PHY2061 Enriched Physics 2 Lecture Notes Gauss’ and Stokes Theorem D. Acosta Page 1 11/15/2006 Vector Calculus Theorems Disclaimer: These lecture notes are not meant to replace the course textbook. The content may be incomplete. Some topics may be unclear. These notes are only meant to be a study aid and a supplement to your own notes. Please report any inaccuracies to the professor. Gauss’ Theorem (Divergence Theorem) Consider a surface S with volume V. If we divide it in half into two volumes V1 and V2 with surface areas S1 and S2, we can write: 12SSSdddΦ= ⋅ = ⋅ + ⋅∫∫∫EA EA EAvvv since the electric flux through the boundary D between the two volumes is equal and opposite (flux out of V1 goes into V2). Now let’s continue this process of dividing the original volume into a great number of infinitesimal volumes, each cubic in shape: iiSSddΦ= ⋅ = ⋅∑∫∫EA EAvv Consider now one of these small cubic volumes. Consider one corner of this cube at position()000,,xyz. The length of each side is ,,xyzΔΔΔ, and each face is perpendicular to one of the coordinate axes. S1 V1 D S2 V2PHY2061 Enriched Physics 2 Lecture Notes Gauss’ and Stokes Theorem D. Acosta Page 2 11/15/2006 We are interested in computing the flux passing through this small volume. The flux through the top and bottom faces will only depend on yE since 0d⋅=EA for the other 4 faces. Since the cube is infinitesimal, we can do a Taylor expansion of the field about ()000,,xyz and find the y component of the field at the center of the bottom face ,,22xzxyzΔΔ⎛⎞++⎜⎟⎝⎠: ()000 000bottom000 ,, ,,,,22yyyy xyz xyzEExzEExyzxz∂∂ΔΔ⎛⎞ ⎛⎞=+ + +⎜⎟ ⎜⎟∂∂⎝⎠ ⎝⎠" Similarly for the top face ,,22xzxyyzΔΔ⎛⎞++Δ+⎜⎟⎝⎠ we have: ()000 000 000top000 ,, ,, ,,,,22yyyyy xyz xyz xyzEEExzEExyz yxzy∂∂∂ΔΔ⎛⎞ ⎛⎞=+ + +Δ⎜⎟ ⎜⎟∂∂∂⎝⎠ ⎝⎠" So the net flux between top and bottom is: top bottomyyxzE xzEΦ=ΔΔ−ΔΔtop-bottom The negative sign arises because the electric field points into one surface (chosen to be the bottom) and out of the other (top). Thus, ()000,,xyz x y z yΔ zΔxΔ EPHY2061 Enriched Physics 2 Lecture Notes Gauss’ and Stokes Theorem D. Acosta Page 3 11/15/2006 ()()000top-bottom000top-bottom,,22,,22yyyyyyyyEEExzExyz yxzyxzEExzExyzxzExyzy∂∂∂⎡⎤ΔΔ⎛⎞ ⎛⎞+++Δ⎜⎟ ⎜⎟⎢⎥∂∂∂⎝⎠ ⎝⎠⎢⎥Φ=ΔΔ⎢⎥∂∂ΔΔ⎛⎞ ⎛⎞−−−⎢⎥⎜⎟ ⎜⎟∂∂⎝⎠ ⎝⎠⎣⎦∂Φ=ΔΔΔ∂ We can apply the same procedure to the other two pairs of sides: front-backleft-rightxzExyzxExyzz∂Φ=ΔΔΔ∂∂Φ=ΔΔΔ∂ So the total flux passing through this infinitesimal volume is: iyxziSiEEEdxyzxyzdV∂⎛⎞∂∂Φ= ⋅ =ΔΔΔ + +⎜⎟∂∂∂⎝⎠Φ= ∇⋅∫EAEv Here we have introduced the volume element dV x y z=ΔΔΔ, and the divergence operator: div yxzEEExyz∂∂∂∇⋅ ≡ = + +∂∂∂EE Where the gradient operator is defined by: ˆˆˆxyz∂∂∂∇=++∂∂∂xyz Now we can sum the contributions from all infinitesimal volumes comprising the full volume V: (),, iiiiiiiSiixyzddV E dV EΦ= Φ = ⋅=∇⋅→∇⋅∑∑∫∑∫EAv SVddV⇒⋅= ∇⋅∫∫EA Ev This forms Gauss’ Theorem, or the Divergence Theorem. It states that the surface integral of d⋅EA can be related to the volume integral of ∇⋅E .PHY2061 Enriched Physics 2 Lecture Notes Gauss’ and Stokes Theorem D. Acosta Page 4 11/15/2006 Differential form of Gauss’ Law So how does Gauss’ Theorem relate to what we have learned in electromagnetism? Consider Gauss’ Law, which relates the electric flux through a closed surface to the net enclosed charge: enc0SqdεΦ= ⋅ =∫EAv We can relate the enclosed charge to the volume integral of the differential charge density per unit volume, ρ, in the enclosed volume: encVqdVρ=∫ And by Gauss’ Theorem, we can relate the surface integral of d⋅EA to the volume integral of ∇⋅E , giving us: 01 VVdV dVρε∇⋅ =∫∫E Now there is nothing special about what volume V we choose, so the above expression must hold for all volumes. That means that the integrands must be equal: 0ρε∇⋅ =E This is the differential form of Gauss’ Law. It holds for every point in space. When combined with further differential laws of electromagnetism (see next section), we can derive a differential equation for electromagnetic waves. For example, consider a constant electric field: 0ˆE=Ex. It is easy to see that the divergence of E will be zero, so the charge density ρ=0 everywhere. Thus, the total enclosed charge in any volume is zero, and by the integral form of Gauss’ Law the total flux through the surface of that volume must be zero. On the other hand, if 0ˆˆ'Ex E=+Exy, then 000xEEExρε∂∇⋅ = = ⇒ =∂E Note: If one integrates the obtained charge density within an certain enclosed volume, you will get exactly the same amount of enclosed charge as if you used the integral form of Gauss’ Law (try it, it really works!)PHY2061 Enriched Physics 2 Lecture Notes Gauss’ and Stokes Theorem D. Acosta Page 5 11/15/2006 Stokes Theorem Consider the line integral of a vector function around a closed curve C: CdΓ= ⋅∫Fsv This integral is called the “circulation”. We use the right-hand rule to define the direction of the area vector (perpendicular to the surface) with respect to the integration direction (counter-clockwise in this case). Now suppose we subdivide this surface into two regions, and calculate the circulation of each around closed curves C1 and C2: 1212 CCddΓ= ⋅ Γ= ⋅∫∫Fs Fsvv It should be clear that in both cases, line segment AB is traversed in opposite directions, so the contribution to Γ1 is equal and opposite to Γ2. Thus: 12Γ=Γ +Γ We can continue subdividing the surface into N subregions and we’ll get: 1Nii=Γ= Γ∑ Let’s let these regions become infinitesimal is size, and calculate the circulation for just one infinitesimal area aligned in the x-y plane. C1 A B C2 ds F C APHY2061 Enriched Physics 2 Lecture Notes Gauss’ and Stokes Theorem D. Acosta Page 6 11/15/2006 To find the circulation for this infinitesimal contour, let’s Taylor expand F about the bottom left corner at ()00,xy to find the value at the center of each of the 4 line segments: ()()()()0000000000
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