# How is the coefficient of the kinetic term in a lagrangian determined?

This question comes from Srednicki's textbook "Quantum Field Theory". On page 517, the lagrangian for scalar electrodynamics is given by
$$\mathcal{L} = -(D^{\mu}\varphi)^{\dagger} D_{\mu}\varphi - V(\varphi) - \frac{1}{4} F^{\mu\nu} F_{\mu\nu} , \tag{85.1}$$ where $\varphi$ is a complex scalar field and $D_{\mu} = \partial_{\mu} -igA_{\mu}$.

On page 523, the lagrangian for a general nonabelian gauge theory is given by $$\mathcal{L} = -\frac{1}{2} D^{\mu}\phi D_{\mu}\phi - V(\phi) - \frac{1}{4} F^{a\mu\nu} F^{a}_{\mu\nu} \tag{86.1}$$ where $\phi_{i}$ are real scalar fields; and $$(D_{\mu}\phi)_{i} = \partial_{\mu}\phi_{i} - ig_{a} A_{\mu}^{a}(\mathcal{T}^{a})_{ij} \phi_{j} \tag{86.2}$$ is the covariant derivative, and the adjoint index $a$ runs over all generators of all gauge groups.

My question is: Why is there -1 in front of the kinetic term in eq. (85.1), while there is $-\frac{1}{2}$ in the same place in eq. (86.1)? In general, how is the coefficient determined?

Normalization.

The action (and the Lagrangian) of a theory are defined up to a normalization factor (and other things, like total derivatives, but let us focus on multiplicative factors). There is no $\textit{correct}$ way to choose this normalization, you just have to stick with one and be coherent.

Suppose you have a real scalar field $\phi$: a Lagrangian for such a field will be $$-\frac12\partial_\mu\phi\partial^\mu\phi-V(\phi).$$ When you derive w.r.t. $\partial_\mu\phi$, you have two fields to derive, and the factor of $\frac12$ gets cancelled in the motion equations: $$-\partial_\mu\partial^\mu\phi=-\frac{\partial V}{\partial \phi}.$$ With this normalization, the LHS has no multiplicative factors, just the Laplacian of $\phi$.

Now let $\varphi$ be a complex scalar field. You interpret $\varphi$ and $\varphi^*$ to be two independent fields: when you vary with respect to one, you do not vary with respect to the other. A typical Lagrangian is now $$-\partial_\mu\varphi\partial^\mu\varphi^*-V(\varphi,\varphi^*),$$ and motion equations are obtained by deriving w.r.t. $\varphi$ and $\varphi^*$: $$-\partial_\mu\partial^\mu\varphi^*=-\frac{\partial V}{\partial\varphi},\\ -\partial_\mu\partial^\mu\varphi=-\frac{\partial V}{\partial\varphi^*}.$$ You do not have the factor $2$ in front of the kinetic term, as you do not need it to properly normalize the LHS of the equation. Now, try to plug $$\varphi=\frac{\phi_1+i\phi_2}{\sqrt 2}$$ and analogous for $\varphi^*$ and see what happens.

About the sign: if you follow Srednicki, you're using $(-,+,+,+)$ as metric tensor. The minus in the front of the kinetic term is needed to have a definite kinetical energy. Remember that you can interpret the action as $S=\int(T-V)dt$: the terms with two time derivatives compose the kinetic energy, that enters with its sign in the action. The kinetic term must be a square of something, to have positive kinetic energy. To find it, we can write $$-\partial_\mu\phi\partial^\mu\phi=(\partial_0\phi)^2-(\partial_i\phi)^2$$ Now, the terms in the parentheses are squares, so they are positive. The first parenthesis enters in the kinetical energy, and it's positive as we wanted it to. Notice that, if you had the metric $(+,-,-,-)$, you would have needed to change the sign of the first term in your Lagrangians: you always want a positive kinetic energy.

1. Usually the coefficient in front of the time derivative squared is normalized to a $\frac{1}{2}$.

2. The minus signs "$-$" in eqs. (85.1) & (86.1) are because of $(-,+,+,+)$ Minkowski sign convention.

3. The "$1$" in eq. (85.1) is because the field $\phi=\frac{1}{\sqrt{2}}(\phi_1+i\phi_2)$ is complex. In contrast, the field $\phi$ in eq. (86.1) is real.

4. See e.g. this Phys.SE post for an explanation of the $-\frac{1}{4}$ in front of the ${\rm Tr}F^2$ term.

5. More generally, a monomial of the form $\prod_{i=1}^n X_i^{m_i}$ in the Lagrangian density is often normalized as $\prod_{i=1}^n \frac{X_i^{m_i}}{m_i!}$. Here $m_i\in \mathbb{N}_0$ is the multiplicity of $X_i$.