Jackson’s Laplacian in spherical Coordinates [Proof]

If you took a look at one of the previous posts on how to remember the Laplacian in different forms by using a metric,  you will notice that the form of  the Laplacian that we get is:

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

But in Jackson’s Classical Electrodynamics, III edition he notes the following:

20190218_135719

This is an interesting form of the Laplacian that perhaps not everyone has encountered. This can obtained from the known form by making the substitution u = r \psi and simplifying. The steps to which have been outlined below:

20190218_135737_120190218_135743_1

 

 

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Einstein’s field equation from the Action principle

In this post, we will derive the Einstein’s field equation from the least action principle. Let’s just construct an action with the Ricci scalar, Lagrangian and an invariant space-time volume element( \sqrt{g} d^{4}x ; g- metric) : All these quantities  have the same value in all frames of reference.

S = \int R \sqrt{g} d^{4}x  - \int \mathcal{L} \sqrt{g} d^{4}x

\delta S = \int \delta ( R \sqrt{g} )  d^{4}x  - \int \delta  ( \mathcal{L} \sqrt{g} ) d^{4}x  = 0

\delta S = \int \left( \delta ( R  )\sqrt{g} + \frac{\delta g}{2 \sqrt{g}} R  \right) d^{4}x   - \int \frac{1}{\sqrt{g}} \frac{ \delta  ( \mathcal{L} \sqrt{g} )}{ \delta g^{\mu \nu}}  \sqrt{g}  \delta g^{\mu \nu} d^{4}x  

 

Let’s call \int \frac{1}{\sqrt{g}} \frac{ \delta  ( \mathcal{L} \sqrt{g} )}{ \delta g_{\mu \nu}}  = T_{\mu \nu}    and also subsitute  R = R_{\mu \nu} g^{\mu \nu} in the above equation.

\delta S = \int \left( \delta ( R_{\mu \nu} g^{\mu \nu}  )\sqrt{g} + \frac{\delta g}{2 \sqrt{g}} R \right)   d^{4}x  - \int T_{\mu \nu}\sqrt{g} \delta  g^{\mu \nu}   d^{4}x  

\delta S = \int \left(R_{\mu \nu} \delta (  g^{\mu \nu}  )\sqrt{g}   + \delta(R_{\mu \nu})   g^{\mu \nu} \sqrt{g} + \frac{\delta g}{2 \sqrt{g}} R  - T_{\mu \nu}\sqrt{g} \delta  g^{\mu \nu}   \right)   d^{4}x    .

To simplify further we employ Jacobi’s formula ** \delta g =  -g g_{\mu \nu} \delta g^{\mu \nu} and also note that the boundary terms that arise out of \delta(R_{\mu \nu})  vanish according to Stoke’s theorem.

\delta S = \int \left(R_{\mu \nu} - \frac{ g^{\mu \nu}}{2} R  - T_{\mu \nu} \right) \sqrt{g} \delta (  g^{\mu \nu} ) d^{4}x  = 0   .

This tells us that under a variation of the metric if we would like a stationary action then it has to be true that :

R_{\mu \nu} - \frac{ g_{\mu \nu}}{2} R   =  T_{\mu \nu} 

where R_{\mu \nu} – Riemann Tensor, R – Ricci scalar, T_{\mu \nu} – Energy-momentum Tensor and   g_{\mu \nu} – Metric Tensor

This is the Einstein’s field equation. Inserting the constants at the right places yields the more familiar form:

R^{\mu \nu} - \frac{ g^{\mu \nu}}{2} R   = \frac{ 8 \pi G}{c^4} T^{\mu \nu} 

What would happen ?

As a fun little exercise you can start off with the invariant volume given by the inverse metric instead. i.e \sqrt{g^{-1}} d^{4}x   and you will obtain the same equation. BUT, you should be careful when employing the Jacobi’s formula else you might ending up with a extra minus sign with the Ricci scalar term.

Some more fun questions to ponder:

  •  Why is \sqrt{g} d^{4}x called the  invariant space-time volume element ?
  •  Why is the action defined as S = \int R -  L ? Why not any other way?
  •  How do you prove the Jacobi’s formula ?

 

Remembering the Laplacian in different coordinate systems

When one is learning about Laplace and Poisson equations it can be frustrating to remember its form in different coordinate systems.  But when one is introduced to four vectors, special relativity and so on, here is a simple way to remember the Laplacian in any coordinate system.

