# Lagrange formalism in field theory

I recently had a discussion with a friend of mine who is like me studying physics. And we might got used to a misconception about the Lagrange-Formalism in field theory. In common field theory books one just states that the action is given by $$\mathcal S[\phi] = \int \mathcal L (\phi(x), \partial_\mu\phi(x),x)\, \text d ^4 x$$ where $$\phi$$ is some field and $$x$$ is a point in space time. In classical mechanics we learn that the action is given by $$\mathcal S[q] = \int_a^b L(q(t), \dot{q}(t),t)\, \text d t$$ This functional maps a trajectory $$t\mapsto q(t)$$ to a scalar $$\mathcal S$$ where the path which is taken by the object is the one wich yields the minimal value of $$\mathcal S$$. So in field theory I always thought we would do something quite similar. We would search a field configuration $$\phi(x)$$ which minimizes the value computed by $$\mathcal S[\phi]$$. To do so we would integrate over all space time $$\mathbb R \times \mathbb R ^3$$. Or would we rather integrate over all space $$\mathbb R ^3$$ and then over a time interval $$[a, b]$$, since we need some kind of boundary of the field configuration to later perform the variation of $$\mathcal S$$? Or do i get everything wrong and we really describe the propagation of a field $$\phi$$ from one point in space-time to another?

I am actually very confused about a thing which i thought i understood pretty well and i am very thankful for every insight. Once again i am very shocked that one tends to get used to formalism even it is not understood deeply.

• Classically, the action is equivalent to an equation of motion for $\phi$. However, we should define the domain that we want to take the infimum (i.e. boundary condition). As an example, we can think of having $\phi$ fixed at all "space" points for given "times" $t_i$ and $t_f$ (that implicitly implies zero otherwise). Thus, the integral is understood as a functional over "this" set of functions. It also fixes the domain of integration (which can be said is always the whole space): an interval in "time" and all space, since the other parts are zero (you could define other bound. cond.). May 9, 2020 at 13:22

If you already know point mechanics, then one way to build intuition for field theory is to consider the space variable $$\vec{x}$$ of spacetime as a continuous index $$j$$. Even better: discretize space $$\mathbb{R}^3$$ with some lattice parameter $$a$$ that eventually is taken to zero. (But keep the time interval $$[t_i,t_f]$$ continuous.) Then the field $$\phi(t,\vec{x})=q^j(t)$$ becomes like infinitely many point particles. Here space derivatives $$\partial/\partial \vec{x}$$ are replaced by corresponding discretized derivatives, space integrals with corresponding lattice sums, etc. Then the field theoretic action $$S[\phi] =S[q]$$ becomes like a point particle action. For an enlightening example, see e.g. H. Goldstein, Classical mechanics, section 13.1.
The action $$S[\phi]$$ is a functional of the map $$\phi:{\mathbb R}^4\to {\mathbb R}$$, which we seek to make stationary under variations $$\delta \phi$$ that vanish at infinity.
We can think of as a history $$\phi(x,t):{\mathbb R}^3\times {\mathbb R}\to {\mathbb R}$$. In principle we should do the same as in classical mechanics: choose a beginning and end time $$t_1$$ and $$t_2$$ and make variations such that $$\delta \phi(x, t_1)= \delta \phi(x,t_2)=0$$ for all $$x\in {\mathbb R}^3$$. In practice we never mention that we are doing this, but instead just ignore any integrated out parts when computing the variation. The variations at spatial infinity are not required to be zero from the classical viewpoint and they can give rise boundary conditions on $$\phi$$.
In field theory, action us defined as $$S[x] = \int L(φ_a, ∂_µφ_a)\, \text dt$$ = $$\int L'd^3x dt$$ = $$\int L'd^4x$$, where L is Lagrangian and L' is Lagrangian density. Then we vary action and require that $$\delta S=0$$ by the principle of least action. This variation is taken under the boundary condition that $$\delta \phi(x, t_1)= \delta \phi(x,t_2)=0$$. It leads to familiar Euler Lagrange equations which is same in the form as that in classical mechanics with an additional change that the independent parameter is not position anymore but fields.