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.

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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.

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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.

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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.

 

 

 

 

 

 

 

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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)

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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

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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)  

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!

 

 

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 😀

 

On the direction of the cross product of vectors

One of my math professors always told me:

Understand the concept and not the definition

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 :

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.

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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.

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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 ?

crossproduct

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 ?

cross-product-simple

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!