Continuum Mechanics PDF

Continuum Mechanics It is one of the applied mathematics courses that is relied upon to study some branches of mathematics Continuum Mechanics Elasticity...

Continuum Mechanics It is one of the applied mathematics courses that is relied upon to study some branches of mathematics Continuum Mechanics Elasticity Fliud mechanics In physics, all matter is made up of particles, atoms, and other small things. If we look at it with a microscope, these materials are not continuous (disconnected). But if we look at them with the naked eye, we can consider them as continuous materials. In many issues in daily life we deal with parts of matter, these parts are very large compared to small particles, and therefore the mechanics of cohesion deals with them as continuous materials. Continuum mechanics also studies the behavior of fluids (liquids or gases), as well as the behavior of solid bodies by the naked eye. This science also neglects the nature of matter itself and treats it as being uniformly distributed in the body that is dealt with (i.e. continuous), and therefore it is possible to define the quantities that we are dealing with (velocity, density, displacement....etc) as continuous functions with respect to position. Page | 1  Tensors In cartesian coordinate, if 𝑎 = 𝑎𝑥 𝑖 + 𝑎𝑦 𝑗 + 𝑎𝑧 𝑘 and 𝑏 = 𝑏𝑥 𝑖 + 𝑏𝑦 𝑗 + 𝑏𝑧 𝑘 then 𝑎 ⋅ 𝑏 = 𝑎𝑥 𝑏𝑥 + 𝑎𝑦 𝑏𝑦 + 𝑎𝑧 𝑏𝑧 , (1) We can use a simpler method to express the vectors by replacing 𝑥, 𝑦, 𝑧 with 𝑥1 , 𝑥2 , 𝑥3 and replacing the base vectors 𝑖 , 𝑗 , 𝑘 by 𝑒1 , 𝑒2 , 𝑒3 3 𝑥 = 𝑥1 𝑒1 + 𝑥2 𝑒2 + 𝑥3 𝑒3 = ∑ 𝑥𝑖 𝑒𝑖 , (2) 𝑖=1 and if we agree to remove the summation symbol ∑ then 𝑥 = 𝑥𝑖 𝑒𝑖 , 𝑖 = 1,2,3 So, if the index is repeated then this mean that we have a summation from 1 to 3. For example, if 𝑎 = 𝑎1 𝑒 1 + 𝑎2 𝑒 2 + 𝑎3 𝑒 3 and 𝑏 = 