Cooking up a Lorentz invariant Lagrangian

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) 


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 .

** There is a very important reason why this quantity did not work out.  This post was inspired in part by Micheal Brown’s answer on stackexchange . I request the interested reader to check that post for a detailed explanation.

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) 


\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)