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NPTEL – Physics – Mathematical Physics - 1 Lecture 5 Gauss’s divergence theorem Let V be a closed bounded region in space whose boundary is S. Let 𝐴⃗(𝑥, 𝑦, 𝑧) is a vector that is continuous and has a continuous first partial derivatives in V, then 0 0 ⃗⃗. ⃗⃗⃗⃗⃗ 𝐴) 𝑑𝑣 where 𝑛̂ is an outward drawn normal to the elemental surface ds. ∫𝑆 𝐴⃗ . n̂ 𝑑𝑠 = ∫𝑉 (∇ Proof of the divergence theorem Imagine a volume v is split up into a number if parallelepipeds. For each parallelepiped, ⃗⃗. 𝐴⃗ 𝑑𝑣 ∑ 𝐴⃗ . 𝑑𝑆⃗ =∇ (Refer to the physical interpretation of divergence where 𝑣⃗ is replaced by 𝐴⃗ ) Now 𝐴⃗. 𝑑𝑠⃗ terms cancel pair wise for all interior surfaces. Only contribution from the interior faces survive. When we sum over all the contribution from the tiny parallelepipeds, ⃗⃗. 𝑣⃗ 𝑑𝑣 ∑𝑒𝑥𝑡𝑒𝑟𝑖𝑜𝑟 𝑠𝑢𝑟𝑓𝑎𝑐𝑒𝑠 𝐴⃗ . 𝑑𝑠⃗ = ∑𝑣𝑜𝑙𝑢𝑚𝑒 ∇ Thus 0 0 ∫𝑆 𝐴⃗ . n̂ 𝑑𝑠 = ∫𝑣 ⃗∇⃗. 𝐴⃗ 𝑑𝑣 Example Verify the divergence theorem for 𝐹⃗ = 𝑥 2 x̂ + 𝑦 2 ŷ + 𝑧 2 ẑ over a unit cube (0 ≤ 𝑥, 𝑦, 𝑧 ≤ 1) Joint initiative of IITs and IISc – Funded by MHRD Page 23 of 32 NPTEL – Physics – Mathematical Physics - 1 ⃗∇⃗. ⃗⃗⃗⃗ 𝐹 = 2(𝑥 + 𝑦 + 𝑧) 0 ⃗⃗. 𝐹⃗ 𝑑𝑣 = 2(𝑥 + 𝑦 + 𝑧)𝑑𝑥𝑑𝑦𝑑𝑧 = 3 ∫𝑣 ∇ Now there are six faces of the cube 0 0 ∫ 𝐹⃗ . n̂ 𝑑𝑠 = ∫𝑠 (𝑥=0) 𝐹⃗ . n̂ 𝑑𝑠 + ∫𝑠 1 0 ∫𝑠 5(𝑧=0) 0 𝐹⃗ . n̂ 𝑑𝑠 + ∫𝑠 6(𝑧=1) 2(𝑥=1) 0 𝐹⃗ . n̂ 𝑑𝑠 + ∫𝑠 3(𝑦=0) 0 𝐹⃗ . n̂ 𝑑𝑠 + ∫𝑠 4(𝑦=1) ⃗⃗⃗⃗ 𝐹. n̂ 𝑑𝑠 + 𝐹⃗ . n̂ 𝑑𝑠 0 1 1 For 𝑠1 , n̂ = − x̂ , So ∫𝑠 𝐹⃗ . n̂ 𝑑𝑠 = ∫𝑦=0 0 ∫𝑧=0 −𝑥 2 𝑑𝑠 = 0 . 1 0 For 𝑠2 n̂ = x̂ ∫𝑠 𝐹⃗ . ̂𝑛 𝑑𝑠 = 1 2 Only 𝑠4 and 𝑠6 are the other surfaces contribution. Total =3. Example of divergence Theorem in Physics The applicability of the concept of divergence can be explained through the celebrated Gauss’s law of electrostatics. Consider the electric field due to a point charge. The field lines move out radially in all directions and according to Coulomb’s law, the field intensity E, falls off with the distance r as 1 . 𝑟2 Now if the aim is to calculate the flux of this field through a given surface that encloses the charge. It is 0 easy to argue that the surface integral ∫ 𝐸⃗ . 𝑑𝑠 will be a constant which is independent of the distance at 𝑠 which the surface is located from the point charge, as the surface grows as 𝑟 2 with field going down as 1 , r2 making the above surface integral independent of r. according to Gauss’s law of electrostatics, 0 𝑞 ∫𝑠 𝐸⃗ . 𝑑𝑠 = 𝜀 where 𝜀0 is the permittivity of the medium. 0 Applying the divergence theorem, 0 0 ⃗⃗. 𝐸⃗⃗ )𝑑𝑣 = ∫ 𝐸⃗ . 𝑑𝑠 = ∫(∇ 𝑠 Joint initiative of IITs and IISc – Funded by MHRD 𝑣 𝑞 𝜀0 Page 24 of 32 NPTEL – Physics – Mathematical Physics - 1 Instead of a point charge, the above equation will hold for a given uniform volume charge density, 𝜌 which is given by, 0 ∫𝑣 𝜌𝑑𝑣 = 𝑞𝑒𝑛𝑐 where the volume v is enclosed by the surface s. And the above form of Gauss’s law of electrostatics law takes, 0 ⃗⃗. 𝐸⃗⃗ )𝑑𝑣 = 𝑞𝑒𝑛𝑐 = 1 ∫0 𝜌𝑑𝑣 ∫𝑣 (∇ 𝜀 𝜀 𝑣 0 0 𝜌 ⃗⃗. 𝐸⃗⃗ − Thus ∫𝑣 (∇ 𝜀0 0 )𝑑𝑣 = 0 For an arbitrary volume, ⃗⃗. 𝐸⃗⃗ − ∇ ⃗⃗ = Hence ⃗∇⃗. 𝐸 𝜌 𝜀0 =0 𝜌 𝜀0 1 Thus the divergence of the electric field is nonzero (as it should be as E diverges as 𝑟2 )and is given by The above relation is also known as one of the Maxwell’s equations in Electrodynamics. Joint initiative of IITs and IISc – Funded by MHRD Page 25 of 32 𝜌 𝜀0 .