𝑏1 𝑒 1 + 𝑏2 𝑒 2 + 𝑏3 𝑒 3 then: 𝑎 ⋅ 𝑏 = 𝑎1 𝑏1 + 𝑎2 𝑏2 + 𝑎3 𝑏3 = 𝑎𝑖 𝑏𝑖 , 𝑖 = 1,2,3 Also, if 𝜙 is a scalar function, then 𝜕 𝜕 𝜕 𝜕 𝜕 𝜕 𝜕𝜙 𝜕𝜙 𝜕𝜙 ∇𝜙 = ( , , )𝜙 = ( 𝑖 + 𝑗 + 𝑘) 𝜙 = 𝑖+ 𝑗+ 𝑘 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕 𝜕 𝜕 𝜕𝜙 𝜕𝜙 𝜕𝜙 =( 𝑒1 + 𝑒2 + 𝑒3 ) 𝜙 = 𝑒1 + 𝑒2 + 𝑒 𝜕𝑥1 𝜕𝑥2 𝜕𝑥3 𝜕𝑥1 𝜕𝑥2 𝜕𝑥3 3 3 𝜕𝜙 𝜕𝜙 =∑ 𝑒𝑖 = 𝑒𝑖 𝜕𝑥𝑖 𝜕𝑥𝑖 𝑖−1 Note that 1. Any index exists only once in product is called free index and we do not apply any summation over it. Page | 2  2. Any index repeated twice in product is called repeated index and we apply a summation over it from 1 to 3. 3. No index can be repeated more than two times in product Examples: 3 3 𝑎𝑖𝑗 𝑥𝑖 𝑥𝑗 = ∑ ∑ 𝑎𝑖𝑗 𝑥𝑖 𝑥𝑗 , 𝑖, 𝑗 = 1,2,3 (9 terms) 𝑖=1 𝑗=1 3 3 3 𝑎𝑖𝑗𝑘 𝑥𝑖 𝑥𝑗 𝑥𝑘 = ∑ ∑ ∑ 𝑎𝑖𝑗𝑘 𝑥𝑖 𝑥𝑗 𝑥𝑘 , 𝑖, 𝑗, 𝑘 = 1,2,3 (27 terms) 𝑖=1 𝑗=1 𝑘=1 Kronecker delta (𝜹𝒊𝒋 ) 1 𝑖𝑓 𝑖=𝑗 𝛿𝑖𝑗 = [ 0 𝑖𝑓 𝑖≠𝑗 𝛿11 = 𝛿22 = 𝛿33 = 1 , 𝛿12 = 𝛿13 = 𝛿23 = 𝛿21 = 𝛿31 = 𝛿32 = 0 𝛿𝑖𝑗 sometimes called “substitution operator”. Example: show that 𝜹𝒊𝒌 𝒂𝒌 = 𝒂𝒊 Solution: 𝛿𝑖𝑘 𝑎𝑘 = 𝛿𝑖1 𝑎1 + 𝛿𝑖2 𝑎2 + 𝛿𝑖3 𝑎3 Since 𝑖 can be 1,2, or 3 then 𝑖𝑓 𝑖 = 1 ⇒ 𝛿𝑖1 = 1 , 𝛿𝑖2 = 𝛿𝑖3 = 0 ∴ 𝛿1𝑘 𝑎𝑘 = 𝑎1 𝑖𝑓 𝑖 = 2 ⇒ 𝛿𝑖2 = 1 , 𝛿𝑖1 = 𝛿𝑖3 = 0 ∴ 𝛿2𝑘 𝑎𝑘 = 𝑎2 𝑖𝑓 𝑖 = 3 ⇒ 𝛿𝑖3 = 1 , 𝛿𝑖1 = 𝛿𝑖2 = 0 ∴ 𝛿3𝑘 𝑎𝑘 = 𝑎3 ∴ 𝛿𝑖𝑘 𝑎𝑘 = 𝑎𝑖 So, the operator 𝛿𝑖𝑘 affects 𝑎𝑘 by replacing the index 𝑘 with the index 𝑖. Page | 3 Notes: 1. The repeated index sometimes called “dummy index” because we can always replace it with another index for example 𝑎𝑖 𝑏𝑖 = 𝑎𝑘 𝑏𝑘 = 𝑎𝑗 𝑏𝑗 = 𝑎1 𝑏1 + 𝑎2 𝑏2 + 𝑎3 𝑏3 also 𝑎𝑖𝑗 𝑥𝑗 = 𝑎𝑖𝑘 𝑥𝑘 𝜕𝑥𝑖 1 𝑖𝑓 𝑖 = 𝑘 2. ={ 𝜕𝑥𝑘 0 𝑖𝑓 𝑖≠𝑘 where 𝑥1 , 𝑥2 , 𝑥3 are independent variables, i.e. 𝜕𝑥𝑖 = 𝛿𝑖𝑘 , 𝑒 𝑖 ⋅ 𝑒 𝑘 = 𝛿𝑖𝑘 𝜕𝑥𝑘 𝛿𝑖𝑖 = 𝛿𝑗𝑗 = 𝛿𝑘𝑘 = 𝛿11 + 𝛿22 + 𝛿33 = 3 Example: Show that 𝝏𝒓 𝒙𝒌 = 𝝏𝒙𝒌 𝒓 Solution: Since we can write the norm of the position vector as 𝑟 2 = 𝑟 ⋅ 𝑟 = 𝑥 ⋅ 𝑥 = 𝑥𝑖 𝑥𝑖 differentiate both sides w.r.t. 