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Quantum Field Theory lecture by Tong (Cambridg)

  1. Apr 15, 2012 #1
    I am struggling with equation 1.5 in Tong's QFT course. I try to understand/explain it in strict calculus, i.e. without physics shortcuts like "small variations". I guess in the full blown explanation, [itex]\delta S[/itex] is a total derivative.

    To be specific, with total derivative I mean the linear map that best approximates a given function [itex]f[/itex] at a given point. For [itex]f:ℝ\toℝ[/itex] we have [itex]D(f,x_0):ℝ\toℝ[/itex], i.e. [itex]D(f,x_0)(h) \in ℝ[/itex]. Often it is also denoted as just [itex]\delta f[/itex].

    In terms of total derivative, I wonder if the following simplification of Tong's equation 1.5 still catches what is going in in mathematical terms (no longer physical). Let [itex]S_{a,b}(f) = \int_a^b f(x)dx[/itex] a functional that maps functions [itex]f[/itex] to the real line. Then [itex]D(S_{a,b},f) = \delta S_{a,b}[/itex] should be well defined given any necessary smoothness conditions. In particular [itex]D(S_{a,b},f)[/itex] maps functions [itex]h[/itex] of the same type of [itex]f[/itex] to real numbers. Because the integral is linear, so my hunch, its best linear approximation should be itself. InTong's course, equation 1.5, first line, I find what I understand to be

    [tex]\delta \int_a^b f(x) dx = \int_a^b \delta f dx[/tex]

    Can anyone explain how the algebraic types on the left and on the right would match up? My interpretation is, that on the left I have a the total derivative of a functional, which itself should be a functional, written explicitly as [itex]D(S_{a,b},f)[/itex]. On the right I have the integral over, hmm, the total derivative of [itex]f[/itex], where I don't see how this could be a functional?

