## Cooking up a Lorentz invariant Lagrangian (Part-I)

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

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’).

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

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

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

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.

## Euler-Lagrange equations – Field Theory

Consider a function $\phi(x)$ where $x$ is a point in spacetime. If we assume that the Lagrangian is dependent on $\phi(x)$ and its derivative $\partial_{\mu} \phi(x)$. The action is then given by,

$S = \int d^{4}x \mathcal{L}(\phi(x) , \partial_{\mu} \phi(x))$

According to the principle of least action, we have:

$\delta S = 0 = \int d^{4}x \delta \mathcal{L}(\phi(x) , \partial_{\mu} \phi(x))$

$\delta \mathcal{L}(\phi(x) , \partial_{\mu} \phi(x)) = \frac{\partial \mathcal{L}}{\partial \phi} \delta \phi + \frac{\partial \mathcal{L}}{\partial( \partial_{\mu} \phi)} \delta(\partial_{\mu} \phi)$

$\delta \mathcal{L}(\phi(x) , \partial_{\mu} \phi(x)) = \frac{\partial \mathcal{L}}{\partial \phi} \delta \phi + \frac{\partial \mathcal{L}}{\partial( \partial_{\mu} \phi)} \partial_{\mu} (\delta\phi)$

According to product rule,

$\frac{\partial \mathcal{L}}{\partial( \partial_{\mu} \phi)} \partial_{\mu} (\delta\phi) + \partial_{\mu} \left( \frac{\partial \mathcal{L}}{\partial( \partial_{\mu} \phi)} \right) \delta\phi = \partial_{\mu} \left( \frac{\partial \mathcal{L}}{\partial( \partial_{\mu} \phi)} \delta\phi \right)$

Therefore,

$\delta \mathcal{L}(\phi(x) , \partial_{\mu} \phi(x)) = \left( \frac{\partial \mathcal{L}}{\partial \phi} - \partial_{\mu} \left( \frac{\partial \mathcal{L}}{\partial( \partial_{\mu} \phi)} \right) \right)\delta \phi + \partial_{\mu} \left( \frac{\partial \mathcal{L}}{\partial( \partial_{\mu} \phi)} \delta\phi \right)$

$\delta S = \int d^{4}x \left[ \left( \frac{\partial \mathcal{L}}{\partial \phi} - \partial_{\mu} \left( \frac{\partial \mathcal{L}}{\partial( \partial_{\mu} \phi)} \right) \right)\delta \phi + \partial_{\mu} \left( \frac{\partial \mathcal{L}}{\partial( \partial_{\mu} \phi)} \delta\phi \right) \right]$

The second term in that equation is zero because the end points of the path are fixed i.e

This then leads to the Euler-Lagrange equation for a field as :

$\frac{\partial \mathcal{L}}{\partial \phi} = \partial_{\mu} \left( \frac{\partial \mathcal{L}}{\partial( \partial_{\mu} \phi)} \right)$

## 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  $|v_{1}> , |v_{2}> , \hdots |v_{n}>$.

$|v_{1}^{'} > = |v_{1}>$

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 $ = 0$ .

$ = - \lambda $

$0 = - \lambda$

$\lambda = $

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

$|v_{2}^{'} > = |v_{2}> - |v_{1}>$

which upon normalization looks like so:

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

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} = $

$\lambda_{2} = $

Therefore we get,

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

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

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

Generalizing, we obtain:

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

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

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!

## Parallax method, 61-Cygni and the Hipporcas mission

It is trivial for most astronomy textbooks to illustrate the parallax method as follows:

This is absolutely fascinating, but it was really hard to find actual images of stars in books that illustrate this.

This is the proper motion of 61-Cygni, a binary star system over a span of couple of years.

But Bessel discovered that in addition to this proper motion, 61-Cygni also wobbled a little bit from side to side because of the parallax during observation.

The following is a plot of the motion of 61 Cygni – A which beautifully  elucidates the proper motion and the effect of parallax (i.e the wiggle of the blue line with respect to the mean free path)

In addition, if you would like to actually play around with data for yourself, the The Hipparcos Space Astrometry Mission might interest you a lot. The mission was Launched in August 1989 and successfully observed the celestial sphere for 3.5 years before operations ceased in March 1993 employ

The documentation and the catalogue are fairly clear ,  instructive and easy to use. Have fun!

## 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!

## Fibonacci sequence in the hiding

What ?!! There exists such an elegant decimal representation of the Fibonacci sequence? Well yes! and the only thing that you need to know to prove this is that if the Fibonacci numbers were the coefficients to a power series expansion, then the Fibonacci generating function is given as follows:

$1x + 1x^{2} + 2x^{3} + 3x^{4} + 5x^{5} + \hdots = \frac{x}{1-x-x^{2}}$

Subsituting the value of $x = \frac{1}{10}$, we get :

$\frac{1}{10} + \frac{1}{10}^{2} + 2(\frac{1}{10})^{3} + 3(\frac{1}{10})^{4} + 5(\frac{1}{10})^{5} + \hdots = \frac{\frac{1}{10}}{1-\frac{1}{10}-\frac{1}{10}^{2}}$

$0.1 + 0.01 + 0.002 + 0.0003 + 0.00005 + \hdots = \frac{10}{89}$

$0.01 + 0.001 + 0.0002 + 0.00003 + 0.000005 + \hdots = \frac{1}{89}$

Proved. 😀

## Prof.Ghrist at his best!

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.

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 😀