𝑥𝑘 𝜕𝑟 𝜕𝑥𝑖 𝜕𝑥𝑖 𝜕𝑥𝑖 2𝑟 = 𝑥𝑖 + 𝑥𝑖 = 2𝑥𝑖 = 2𝑥𝑖 𝛿𝑖𝑘 = 2𝑥𝑘 𝜕𝑥𝑘 𝜕𝑥𝑘 𝜕𝑥𝑘 𝜕𝑥𝑘 𝜕𝑟 𝜕𝑥𝑖 𝑟 = 𝑥𝑖 = 𝑥𝑖 𝛿𝑖𝑘 = 𝑥𝑘 𝜕𝑥𝑘 𝜕𝑥𝑘 𝜕𝑟 𝑥𝑘 = 𝜕𝑥𝑘 𝑟 Example: Find the value of (𝑖) 𝛿𝑖𝑘 𝛿𝑖𝑘 (𝑖𝑖) 𝛿𝑖𝑘 𝛿𝑖𝑚 Page | 4 Solution: (𝑖) 𝛿𝑖𝑘 𝛿𝑖𝑘 = 𝛿𝑖𝑖 = 𝛿11 + 𝛿22 + 𝛿33 = 1 + 1 + 1 = 3 (𝑖𝑖) 𝛿𝑖𝑘 𝛿𝑖𝑚 = 𝛿𝑘𝑚 Direction cosines: The direction cosine of a vector 𝑥 ′ are cos 𝛼 , cos 𝛽 , cos 𝛾 , where 𝛼, 𝛽, 𝛾 are the angles between the vector 𝑥 ′ and the positive coordinates of 𝑥1 , 𝑥2 , 𝑥3. 𝑥3 𝑒Ԧ3 𝑒𝑖ƴ 𝑥ƴ 𝜸 𝜷 𝑒Ԧ1 𝜶 𝑥2 𝑒Ԧ2 𝑥1 ∴ cos 𝛼 = 𝑒1′ ⋅ 𝑒1 = ℓ11 where 𝑒1′ is the unit vector in the direction of 𝑥1′ and 𝑒1 is the unit vector in the direction of 𝑥1. Also cos 𝛽 = 𝑒 ′1 ⋅ 𝑒 2 = ℓ12 cos 𝛾 = 𝑒 ′1 ⋅ 𝑒 3 = ℓ13 ∴ ℓ1𝑗 = 𝑒 ′1 ⋅ 𝑒 𝑗 Similarly, ℓ2𝑗 = 𝑒 ′2 ⋅ 𝑒 𝑗 , ℓ3𝑗 = 𝑒 ′3 ⋅ 𝑒 𝑗 ∴ ℓ𝑖𝑗 = 𝑒 ′𝑖 ⋅ 𝑒 𝑗 Transformation law: Suppose that we have a coordinate system with perpendicular axes and has unit vectors 𝑒 1 , 𝑒 2 , 𝑒 3 as shown in the figure. Page | 5 If we rotate these axes around the origin 𝑂 such that they are parallel to other unit vectors 𝑒 ′1 , 𝑒 ′2 , 𝑒 ′3 then the position vector 𝑥 of a fixed point 𝑃 for the old axes is 𝑥 = (𝑥1 , 𝑥2 , 𝑥3 ) = 𝑥𝑗 𝑒𝑗 (1) and for the new axes 𝑥 = (𝑥1ƴ , 𝑥2ƴ , 𝑥3ƴ ) = 𝑥𝑖′ 𝑒 ′𝑖 (2) 𝑥3 𝑥ƴ 3 𝑝 𝑥 𝑥ƴ 2 𝑒Ԧ3 𝑒Ԧ3′ 𝑒Ԧ2′ 𝑥2 𝑒Ԧ1 𝑒Ԧ2 𝑒Ԧ1′ 𝑥1 𝑥ƴ 1 If we want to express (𝑥1ƴ , 𝑥2ƴ , 𝑥3ƴ ) in terms of (𝑥1 , 𝑥2 , 𝑥3 ) and (𝑒 ′1 , 𝑒 ′2 , 𝑒 ′3 ) in terms of (𝑒 1 , 𝑒 2 , 𝑒 3 ), then from (2) 𝑥𝑖′ = 𝑥 ∙ 𝑒 ′𝑖 = 𝑥𝑗′ 𝑒𝑗′ ⋅ 𝑒 ′𝑖 (from 2) 𝑥𝑖′ = 𝑥𝑗 𝑒 𝑗 ⋅ 𝑒 ′𝑖 (from 1) 𝑥𝑖′ = ( 𝑒 𝑗 ⋅ 𝑒 ′𝑖 ) 𝑥𝑗 = ℓ𝑖𝑗 𝑥𝑗 (3) Note that for any vector 𝑢: (𝑢 ⋅ 𝑒 𝑖 ) 𝑒 𝑖 = (𝑢𝑗 𝑒 𝑗 ⋅ 𝑒 𝑖 ) 𝑒 𝑖 = 𝑢𝑗 𝛿𝑖𝑗 𝑒 𝑖 = 𝑢𝑖 𝑒 𝑖 = 𝑢 𝑒 ′𝑖 = (𝑒 ′𝑖 ⋅ 𝑒 𝑗 ) 𝑒 𝑗 = ℓ𝑖𝑗 𝑒 𝑗 (4) So, we obtained a formula to transfer vectors when rotated as in (3) and a formula to transfer unit vectors as in (4). These rules are important in studying tensors. Page | 6  ∴ ℓ𝑖𝑘 𝑒 ′𝑖 = ℓ𝑖𝑘 ℓ𝑖𝑗 𝑒 𝑗 (from 4) ℓ𝑖𝑘 𝑒 ′𝑖 = 𝛿𝑗𝑘 𝑒 𝑗 = 𝑒 𝑘 (5) so, from (4) and (5) 𝑒 ′𝑖 = ℓ𝑖𝑗 𝑒 𝑗 , 𝑒 𝑖 = ℓ𝑗𝑖 𝑒 ′𝑗 (6) Also, for any vector 𝑥 𝑥 = 𝑥𝑖 𝑒 𝑖 = 𝑥𝑖 ℓ𝑗𝑖 𝑒 ′𝑗 (from 6𝑖𝑖 ) (7) and 𝑥 = 𝑥𝑗′ 𝑒 ′𝑗 = 𝑥 ′𝑗 ℓ𝑗𝑖 𝑒 𝑖 (from 6𝑖𝑖 ) (8) From (7) and (8) 𝑥𝑖 𝑒 𝑖 = 𝑥 ′𝑗 𝑒 ′𝑗 𝑥𝑖 𝑒 𝑖 ⋅ 𝑒 𝑘 = 𝑥𝑗′ 𝑒 ′𝑗 ⋅ 𝑒 𝑘 ⇒ 𝑥𝑖 𝛿 𝑖𝑘 = 𝑥𝑗′ 𝑒 ′𝑗 ⋅ 𝑒 𝑘 𝑥𝑘 = 𝑥 ′𝑗 ℓ𝑗𝑘 (9) From (3) and (9) 𝑥𝑖′ = ℓ𝑖𝑗 𝑥𝑗 , 𝑥𝑖 = ℓ𝑗𝑖 𝑥𝑗′ (10) The vector 𝑟 has components 𝑥𝑖 or 𝑥 ′𝑖 according to the axes used where, 𝑥𝑖 = 𝑙𝑗𝑖 𝑥 ′𝑖 , 𝑥𝑖′ = 𝑙𝑖𝑗 𝑥𝑗 Here, the vector is called a tensor of order one. Rotation Matrix: The nine quantities 𝑙𝑖𝑗 can be represented as a matrix called the rotation matrix. i.e. 𝑙11 𝑙12 𝑙13 𝐿 = [𝑙𝑖𝑗 ] = [𝑙21 𝑙22 𝑙23 ] 𝑙31 𝑙32 𝑙33 ′ Since 𝑙𝑖𝑘 𝑙𝑘𝑗 = 𝑙𝑖𝑘 𝑙𝑗𝑘 = 𝑙𝑖𝑘 𝑒 ′𝑗 ⋅ 𝑒 𝑘 = 𝑒 ′𝑗 ⋅ 𝑙𝑖𝑘 𝑒 𝑘 = 𝑒 ′𝑗 ⋅ 𝑒 ′𝑖 (from 4) ′ ∴ 𝑙𝑖𝑘 𝑙𝑘𝑗 = 𝛿𝑖𝑗 (11) and 𝑖𝑓 𝐼 = [𝑆𝑖𝑗 ], then from (11) 𝐿 𝐿𝑇 = 𝐼 Page | 7  ∵ det 𝐿 = det 𝐿𝑇 ∴ (det 𝐿)2 = 1 ⇒ det 𝐿 = ±1 (12) ∵ 𝐿 𝐿𝑇 = 𝐼 ⇒ 𝐿𝑇 𝐿 = 𝐼 ∴ 𝑙𝑗𝑖′ 𝑙𝑖𝑘 = 𝑙𝑖𝑗 𝑙𝑖𝑘 = 𝛿𝑗𝑘 i.e., the rotation matrix must satisfy equation (12) and we say that its orthogonal matrix. We say that 𝐿 is proper orthogonal if det 𝐿 = +1. This mean that we have a rotation only. We say that 𝐿 is improper orthogonal if det 𝐿 = −1, it means that we have a rotation and a reflection. In the previous case 𝑒 1 × 𝑒 2 = 𝑒 3 and in the latter case 𝑒 1 × 𝑒 2 = −𝑒 3. Note that when the axes 𝑥𝑖 are rotated about the origin to the near axes 𝑥 ′𝑖 , the distance |𝑂𝑃| remains invariant, and we call the distance scalar quantity or a tensor of order zero. Example: prove that the OP is uncharged under the effect of rotation. Solution: ⃗⃗⃗⃗⃗Ԧ = 𝑥𝑖 and after rotation 𝑂𝑃 Let 𝑂𝑃 ⃗⃗⃗⃗⃗Ԧ′ = 𝑥𝑖′ , |𝑂𝑃 ⃗⃗⃗⃗⃗Ԧ′| = 𝑥𝑖′ 𝑥𝑖′ ∵ 𝑥𝑖′ 𝑥𝑖′ = (𝑙𝑖𝑗 𝑥𝑗 )(𝑙𝑖𝑘 𝑥𝑘 ) = 𝑙𝑖𝑗 𝑙𝑖𝑘 𝑥𝑗 𝑥𝑘 = 𝛿𝑗𝑘 𝑥𝑗 𝑥𝑘 = 𝑥𝑘 𝑥𝑘 ∴ 𝑥1′ 2 + 𝑥2′ 2 + 𝑥3′ 2 = 𝑥12 + 𝑥22 + 𝑥32 ∴ |𝑂𝑃| does not change under the effect of