    Any hints appreciated.
  2. jcsd
  3. Apr 15, 2012 #2


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  4. Apr 16, 2012 #3


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    Mathematicians always struggle to understand the meaning of the symbol [itex]\delta[/itex] which often used by physicists. I found that puzzling, after all the “calculus of variations” is a very old mathematical discipline.
    With the help of [itex]\delta[/itex], physicists can do in two lines what mathematicians would do in two pages! This is why physicists are so keen on using [itex]\delta[/itex]. To demonstrate this to you, I (pretending to be a mathematician) will derive the Euler-Lagrange equations the same way some mathematicians do it using the calculus of variations.
    First, we need to define the Lagrange function for a relativistic classical field. Let [itex]\bar{D}\subset \mathbb{R}^{4}[/itex] be a bounded region corresponding to a bounded space-time region [itex]\bar{U}\subset M^{4}[/itex]. [itex]D[/itex] is an open subset of [itex]\mathbb{R}^{4}[/itex] and [itex]U \subseteq M^{4}[/itex]. Let [itex]f: \ \bar{D} \rightarrow \mathbb{R}^{s}[/itex] be a twice differentiable function, i.e., [itex]f \in \mathcal{l}^{2}(\bar{D},\mathbb{R}^{s})[/itex]. Let [itex]\phi^{A}\equiv \pi^{A}\circ f[/itex] where [itex]\pi^{A}: \ D \rightarrow \mathbb{R}[/itex] are some finite set of projection mappings, and [itex]A[/itex] is a multi-index taking values in [itex]\{0,1,2, … ,s\}[/itex]; [itex]A=0[/itex] is a scalar [itex]\phi[/itex](i.e., no index), [itex]A=1[/itex] is a vector; [itex]\phi^{a}, \ a = 0, 1,.., 3[/itex] (i.e.,one index), [itex]A=2[/itex] is rank two tensor; [itex]\phi^{ab}[/itex], … . Since [itex]\phi^{A}: \ D \rightarrow \mathbb{R}[/itex], we can associate [itex]y^{A} = \phi^{A}(x)[/itex] with components of a Minkowski tensor field, and [itex]y^{A}_{a}= \phi^{A}_{,a}[/itex] with the derivatives of those components with respect to coordinates [itex]x^{a}[/itex].
    Let the Lagrange function [itex]\mathcal{L}: \bar{D} \times \mathbb{R}^{s}\times \mathbb{R}^{4s}\rightarrow \mathbb{R}[/itex], and assume that it is twice differentiable. It’s real value is denoted by [itex]\mathcal{L}(x,y^{A},y^{A}_{a})[/itex]. Let the action functional [itex]S[/itex] be the totally differentiable function [itex]S: \ \mathcal{l}^{2}(\bar{D};\mathbb{R}^{s}) \rightarrow \mathbb{R}[/itex] such that
    [tex]S(f) = \int_{\bar{D}}d^{4}x \ \mathcal{L}(x,y^{A},y^{A}_{a})|_{N},[/tex]
    with [itex]N[/itex] is the submanifold [itex]y^{A}=\phi^{A} (x), \ y^{A}_{a}=\partial_{a}\phi^{A} (x)[/itex]. A new function [itex]f + \epsilon h[/itex], with the same boundary value as [itex]f[/itex], can be defined by
    [tex]y^{A} = \phi^{A}(x) + \epsilon h^{A}(x)[/tex]
    [tex]h^{A}(x)|_{\partial D} = 0.[/tex]
    It is assumed that [itex]h^{A}[/itex] is twice differentiable, [itex]\epsilon \ll 1[/itex] is a small positive number and [itex]\partial D[/itex] is the boundary of the domain [itex]D[/itex]. The “variation” of the action integral is then defined by
    \delta S(f) = S(f + \epsilon h) - S(f) = \int_{\bar{D}} d^{4}x ( \mathcal{L}(x,y,y_{a})|_{N'} - \mathcal{L}(x,y,y_{a})|_{N}),
    where [itex]N'[/itex] is the submanifold [itex]y = \phi (x) + \epsilon h(x), \ y_{a}= \partial_{a}\phi + \epsilon \partial_{a}h[/itex].
    Expanding [itex]\mathcal{L}|_{\bar{N}}[/itex] in powers of [itex]\epsilon[/itex] leads to
    \delta S(f) = \epsilon \int_{\bar{D}}d^{4}x \{ h^{A}(x) \frac{\partial \mathcal{L}}{\partial y^{A}}|_{N} + \partial_{a}h^{A}(x) \frac{\partial \mathcal{L}}{\partial y^{A}_{a}}|_{N}\} + \mathcal{O}(\epsilon^{2}).
    Now, integrating by part and using Gauss’s theorem allow us to write
    \delta S(f) = \epsilon \int_{\bar{D}}d^{4}x \ h^{A}(x) E_{A}(x) + \epsilon \int_{\partial D}d\Sigma^{a}\ h^{A}(x) \frac{\partial \mathcal{L}}{\partial y^{A}_{a}}|_{N} + \mathcal{O}(\epsilon^{2}), \ \ (1)
    E_{A}(x) \equiv \frac{\partial \mathcal{L}}{\partial y^{A}}|_{N} - \frac{d}{dx^{a}}\left( \frac{\partial \mathcal{L}}{\partial y^{A}_{a}}|_{N}\right).
    Since [itex]h^{A}(x)|_{\partial D}= 0[/itex], the second integral in eq(1) vanishes. To pick the “critical” value of the action integral [itex]S(f)[/itex], we must demand that
    [tex]\lim_{\epsilon \rightarrow 0^{+}}\left( \frac{\delta S(f)}{\epsilon}\right) = 0.[/tex]
    Thus we find that
    [tex]\int_{\bar{D}}d^{4}x \ h^{A}(x) E_{A}(x) = 0,[/tex]
    for all [itex]h^{A}(x)[/itex] that vanishes on the boundary [itex]\partial D[/itex]. This implies that
    [tex]E_{A}(x) = 0,[/tex]
    must hold in the domain [itex]D \subset \mathbb{R}^{4}.[/itex]

    Now, going back to be a physicist again, I have to say that the above derivation is a total waste of space and time.

    Last edited: Apr 16, 2012
  5. May 18, 2012 #4
    Thanks samalkhaiat, it is pity that I found your derivation only. I seem to have missed the notification. It certainly helps me to understand the topic better.

    You are certainly right in that for physics it is often practical to use shortcut notation. Personally I prefer to first read and understand it the mathematical background before then moving forward to shorthand notations. As an example, when I first saw the Einstein summation convention over equals indexes without further explanation, I was completely lost of what to do with those unbound variables.

    In the meantime I came across the wonderful book "Emmy Noether's wonderful theorem" by Neuenschwander. It contains just the right dose of hints about things like summation over double indexes or which the free variables are etc that allows to bridge the gap from pure math knowledge to physics math notation. Further he manages to explain the Lagrangian without lone deltas. Really nice book.
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