\nabla^2 \phi = \frac{1}{\sqrt{|g|}} \frac{\partial}{\partial x^{j}} \left(\sqrt{|g|} g^{ij}  \frac{\partial \phi}{\partial x^{j}} \right)

where we g^{ij} is the inverse of the metric g_{ij} , |g| is the determinant of the metric . And one specifies the coordinate system by mentioning the form of the metric. Let’s look at how this works out:

Cartesian Coordinates

x^{j} = (x,y,z)

g_{ij} = \begin{bmatrix} 1  &  0 &  0 \\  0 & 1 & 0 \\ 0 & 0 & 1  \end{bmatrix}

g^{ij} = \begin{bmatrix} 1  &  0 &  0 \\  0 & 1 & 0 \\ 0 & 0 & 1  \end{bmatrix}

|g_{ij}| = 1

\nabla^2 \psi = \frac{1}{1} \frac{\partial}{\partial x^{j}} \left(\sqrt{1} g^{ij}  \frac{\partial \psi}{\partial x^{j}} \right)  =  \frac{\partial^2 \psi}{\partial x^2} + \frac{\partial^2 \psi}{\partial y^2} + \frac{\partial^2 \psi}{\partial z^2}

 

Cylindrical Coordinates

x^{j} = (r,\phi,z)

g_{ij} = \begin{bmatrix} 1  &  0 &  0 \\  0 & r^2 & 0 \\ 0 & 0 & 1  \end{bmatrix}

g^{ij} = \begin{bmatrix} 1  &  0 &  0 \\  0 & \frac{1}{r^2} & 0 \\ 0 & 0 & 1  \end{bmatrix}

|g_{ij}| = r^2

\nabla^2 \psi = \frac{1}{r} \frac{\partial}{\partial x^{j}} \left(r g^{ij}  \frac{\partial \phi}{\partial x^{j}} \right)   

\nabla^2 \psi = \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial \psi}{\partial r} \right) + \frac{1}{r} \frac{\partial}{\partial \phi} \left( \frac{r}{r^2} \frac{\partial \psi}{\partial \phi} \right)   + \frac{1}{r} \frac{\partial}{\partial z} \left( r \frac{\partial \psi}{\partial z} \right)  

Noting that \frac{\partial r}{\partial \phi}  = 0 = \frac{\partial r}{\partial z}  because they are independent variables, we get

\nabla^2 \psi = \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial \psi}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2 \psi}{\partial \phi^2}   + \frac{\partial^2 \psi}{\partial z^2}   

 

Spherical Coordinates

x^{j} = (r,\phi,\theta)

g_{ij} = \begin{bmatrix} 1  &  0 &  0 \\  0 & r^2 & 0 \\ 0 & 0 & r^2 \sin^2(\theta)  \end{bmatrix}

g^{ij} = \begin{bmatrix} 1  &  0 &  0 \\  0 & \frac{1}{r^2} & 0 \\ 0 & 0 & \frac{1}{r^2 \sin^2(\theta)}  \end{bmatrix}

|g_{ij}| = r^4  \sin^2(\theta)

Following the same approach as the Cylindrical and Cartesian coordinates, we get the following form for the Laplacian in Spherical coordinates,

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

 

 

 

 

 

 

 

 

 

 

 

 

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!

 

 

 

Ansatz to Gram-Schmidt Orthonormalization

The Gram–Schmidt process is a method for orthonormalising a set of vectors in an inner product space and the trivial way to remember this is through an ansatz :

Let |v_{1}> , |v_{2}> , \hdots |v_{n}>    be a set of normalized basis vectors but we would also like to make them orthogonal.  We will call |v_{1}^{'}> , |v_{2}^{'}> , \hdots |v_{n}^{'}>  be the orthonormalized set of basis vectors formed out  |v_{1}> , |v_{2}> , \hdots |v_{n}>  .

Let’s start with the first vector:

|v_{1}^{'} > = |v_{1}> 

Now we construct a second vector |v_{2}^{'}> out of |v_{1}^{'}> and |v_{2}> :

|v_{2}^{'} > = |v_{2}> - \lambda |v_{1}^{'}>

But what must be true of |v_{2}^{'} > is that  |v_{1}^{'}> and |v_{2}^{'}> must be orthogonal i.e <v_{1}^{'}|v_{2}^{'}> = 0 .

<v_{1}^{'}|v_{2}^{'} > = <v_{1}^{'}|v_{2}> - \lambda <v_{1}^{'}|v_{1}^{'}>

0 = <v_{1}^{'}|v_{2}> - \lambda 

\lambda = <v_{1}^{'} | v_{2}>

Therefore we get the following expression for v_{2}^{'} ,

|v_{2}^{'} > = |v_{2}> -  <v_1^{'} | v_{2} >|v_{1}>

which upon normalization looks like so:

|v_{2}^{'} > = \frac{|v_{2}^{'} >}{<v_{2}^{'} |v_{2}^{'} > }

 

 