rotation. Cartesian tensors Suppose that 𝑉 is a vector space. A tensor 𝑇 of order two is a linear mapping 𝑇: 𝑉 → 𝑉. If we have unit basis vectors which are perpendicular and orthonormal in 𝑉 then 𝑇 is called a cartesian tensor. Suppose that 𝑢, 𝑣 ∈ 𝑉 such that 𝑢 = 𝑇𝑣 Page | 8 In case we have a unit basis vectors {𝑒𝑖 } then 𝑢𝑖 = 𝑇𝑖𝑗 𝑣𝑗 (13) and if we have a unit basis vector {𝑒𝑖′ } then 𝑢𝑝′ = 𝑇𝑝𝑞 ′ 𝑣𝑞′ (14) since 𝑢 and 𝑣 are vectors then their components transform according to (12) 𝑖. 𝑒. 𝑢𝑝′ = ℓ𝑝𝑖 𝑢𝑖 , 𝑣𝑞′ = ℓ𝑞𝑗 𝑣𝑗 substitute in (14) 𝑢𝑝′ = 𝑇𝑝𝑞 ′ 𝑣𝑞′ ⇒ ′ ℓ𝑝𝑖 𝑢𝑖 = 𝑇𝑝𝑞 ℓ𝑞𝑗 𝑣𝑗 using (13) ′ ∴ ℓ𝑝𝑖 𝑇𝑖𝑗 𝑣𝑗 = 𝑇𝑝𝑞 ℓ𝑞𝑗 𝑣𝑗 ′ ∴ ( ℓ𝑝𝑖 𝑇𝑖𝑗 − 𝑇𝑝𝑞 ℓ𝑞𝑗 ) 𝑣𝑗 = 0 and since this is true for all 𝑣 ∈ 𝑉, then ′ (ℓ𝑝𝑖 𝑇𝑖𝑗 − 𝑇𝑝𝑞 ℓ𝑞𝑗 ) = 0 ′ ∴ ℓ𝑝𝑖 𝑇𝑖𝑗 = 𝑇𝑝𝑞 ℓ𝑞𝑗 ′ 𝑖. 𝑒. ℓ𝑝𝑖 𝑇𝑖𝑗 = 𝑇𝑝𝑟 ℓ𝑟𝑗 multiply both sides by ℓ𝑞𝑗 ℓ𝑞𝑗 ℓ𝑝𝑖 𝑇𝑖𝑗 = ℓ𝑞𝑗 ℓ𝑟𝑗 𝑇𝑝𝑟′ = 𝛿𝑞𝑟 𝑇𝑝𝑟 ′ ′ = 𝑇𝑝𝑞 ′ (15) ∴ 𝑇𝑝𝑞 = ℓ𝑝𝑖 ℓ𝑞𝑗 𝑇𝑖𝑗 This is the transformation law for the components of a cartesian tensor of order two (a 𝐶𝑇(2)) where we transform from unit basis vectors {𝑒𝑖 } to unit basis vectors {𝑒𝑖′ }. ′ Note that ℓ𝑝𝑖 ℓ𝑞𝑗 𝑇𝑝𝑞 = ℓ𝑝𝑖 ℓ𝑞𝑗 (ℓ𝑝𝑟 ℓ𝑞𝑠 𝑇𝑟𝑠 ) (𝑓𝑟𝑜𝑚 15) = ℓ𝑝𝑖 ℓ𝑝𝑟 ℓ𝑞𝑗 ℓ𝑞𝑠 𝑇𝑟𝑠 = 𝛿𝑖𝑟 𝛿𝑗𝑠 𝑇𝑟𝑠 = 𝑇𝑖𝑗 Page | 9 The definition of a 𝐶𝑇(2) can be generalized to a 𝐶𝑇(𝑛) with components 𝑇 𝑖𝑗𝑘... ⏟ 𝑛 indices and can be transformed according to ′ (16) 𝑇𝑝𝑞𝑟… = ℓ𝑝𝑖 ℓ𝑞𝑗 ℓ𝑟𝑘... 𝑇𝑖𝑗𝑘… when we change the basis vectors {𝑒𝑖 } to {𝑒𝑖′ }. Examples: 1. Any scalar quantity is a cartesian tensor of order zero i.e. 𝐶𝑇(0), e.g. 𝜙 ′ = 𝜙 for any scalar quantity 𝜙. 2. Any vector 𝑣 is a cartesian tensor of order one i.e. 𝐶𝑇(1) e.g. 𝑣𝑝′ = ℓ𝑝𝑖 𝑣𝑖 3. The tensor product of any two vectors 𝑢 , 𝑣 is defined by 𝑢⨂𝑣 with components (𝑢⨂𝑣)𝑖𝑗 = 𝑢𝑖 𝑣𝑗. This is a cartesian tensor of order 2 i.e 𝐶𝑇(2) because 𝑢𝑖′ 𝑣𝑗′ = (ℓ𝑖𝑝 𝑢𝑝 )(ℓ𝑗𝑞 𝑣𝑞 ) = ℓ𝑖𝑝 ℓ𝑗𝑞 𝑢𝑝 𝑣𝑞 This product is sometimes called dyadic product. 