Feynman’s trick of parametric integration applied to Laplace Transforms

Parametric Integration is an Integration technique that was popularized by Richard Feynman but was known since Leibinz’s times. But this technique rarely gets discussed beyond a niche set of problems mostly in graduate school in the context of Contour Integration.

A while ago, having become obsessed with this technique I wrote this note on applying it to Laplace transform problems  and it is now public for everyone to take a look.

( Link to notes on Google Drive ) 

I would be open to your suggestions, comments and improvements on it as well. Cheers!

Using Complex numbers in Classical Mechanics

When one is solving problems on the two dimensional plane and you are using polar coordinates, it is always a challenge to remember what the velocity/acceleration in the radial and angular directions (v_r , v_{\theta}, a_r, a_{\theta} ) are. Here’s one failsafe way using complex numbers that made things really easy :

z = re^{i \theta}

\dot{z} = \dot{r}e^{i \theta} + ir\dot{\theta}e^{i \theta} = (\dot{r} + ir\dot{\theta} ) e^{i \theta}

From the above expression, we can obtain v_r = \dot{r} and v_{\theta} = r\dot{\theta}

\ddot{z} =  (\ddot{r} + ir\ddot{\theta} + i\dot{r}\dot{\theta} ) e^{i \theta}   + (\dot{r} + ir\dot{\theta} )i \dot{\theta} e^{i \theta} 

\ddot{z} =  (\ddot{r} + ir\ddot{\theta} + i\dot{r}\dot{\theta}  + i  \dot{r} \dot{\theta} - r\dot{\theta}\dot{\theta} )e^{i \theta} 

\ddot{z} =  (\ddot{r} - r(\dot{\theta})^2+ i(r\ddot{\theta} + 2\dot{r}\dot{\theta} ) )e^{i \theta} 

From this we can obtain a_r = \ddot{r} - r(\dot{\theta})^2 and a_{\theta} = (r\ddot{\theta} + 2\dot{r}\dot{\theta}) with absolute ease.

Something that I realized only after a mechanics course in college was done and dusted but nevertheless a really cool and interesting place where complex numbers come in handy!

 

 

Beautiful Proofs(#3): Area under a sine curve !

So, I read this post on the the area of the sine curve some time ago and in the bottom was this equally amazing comment :

Screenshot from 2017-06-07 00:19:11

\sum sin(\theta)d\theta =   Diameter of the circle/ The distance covered along the x axis starting from 0 and ending up at \pi.

And therefore by the same logic, it is extremely intuitive to see why:

\int\limits_{0}^{2\pi} sin/cos(x) dx = 0

Because if a dude starts at 0 and ends at 0/ 2\pi/ 4\pi \hdots, the effective distance that he covers is 0.

Circle_cos_sin.gif

If you still have trouble understanding, follow the blue point in the above gif and hopefully things become clearer.

 

The generalized product rule ( Leibniz Formula )

The generalized product rule at the face of it not intuitive and that’s probably why a lot of people make mistakes when asked to differentiate functions n-times.

If f and g are functions that are continuous and differentiable everywhere, then:

image

Now, we would like to find out a generalized expression for the n-th
derivative of fg.

In order to arrive at that formulation lets calculate a
few derivatives to see whether we can find a pattern:

image
image
image
image

.

.

.

.

You must have noticed a pattern in the above expressions. The coefficients seem are the one in the binomial expansion of (x+y)n

Therefore we can write the expression for the n-derivative of fg as the following:

image

where (i) means to differentiate i-times. This is also known as Leibniz Formula.

Have a good one!

On the direction of the cross product of vectors

One of my math professors always told me:

Understand the concept and not the definition

image

A lot of times I have fallen into this pitfall where I seem to
completely understand how to methodically do something without actually
comprehending what it means.

And only after several years after I first
encountered the notion of cross products did I actually understand what
they really meant. When I did, it was purely ecstatic!

Why on earth is the direction of cross product orthogonal ? Like seriously…

I mean this is one of the burning questions regarding the cross
product and yet for some reason, textbooks don’t get to the bottom of
this. One way to think about this is :

It is modeling a real life scenario!!

The scenario being :

image

When you try to twist a screw (clockwise screws being the convention)
inside a block in the clockwise direction like so, the nail moves down
and vice versa.

image

i.e When you move from the screw from u to v, then the direction of the cross product denotes the direction the screw will move..

That’s why the direction of the cross product is orthogonal. It’s really that simple!

Another perspective

Now that you get a physical feel for the direction of the cross product, there is another way of looking at the direction too:

Displacement is a vector. Velocity is a vector.
Acceleration is a vector. As you might expect, angular displacement,
angular velocity, and angular acceleration are all vectors, too.

But which way do they point ?

image

Let’s take a rolling tire. The velocity vector of every point in the
tire is pointed in every other direction.

BUT every point on a rolling
tire has to have the same angular velocity – Magnitude and Direction.

How can we possibly assign a direction to the angular velocity ?

image

Well, the only way to ensure that the direction of the angular velocity
is the same for every point is to make the direction of the angular
velocity perpendicular to the plane of the tire.

Problem solved!

Why is the area under one hump of a sine curve exactly 2?

Girls' Angle

blog_073013_02

I was talking with a student recently who told me that he always found the fact that $latex int_0^{pi} sin x , dx = 2$ amazing. “How is it that the area under one hump of the sine curve comes out exactly 2?” He asked me if there is an easy way to see that, or is it something you just have to discover by doing the computation.

If you’ve wondered about this too, perhaps you’ll find the following of interest.

