Einstein’s field equation from the Action principle

January 21, 2019January 27, 2019 / 153armstrong / Leave a comment

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

January 17, 2019January 17, 2019 / 153armstrong / 1 Comment

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

December 13, 2018 / 153armstrong / 1 Comment

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!

Cooking up a Lorentz invariant Lagrangian (Part-II)

December 8, 2018December 8, 2018 / 153armstrong / Leave a comment

In a previous post, we took a look at Lorentz Invariant scalar fields by using Temperature as an example and derived some Lorentz invariant quantities. In this post we will do the same thing but by being not so rigorous.

20181208_000628

where $dx^{\mu}$ and $dx^{\mu ' }$ are small infinitesimal change in coordinates.

But by virtue of series expansion we can express the second expression in the following way and find another Lorentz invariant quantity for scalar fields.

20181208_000946

By a similar analysis, we can obtain

$\partial^{\beta} \phi \ dx_{\beta} = \partial^{\beta^{'}} \phi^{'} \ dx_{\beta^{'}}$

Using the above and the invariant distance ( $ds^{2} = dx^{\mu}{dx_{\mu}} = dx^{\mu^{'}}{dx_{\mu^{'}}} = C_{3}$ ) we find that $\partial^{\beta} \phi \partial_{\beta} \phi$ is also invariant under the Lorentz Transformation.

20181208_030246

But what does this quantity physically represent and why is it important in physics ? That’s the question we will address in the next post of this mini-series.

A note on Wave-functions and Fourier Transforms

December 3, 2018December 3, 2018 / 153armstrong / Leave a comment

In quantum mechanics you can denote the wave-function in the position or the momentum basis. Written in the momentum basis, it would look something like:

$|\psi(x)> = a_0 |p_0> + a_1 |p_1> + \hdots$

$|\psi(x)> = \sum\limits_{n} a_n |p_n>$

But momentum is a continuous variable and it varies from $-\infty$ to $\infty$ .

unnamed

Therefore changing to the integral representation we get that:

$|\psi(x)> = \int\limits_{-\infty}^{\infty} dp \ a_n(p) |p>$

But $a_n(p)$ is just the projection of the momentum vector on the wavefunction:

$|\psi(x)> = \int\limits_{-\infty}^{\infty} dp <p|\psi(x)>|p>$

We are also aware from our knowledge of Fourier Transform* that the wave function written in momentum space is given as :

$|\psi(x)> = \frac{1}{\sqrt{2 \pi \hbar}} \int\limits_{-\infty}^{\infty} dp \ \tilde{\psi}(p) |e^{\frac{ipx}{\hbar}}>$

Comparing both the above equations if we take the momentum basis as $|p> = e^{\frac{ipx}{\hbar}}$ , then:

$<p|\psi(x)> = \frac{1}{\sqrt{2 \pi \hbar}} \tilde{\psi}(p)$

We can perform a similar analysis by expanding the wavefunction about the position basis and get

$<x| \tilde{\psi}(p)> = \frac{1}{\sqrt{2 \pi \hbar}} \psi(x)$

** Where does the $\frac{1}{\sqrt{2 \pi \hbar}}$ in the Fourier Transform come from ?

We know from Fourier transform is defined as follows:

$|\psi(x)> = \frac{1}{\sqrt{2 \pi}} \int\limits_{-\infty}^{\infty} dk \ \psi_{1}(k) |e^{ikx}>$

Plugging in $p = \hbar k$ and rewriting the above equation we get,

$|\psi(x)> = A \int\limits_{-\infty}^{\infty} dp \ \tilde{\psi}(p) |e^{\frac{ipx}{\hbar}}>$

We find that from

$\int\limits_{-\infty}^{\infty} dx <\psi(x)| \psi(x)> = 1$

that the normalization constant is not $\frac{1}{\sqrt{2 \pi}}$ but $\frac{1}{\sqrt{2 \pi \hbar}}$ . Therefore,

$|\psi(x)> = \frac{1}{\sqrt{2 \pi \hbar}} \int\limits_{-\infty}^{\infty} dp \ \tilde{\psi}(p) |e^{\frac{ipx}{\hbar}}>$

Cooking up a Lorentz invariant Lagrangian (Part-I)

November 23, 2018December 5, 2018 / 153armstrong / 1 Comment

Let’s consider a scalar field, say temperature of a rod varying with time i.e $T(x,t)$ . (something like the following)

w1hldkr

We will take this setup and put it on a really fast train moving at a constant velocity $v$ (also known as performing a ‘Lorentz boost’).

train-animated-gif-3

Now the temperature of the bar in this new frame of reference is given by $T^{'}(x^{'}, t^{'})$ where,

$x^{'}(x,t) = \gamma \left( x - vt \right)$

$t^{'}(x,t) = \gamma \left( t - \frac{v x}{c^{2}} \right)$

Visualizing the temperature distribution of the rod under a Length contraction.