4. The Kronecker delta 𝛿𝑖𝑗 is a cartesian tensor of order 2, 𝐶𝑇(2) because ′ ′ ′ 𝛿𝑝𝑞 = 𝑒⏟𝑝 ∙ 𝑒 𝑞 = (ℓ ⏟ 𝑝𝑖 𝑒 𝑖 ) ∙ (ℓ𝑞𝑗 𝑒 𝑗 ) = ℓ𝑝𝑖 ℓ𝑞𝑗 ( 𝑒 𝑖 ∙ 𝑒 𝑗 ) = ℓ𝑝𝑖 ℓ𝑞𝑗 𝛿𝑖𝑗 𝑓𝑟𝑜𝑚(5) 𝑓𝑟𝑜𝑚(8𝑖) 5. The stress tensor 𝜏𝑖𝑗 is a cartesian tensor of order two (𝑖. 𝑒. 𝐶𝑇(2)) where ′ 𝜏𝑝𝑞 = ℓ𝑝𝑖 ℓ𝑞𝑗 𝜏𝑖𝑗 6. The permutation tensor 𝜖 is a cartesian tensor of order three (𝐶𝑇(3)). Note that 𝑒𝑣𝑒𝑛 𝑝𝑒𝑟𝑚𝑢𝑡𝑎𝑡𝑖𝑜𝑛 → 𝜖123 = 𝜖312 = 𝜖231 = 1 𝑜𝑑𝑑 𝑝𝑒𝑟𝑚𝑢𝑡𝑎𝑡𝑖𝑜𝑛 → 𝜖132 = 𝜖213 = 𝜖321 = −1 Page | 10  The rest are zeros, i.e. 𝑖𝑓 𝑖 = 𝑗, 𝑜𝑟 𝑗 = 𝑘 𝑜𝑟 𝑘 = 𝑖 ⏟ 𝑛𝑜 𝑝𝑒𝑟𝑚𝑢𝑡𝑎𝑡𝑖𝑜𝑛 Note: In example (3) we have to notice that the cartesian tensor of order two cannot be represented as a dyadic product, i.e. for any 𝑇, then 𝑇 cannot be written as 𝑢 ⊗ 𝑣 for any two vectors 𝑢, 𝑣. In other words, there are no 𝑢, 𝑣 such that 𝑇𝑖𝑗 = 𝑢𝑖 𝑣𝑗. In fact, we can write. 𝑇 = 𝑇𝑖𝑗 (𝑒 𝑖 ⊗ 𝑒 𝑗 ) or 𝑇 = 𝑇𝑖𝑗′ (𝑒 ′𝑖 ⊗ 𝑒 ′𝑗 ) for any basis vectors. Representing CT(2)'s with matrices: Tensor equations are equations independent of any basis, and when we choose the basis of {𝑒⃑𝑖 } then 𝐶𝑇(2) has elements 𝑇𝑖𝑗 with respect to {𝑒⃑𝑖 }. And if we define the following matrices: 𝑇 = [𝑇𝑖𝑗 ] , 𝐿 = [ℓ𝑖𝑗 ] , 𝑇 ′ = [𝑇𝑖𝑗′ ] then from the transformation law (15) 𝑇̀𝑝𝑞 = ℓ𝑝𝑖 ℓ𝑞𝑗 𝑇𝑖𝑗 which is written in matrix form as: 𝑇 ′ = 𝐿 𝑇 𝐿𝑇 multiplying from the left by 𝐿𝑇 we get: 𝐿𝑇 𝑇 ′ = 𝑇 𝐿𝑇 multiplying from the left by 𝐿 we get: 𝐿𝑇 𝑇 ′ 𝐿 = 𝑇 𝑜𝑟 𝑇 = 𝐿𝑇 𝑇 ′ 𝐿 By components Page | 11  𝑇̀𝑝𝑞 = ℓ𝑝𝑖 ℓ𝑞𝑗 𝑇𝑖𝑗 multiplying by ℓ𝑝𝑟 ℓ𝑞𝑠 we find that: ℓ𝑝𝑟 ℓ𝑞𝑠 𝑇 ′ 𝑝𝑞 = ℓ𝑝𝑟 ℓ𝑞𝑠 ℓ𝑝𝑖 ℓ𝑞𝑗 𝑇𝑖𝑗 ℓ𝑝𝑟 ℓ𝑞𝑠 𝑇 ′ 𝑝𝑞 = 𝛿𝑟𝑖 𝛿𝑠𝑗 𝑇𝑖𝑗 ℓ𝑝𝑟 ℓ𝑞𝑠 𝑇 ′ 𝑝𝑞 = 𝑇𝑟𝑠 ∴ 𝑇 = 𝐿𝑇 𝑇𝐿 Operations on tensors: Suppose that we have a scalar quantity 𝜙 and two tensors 𝑇 , 𝑆 which are of order 𝑛 i.e 𝐶𝑇(𝑛), then 𝜙𝑇 , 𝑇+𝑠 are 𝐶𝑇(𝑛). Also if 𝑇 is a 𝐶𝑇(𝑛) and 𝑆 is a 𝐶𝑇(𝑚) then 𝑈 = 𝑆 𝑇 is defined as 𝑈 𝑖𝑗𝑘…𝑝𝑞𝑟𝑠… ⏟ =𝑆 𝑖𝑗𝑘…. ⏟ 𝑇 𝑞𝑟𝑠…. ⏟ (𝑛+𝑚)𝑖𝑛𝑑𝑖𝑐𝑒𝑠 𝑚 𝑖𝑛𝑑𝑖𝑐𝑒𝑠 𝑛 𝑖𝑛𝑑𝑖𝑐𝑒𝑠 Which is a 𝐶𝑇(𝑛 + 𝑚). Contraction of tensors: Suppose that 𝑇 is a 𝐶𝑇(𝑛) with components 𝑇𝑖𝑗𝑘…𝑝𝑞… for the basis vectors. If we put 𝑃 = 𝑘 and add with repeat to 𝑘 then we say that the tensor 𝑇 is contracted and we obtain a 𝐶𝑇(𝑛 − 2). ′ 𝑇𝑎𝑏𝑐…𝑔ℎ… = ℓ𝑎𝑖 ℓ𝑏𝑗 ℓ𝑐𝑘 … ℓ𝑔𝑝 ℓℎ𝑞 … 𝑇𝑖𝑗𝑘…𝑝𝑞… let 𝑔 = 𝑐 then ′ 𝑇𝑎𝑏𝑐…𝑐ℎ… = ℓ𝑎𝑖 ℓ𝑏𝑗 ℓ ⏟𝑐𝑘 … ℓ𝑐𝑝 ℓℎ𝑞 … 𝑇𝑖𝑗𝑘…𝑝𝑞… 𝛿𝑘𝑝 ′ ∴ 𝑇𝑎𝑏𝑐…𝑐ℎ… = ℓ𝑎𝑖 ℓ𝑏𝑗 … ℓℎ𝑞 … 𝑇𝑖𝑗𝑘…𝑘𝑞… Page | 12 Examples: ′ 1. 𝑇 is a 𝐶𝑇(2) and its components are transformed according to 𝑇𝑝𝑞 = ℓ𝑝𝑖 ℓ𝑞𝑗 𝑇𝑖𝑗. let 𝑝 = 𝑞 then ′ 𝑇𝑞𝑞 = ℓ𝑞𝑖 ℓ𝑞𝑗 𝑇𝑖𝑗 = 𝛿𝑖𝑗 𝑇𝑖𝑗 = 𝑇𝑖𝑖 = 𝑇𝑞𝑞 i.e. 𝑇𝑞𝑞 is a 𝐶𝑇(0) which is a scalar quantity (invariant) and we call it trace of 𝑇 and denoted by 𝑡𝑟(𝑇) which is the sum of the diagonal components of 𝑇. Note that 𝑡𝑟(𝑇 𝑛 ) is also a scalar quantity. 2. The dyadic product 𝑢 ⊗ 𝑣 becomes a scalar quantity by contraction (𝑢 ⊗ 𝑣)𝑖𝑗 = 𝑢𝑖 𝑣𝑗 𝑖=𝑗 ⃗⃗⃗⃗⃗⃗⃗⃗⃗Ԧ (𝑢 ⊗ 𝑣)𝑖𝑖 = 𝑢𝑖 𝑣𝑖 = 𝑢 ∙ 𝑣 which is a scalar quantity i.e. 𝐶𝑇(0), that is, the dyadic product of two vectors is transformed by contraction into the inner product of these two vectors. 3. If 𝑇 is a 𝐶𝑇(4) then ′ 𝑇𝑝𝑞𝑟𝑠 = ℓ𝑝𝑖 ℓ𝑞𝑗 ℓ𝑟𝑘 ℓ𝑠𝑚 𝑇𝑖𝑗𝑘𝑚 contract with 𝑟 = 𝑠 ′ ∴ 𝑇𝑝𝑞𝑠𝑠 = ℓ𝑝𝑖 ℓ𝑞𝑗 ℓ𝑠𝑘 ℓ𝑠𝑚 𝑇𝑖𝑗𝑘𝑚 = ℓ𝑝𝑖 ℓ𝑞𝑗 𝛿𝑘𝑚 𝑇𝑖𝑗𝑘𝑚 = ℓ𝑝𝑖 ℓ𝑞𝑗 𝑇𝑖𝑗𝑘𝑘 𝑇𝑖𝑗𝑘𝑘 is a 𝐶𝑇(2). Symmetric and anti-symmetric tensors: We say that the cartesian tensor of order two (𝐶𝑇(2)) is symmetric if 𝑇 𝑇 = 𝑇 and we say that it is skew symmetric 𝑇 ⏟𝑖𝑗 = 𝑇𝑗𝑖 𝑇 ⏟𝑖𝑗 = −𝑇𝑗𝑖 If 𝑇 𝑇 = −𝑇 𝑖. 𝑒. , 6 independent variables 3 independent variables Any 𝐶𝑇(2) can be written as the sum of its symmetric part with its skew- symmetric part. 1 1 𝑇 = (𝑇 + 𝑇 𝑇 ) + (𝑇 − 𝑇 𝑇 ) 2 2 Page | 13 Isotropic tensors: If the components of a cartesian tensor do not change under the effect of arbitrary rotation of the basis vectors, then this tensor is isotropic. i.e. 