View original post 162 more words

Solving the Laplacian in Spherical Coordinates (#1)

In this post, let’s derive a general solution for the Laplacian in Spherical Coordinates. In future posts, we shall look at the application of this equation in the context of Fluids and Quantum Mechanics.

sph_coor

x = rsin\theta cos\phi
y = rsin\theta cos\phi
z = rcos\theta

where

0 \leq r < \infty
0 \leq \theta \leq \pi
0 \leq \phi < 2\pi

The Laplacian in Spherical coordinates in its ultimate glory is written as follows:

\nabla ^{2}f ={\frac {1}{r^{2}}}{\frac {\partial }{\partial r}}\left(r^{2}{\frac {\partial f}{\partial r}}\right)+{\frac {1}{r^{2}\sin \theta }}{\frac {\partial }{\partial \theta }}\left(\sin \theta {\frac {\partial f}{\partial \theta }}\right)+{\frac {1}{r^{2}\sin ^{2}\theta }}{\frac {\partial ^{2}f}{\partial \phi ^{2}}} = 0

To solve it we use the method of separation of variables.

f = R(r)\Theta(\theta)\Phi(\phi)

Plugging in the value of f into the Laplacian, we get that :

\frac{\Theta \Phi}{r^2} \frac{d}{dr} \left( r^2\frac{dR}{dr} \right) + \frac{R \Phi}{r^2 sin \theta} \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{d\theta} \right) + \frac{\Theta R}{r^2 sin^2 \theta} \frac{d^2 \Phi}{d\phi^2} = 0

Dividing throughout by R\Theta\Phi and multiplying throughout by r^2, further simplifies into:

\underbrace{ \frac{1}{R} \frac{d}{dr} \left( r^2\frac{dR}{dr} \right)}_{h(r)} + \underbrace{\frac{1}{\Theta sin \theta} \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{d\theta} \right) + \frac{1}{\Phi sin^2 \theta} \frac{d^2 \Phi}{d\phi^2}}_{g(\theta,\phi)} = 0

It can be observed that the first expression in the differential equation is merely a function of r and the remaining a function of \theta and \phi only. Therefore, we equate the first expression to be \lambda = l(l+1) and the second to be -\lambda = -l(l+1). The reason for choosing the peculiar value of l(l+1) is explained in another post.

\underbrace{ \frac{1}{R} \frac{d}{dr} \left( r^2\frac{dR}{dr} \right)}_{l(l+1)} + \underbrace{\frac{1}{\Theta sin \theta} \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{d\theta} \right) + \frac{1}{\Phi sin^2 \theta} \frac{d^2 \Phi}{d\phi^2}}_{-l(l+1)} = 0 (1)

 

The first expression in (1) the Euler-Cauchy equation in r.

\frac{d}{dr} \left( r^2\frac{dR}{dr} \right) = l(l+1)R

The general solution of this has been in discussed in a previous post and it can be written as:

R(r) = C_1 r^l + \frac{C_2}{r^{l+1}}

 

The second expression in (1) takes the form as follows:

\frac{sin \theta}{\Theta} \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{dr} \right)+ l(l+1)sin^2 \theta + \frac{1}{\Phi} \frac{d^2 \Phi}{d\phi^2} = 0

The following observation can be made similar to the previous analysis

\underbrace{\frac{sin \theta}{\Theta} \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{dr} \right)+ l(l+1)sin^2 \theta }_{m^2} + \underbrace{\frac{1}{\Phi} \frac{d^2 \Phi}{d\phi^2}}_{-m^2} = 0 (2)

 

The first expression in the above equation (2) is the Associated Legendre Differential equation.

\frac{sin \theta}{\Theta} \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{dr} \right)+ l(l+1)sin^2 \theta = m^2

sin \theta \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{dr} \right)+ \Theta \left( l(l+1)sin^2 \theta - m^2 \right) = 0

The general solution to this differential equation can be given as:
\Theta(\theta) = C_3 P_l^m(cos\theta) + C_4 Q_l^m(cos\theta)

 

The solution to the second term in the equation (2) is a trivial one:

\frac{d^2 \Phi}{d\phi^2} = m^2 \Phi
\Phi(\phi) = C_5 e^{im\phi} + C_6 e^{-im\phi}

 

Therefore the general solution to the Laplacian in Spherical coordinates is given by:

R\Theta\Phi = \left(C_1 r^l + \frac{C_2}{r^{l+1}} \right) \left(C_3 P_l^m(cos\theta) + C_4 Q_l^m(cos\theta \right) \left(C_5 e^{im\phi} + C_6 e^{-im\phi}\right)

fuckyeahphysica:

The simple harmonic oscillator

Anonymous asked: Please explain the intuition of solving the SHM equation.

Okay Anon! Here you go, this is my rendition.

The problem

You have a mass suspended on a spring. We want to know where the mass will be at any instant of time.

Describe the motion of the mass

image

The physical solution

Now before we get on to the math, let us first visualize the motion by attaching a spray paint bottle as the mass.

image

Oh, wait that seems like a function that we are familiar with – The sinusoid.

image

Without even having to write down a single equation, we have found out the solution to our problem. The motion that is traced  by the mass is a sinusoid.

But what do I mean by a sinusoid ?

If you took the plotted paper and tried to create that function with the help of sum of polynomials i.e x, x2, x3 … Now you this what it would like :

image

By taking an infinite of these polynomial sums you get the function Since this series of polynomial occurs a lot, its given the name – sine.

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I hope this shed some light on the intuition of the SHM equation. Have fun!

A note on the cosine

For people asking about the cosine and how to think about it . Cosine is merely sine pushed to the side.

The analogy works the same way 😀

Have a good one!