^4-x^2-3)}} \Big)^2$$
$$= \frac {8x^3-2x}{3} \frac {1}{(2x^4-x^2-3)^{\frac 23}}$$
$$= \frac {8x^3-2x}{3(2x^4-x^2-3)^{\frac 23}}$$

[Edited on 24-5-2016 by aga]

blogfast25 - 24-5-2016 at 07:00

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!

$$(x+1).-1(x+1)^{-2}=-(x+1)^{-1}=\frac{1}{1-x}$$
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]

aga - 24-5-2016 at 08:07

(blush)
Yes, it does seem a lot clearer now.

Basically, if something happens to x (e.g. x+1, x^2, 3x+5 etc) then something happens to the result of that, then you need to use the chain rule.

Still not comfortable with dg/dx etc but it feels like it's coming together.

Thanks again for taking the time to check these answers !

blogfast25 - 24-5-2016 at 12:05

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

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

Enjoy!

aga - 24-5-2016 at 12:38

if the same brain can 'hear' the sound of that brain imploding, i just heard it.
aga - 24-5-2016 at 13:13

OK. You've cited Limit Theorem about 3 times now.

What's Limit Theorem all about ?

blogfast25 - 24-5-2016 at 14:52

 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]

blogfast25 - 30-5-2016 at 07:28

Ooopsie. Double post deleted.

[Edited on 30-5-2016 by blogfast25]

### Second order DEs (continued) (II):

blogfast25 - 30-5-2016 at 07:29

In the previous post we saw that for a second order DE of the type:
$$ay''+by'+cy=0\:\text{, where: }y=f(x)$$
... a Chracteristic Equation (CE) can be defined as:
$$a\lambda^2+b\lambda +cy=0$$
The roots of this quadratic equation will be used to determine a full solution of the DE.

The two roots of the CE are determined from:
$$\lambda_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
We can now distinguish several cases:

1. Two Real and distinct roots:

If:
$$b^2-4ac>0$$
Then both roots are Real and distinct and the general solution of the DE is given by:
$$y=c_1e^{\lambda_1 x}+c_2e^{\lambda_2 x}$$
Where c1 and c2 are two integration constants, to be determined from boundary conditions.

An example was shown in the previous post.

2. Two Complex and distinct roots:

If:
$$b^2-4ac<0$$
Then note that the square root of that expression becomes Complex (imaginary):
$$\sqrt{b^2-4ac}=\sqrt{(4ac-b^2) \times -1}=\sqrt{4ac-b^2}\sqrt{-1}$$
$$i=\sqrt{-1}$$
$$\beta =\frac{-b}{2a},\gamma = \frac{\sqrt{4ac-b^2}}{2a}$$
Then the Complex and distinct roots can be written as:
$$\lambda_{1,2} = \beta \pm \gamma i$$
We can then insert these roots into the general solution as above but that yields a complex function.

But using Euler's Formula and with some reworking (not shown here) we then obtain a Real solution:

$$y=c_1e^{\beta x}\cos \gamma x+c_2e^{\beta x}\sin \gamma x$$
Where c1 and c2 are two integration constants, to be determined from boundary conditions.
<hr>
Two more cases to follow:
$$b^2-4ac=0\:\text{and }b=0$$

[Edited on 30-5-2016 by blogfast25]

aga - 30-5-2016 at 09:17

Gulp !
blogfast25 - 30-5-2016 at 10:30

 Quote: Originally posted by aga Gulp !

Just try it for:

$$y''-4y'+5y=0\:\text{where: }y=f(x)$$

The CE has two complex roots. No boundary conditions have been given with the DE, so you don't need to determine values for the integration constants.

