Electricity and Magnetism: Gradient, Divergence, and CurlOutline 1 We introduce vector differential operations of gradient, divergence, and curl 2 Gauss’s theorem and Stokes theorem are discussed Learning Objectives 1 To understand the physical meaning of vector differential operators gradient, divergence and curl 2 To learn to evaluate the gradient, divergence, and curl of given functions 3 To understand theorems involving these operators, particularly Gauss’s and Stokes theorems Gradient of a Scalar Function Consider a scalar function T (x, y , z) We want to compute the change in T , as we move from initial coordinates r ≡ (x, y , z) infinitesimally to the new position r + dr ≡ (x + dx, y + dy , z + dz) Using Taylor expansion (for multi-variables), and retaining terms up to first order T (r+dr) = T (r)+dx ∂T ∂T ∂T +dy +dz +higher order terms ∂x ∂y ∂z Or, to the first order terms, T (r + dr) = T (r) + dr · ∇T Where dr = dxˆi + dy ˆj + dz kˆ ∂Tˆ ∂T ˆ ∂T ˆ k ∇T = i+ j+ ∂x ∂y ∂z where the vector ∇T defined above is called the gradient of scalar field T .Gradient contd. we conclude dT = dr · ∇T .. and is itself a vector quantity Let us examine ∇T a bit more . Thus ∇T defines the rate of change of the scalar field with respect to the spatial coordinates. Defining T (r + dr) = T (r) + dT .. we can write the change in the scalar field dT as dT = dr · ∇T = |dr||∇T | cos θ Let us consider two possibilities: dr is along a constant T surface dr is in an arbitrary direction .Physical Meaning of Gradient Let us plot the constant surfaces of a given scalar field T As per the figure. Thus the direction of ∇T is also the direction of maximum change in the scalar function T .e. it is obvious that the magnitude of the maximum possible change in T is dTmax = |dr||∇T |.Gradient. physical meaning. .. This means |dr||∇T | cos θ = 0 =⇒ cos θ = 0 Thus the direction of ∇T at a given point r is always perpendicular to the constant T surface passing through that point Let us consider dr to be in an arbitrary direction Then from dT = |dr||∇T | cos θ . i. when cos θ = 1... If dr is along a constant T surface then dT = 0. Thus. if one moves in the direction of ∇T .. at a given point r. y . z) = x 2 y + y 2 z + z 2 x + 2xyz .Gradient continued. maximum change in T will take place This property of gradient is used in optimization problems involving location of maxima/minima of scalar functions Examples: 1 Let us consider a scalar function T = r2 = x2 + y2 + z2 It is easy to see that ∇T == 2 ∂Tˆ ∂T ˆ ∂T ˆ i+ j+ k = 2xˆi + 2y ˆj + 2z kˆ = 2r ∂x ∂y ∂z Consider Φ(x.. the gradient operator can be denoted as ∂ ˆ ∂ ˆ ∂ ˆ ∇≡ i+ j+ k ∂x ∂y ∂z In curvilinear coordinates the gradient operator has more complicated forms ∂ ∂ ˆ ∂ ˆ ρˆ + θ+ k ∂ρ ρ∂ θ ∂z ∂ ∂ ˆ ∂ Spherical ∇ ≡ ˆr + θ+ φˆ ∂r r∂θ r sin θ ∂ φ Cylindrical ∇ ≡ . in Cartesian coordinates.Gradient calculation Clearly ∂ Φˆ ∂ Φ ˆ ∂ Φ ˆ i+ j+ k ∂x ∂y ∂z = (2xy + 2yz + z 2 )ˆi + (2yz + 2xz + x 2 )ˆj + (2zx + 2xy + y 2 )kˆ ∇Φ = Thus. Some Properties of Gradient 1 ∇(U + V ) = ∇U + ∇V 2 ∇(UV ) = U∇V + V ∇U 3 ∇(V n ) = nV n−1 ∇V . z) − Fy (x. R flux Φ = F · dS over the six surfaces of a cuboid shown below Clearly. y . y + dy . outward flux through the shaded planes Φy = [Fy (x. z) = Fy (x. y . z) + dy So that Φy = ∂ Fy dxdydz ∂y ∂ Fy . z)] dxdz Using first order Taylor expansion Fy (x.Divergence of a Vector Field ˆ and evaluate its Let us consider a vector field F = Fxˆi + Fy ˆj + Fz k. ∂y . y + dy . Thus we get for the entire cuboid F · dS = (∇ · F)dV . we obtain ∂ Fx ∂ Fy ∂ F z + + dV . where ∇·F = ∂ Fx ∂ Fy ∂ Fz + + ∂x ∂y ∂z is called the divergence of the vector field F. dΦ = ∂x ∂y ∂z where dV = dxdydz is the volume of the cuboid. If we similarly calculate the flux through all the remaining faces of the cuboid and add it to obtain the total flux d Φ. .Divergence contd. .Divergence . .. Note that divergence can be seen as a dot product of the ˆ and the vector field operator ∇ ≡ ∂∂x ˆi + ∂∂y ˆj + ∂∂z k. ˆ F = Fxˆi + Fy ˆj + Fz k. which is called Gauss’s Theorem or Divergence Theorem I S F · dS = Z ∇ · FdV . .Divergence Theorem Divergence Theorem (Gauss’s Theorem): Flux of a vector field F calculated over the surface of an arbitrarily shaped volume satisfies the following result. (1) V that is the flux of a vector field over the surface enclosing a given volume is equal to the volume integral of its divergence over the given volume. because contribution from those surfaces of the cuboids which are inside the volume will cancel due to opposite orientations for the adjacent cuboids. 