Why do these terms vanish when applying Euler-Lagrange to the Dirac Lagrangian?

[JohnH]

TL;DR

Looking for a proof of
∂(∂L/∂[overline]Ψ[\overline])=0
and
∂L/∂Ψ=-[overline]Ψ[\overline]m
and
∂L/∂(∂[SUB]μ[/SUB]Ψ)=i[overline]Ψ[\overline]γ[SUP]μ[/SUP]

I'm having trouble following a proof of what happens when the Dirac Lagrangian is put into the Euler-Lagrange equation. This is the youtube video: and you can skip to 2:56 and pause to see all the math laid out. I understand the bird's eye results of the Dirac Lagrangian having an equation of motion that is either the Dirac equation or a [overline]Ψ[\overline] version of the Dirac equation, but I'm unclear about why, when putting the Dirac Lagrangian into the Euler-Lagrange equation, the following pieces are:

∂(∂L/∂[overline]Ψ[\overline])=0
and
∂L/∂Ψ=-[overline]Ψ[\overline]m
and
∂L/∂(∂_μΨ)=i[overline]Ψ[\overline]γ^μ

Thanks for all replies.

 

[div_grad]

You get it by direct computation. The equations of motion for a field $\psi$ described by some Lagrangian $\mathcal{L}$ are the Euler-Lagrange equations $$\frac{\partial\mathcal{L}}{\partial\psi} - \frac{\partial}{\partial x^\mu}\frac{\partial\mathcal{L}}{\partial(\partial_\mu\psi)}=0 \rm{.}$$ In units of $\hbar=c=1$, the Dirac Lagrangian can be written as $$\mathcal{L} = \bar{\psi}\left(\mathrm{i}\gamma^\mu\partial_\mu - m\right)\psi = \bar{\psi}\mathrm{i}\gamma^\mu\partial_\mu\psi - m\bar{\psi}\psi \rm{.}$$ Now, you can plug this Lagrangian into the equations of motion above in which you differentiate with respect to $\psi$ - in this way you will obtain the equations of motion for the field $\bar{\psi}$. Alternatively, you can differentiate with respect to $\bar{\psi}$ in the Euler-Lagrange equations, thereby obtaining the standard Dirac equation (i.e., the equations of motion for the field $\psi$).

Observe that in the Lagrangian there is no term of the form $\partial_\mu\bar{\psi}$, hence you get $$\frac{\partial\mathcal{L}}{\partial(\partial_\mu\bar{\psi})}=0 \rm{,}$$ which shows you that $\frac{\partial\mathcal{L}}{\partial\bar{\psi}}=0$ - this is one of the things you asked about. You can work out the rest of the identities from your Question yourself in the same manner.