[Edited on 30-5-2016 by blogfast25]

### Ct'ued from above:

blogfast25 - 7-6-2016 at 06:40

For:
$$ay''+by'+cy=0$$
3. Repeat Real roots if:
$$b^2-4ac=0$$
$$\lambda_{1,2}=\frac{-b}{2a}=\lambda$$
$$\implies y=c_1e^{\lambda x}+c_2xe^{\lambda x}$$
Example/exercise:
$$y''+2y'+y=0$$
$$b^2-4ac=2^2-4.1.1=0$$
$$\implies \lambda=-\frac{b}{2a}=-\frac22=-1$$
Solution:
$$y=c_1e^{-x}+c_2xe^{-x}$$

Special case: b = 0
$$b=0 \implies \beta =0, \gamma = \sqrt{\frac{c}{a}}$$
a. Real roots:
$$\implies y=c_1e^{\lambda x}+c_2xe^{\lambda x}$$
b. Complex roots:
$$\implies y=c_1\cos \gamma x+c_2\sin \gamma x$$
<hr>
Coming up: a partial differential equation that makes music!

### Music notes!

blogfast25 - 9-6-2016 at 06:53

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]

aga - 9-6-2016 at 09:07

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.

[Edited on 9-6-2016 by aga]

blogfast25 - 9-6-2016 at 10:01

 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.

Yes. Trust me, it gets more beautiful as we get on with this problem.

The 'hieroglyphs' could be avoided by a notation we used above, for example:

$$\frac{\partial^2u}{\partial x^2}=u_{xx}$$

The wave equation then becomes:

$$y_{tt}=c^2y_{xx}$$
But that doesn't have the same allure. It is much more practical though...

[Edited on 9-6-2016 by blogfast25]

aga - 9-6-2016 at 14:12

Aspersions cast i not, young 25 of blogfastingness.

The drawn glyphs admire i, and that muchly.

Of great beauty they are.

### Music notes! (Ct'nued)

blogfast25 - 10-6-2016 at 08:28

So far we obtained:
$$\frac{1}{c^2h}\frac{d^2h}{dt^2}=\frac{1}{\phi}\frac{d^2\phi}{dx^2}$$
This is really a bit of a strange animal. h(t) and φ(x) are obviously not the same (they’re not even dependent on the same variable!). This can only make sense mathematically if both derivative equations (each side of the equation) equals the same number:
$$\frac{1}{c^2h}\frac{d^2h}{dt^2}=\frac{1}{\phi}\frac{d^2\phi}{dx^2}=-\lambda$$
Here λ is called the separation constant. The negative sign is for convenience (in some other PDEs a positive sign is needed, it’s case dependent) Note that we don’t know its value yet though.
<hr>
We can now split the PDE into two parts (two DEs). First part:
$$\frac{1}{c^2h}\frac{d^2h}{dt^2}=-\lambda$$
Which reworks to:
$$\frac{d^2h}{dt^2}+c^2\lambda h=0$$
Second part:
$$\frac{1}{\phi}\frac{d^2\phi}{dx^2}=-\lambda$$
Which reworks to:
$$\frac{d^2\phi}{dx^2}+\lambda \phi=0$$
For the latter we also have the boundary conditions:
$$\phi(0)=0\:\text{and }\phi(L)=0$$
The DE for φ(x) is a very simple one of the type we’ve already solved higher up (see e.g. the Pi1DB problem) and the solutions are:
$$\lambda_n=\Big(\frac{n\pi}{L}\Big)^2\:\text{so }\phi_n(x)=A\sin\Big(\frac{n\pi x}{L}\Big)\:\text{for: }n=1,2,3,...$$
The λ<sub>n</sub> are called eigenvalues and the φ<sub>n</sub> eigenfunctions of the DE. So now we have values for λ (but not for A)!
<hr>
And of course that means we can use these eigenvalues in the DE for h(t):
$$\frac{d^2h}{dt^2}+\Big(\frac{n\pi c}{L}\Big)^2h=0$$
This is a straightforward 2nd order linear DE and its CE has two repeat complex roots and general solution:
$$h(t)=c_1\cos\Big(\frac{n\pi ct}{L}\Big)+c_2\sin\Big(\frac{n\pi ct}{L}\Big)$$
Here we have a useful initial condition:
$$\frac{\partial y(x,0)}{\partial t}=0 \implies \frac{dh(0)}{dt}=0$$
$$\frac{dh}{dt}=-c_1\frac{n\pi c}{L}\sin\Big(\frac{n\pi ct}{L}\Big)+c_2\frac{n\pi c}{L}\cos\Big(\frac{n\pi ct}{L}\Big)$$
$$\frac{dh(0)}{dt}=0 \implies c_2=0$$
So:
$$h(t)=c_1\cos\Big(\frac{n\pi ct}{L}\Big)$$
<hr>
Now remember that we assumed (in German mathspeak: our 'Ansatz'):
$$y(x,t)=\phi(x)h(t)$$
Which gives us a set of solutions:
$$y_n(x,t)=A_n\cos\Big(\frac{n\pi ct}{L}\Big)\sin\Big(\frac{n\pi x}{L}\Big)\:\text{with }A_n=Ac_1$$

$$\text{For: }n=1,2,3,...$$
The Superposition Principle:

