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!

 

 

 

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Cooking up a Lorentz invariant Lagrangian (Part-II)

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

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)

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) 

lorentz_boost
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.

 

 

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

20181118_210223

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}>  .

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):

 

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.



61 Cygni showing proper motion at one year intervals

 

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)

         Source

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!