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Gradient theorem

From Wikipedia, the free encyclopedia
Evaluates a line integral through a gradient field using the original scalar field
Part of a series of articles about
Calculus
abf(t)dt=f(b)f(a){\displaystyle \int _{a}^{b}f'(t)\,dt=f(b)-f(a)}

Thegradient theorem, also known as thefundamental theorem of calculus for line integrals, says that aline integral through agradient field can be evaluated by evaluating the original scalar field at the endpoints of the curve. The theorem is a generalization of thesecond fundamental theorem of calculus to any curve in a plane or space (generallyn-dimensional) rather than just the real line.

Ifφ :URnR is adifferentiable function andγ adifferentiable curve inU which starts at a pointp and ends at a pointq, then

γφ(r)dr=φ(q)φ(p){\displaystyle \int _{\gamma }\nabla \varphi (\mathbf {r} )\cdot \mathrm {d} \mathbf {r} =\varphi \left(\mathbf {q} \right)-\varphi \left(\mathbf {p} \right)}

whereφ denotes the gradient vector field ofφ.

The gradient theorem implies that line integrals through gradient fields arepath-independent. In physics this theorem is one of the ways of defining aconservative force. By placingφ as potential,φ is aconservative field.Work done by conservative forces does not depend on the path followed by the object, but only the end points, as the above equation shows.

The gradient theorem also has an interesting converse: any path-independent vector field can be expressed as the gradient of ascalar field. Just like the gradient theorem itself, this converse has many striking consequences and applications in both pure and applied mathematics.

Proof

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Ifφ is adifferentiable function from someopen subsetURn toR andr is a differentiable function from some closedinterval[a,b] toU (Note thatr is differentiable at the interval endpointsa andb. To do this,r is defined on an interval that is larger than and includes[a,b].), then by themultivariate chain rule, thecomposite functionφr is differentiable on[a,b]:

ddt(φr)(t)=φ(r(t))r(t){\displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}(\varphi \circ \mathbf {r} )(t)=\nabla \varphi (\mathbf {r} (t))\cdot \mathbf {r} '(t)}

for allt in[a,b]. Here the denotes thedot product.

Now suppose the domainU ofφ contains the differentiable curveγ with endpointsp andq. (This isoriented in the direction fromp toq). Ifrparametrizesγ fort in[a,b] (i.e.,r representsγ as a function oft), then

γφ(r)dr=abφ(r(t))r(t)dt=abddtφ(r(t))dt=φ(r(b))φ(r(a))=φ(q)φ(p),{\displaystyle {\begin{aligned}\int _{\gamma }\nabla \varphi (\mathbf {r} )\cdot \mathrm {d} \mathbf {r} &=\int _{a}^{b}\nabla \varphi (\mathbf {r} (t))\cdot \mathbf {r} '(t)\mathrm {d} t\\&=\int _{a}^{b}{\frac {d}{dt}}\varphi (\mathbf {r} (t))\mathrm {d} t=\varphi (\mathbf {r} (b))-\varphi (\mathbf {r} (a))=\varphi \left(\mathbf {q} \right)-\varphi \left(\mathbf {p} \right),\end{aligned}}}

where thedefinition of a line integral is used in the first equality, the above equation is used in the second equality, and thesecond fundamental theorem of calculus is used in the third equality.[1]

Even if the gradient theorem (also calledfundamental theorem of calculus for line integrals) has been proved for a differentiable (so looked as smooth) curve so far, the theorem is also proved for a piecewise-smooth curve since this curve is made by joining multiple differentiable curves so the proof for this curve is made by the proof per differentiable curve component.[2]

Examples

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Example 1

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SupposeγR2 is the circular arc oriented counterclockwise from(5, 0) to(−4, 3). Using thedefinition of a line integral,

γydx+xdy=0πtan1(34)((5sint)(5sint)+(5cost)(5cost))dt=0πtan1(34)25(sin2t+cos2t)dt=0πtan1(34)25cos(2t)dt = 252sin(2t)|0πtan1(34)=252sin(2π2tan1(34))=252sin(2tan1(34)) = 25(3/4)(3/4)2+1=12.{\displaystyle {\begin{aligned}\int _{\gamma }y\,\mathrm {d} x+x\,\mathrm {d} y&=\int _{0}^{\pi -\tan ^{-1}\!\left({\frac {3}{4}}\right)}((5\sin t)(-5\sin t)+(5\cos t)(5\cos t))\,\mathrm {d} t\\&=\int _{0}^{\pi -\tan ^{-1}\!\left({\frac {3}{4}}\right)}25\left(-\sin ^{2}t+\cos ^{2}t\right)\mathrm {d} t\\&=\int _{0}^{\pi -\tan ^{-1}\!\left({\frac {3}{4}}\right)}25\cos(2t)\mathrm {d} t\ =\ \left.{\tfrac {25}{2}}\sin(2t)\right|_{0}^{\pi -\tan ^{-1}\!\left({\tfrac {3}{4}}\right)}\\[.5em]&={\tfrac {25}{2}}\sin \left(2\pi -2\tan ^{-1}\!\!\left({\tfrac {3}{4}}\right)\right)\\[.5em]&=-{\tfrac {25}{2}}\sin \left(2\tan ^{-1}\!\!\left({\tfrac {3}{4}}\right)\right)\ =\ -{\frac {25(3/4)}{(3/4)^{2}+1}}=-12.\end{aligned}}}