To put together a comprehensive solution we need to use the Superposition Principle (SP). There’s nothing new about it and we’ve used the SP several times before when solving second order linear DEs.

Formally the principle is:
<hr>
$$\text{If }(y_1, y_2,...,y_i,...,y_n)\:\text{are solutions to a linear, homogeneous DE, then }\displaystyle\sum_{i=1}^{n}(c_iy_i)\:\text{,where }c_i\:\text{are arbitrary constants, is also a solution to the DE.}$$
<hr>
Looking at the expression for y<sub>n</sub>(x,t), for each value of n we have a particular solution of the PDE, so in accordance with the SP these have to be combined to:
$$y(x,t)=\displaystyle\sum_{n=1}^{\infty}A_n\cos\Big(\frac{n\pi ct}{L}\Big)\sin\Big(\frac{n\pi x}{L}\Big)$$

Scary as it looks, we’ll soon see what it means with some function analysis!

[Edited on 10-6-2016 by blogfast25]

aga - 10-6-2016 at 12:04

Scary ? Terrifying more like !

There's gonna be chain rules everywhere, probably whips and gimp suits too.

blogfast25 - 10-6-2016 at 13:54

 Quote: Originally posted by aga Scary ? Terrifying more like ! There's gonna be chain rules everywhere, probably whips and gimp suits too.

In fact there's almost nothing we've done here that we haven't seen in action before in the thread.

That's the beauty of Separation of Variables (SoV) of simple PDEs: you don't need to know much more than what's needed for simple ODEs.

The solution of the Schrödinger equation applied to the Hydrogen atom also uses SoV. The assumed solution is then:

$$\psi(r,\theta,\phi)=R(r)Y(\theta,\phi)$$

... with R and Y resp. the radial and angular parts of the wave function.

aga - 10-6-2016 at 14:05

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 ?
blogfast25 - 10-6-2016 at 14:41

 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)

blogfast25 - 11-6-2016 at 12:00

Frequency spectrum:

The main purpose of a vibrating string is of course to generate sound. The string ‘bangs’ against the air, an elastic medium, and sound waves are generated. These our ears receive, turn the waves back to oscillations and the resulting electrical impulses are perceived by our brain as sound (here a music note).

We’ve yet to determine the coefficients A<sub>n</sub>but let’s first look and first time-dependent term in the series, cos(nπct/L).

In wave/oscillation mechanics we write:

$$\cos\Big(\frac{n\pi ct}{L}\Big)=\cos\omega t$$
Where ω is the angular velocity and f the frequency of oscillation:
$$\omega=2\pi f=\frac{n\pi c}{L}$$
$$f=\frac{nc}{2L}$$
$$\frac{T}{\rho}=c^2$$
$$f=\frac{n}{2L}\sqrt{\frac{T}{\rho}}$$
We can see that the vibrating string doesn’t have one single frequency but a frequency spectrum. If for n=1 we call:
$$f_1=\frac{1}{2L}\sqrt{\frac{T}{\rho}}$$
... the fundamental frequency (which defines the note), then the frequency spectrum is:
$$f_1, 2f_1, 3f_1,...$$
... where f<sub>2</sub>, f<sub>3</sub>,... are the so-called harmonics.

Note also:

• Shorter string length gives higher pitch
• Higher string tension gives higher pitch
• Higher density strings give lower pitch

And next we’ll see not all the frequencies are equally represented in the spectrum.

