Tensorial Propagators (Part II)

In Part I, we found the central issue when dealing with tensorial propagators $latex -$ they are usually non-invertible. So then how do physicists obtain these propagators? In order to obtain these propagators, they usually exploit gauge invariance. Here, we will discuss what gauge invariance is and continue with our analysis of the photon propagator.

Before we employ gauge invariance, we should have a discussion on what gauge invariance is and the origins of its usefulness.

If a Lagrangian is invariant under continuous group of local transformations, then the gauge fields included in the Lagrangian are said to have a gauge invariance.

Let’s break down this statement and build it up piece by piece. We will first study an example of how an invariant quantity can arise from a continuous group of local transformations. Then we will inspect how this may hold true for Lagrangians using the Noether’s Theorem. Finally, we will connect our investigations to gauge fields.


Continuous Groups and Invariant Quantities

An example (see figure below) of an invariant quantity under a continuous group of local transformations is simply the length, or modulus, of a vector.

Active_Rotation
Active rotation of a vector, A, which clearly preserves the length of the vector. Image credit: Mathematical Methods of Physics: Tutorial, 5th edition, Richard J. Jacob, Arizona State University.

To elaborate on the figure, the vector, $latex \textbf{A}$, is transformed by the active rotation matrix, given by

$latex \displaystyle \text{R}(\varphi) = \left[ \begin{array}{cc} \cos(\varphi) & \sin(\varphi) \\ -\sin(\varphi) & \cos(\varphi) \end{array} \right].$

Recall that $latex \displaystyle \text{R}(\varphi)\in \text{SU}(2)$, that is, $latex \displaystyle \det(\text{R}(\varphi))=1$, $latex \displaystyle \text{R}(\varphi)$ is a $2\times2$ matrix, and $latex \displaystyle \text{R}^\dagger\text{R}=\mathbb{I}$. Here, the continuous group of local transformations is, in fact, represented by  $latex \text{R}(\varphi)$. While it may not be immediately obvious that the length should remain invariant upon rotation, we can quickly perform a check by inspecting and comparing the modulus of both $latex \textbf{A}^\prime$ and $latex \textbf{A}$ as shown below;

$$\begin{align*} \|\textbf{A}^\prime\| = \sqrt{(\mathbf{A}^\prime)^\dagger(\textbf{A}^\prime)} &= \sqrt{(\text{R}\textbf{A})^\dagger(\text{R}\textbf{A})} = \sqrt{\textbf{A}^\dagger (\text{R}^\dagger \text{R}) \textbf{A}} = \sqrt{\textbf{A}^\dagger\mathbb{I}\textbf{A}}\\ &\therefore  \|\textbf{A}^\prime\| = \|\textbf{A}\|. \end{align*}$$

Thus we have shown how the length of a vector may be invariant under rotations.


Lagrangians and Noether’s Theorem

To extend the above example to Lagrangians, we consider Noether’s Theorem formally defined as;

If a system, or its Lagrangian Density, has a continuous symmetric property, then there exists an associated conservation law.

This statement[1] is practically identical to the above statement about gauge invariance, albeit a little easier to understand. There is enough substance to Noether’s Theorem that it deserves its own detailed explanation — perhaps in the future. For now, however, we will consider an example of how a continuous symmetric property can bring forth a conserved quantity by studying the simplest case; the free-particle Lagrangian. The free particle Lagrangian is given by

$latex \displaystyle L(q,\dot{q},t)=\frac{1}{2}m\dot{q}^2$,

then the derivatives of the Lagrangian become

$latex \displaystyle \frac{\partial L}{\partial q} = \frac{\partial}{\partial q}\left(\frac{1}{2} m\dot{q}^2\right) = 0\text{, while } \frac{\partial L}{\partial \dot{q}} =\frac{\partial}{\partial \dot{q}}\left(\frac{1}{2} m\dot{q}^2\right) = m\dot{q}.$

Then the Euler-Lagrange equations become

$latex \displaystyle \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right) = \boxed{\frac{d}{dt}\left(m\dot{q}\right)=0}=\frac{\partial L}{\partial q}.$

The boxed part can be immediately integrated to read $latex m\dot{q} = k$, for some real number $latex k$. Therefore $latex p\equiv m\dot{q}$ is a conserved quantity. To summarize, $latex q(t)$ being a cyclic variable, as the Lagrangian does not explicitly depend on it, had a corresponding conservation law, specifically, the conservation of momentum. Now, to find the corresponding symmetry for the above case, recall that by continuous symmetry we mean any infinitesimal transformation of the Lagrangian’s coordinates such that it leaves the value of the Lagrangian unchanged. As the position coordinate, $latex q(t)$, here is the cyclic variable, we will transform $latex q(t)\to q^\prime(t)=q(t)+\epsilon$ and check the variation, or change, in the Lagrangian;

$latex \displaystyle \begin{aligned}\delta L &\equiv L(q(t)+\epsilon,\dot{q}(t),t) -L(q(t),\dot{q}(t),t)\\ &\approx L(q(t),\dot{q}(t),t) + \epsilon\frac{\partial L}{\partial q} – L(q(t),\dot{q}(t),t) \\ &=\epsilon\frac{\partial L}{\partial q} \end{aligned}.$

Since $latex q(t)$ is a cyclic variable, the derivative of the Lagrangian with respect to $latex q(t)$ vanishes and thus we have shown that the continuous symmetry of a cyclic variable not only produces an invariant quantity, which in this case is the Lagrangian, but also has an associated conservation law.


Before we continue onto gauge fields, and in order to hammer in the analogy, we will link the two examples in the table below.

Matrix Example Lagrangian Counterpart
$latex \varphi$ is the continuous variable $latex q(t)$ is a cyclic variable
$latex R(\varphi)$ belongs to a continuous group $latex q(t)$ has a continuous symmetry
The length is invariant The Lagrangian is invariant

With these examples fleshed out, we will now turn our focus to how gauge fields display this invariance and how to obtain the photon propagator in Part III.

It would be silly to imply that the conservation of momentum is akin to preservation of the Pythagorean theorem in flat spacetime, but we will leave this after thought here because it is fun to think about.

[1] The literature on Noether’s Theorem is both vast and technical, but, in my opinion, Goldstein’s Classical Mechanics provides a detailed, student-friendly, pedagogical treatment.


Credit: Image found on Wikipedia. Author is Sanpaz.

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