This result can be obtained much more simply by noticing that the functionf(x,y)=xy{\displaystyle f(x,y)=xy} has gradientf(x,y)=(y,x){\displaystyle \nabla f(x,y)=(y,x)}, so by the Gradient Theorem:

γydx+xdy=γ(xy)(dx,dy) = xy|(5,0)(4,3)=4350=12.{\displaystyle \int _{\gamma }y\,\mathrm {d} x+x\,\mathrm {d} y=\int _{\gamma }\nabla (xy)\cdot (\mathrm {d} x,\mathrm {d} y)\ =\ xy\,|_{(5,0)}^{(-4,3)}=-4\cdot 3-5\cdot 0=-12.}

Example 2

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For a more abstract example, supposeγRn has endpointsp,q, with orientation fromp toq. Foru inRn, let|u| denote theEuclidean norm ofu. Ifα ≥ 1 is a real number, then

γ|x|α1xdx=1α+1γ(α+1)|x|(α+1)2xdx=1α+1γ|x|α+1dx=|q|α+1|p|α+1α+1{\displaystyle {\begin{aligned}\int _{\gamma }|\mathbf {x} |^{\alpha -1}\mathbf {x} \cdot \mathrm {d} \mathbf {x} &={\frac {1}{\alpha +1}}\int _{\gamma }(\alpha +1)|\mathbf {x} |^{(\alpha +1)-2}\mathbf {x} \cdot \mathrm {d} \mathbf {x} \\&={\frac {1}{\alpha +1}}\int _{\gamma }\nabla |\mathbf {x} |^{\alpha +1}\cdot \mathrm {d} \mathbf {x} ={\frac {|\mathbf {q} |^{\alpha +1}-|\mathbf {p} |^{\alpha +1}}{\alpha +1}}\end{aligned}}}

Here the final equality follows by the gradient theorem, since the functionf(x) = |x|α+1 is differentiable onRn ifα ≥ 1.

Ifα < 1 then this equality will still hold in most cases, but caution must be taken ifγ passes through or encloses the origin, because the integrand vector field|x|α − 1x will fail to be defined there. However, the caseα = −1 is somewhat different; in this case, the integrand becomes|x|−2x = ∇(log |x|), so that the final equality becomeslog |q| − log |p|.

Note that ifn = 1, then this example is simply a slight variant of the familiarpower rule from single-variable calculus.

Example 3

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Suppose there arenpoint charges arranged in three-dimensional space, and thei-th point charge haschargeQi and is located at positionpi inR3. We would like to calculate thework done on a particle of chargeq as it travels from a pointa to a pointb inR3. UsingCoulomb's law, we can easily determine that theforce on the particle at positionr will be

F(r)=kqi=1nQi(rpi)|rpi|3{\displaystyle \mathbf {F} (\mathbf {r} )=kq\sum _{i=1}^{n}{\frac {Q_{i}(\mathbf {r} -\mathbf {p} _{i})}{\left|\mathbf {r} -\mathbf {p} _{i}\right|^{3}}}}

Here|u| denotes theEuclidean norm of the vectoru inR3, andk = 1/(4πε0), whereε0 is thevacuum permittivity.

LetγR3 − {p1, ...,pn} be an arbitrary differentiable curve froma tob. Then the work done on the particle is

W=γF(r)dr=γ(kqi=1nQi(rpi)|rpi|3)dr=kqi=1n(Qiγrpi|rpi|3dr){\displaystyle W=\int _{\gamma }\mathbf {F} (\mathbf {r} )\cdot \mathrm {d} \mathbf {r} =\int _{\gamma }\left(kq\sum _{i=1}^{n}{\frac {Q_{i}(\mathbf {r} -\mathbf {p} _{i})}{\left|\mathbf {r} -\mathbf {p} _{i}\right|^{3}}}\right)\cdot \mathrm {d} \mathbf {r} =kq\sum _{i=1}^{n}\left(Q_{i}\int _{\gamma }{\frac {\mathbf {r} -\mathbf {p} _{i}}{\left|\mathbf {r} -\mathbf {p} _{i}\right|^{3}}}\cdot \mathrm {d} \mathbf {r} \right)}