### Music notes! (Ct'nued)

blogfast25 - 12-6-2016 at 11:05

The Amplitude Spectrum:

We now need to derive values for A<sub>n</sub> 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<sub>n</sub> 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<sub>n</sub> should be seen as amplitudes, associated with each frequency and they vary inversely with n<sup>2</sup>:
$$f_1,2f_1,3f_2,4f_1,...$$
$$A_1,\frac{A_1}{4},\frac{A_1}{9},\frac{A_1}{16},...$$
So the amplitudes diminish very quickly with increasing values of n.

We can also see that the A<sub>n</sub> increase with y<sub>0</sub> 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]

aga - 12-6-2016 at 12:14

 Quote: Originally posted by blogfast25 This isn’t a particularly difficult integral but it is tedious and I’ll only present the final result: $$A_n=\frac{4ay_0}{n^2\pi^2}\sin\Big(\frac{n\pi }{a}\Big)$$

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 ...

blogfast25 - 12-6-2016 at 13:08

 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 ...

Lots of paper, yes. You end up with like 6 individual pieces you need to put back together again! But it requires only listed integrals (see cheat sheet). More lots of algebra than calculus, really...

### Music notes! (Ct'nued)

blogfast25 - 13-6-2016 at 09:54

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<sub>n</sub> 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

blogfast25 - 15-6-2016 at 08:12

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<sup>-2</sup> s<sup>-1</sup>, D the diffusion coefficient (aka the diffusivity) (in m<sup>2</sup> s<sup>-1</sup> 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)

blogfast25 - 16-6-2016 at 08:51

So far we obtained the fully separated DEs:
$$X''+\lambda X=0\:\text{and: }T'+\lambda DT=0$$
1. Solving the left hand side first:
$$X''+\lambda X=0$$
$$X=c_1\sin(\sqrt{\lambda}x)+c_2\cos(\sqrt{\lambda}x)$$
With the first boundary condition:
$$u_x(0,t)=X'(0)T(t)=0 \implies X'(0)=0$$
$$X'=c_1\sqrt{\lambda}\cos(\sqrt{\lambda}x)-c_2\sqrt{\lambda}\sin(\sqrt{\lambda}x)=0$$
$$x=0 \implies c_1=0$$
Which gives us:
$$X=c_2\cos(\sqrt{\lambda}x)$$
With the second boundary condition:
$$u_x(L,t)=X'(L)T(t)=0 \implies X'(L)=0$$
$$X’(L)=-c_2\sqrt{\lambda}\sin(\sqrt{\lambda}L)=0$$
$$\implies \sqrt{\lambda}L=n\pi$$
$$\lambda=\Big(\frac{n\pi}{L}\Big)^2\:\text{for: }n=0,1,2,3,...$$
$$X=c_2\cos\Big(\frac{n\pi x}{L}\Big)$$
Note that the series n=0,1,2,3,... includes the zeroth term because X is of the cosine form (a consequence of the boundary condition). Had X been of the sine form then n=0 would be a trivial solution, making u(x,t) zero over the entire x-domain. But cos 0 is not zero (it’s one). For that reason the n=0 term has to be included in the cosine series.

2. Solving the right hand equation:
$$T'+\lambda DT=0$$
$$T=Ae^{-\lambda Dt}=Ae^{-\Big(\frac{n\pi}{L}\Big)^2Dt}\:\text{for: }n=0,1,2,3,...$$
So with u(x,t)=T(t)X(x):
$$u_n=A_ne^{-\Big(\frac{n\pi}{L}\Big)^2Dt}\cos\Big(\frac{n\pi x}{L}\Big)$$
And with the Superposition Principle:
$$u(x,t)=A_0+\displaystyle \sum_{n=1}^{\infty}A_ne^{-\Big(\frac{n\pi}{L}\Big)^2Dt}\cos\Big(\frac{n\pi x}{L}\Big)$$
The term A<sub>0</sub> stems from the fact that for n=0 both the exponential factor and the cosine factor of the zeroth term equal one (1). That way the series can be started at n=1.

In the so far unused initial condition u(x,0)=f(x), the exponential factors all become 1 (e0=1), so we get:

$$u(x,0)=f(x)=A_0+\displaystyle \sum_{n=1}^{\infty}A_n\cos\Big(\frac{n\pi x}{L}\Big)$$
As in the vibrating spring problem, the coefficients A<sub>n</sub> are determined from:
$$A_n=\frac{2}{L}\int_0^Lf(x)\cos\Big(\frac{n\pi x}{L}\Big)dx$$
<hr>
Next we’ll use two realistic initial dye distributions (u(x,0)=f(x)) and do some plotting of u(x,t)!

