Monday, June 2, 2014

lagrangian formalism - Does it make sense to speak in a total derivative of a functional? Part I



I would like to consider the problem of the total derivative of a given functional L[ϕ(x,y,z,t),ϕx(x,y,z,t),ϕy(x,y,z,t),ϕz(x,y,z,t),ϕt(x,y,z,t),x,y,z,t],

where all variable are independent each other.


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=fxdx+fydy,

while the total differential of the functions x and y with respect to u and v are given by dx=gudu+gvdv,dy=hudu+hvdv.


Now, let us replacing (I.3) and (I.4) in (I.2), such that now we have dz=(fxgu+fyhu)du+(fxgv+fyhv)dv.


Thus, knowing that the total differential of z with respect to u and v is given by dz=fudu+fvdv,

we can, by direct comparison, to conclude that zu=fxgu+fyhu,zv=fxgv+fyhv.
And here arises my first question: Does it make sense to speak of the total derivative of z in relation both of the two variables u and v?


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=fududu+fvdvdu=fuanddzdv=fududv+fvdvdv=fv.

If the answer will be no, so the notation dz/du and dz/dv cannot be used and we can only speak in the validity of the equations (I.7) and (I.8). Here, zufuandzvfv.


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),

where the set of prime coordinates are independent of each other. Similarly, the set of coordinates without prime are also independent of each other. Thus, such that the total differential are dx=fxdx+fydy+fzdz+ftdt,dy=gxdx+gydy+gzdz+gtdt,dz=hxdx+hydy+hzdz+htdt,dt=wxdx+wydy+wzdz+wtdt,
and so, we have found to case of x, for example, that dxdx=fx,dxdy=fydxdz=fz,anddxdt=ft.
And again, we ask ourselves: is it valid to use the d/dx, d/dy, d/dz and d/dt notation, since the function x has a dependence on the variables x, y, z and 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: dxdt=fxdxdt+fydydt+fzdzdt+ft,

where, by simplicity, we have consider only the total derivative to x. Obviously, that y, z and t have analog equations. If x, y, z and t are not explicitly dependents of t variable, so, of course, ft=gt=ht=wt=0.
We also point out that Eq. (I.10) can be rewritten as dx=(fxdxdt+fydydt+fzdzdt+ft)dt=dfdtdt.


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=Lϕdϕ+L(iϕ)d(iϕ)+Lxdx+Lydy+Lzdz+Ltdt.

Here, we can immediately think in the total derivative as (I will do the exposition only the x variable.) dLdx=Lϕϕx+L(iϕ)(iϕ)x+Lx,
once that x, y and z are independents each other. However, if remember that dϕ=ϕxdx+ϕydy+ϕzdz+ϕtdt,d(iϕ)=(iϕ)xdx+(iϕ)ydy+(iϕ)zdz+(iϕ)tdt,
we can, instead of we immediately write equation (I.13), rewritten the equation (I.12) as dL=(Lϕϕx+L(iϕ)(iϕ)x+Lx)dx+(Lϕϕy+L(iϕ)(iϕ)y+Ly)dy+(Lϕϕz+L(iϕ)(iϕ)z+Lz)dz+(Lϕϕt+L(iϕ)(iϕ)t+Lt)dt
Here's the dilemma! Since ϕ and iϕ are functions of the variables x, y, z and t, and, in addition, the functional L itself depends explicitly on these same variables, we could then think that the functional L is implicitly a function of the variables x, y, z and t, and, therefore, L=L(x,y,z,t). If so, then the total implicit'' the total differential of L would be given by dL=Lxdx+Lydy+Lzdz+Ltdt
But this is not right since it contradicts the Eq. (I.12)! Based on this contradiction, I ask: who are the terms in parentheses in Eq. (I.14)? Is it possible to speak in a total derivative of the functional L?



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





  1. Consider for simplicity a single real scalar field ϕ:R4R

    on a 4-dimensional spacetime R4. The Lagrangian density L: R×R4×R4    R
    is 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.




  2. The integrand ϕL:R4R

    of the action functional S[ϕ] := R4d4x (ϕL)(x)
    is the pullback x    (ϕL)(x) := L(ϕ(x),ϕ(x),x)
    of the Lagrangian density L by the field ϕ.





  3. The derivative x    d(ϕL)(x)dxμ

    of the pullback (E) is by definition the total derivative [wrt. the spacetime coordinate xμ].




  4. Be aware that physics texts usually don't bother to spell out the difference between the Lagrangian density L and its pullback ϕL, either in words or notation. It is implicitly understood.




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