I would like to consider the problem of the total derivative of a given functional L[ϕ(x,y,z,t),∂ϕ∂x(x,y,z,t),∂ϕ∂y(x,y,z,t),∂ϕ∂z(x,y,z,t),∂ϕ∂t(x,y,z,t),x,y,z,t],
However, before expressing my inquiry about the problem itself, I will make a brief preamble as motivation, or warm-up, as you wish. All the exhibition considered here takes into account that all functions are continuous and differentiable in any order, that is, they are all C∞ class.
Let us consider the case in which z is a function of two variables x and y, say z=f(x,y), while x and y, in turn, are functions of two variables u and v, so that x=g(u,v) and y=h(u,v). Then z becomes a function of u and v, namely, z=f(g(u,v),h(u,v))=f(u,v). Here, we consider u and v as independents variables.
As we know, the total differential of the function z=f(x,y) with respect to x and y is given by dz=∂f∂xdx+∂f∂ydy,
Now, let us replacing (I.3) and (I.4) in (I.2), such that now we have dz=(∂f∂x∂g∂u+∂f∂y∂h∂u)du+(∂f∂x∂g∂v+∂f∂y∂h∂v)dv.
Thus, knowing that the total differential of z with respect to u and v is given by dz=∂f∂udu+∂f∂vdv,
If the answer will be yes, and I think that this to be the answer, so, from Eq. (I.6), it is valid that dzdu=∂f∂ududu+∂f∂vdvdu=∂f∂uanddzdv=∂f∂ududv+∂f∂vdvdv=∂f∂v.
The situation is similar when we are considering coordinates transformation of type: x′=f(x,y,z,t),y′=g(x,y,z,t),z′=h(x,y,z,t),t′=w(x,y,z,t),
To finalize this preamble, which already is very long and tiresome, let us consider that the variables x, y and z have dependence with t, i.e., we have x(t), y(t) and z(t), so we can write: dx′dt=∂f∂xdxdt+∂f∂ydydt+∂f∂zdzdt+∂f∂t,
After this exhaustive exposition, I want to turn back in the original problem of the functional (I.1), whose total differential is given by dL=∂L∂ϕdϕ+∂L∂(∂iϕ)d(∂iϕ)+∂L∂xdx+∂L∂ydy+∂L∂zdz+∂L∂tdt.
To conclude, I would like to justify this exposition, and its inquiries, saying that the problem arises when I try to derive the Noether theorem. In a certain passage, similar terms arose, suggesting the use of a total derivative. However, I was unsure whether such a procedure would be correct or valid.
See Does it make sense to speak in a total derivative of a functional? Part II to additional motivation.
Answer
Consider for simplicity a single real scalar field ϕ:R4→R
on a 4-dimensional spacetime R4. The Lagrangian density L: R×R4×R4 → Ris a differentiable function. We can construct partial derivatives of the Lagrangian density L wrt. any of its 1+4=4=9 arguments. See also this & this related Phys.SE posts.The integrand ϕ∗L:R4→R
of the action functional S[ϕ] := ∫R4d4x (ϕ∗L)(x)is the pullback x ↦ (ϕ∗L)(x) := L(ϕ(x),∂ϕ(x),x)of the Lagrangian density L by the field ϕ.The derivative x ↦ d(ϕ∗L)(x)dxμ
of the pullback (E) is by definition the total derivative [wrt. the spacetime coordinate xμ].Be aware that physics texts usually don't bother to spell out the difference between the Lagrangian density L and its pullback ϕ∗L, either in words or notation. It is implicitly understood.
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