[Edited on 17-6-2016 by blogfast25]

Richard3050 - 16-6-2016 at 17:26

Chemistry! For beginners, with a ‘no elements’ approach!
blogfast25 - 16-6-2016 at 17:46

 Quote: Originally posted by Richard3050 Chemistry! For beginners, with a ‘no elements’ approach!

### Modelling diffusion (Ct'nued)

blogfast25 - 18-6-2016 at 08:26

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 u<sub>0</sub> over a narrow length a, centred about L/2, like this:

Now we calculate the coefficients A<sub>n</sub>:

$$A_n=\frac{2}{L}\int_0^Lf(x)\cos\Big(\frac{n\pi x}{L}\Big)dx$$
$$A_n=\frac{2}{L}\int_{\frac{L-a}{2}}^{\frac{L+a}{2}}u_0\cos\Big(\frac{n\pi x}{L}\Big)dx$$
$$A_n=\frac{2u_0}{n\pi}\big[\sin\Big(\frac{n\pi x}{L}\Big)\Big]_{\frac{L-a}{2}}^{\frac{L+a}{2}}$$
$$A_n=\frac{2u_0}{n\pi}\Big[\sin\Big(\frac{n\pi}{2}+\frac{n\pi a}{2L}\Big)-\sin\Big(\frac{n\pi}{2}-\frac{n\pi a}{2L}\Big)\Big]$$
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<sub>r</sub>:
$$t_r=\frac{D}{L^2}t$$
$$\implies t=\frac{L^2}{D}t_r$$
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<sup>-9</sup> m<sup>2</sup> s<sup>-1</sup>!

Assuming a tube length of 10 cm (0.1 m), then t= 1 x 10<sup>7</sup> x t<sub>r</sub> = 16 weeks x t<sub>r</sub>! 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<sub>r</sub>.

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 u<sub>0</sub>, then calculation of A<sub>n</sub> 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<sub>r</sub>.

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

blogfast25 - 23-6-2016 at 08:54

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<sub>2,2</sub> for a particle in a rectangular box (with zero potential energy, length twice the width):

The probability density ψ<sup>2</sup><sub>2,2</sub> 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)

blogfast25 - 24-6-2016 at 10:25

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]

aga - 24-6-2016 at 12:02

Polar co-ordinates.

blogfast25 - 24-6-2016 at 13:12

 Quote: Originally posted by aga Polar co-ordinates. Now we're talking business ...

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

aga - 24-6-2016 at 13:29

Astonishingly, i have !

I was introduced to them as a convenient way to plot shapes on a computer screen, then rotate those shapes.

blogfast25 - 24-6-2016 at 14:10

Grab a piece of chalk, the blackboard is yours!
aga - 24-6-2016 at 14:56

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]

blogfast25 - 24-6-2016 at 16:24

That wasn't a bad lecture, actually.

I like these plots: cycloids:

http://curvebank.calstatela.edu/cycloidmaple/cycloid.htm

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

blogfast25 - 25-6-2016 at 08:31

3. Ansatz and separation of variables:

We assume u can be written as the product of two functions, one only dependent on r, the other only dependent on θ:
$$u(r,\theta)=R(r)Y(\theta)$$
Then we insert the assumed form of u into the PDE. This looks frightening (?) but because of the nature of u, it’s very easy. The first term e.g. becomes:
$$\frac{\partial^2}{\partial r^2}u(r,\theta)=\frac{\partial^2}{\partial r^2}(R(r)Y(\theta))=Y(\theta)\frac{d^2R}{dr^2}=Y(\theta)R''$$
Ploughing on with the other terms we get:
$$YR''+\frac{1}{r}YR'+\frac{1}{r^2}RY''+k^2RY=0$$
Divide both sides by RY:
$$\frac{R''}{R}+\frac{1}{r}\frac{R'}{R}+\frac{1}{r^2}\frac{Y''}{Y}+k^2=0$$
Multiply both sides with r<sup>2</sup>:
$$r^2\frac{R''}{R}+r\frac{R'}{R}+\frac{Y''}{Y}+r^2k^2=0$$
Bringing the Y based term to the right and introducing a separation constant m<sup>2</sup>:
$$r^2\frac{R''}{R}+r\frac{R'}{R}+r^2k^2=-\frac{Y''}{Y}=m^2$$
So we obtain two ODEs, one in R(r):
$$r^2R''+rR'+\Big(r^2k^2-m^2\Big)R= m^2$$
... and one in Y(θ):
$$\frac{Y''}{Y}=-m^2$$
The value of the separation constant m<sup>2</sup> is yet to be determined.