Temperature is a scalar field and therefore irrespective of which frame of reference you are on, the temperature at each point on the rod will remain the same on both the frames i.e

$T^{'}(x^{'}, t^{'})= T(x, t)$

Therefore we can say that Temperature (a scalar field) is Lorentz invariant. Now what other quantities can we make from T that would also be Lorentz invariant ?

Is $\nabla . T^{'}(x^{'}, t^{'})= \nabla . T(x, t) ?$

Well, let’s give it a try:

$\frac{\partial T}{\partial x} = \frac{\partial T^{'}}{\partial x^{'}} \frac{\partial x^{'}}{\partial x} + \frac{\partial T}{\partial t^{'}} \frac{\partial t^{'}}{\partial x}$

$\frac{\partial T}{\partial x} = \frac{\partial T^{'}}{\partial x^{'}} \gamma - \frac{\partial T}{\partial t^{'}} \gamma v$

$\frac{\partial T}{\partial x} = \gamma \left( \frac{\partial T^{'}}{\partial x^{'}} - \frac{\partial T}{\partial t^{'}} v \right)$

——–

$\frac{\partial T}{\partial t} = \frac{\partial T^{'}}{\partial x^{'}} \frac{\partial x^{'}}{\partial t} + \frac{\partial T^{'}}{\partial t^{'}} \frac{\partial t^{'}}{\partial t}$

$\frac{\partial T}{\partial t} = - \frac{\partial T^{'}}{\partial x^{'}} \gamma v + \frac{\partial T^{'}}{\partial t^{'}} \gamma$

$\frac{\partial T}{\partial t} = \gamma \left( - \frac{\partial T^{'}}{\partial x^{'}} v + \frac{\partial T^{'}}{\partial t^{'}} \right)$

Clearly,

$\frac{\partial T}{\partial x} + \frac{\partial T}{\partial t} \neq \frac{\partial T^{'}}{\partial x^{'}} + \frac{\partial T^{'}}{\partial t^{'}}$

But just for fun let’s just square the terms and see if we can churn something out of that:

$\left( \frac{\partial T}{\partial x} \right)^{2} = \gamma^{2} \left( \frac{\partial T^{'}}{\partial x^{'}} - \frac{\partial T^{'}}{\partial t^{'}} v \right)^{2}$

$\left( \frac{\partial T}{\partial t} \right)^{2} = \gamma^{2} \left( - \frac{\partial T^{'}}{\partial x^{'}} v + \frac{\partial T^{'}}{\partial t^{'}} \right) ^{2}$

We immediately notice that:

$\left( \frac{\partial T}{\partial t} \right)^{2} - \left( \frac{\partial T}{\partial x} \right)^{2} = \gamma^{2} \left[ \left( - \frac{\partial T^{'}}{\partial x^{'}} v + \frac{\partial T^{'}}{\partial t^{'}} \right) ^{2} - \left( \frac{\partial T^{'}}{\partial x^{'}} - \frac{\partial T^{'}}{\partial t^{'}} v \right)^{2} \right]$

$\left( \frac{\partial T}{\partial t} \right)^{2} - \left( \frac{\partial T}{\partial x} \right)^{2} = \left( \frac{\partial T^{'}}{\partial t^{'}} \right)^{2} - \left( \frac{\partial T^{'}}{\partial x^{'}} \right)^{2}$

Therefore in addition to realizing that $T$ is Lorentz invariant, we have also found another quantity that is also Lorentz invariant. This quantity is also written as $\partial_{\mu} T \partial^{\mu} T$ .

Deeper meaning

$S = \int \mathcal{L}( \phi, \partial_{\mu} \phi) dt$

We know that nature is relativistic and when we are are cooking up a Lagrangian for a theory, we better make sure that it is Lorentz invariant as well. What the above analysis on scalar fields hints us is that the Lagrangian for such a theory can contain terms like $\partial_{\mu} \phi \partial^{\mu} \phi$ in it as the quantity does not change under a Lorentz transformation.

This discussion finds a deeper ground in Quantum Field Theory. For example if $\phi$ is a scalar field, then a Lorentz invariant Lagrangian could take any of the following possible forms:

$\mathcal{L} = \phi , \partial_{\mu} \phi \partial^{\mu} \phi, \partial_{\mu} \phi \partial^{\mu} \phi - m^{2} \phi^2 , \partial_{\mu} \phi \partial^{\mu} \phi - m^{2} \phi^2 + g \phi^4, \hdots$

All of them keep the action $S$ invariant under a Lorentz transformation.

Ansatz to Gram-Schmidt Orthonormalization

November 18, 2018November 18, 2018 / 153armstrong / Leave a comment

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

April 18, 2018April 18, 2018 / 153armstrong / Leave a comment

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 !

June 7, 2017June 9, 2017 / 153armstrong / Leave a comment

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.

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

June 6, 2017June 6, 2017 / 153armstrong / Leave a comment

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

$z^n = 1$

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