2 for all cuboids. This will lead to the LHS of Eq. RHS of Eq. Perform the sum of Eq.e. .S. 1.H.. L.H. i. R. of the sum will only have the contribution from the external surface S of the volume V .S. 1 Therefore QED.Proof of Divergence Theorem Divergence theorem is easy to prove using the result we proved for an infinitesimal cuboid F · dS = (∇ · F)dV (2) Divide the given volume into a large number of such infinitesimal cuboids. of the sum will simply be the volume integral of ∇ · F over the entire volume V . in cylindrical polar coordinates. in spherical polar coordinates. the divergence is given by ∇·V = 1 ∂ vθ ∂ vz 1 ∂ (ρvρ ) + + ρ ∂ρ ρ ∂θ ∂z For vector V = vr ˆr + vθ θˆ + vφ φˆ. .Divergence in Curvilinear Coordinates For vector V = vρ ρˆ + vθ θˆ + vz zˆ. the divergence is given by ∇·V = 1 ∂ 1 ∂ vφ 1 ∂ 2 (r vr ) + (sin θ vθ ) + 2 r ∂r r sin θ ∂ θ r sin θ ∂ φ Note the different forms of divergence operator in different coordinate systems. so ∇·r = 1 ∂ (r 2 r ) =3 r2 ∂r In cylindrical polar coordinates r = ρ ρˆ + zˆ z . so ∇·r = 1 ∂ (ρρ) ∂ z + = 2+1 = 3 ρ ∂ρ ∂z .Examples of Divergence Calculation Calculate the divergence of vector r = xˆi + y ˆj + z kˆ In Cartesian coordinates ∇·r = ∂x ∂y ∂z + + =3 ∂x ∂y ∂z In spherical polar coordinates r = rˆr . dl = F. z)∆z .dy ˆj ≈ Fy (y . z)∆y Z F.dl = Z F.dz kˆ BC ≈ Fz (y + ∆y . line integral along a infinitesimal rectangular path shown below Clearly I F.Curl of a Vector Field ˆ and evaluate its Let us consider a vector field F = Fxˆi + Fy ˆj + Fz k.dl + Z AB + F.dl = Z F.dl DA Now Z ZAB F.dl F.dl + CD Z Z BC F. .dl = Fy ∆y + Fz ∆z + ∂y AB BC Z Z Similarly one can show (by integrating in AD and DC directions) Z Z ∂ Fy F.dl + F.dl = − ∆Sx ∂y ∂z (3) . Using first order Taylor expansion Fz (y + ∆y .dl + F.. z) + ∂ Fz ∆y ∂y So that ∂ Fz ∆z∆y F.Curl contd. z) = Fz (y .dl = − Fy ∆y + Fz ∆z + ∆z∆y ∂z CD DA By adding all the contributions we obtain I ∂ Fz ∂ Fy F.. is the area of the infinitesimal loop. directed along the x axis. denoted as ∇× .. Where ∆Sx = ∆y ∆z.Curl contd... Let us define a quantity called curl. . . ˆi ˆj kˆ . . . . . ∇ × F = . ∂∂x ∂∂y ∂∂z . . . . . Fx Fy Fz . 3 as I F. Using this we can cast Eq.dl = (∇ × F)x ∆Sx = (∇ × F) · ∆S (4) . then the line integral is equal to the surface integral of the curl of F.Stokes’ Theorem Stokes’ Theorem: If a vector field F is integrated along a closed loop of an arbitrary shape. evaluated over the area enclosed by the loop I F.dl = Z S (∇ × F) · dS . S Note that in the line integral. we get the desired result I F. Upon adding the contribution of all such loops. . 4 will hold.Proof of Stokes’ Theorem Outline of the proof: We can split the area enclosed by the loop into a large number of infinitesimal loops as shown. for each one of which Eq.dl = Z (∇ × F) · dS. the contribution only from the boundary of the loop will survive because the contribution from the internal lines gets canceled from adjacent loops. Curl in Cylindrical Coordinate system Curl in Cylindrical Coordinates: For a vector field A = Aρ ρˆ + Aθ θˆ + Az ˆz 1 ∂ Az ∂ Aθ − ρˆ ∇×A = ρ ∂θ ∂z ∂ Aρ ∂ Az ˆ + − θ ∂z ∂ρ 1 ∂ (ρAθ ) ∂ Aρ ˆz + − ρ ∂ρ ∂φ . Curl in Spherical Polar Coordinate System Curl in Spherical Polar Coordinates: For a vector field A = Ar ˆr + Aθ θˆ + Aφ φˆ ∂ Aθ ∂ 1 ˆr Aφ sin θ − ∇×A = r sin θ ∂ θ ∂φ ∂ 1 1 ∂ Ar + − rAφ θˆ r sin θ ∂ φ ∂r 1 ∂ ∂ Ar ˆ + (rAθ ) − φ r ∂r ∂θ . Examples of Calculation of Curl 1 Calculate the curl of the vector field F = −yˆi + zˆj + x 2 kˆ . . . ˆi ˆj kˆ . . . . . ∇ × F = . ∂∂x ∂∂y ∂∂z . . . . −y z x 2 . using spherical coordinates (r = rˆr ) and cylindrical coordinates (r = ρ ρˆ + zˆ z) . Verify the result of 2. where r = xˆi + y ˆj + z k. = −ˆi + 2xˆj + kˆ 2 3 ˆ Easy to verify that ∇ × r = 0.