[Edited on 25-6-2016 by blogfast25]

aga - 25-6-2016 at 13:40

 Quote: Originally posted by blogfast25 $$\frac{Y''}{Y}=-m^2$$ The value of the separation constant m2 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]

blogfast25 - 25-6-2016 at 14:31

Quote: Originally posted by aga
 Quote: Originally posted by blogfast25 $$\frac{Y''}{Y}=-m^2$$ The value of the separation constant m2 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. x<sub>1</sub>,x<sub>2</sub>, x<sub>3</sub>, x<sub>4</sub>, 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.

wg48 - 27-6-2016 at 09:20

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!

Then you haven't lived WELL if you haven't used Euler angles plus range their first and second derivatives their estimates and time.

But if you want to live really well you need to get in to metrics of relativity and black holes.

morganbw - 27-6-2016 at 10:41

Sorry gentlemen, I do like math and even calculus, I am not extremely well versed and I usually have to do some research to use it meaningfully. I have not read all post in this thread and I am drinking just a little bit but it seems to me that this is a mastrubation post.

Self love screams from these post. Am I wrong? Think about it just a little. Calculus was developed between three and four hundred years ago, this is not new shit. Difficult shit for sure, for most of us mortals, but entirely doable.

And no, I am not able to give the theorem of calculus in its entirety without hitting the literature, but I am able to hit the literature and do pretty damn well. I am no doubt slower that a true mathematician, but tortoise or rabbit the end is the same place.

Please gentlemen, you are just who you are, please inform all of us when you are able to expand on ( an example, Newton ) his math and thoughts are hundreds of years old. ( Can you do it without referencing perhaps Einstein )

No gentlemen, you cannot, because you are not the genius you wished to be. We play with the thoughts of those who came before us, (effing A, I do as well )wishing to show something different/better, but it is not there yet, I do believe that it will be someday, but I am certain that it does not exist in this thread.

blogfast25 - 27-6-2016 at 11:34

@morganbw:

Despite your unnecessary reference to 'mastrubation' [sic], your multiple spelling, grammar and punctuation mistakes, not to mention your conflation between 'thread' and 'post', I'll keep this short and polite.

This is an educational thread for those who wish to learn something 'in house' about calculus and its applications in science. As such nothing new occurs here, which of course is also true for 99 % of what goes on in other SM threads. That's the nature of something educational.

Maybe morgan wants to point to whatever it is that he has done here that hasn't already been done elsewhere and possibly a million times already? Perhaps the many excellent but far, far from new (or unique) syntheses that are performed here by a few members are, by that reasoning, also 'm*st*rbation' and expressions of 'self-love'? (They are not, to be clear)

Finally, if you don't like the subject of this thread or how it's being presented, then don't read it. Inane comments like yours don't help one iota.

As regards:

 Quote: No gentlemen, you cannot, because you are not the genius you wished to be.

Again that's offensive (but it's your right to be so) and wholly incorrect. I'm pretty sure know my limitations better than you know yours.

[Edited on 27-6-2016 by blogfast25]

aga - 27-6-2016 at 11:39

Erm, in this thread I learnt at least a bit of Calculus from bloggers, and for Free !

However the knowledge cums, i'm glad it does, no matter if it is from the enlightening words of morganbw or from Dr Flexi Jerkoff himself.

As it is a Maths thread, erm, perhaps you'd care to share some of your own pearls of wisdom ?

