Math 21a Supplement 2 on Work and EnergyAs mentioned at the end of Section 5.1 in the text, the work done in moving a particle alonga path γ in R3 when a force vector F acts is defined to bew(γ) ≡ ∫γ F•dx(1)This is to say that term ‘work done in moving along γ’ in physics has a specific meaning, the latterbeing the value of (1). (Presumably, this mathematical definition of work meshes well with ourintuitive idea of work done.)Let me remind you (from the definition in the text) that the shorthand on the right side of(1) has the following meaning: Parameterize the path γ by choosing an interval, [a, b], on the line,and a vector valued function x(t) for a ≤ t ≤ b which traces out the path γ. With this done, then theright side of (1) is computed by the ordinary integralw(γ) = ab∫F(x(t))•x´(t) dt .(2)Now, suppose that the path γ is the path that the particle would have travelled were itstrajectory γ given by Newton’s law; thus x(t) obeysm x´´ = F .(2)In this case, if we substitute m x´´ for F in (2) we see thatw(γ) = mab∫ x´´•x´ dt ,(3)which is nothing more thanw(γ) = 2−1 mab∫ddt |x´|2 dt = (2−1 m |x´|2)t=b - (2−1 m |x´|2)t=a.(4)(The final equality in (4) is just the Fundamental Theorem of Calculus.) In any event, (4)demonstrates that the work done by a particle moving according to Newton’s law is given by thedifference between the squared norms of the velocities times the mass divided by 2.This last formula is not unrelated to the notion of conservation of energy . In this regard,suppose that F = ∇V, that is the force is the gradient of a potential. (For example, the gravitationalforce from a point particle of mass M at the origin is the gradient of V = G m M r−1, where r = |x| isthe distance to the origin.) Anyway, if F = ∇V, the energy of a particle moving on the trajectoryx(t) at time t is defined to bee ≡ 2−1 m |x´|2 - V(x).(5)It then turns out that e´ = 0 when the particle moves according to Newton’s law law m x´´ = ∇V ascan be seen by differentiating (5). In this regard, use the Chain rule to writeV(x(t))´ = ∇V•x´,(6)while writing 2−1 m (|x´|2)´ = m x´´•x´ and then substituting ∇V for m x´´ to see that the timederivative of the first term on the right side in (5) just cancels that in the second.The relation of all of this to (4) is that when e is constant, then the difference on the farright side of (4) is the same asw(γ) = V(x(tb)) - V(x(ta)) .(7)Thus, the work done when a particle moves by Newton’s law against a force which is a gradient isgiven by the difference between the values of the potential function at the end and beginning pointsof the motion. (We will see in Section 5.2 of the text that this last conclusion is true even if theparticle ignores Newton’s
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