# Solving the Euler-Lagrange equations for a complex scalar field in which the time derivatives and gradient are separate

This is found at the bottom of page 9 of David Tong's QFT lectures. The Euler-Lagrange equations for the complex scalar field:

$$\mathcal L=\frac{i}{2}(\psi^*\dot\psi-\dot{\psi^*}\psi)-\nabla\psi^*\cdot\nabla\psi-m\psi^*\psi \tag{1.15}.$$

However, to obtain the equation of motion for $$\psi^*$$, we need the following derivatives of the Lagrangian density:

$$\partial_\mu\left(\frac{\partial\mathcal L}{\partial(\partial_\mu\psi^*)}\right),\quad \frac{\partial\mathcal L}{\partial\psi^*}.$$

In the notes, the following derivatives are instead used:

$$\frac{\partial\mathcal L}{\partial\psi^*},\quad \frac{\partial\mathcal L}{\partial\dot{\psi^*}}, \quad \frac{\partial\mathcal L}{\partial\nabla\psi^*}. \tag{1.16}$$

I don't see why we can use these instead (ie. take the derivative of $$\mathcal L$$ wrt. the time derivative and the gradient of $$\psi^*$$ separately and then combine them, which I think is what's being done here).

The double $$\mu$$ index is being summed over, so if you write it out, you get
$$\partial_\mu\left(\frac{\partial\mathcal L}{\partial(\partial_\mu\psi^*)}\right) =\sum_{\mu=0}^3\partial_\mu\left(\frac{\partial\mathcal L}{\partial(\partial_\mu\psi^*)}\right) = \partial_0\left(\frac{\partial\mathcal L}{\partial(\partial_0\psi^*)}\right)+\sum_{i=1}^3\partial_i\left(\frac{\partial\mathcal L}{\partial(\partial_i\psi^*)}\right) =\partial_t\left(\frac{\partial\mathcal L}{\partial(\dot{\psi^*})}\right)+\nabla\left(\frac{\partial\mathcal L}{\partial(\nabla\psi^*)}\right)$$
If you take the Euler-Lagrange equation you have the following term $$\partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\psi^*)}\right)\tag{1}$$ in which the $$\mu$$ index is summed over. This means that the quantity $$(1)$$ has to be thought as $$\partial_0\left(\frac{\partial\mathcal{L}}{\partial(\partial_0\psi^*)}\right)+\partial_1\left(\frac{\partial\mathcal{L}}{\partial(\partial_1\psi^*)}\right)+\partial_2\left(\frac{\partial\mathcal{L}}{\partial(\partial_2\psi^*)}\right)+\partial_3\left(\frac{\partial\mathcal{L}}{\partial(\partial_3\psi^*)}\right)\tag{2}$$ but since $$0\equiv t$$, $$1\equiv x$$, $$2\equiv y$$, $$3\equiv z$$, the last three elements of $$(2)$$ are just the gradient of the derivative of the lagrangian wrt the gradient of $$\psi$$ $$\nabla\left(\frac{\partial\mathcal{L}}{\partial\nabla\psi^*}\right)$$ Moreover, given that $$\partial_0\equiv\partial_t$$ the same can be said to the first term of $$(2)$$ which is going to be $$\partial_t\left(\frac{\partial\mathcal{L}}{\partial\dot{\psi}^*}\right)$$ With this you get your result