Edit:

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

[Edited on 27-6-2016 by aga]

blogfast25 - 27-6-2016 at 11:53

 Quote: Originally posted by aga I'll swap for my secret formula for actually making Aluminium Sulphate xHydrate crystals.

Very slow, controlled evaporation of an Al sulphate solution, with careful nucleation should yield a defined hydrate as first crystals.

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

blogfast25 - 27-6-2016 at 11:59

4. Solving the angular equation:
$$\frac{Y''}{Y}=-m^2$$
$$Y''+m^2 Y=0$$
With two complex roots, +im and –im this has the general solution:
$$Y(\theta)=c_1\cos m\theta + c_2\sin m\theta$$
θ is of course an angle, in radians, and an angle ‘repeats’ itself every 2π. For that reason we can choose either c<sub>1</sub> or<sub>2</sub> to be zero. We choose the latter:
$$c_2=0 \implies Y=c_1\cos m\theta$$
Again, because θ is an angle the following condition must also be met:
$$Y_m(\theta)=Y_m(\theta+2\pi ) \implies \cos(m\theta)=\cos(m(\theta+2\pi))=\cos(m\theta+2m\pi)$$
$$\implies m=0, \pm 1, \pm 2, \pm 3,...$$
The negative values are trivial because:
$$\cos(|m|\theta)=\cos(-|m|\theta)$$
The function needs to be normalised, with the normalisation requirement:
$$\int_0^{2\pi}Y^*Yd\theta=1$$
Y<sup>*</sup> is the complex conjugate of Y. But Y is a Real function, so:
$$Y^*=Y \implies Y^*Y=Y^2$$
$$\int_0^{2\pi}Y^2d\theta=\int_0^{2\pi}c_1^2\cos^2(m\theta) d\theta=1$$
$$\implies c_1=\frac{1}{\sqrt{\pi}}$$
$$Y_m=\frac{1}{\sqrt{\pi}}\cos m\theta$$
There’s one exception, for m=0:
$$m=0 \implies Y_0=c_1$$
$$\int_0^{2\pi}c_1^2d\theta=1 \implies c_1=\frac{1}{\sqrt{2\pi}}$$
$$Y_0=\frac{1}{\sqrt{2\pi}}$$
So we have a fully defined solution for Y(θ), as well as values for m:
$$m=0,1,2,3,...$$

[Edited on 27-6-2016 by blogfast25]

aga - 27-6-2016 at 12:03

Yay ! morganbw's post caused the next page to appear.
it was taking ages for page 18 to render.

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 ?

[Edited on 27-6-2016 by aga]

blogfast25 - 28-6-2016 at 06:41

 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 ?

Sorry, only just saw your edit.

Firstly, both polar coordinates and Cartesian coordinates are completely valid ways to define the position of a point (the position does not change in function of the coordinate system used).

Derivatives (first, second or whatnot) can be used in both systems, the same rules of derivation apply.

A derivative like:

$$\frac{\partial u(r,\theta)}{\partial r}$$

... is the rate of change of the wave function (in polar coordinates) vis-a-vis the r-component of the position (r,θ).

 Quote: In plain english [...]

That's a perfectly legitimate question but hard to answer. Mathematical statements like the Schrödinger equation are jam packed full of logic. But that makes 'unpacking' the parcel into intelligible spoken or written language quite hard.

Incidentally, the Schrödinger equation used here is the time-independent one. It is free of time and therefore cannot be used to determine velocities or angular velocities (both are derivatives to time). For these we need to apply quantum operators on the wave function.

I'll wait for your response before solving the radial part of the SE and then we'll be able to start plotting some functions!

aga - 28-6-2016 at 09:26

OK. It was a silly question really now i'm sober enough to think about it.

It's a bit like asking what a 5-dimension co-ordinate set actually means ...

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

blogfast25 - 28-6-2016 at 11:32

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<sub>m</sub>:
$$m \to J_m(\rho)=J_m(kr)$$
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<sub>m</sub> functions are sophisticated infinite term polynomials and they look like this:

J<sub>0</sub> compared a cosine function, to show the resemblance:

J<sub>3</sub> 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.