Now for eachi, direct computation shows that

rpi|rpi|3=1|rpi|.{\displaystyle {\frac {\mathbf {r} -\mathbf {p} _{i}}{\left|\mathbf {r} -\mathbf {p} _{i}\right|^{3}}}=-\nabla {\frac {1}{\left|\mathbf {r} -\mathbf {p} _{i}\right|}}.}

Thus, continuing from above and using the gradient theorem,

W=kqi=1n(Qiγ1|rpi|dr)=kqi=1nQi(1|api|1|bpi|){\displaystyle W=-kq\sum _{i=1}^{n}\left(Q_{i}\int _{\gamma }\nabla {\frac {1}{\left|\mathbf {r} -\mathbf {p} _{i}\right|}}\cdot \mathrm {d} \mathbf {r} \right)=kq\sum _{i=1}^{n}Q_{i}\left({\frac {1}{\left|\mathbf {a} -\mathbf {p} _{i}\right|}}-{\frac {1}{\left|\mathbf {b} -\mathbf {p} _{i}\right|}}\right)}

We are finished. Of course, we could have easily completed this calculation using the powerful language ofelectrostatic potential orelectrostatic potential energy (with the familiar formulasW = −ΔU = −qΔV). However, we have not yetdefined potential or potential energy, because theconverse of the gradient theorem is required to prove that these are well-defined, differentiable functions and that these formulas hold (see below). Thus, we have solved this problem using only Coulomb's law, the definition of work, and the gradient theorem.

Converse of the gradient theorem

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The gradient theorem states that if the vector fieldF is the gradient of some scalar-valued function (i.e., ifF isconservative), thenF is a path-independent vector field (i.e., the integral ofF over some piecewise-differentiable curve is dependent only on end points). This theorem has a powerful converse:

Theorem IfF is a path-independent vector field, thenF is the gradient of some scalar-valued function.[3]

It is straightforward to show that a vector field is path-independent if and only if the integral of the vector field over every closed loop in its domain is zero. Thus the converse can alternatively be stated as follows: If the integral ofF over every closed loop in the domain ofF is zero, thenF is the gradient of some scalar-valued function.

Proof of the converse

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SupposeU is anopen,path-connected subset ofRn, andF :URn is acontinuous and path-independent vector field. Fix some elementa ofU, and definef :UR byf(x):=γ[a,x]F(u)du{\displaystyle f(\mathbf {x} ):=\int _{\gamma [\mathbf {a} ,\mathbf {x} ]}\mathbf {F} (\mathbf {u} )\cdot \mathrm {d} \mathbf {u} }Hereγ[a,x] is any (differentiable) curve inU originating ata and terminating atx. We know thatf iswell-defined becauseF is path-independent.

Letv be any nonzero vector inRn. By the definition of thedirectional derivative,f(x)v=limt0f(x+tv)f(x)t=limt0γ[a,x+tv]F(u)duγ[a,x]F(u)dut=limt01tγ[x,x+tv]F(u)du{\displaystyle {\begin{aligned}{\frac {\partial f(\mathbf {x} )}{\partial \mathbf {v} }}&=\lim _{t\to 0}{\frac {f(\mathbf {x} +t\mathbf {v} )-f(\mathbf {x} )}{t}}\\&=\lim _{t\to 0}{\frac {\int _{\gamma [\mathbf {a} ,\mathbf {x} +t\mathbf {v} ]}\mathbf {F} (\mathbf {u} )\cdot \mathrm {d} \mathbf {u} -\int _{\gamma [\mathbf {a} ,\mathbf {x} ]}\mathbf {F} (\mathbf {u} )\cdot d\mathbf {u} }{t}}\\&=\lim _{t\to 0}{\frac {1}{t}}\int _{\gamma [\mathbf {x} ,\mathbf {x} +t\mathbf {v} ]}\mathbf {F} (\mathbf {u} )\cdot \mathrm {d} \mathbf {u} \end{aligned}}}To calculate the integral within the final limit, we mustparametrizeγ[x,x +tv]. SinceF is path-independent,U is open, andt is approaching zero, we may assume that this path is a straight line, and parametrize it asu(s) =x +sv for0 <s <t. Now, sinceu'(s) =v, the limit becomeslimt01t0tF(u(s))u(s)ds=ddt0tF(x+sv)vds|t=0=F(x)v{\displaystyle \lim _{t\to 0}{\frac {1}{t}}\int _{0}^{t}\mathbf {F} (\mathbf {u} (s))\cdot \mathbf {u} '(s)\,\mathrm {d} s={\frac {\mathrm {d} }{\mathrm {d} t}}\int _{0}^{t}\mathbf {F} (\mathbf {x} +s\mathbf {v} )\cdot \mathbf {v} \,\mathrm {d} s{\bigg |}_{t=0}=\mathbf {F} (\mathbf {x} )\cdot \mathbf {v} }where the first equality is fromthe definition of the derivative with a fact that the integral is equal to 0 att = 0, and the second equality is from thefirst fundamental theorem of calculus. Thus we have a formula forvf, (one of ways to represent thedirectional derivative) wherev is arbitrary; forf(x):=γ[a,x]F(u)du{\displaystyle f(\mathbf {x} ):=\int _{\gamma [\mathbf {a} ,\mathbf {x} ]}\mathbf {F} (\mathbf {u} )\cdot \mathrm {d} \mathbf {u} } (see its full definition above), its directional derivative with respect tov isf(x)v=vf(x)=Dvf(x)=F(x)v{\displaystyle {\frac {\partial f(\mathbf {x} )}{\partial \mathbf {v} }}=\partial _{\mathbf {v} }f(\mathbf {x} )=D_{\mathbf {v} }f(\mathbf {x} )=\mathbf {F} (\mathbf {x} )\cdot \mathbf {v} }where the first two equalities just show different representations of the directional derivative. According tothe definition of the gradient of a scalar functionf,f(x)=F(x){\displaystyle \nabla f(\mathbf {x} )=\mathbf {F} (\mathbf {x} )}, thus we have found a scalar-valued functionf whose gradient is the path-independent vector fieldF (i.e.,F is a conservative vector field.), as desired.[3]