That might have seemed trivial geometrically, but this process can be generalized for any complete n-dimensional vector space. Let’s continue the Gram – Schmidt for the third vector by choosing |v_{3}^{'} > of the following form and generalizing this process:

|v_{3}^{'} > = |v_{3}> - \lambda_{1} |v_{1}^{'}> - \lambda_{2} |v_{2}^{'}>

The values for \lambda_{1} and \lambda_{1} are found out to be as:

\lambda_{1} =  <v_{1}^{'}|v_{3}>

\lambda_{2}  = <v_{2}^{'}|v_{3}>

Therefore we get,

|v_{3}^{'} > = |v_{3}> - <v_{1}^{'}|v_{3}>|v_{1}^{'}> - <v_{2}^{'}|v_{3}>|v_{2}^{'}> (or)

|v_{3}^{'} > = |v_{3}> -  \sum\limits_{j=1,2} <v_{j}^{'} | v_{3}> |v_{j}^{'}> 

|v_{3}^{'} > = \frac{|v_{2}^{'} >}{<v_{3}^{'} |v_{3}^{'} > }

 

Generalizing, we obtain:

|v_{i}^{'} > = |v_{i}> -  \sum\limits_{j=1,2,...,i-1} <v_{j}^{'} | v_{i}> |v_{j}^{'}> 

|v_{i}^{'} > = \frac{|v_{i}^{'} >}{<v_{i}^{'} |v_{i}^{'} > }

Now although you would never need to remember the above expression because you can derive it off the bat with the above procedure, it is essential to understand how it came out to be.

Cheers!

 

Example (to be added soon):

 

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!

 

 

Prof.Ghrist at his best!

Screenshot from 2017-07-28 10:55:27

To understand why this is true, we must start with the Fundamental Theorem of Vector calculus. If F  is a conservative field ( i.e F = \nabla \phi  ), then

\int\limits_{A}^{B} F.dr = \int\limits_{A}^{B} \nabla\phi .dr = \phi_{A} - \phi_{B}

What this means is that the value is dependent only on the initial and final positions. The path that you take to get from A to B is not important.

Screenshot from 2017-07-28 11:29:46

Now if the path of integration is a closed loop, then points A and B are the same, and therefore:

\int\limits_{A}^{A} F.dr = \int\limits_{A}^{A} \nabla\phi .dr = \phi_{1} - \phi_{1} = 0

Now that we are clear about this, according to Stokes theorem the same integral for a closed region can be represented in another form:

\int_{C} F.dr = \int\int_{A} (\nabla X F) .\vec{n} dA  = 0

From this we get that Curl = \nabla X F = 0 for a conservative field (i.e F = \nabla \phi ). Therefore when a conservative field is operated on by a curl operator (\nabla X ), it yields 0.

Bravo Prof.Ghrist! Beautifully said 😀

 

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.

 

nth roots of unity : A geometric approach

When one is dealing with complex numbers, it is many a times useful to think of them as transformations. The problem at hand is to find the nth roots of unity. i.e

z^n = 1

Multiplication as a Transformation

Multiplication in the complex plane is mere rotation and scaling. i.e

z_{1} = r_{1}e^{i\theta_{1}}, z_{2} = r_{2}e^{i\theta_{2}} 

z_{1}z_{2} = \underbrace{r_{1} r_{2}}_{scaling} \underbrace{e^{i(\theta_{1} + \theta_{2})}}_{rotation}

Now what does finding the n roots of unity mean?

If you start at 1 and perform n equal rotations( because multiplication is nothing but rotation + scaling ), you should again end up at 1.

We just need to find the complex numbers that do this.i.e

z^n = 1

\underbrace{zz \hdots z}_{n} = 1

z = re^{i\theta}

r^{n}e^{i(\theta + \theta + \hdots \theta)} = 1e^{2\pi k i}

r^{n}e^{in\theta} =1e^{2\pi k i}

This implies that :

\theta = \frac{2\pi k}{n}, r = 1

And therefore :

z = e^{\frac{2\pi k i}{n}}

Take a circle, slice it into n equal parts and voila you have your n roots of unity.

main-qimg-6da134e3e5735a9fb92355d53f95e4ed

Okay, but what does this imply ?

Multiplication by 1 is a 360^o/0^o rotation.

drawing

When you say that you are multiplying a positive real number(say 1) with 1 , we get a number(1) that is on the same positive real axis.

Multiplication by (-1) is a 180^o rotation.

drawing-1

When you multiply a positive real number (say 1) with -1, then we get a number (-1) that is on the negative real axis

The act of multiplying 1 by (-1) has resulted in a 180o transformation. And doing it again gets us back to 1.

Multiplication by i is a 90^o rotation.

drawing-2

Similarly multiplying by i takes 1 from real axis to the imaginary axis, which is a 90o rotation.

This applies to -i as well.

That’s about it! – That’s what the nth roots of unity mean geometrically. Have a good one!