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<sub>m,n</sub>, where n designates the n<sup>th</sup> zero of J<sub>m</sub>:
$$k_{m,n}R=z_{m,n}$$
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)

blogfast25 - 30-6-2016 at 09:06

6. Quasi-solutions of the radial equation - plots:

Two cases need to be considered:

1. m = 0
$$R(r)=c_1\cos(kr)$$
$$R(R)=c_1\cos(kR)=0 \implies kR=\frac{n\pi}{2}$$
$$n=1,3,5,7,...$$
$$k=\frac{n\pi}{2R}$$
$$R_0(r)=c_1\cos\Big(\frac{n\pi r}{2R}\Big)$$
Normalisation condition to obtain c<sub>1</sub>:
$$\int_0^Rc_1^2\cos^2\Big(\frac{n\pi r}{2R}\Big)rdr \implies c_1=\frac{2}{R}$$
$$R_0(r)=\frac{2}{R}\cos\Big(\frac{n\pi r}{2R}\Big)$$
2. m = 1,2,3,...
$$R(r)=c_2\sin(kr)$$
$$R(R)=c_2\sin(kR)=0 \implies kR=n\pi$$
$$n=1,2,3,...$$
$$k=\frac{n\pi r}{R}$$
$$R_m(r)=c_2\sin\Big(\frac{n\pi r}{R}\Big)$$
Normalisation condition to obtain c<sub>2</sub>:

$$\int_0^Rc_2^2\sin^2\Big(\frac{n\pi r}{R}\Big)rdr \implies c_2=\frac{2}{R}$$
$$R(r)=\frac{2}{R}\sin\Big(\frac{n\pi r}{R}\Big)$$
Putting it all together:
$$u(r,\theta)=R(r)Y(\theta)$$
m = 0:
$$u_0(r,\theta)\approx \frac{2}{R\sqrt{2\pi}}\cos\Big(\frac{n\pi r}{2R}\Big)$$
$$n=1,3,5,...$$
m = 1,2,3,...
$$u_m(r,\theta)\approx \frac{2}{R\sqrt{\pi}}\sin\Big(\frac{n\pi r}{R}\Big)\cos(m\theta)$$
$$n=1,2,3,...$$
Plotting the functions:

Instead of plotting the values of the wave functions u, the probability density functions will be plotted, for a circular well of R=1 m. As u is a Real function, the probability density function is simply:
$$\rho=u_m^2(r,\theta)$$
(not to be confused with the previously used rho)

Of course, R=1 m is not a realistic value for a quantum system. For a more realistic size of say 1 nm (1x10<sup>-9</sup> m) the vertical axis values would have to be multiplied by 1x10<sup>18</sup> m<sup>-2</sup>.

Left (next post) are the 3D plots, right the contour plots, with the actual boundary indicated as a thick black line. In the contour plots, areas of high probability density are marked as light colours and values close to zero as red.

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

blogfast25 - 30-6-2016 at 09:14

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)

blogfast25 - 30-6-2016 at 09:27

m=2, n=1

m=2, n=2

m=3, n=1

m=3, n=2

wg48 - 26-7-2016 at 16:56

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.

[Edited on 27-7-2016 by wg48]

blogfast25 - 27-7-2016 at 05:42

 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]

wg48 - 27-7-2016 at 10:52

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]

I had no particular orbital in mind. My interest was just general curiosity. In part motivated by how an object with some symmetry can have modes with a different symmetry. I have now sorted that out now.

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.

blogfast25 - 27-7-2016 at 12:59

 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:

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

The roots of the Bessel function (zm,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 zm,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]

wg48 - 27-7-2016 at 14:17

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.

As for ringing a bell: the whealing sounds of bells is caused by the anharmonic relationship between its fundamental and overtones. The fundamental decays faster than some overtones producing a sound that chances frequency, the walling. LOL

Thanks for the offer of help. I may get round to it some time. I am presently improving my attenuation graph and trying to work out how I can get mathcad to calculate masses from a chemical equation. I am stuck at inputting the atomic weights and abbreviations at the moment.

blogfast25 - 27-7-2016 at 14:24

 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.

Chaotic and noise in the sense of inharmonic.

The amplitude spectrum (relative weight of the various frequencies) is indeed determined by the initial condition:

$$u(r,\theta,0)=f(x,y)$$

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