Example of the converse principle

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Main article:Electric potential energy

To illustrate the power of this converse principle, we cite an example that has significantphysical consequences. Inclassical electromagnetism, theelectric force is a path-independent force; i.e. thework done on a particle that has returned to its original position within anelectric field is zero (assuming that no changingmagnetic fields are present).

Therefore, the above theorem implies that the electricforce fieldFe :SR3 is conservative (hereS is someopen,path-connected subset ofR3 that contains acharge distribution). Following the ideas of the above proof, we can set some reference pointa inS, and define a functionUe:SR by

Ue(r):=γ[a,r]Fe(u)du{\displaystyle U_{e}(\mathbf {r} ):=-\int _{\gamma [\mathbf {a} ,\mathbf {r} ]}\mathbf {F} _{e}(\mathbf {u} )\cdot \mathrm {d} \mathbf {u} }

Using the above proof, we knowUe is well-defined and differentiable, andFe = −∇Ue (from this formula we can use the gradient theorem to easily derive the well-known formula for calculating work done by conservative forces:W = −ΔU). This functionUe is often referred to as theelectrostatic potential energy of the system of charges inS (with reference to the zero-of-potentiala). In many cases, the domainS is assumed to beunbounded and the reference pointa is taken to be "infinity", which can be maderigorous using limiting techniques. This functionUe is an indispensable tool used in the analysis of many physical systems.

Generalizations

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Main articles:Stokes' theorem andClosed and exact differential forms

Many of the critical theorems of vector calculus generalize elegantly to statements about theintegration of differential forms onmanifolds. In the language ofdifferential forms andexterior derivatives, the gradient theorem states that

γϕ=γdϕ{\displaystyle \int _{\partial \gamma }\phi =\int _{\gamma }\mathrm {d} \phi }

for any0-form,ϕ, defined on some differentiable curveγRn (here the integral ofϕ over the boundary of theγ is understood to be the evaluation ofϕ at the endpoints ofγ).

Notice the striking similarity between this statement and thegeneralized Stokes’ theorem, which says that the integral of anycompactly supported differential formω over theboundary of someorientable manifoldΩ is equal to the integral of itsexterior derivativedω over the whole ofΩ, i.e.,

Ωω=Ωdω{\displaystyle \int _{\partial \Omega }\omega =\int _{\Omega }\mathrm {d} \omega }

This powerful statement is a generalization of the gradient theorem from 1-forms defined on one-dimensional manifolds to differential forms defined on manifolds of arbitrary dimension.

The converse statement of the gradient theorem also has a powerful generalization in terms of differential forms on manifolds. In particular, supposeω is a form defined on acontractible domain, and the integral ofω over any closed manifold is zero. Then there exists a formψ such thatω = dψ. Thus, on a contractible domain, everyclosed form isexact. This result is summarized by thePoincaré lemma.

See also

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References

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  1. ^Williamson, Richard E.;Trotter, Hale F. (2004).Multivariable mathematics.Pearson Education International (4th ed.). Upper Saddle River, N.J:Pearson Prentice Hall. p. 374.ISBN 978-0-13-067276-6.
  2. ^Stewart, James; Clegg, Dan; Watson, Saleem (2021). "16.3 The Fundamental Theorem for Line Integrals".Calculus (Ninth ed.). Australia ; Boston, MA, USA:Cengage. pp. 1182–1185.ISBN 978-1-337-62418-3.
  3. ^abWilliamson & Trotter 2004, p. 410
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