𝑇 is isotropic if 𝑇𝑖𝑗′ = 𝑇𝑖𝑗 Examples: 1. 𝐶𝑇(0): all scalar quantities are isotropic 2. 𝐶𝑇(1): only the trivial case of the vector 0 is isotropic 3. 𝐶𝑇(2): the product of a scalar quantity with 𝛿𝑖𝑗 is the only isotropic tensor of order two. 4. 𝐶𝑇(3): the product of scalar quantities with 𝜖𝑖𝑗𝑘 are the only isotropic tensors of order three. 5. 𝐶𝑇(4): only the independent tensors which are a linear combination of 𝛿𝑖𝑗 𝛿𝑘𝑚 , 𝛿𝑖𝑘 𝛿𝑗𝑚 , 𝛿𝑖𝑚 𝛿𝑗𝑘. 6. 𝐶𝑇(2𝑛): only the independent tensors which are a linear combination of the product of 𝑛 of 𝛿’s. 7. 𝐶𝑇(2𝑛 + 1): only the independent tensors which are linear combination of the product of (𝑛 − 1) of 𝛿’s with one 𝜖. Advanced topics in tensors: For any pair of vectors 𝑢 , 𝑣 , the dyadic product 𝑢 ⊗ 𝑣 can be defined by the identity (𝑢 ⊗ 𝑣)𝑖𝑗 = 𝑢𝑖 𝑣𝑗 ⇒ (𝑢 ⊗ 𝑣) = 𝑢𝑖 𝑣𝑗 ( 𝑒 𝑖 ⊗ 𝑒 𝑗 ) (𝑢 ⊗ 𝑣) 𝑎 ≡ 𝑢 (𝑣 ⋅ 𝑎) ∀𝑎 ∈ 𝑉 specifically (𝑒 𝑖 ⊗ 𝑒 𝑗 ) 𝑒 𝑘 = 𝑒 𝑖 (𝑒 𝑗 ⋅ 𝑒 𝑘 ) = 𝛿𝑗𝑘 𝑒 𝑖 Generally, if 𝑇 is a 𝐶𝑇(2) Page | 14  𝑇 𝑢 = (𝑇𝑖𝑗 𝑒 𝑖 ⊗ 𝑒 𝑗 ) 𝑢𝑘 𝑒 𝑘 = 𝑇𝑖𝑗 𝑢𝑘 ( 𝑒 𝑖 ⊗ 𝑒 𝑗 ) 𝑒 𝑘 = 𝑇𝑖𝑗 𝑢𝑘 (𝑒 𝑗 ⋅ 𝑒 𝑘 ) 𝑒 𝑖 = 𝑇𝑖𝑗 𝑢𝑘 𝛿𝑗𝑘 𝑒 𝑖 = 𝑇𝑖𝑗 𝑢𝑗 𝑒 𝑖 and 𝑢 𝑇 = 𝑢𝑘 𝑒 𝑘 (𝑇𝑖𝑗 𝑒 𝑖 ⊗ 𝑒 𝑗 ) = 𝑢𝑘 𝑇𝑖𝑗 𝑒 𝑘 ( 𝑒 𝑖 ⊗ 𝑒 𝑗 ) = 𝑢𝑘 𝑇𝑖𝑗 (𝑒 𝑖 ⋅ 𝑒 𝑘 ) 𝑒 𝑗 = 𝑢𝑘 𝑇𝑖𝑗 𝛿𝑖𝑘 𝑒 𝑗 = 𝑢𝑖 𝑇𝑖𝑗 𝑒 𝑗 so 𝑇 𝑒 𝑗 = 𝑇𝑖𝑗 𝑒𝑖 because 𝑇𝑒 𝑗 = (𝑇𝑖𝑘 𝑒 𝑖 ⊗ 𝑒 𝑘 ) 𝑒 𝑗 = 𝑇𝑖𝑗 (𝑒 𝑗 ⋅ 𝑒 𝑘 ) 𝑒 𝑖 = 𝑇𝑖𝑘 𝛿𝑘𝑗 𝑒 𝑖 = 𝑇𝑖𝑗 𝑒 𝑖 Note that: 𝑇 (𝑢 ⊗ 𝑣) = 𝑣 ⊗ 𝑢 Example: Find the value of ∆ = det(𝑎𝑖𝑗 ) using permutation tensor 𝜖 Solution: 𝑎 11 𝑎 12 𝑎 13 ∆ = det(𝑎𝑖𝑗 ) = |𝑎 21 𝑎 22 𝑎 23 | = 𝑎 31 𝑎 32 𝑎 33 𝑎22 𝑎23 𝑎21 𝑎23 𝑎21 𝑎22 = 𝑎11 |𝑎 𝑎 | − 𝑎12 |𝑎 𝑎 | + 𝑎13 |𝑎 𝑎 | 𝜖𝑖𝑗𝑘 𝑎1𝑖 𝑎2𝑗 𝑎3𝑘 32 33 31 33 31 32 = 𝑎11 (𝑎22 𝑎33 − 𝑎32 𝑎23 ) − 𝑎12 (𝑎21 𝑎33 − 𝑎23 𝑎31 ) + 𝑎13 (𝑎21 𝑎32 − 𝑎22 𝑎31 ) ∆ = 𝜖123 𝑎11 𝑎22 𝑎33 + 𝜖132 𝑎11 𝑎23 𝑎32 + 𝜖213 𝑎12 𝑎21 𝑎33 +𝜖231 𝑎12 𝑎23 𝑎31 + 𝜖312 𝑎13 𝑎21 𝑎32 + 𝜖321 𝑎13 𝑎22 𝑎31 = 𝜖𝑖𝑗𝑘 𝑎1𝑖 𝑎2𝑗 𝑎3𝑘 Also ∆= 𝜖𝑖𝑗𝑘 𝑎𝑖1 𝑎𝑗2 𝑎𝑘3 (17) If we replace 1 , 2 , 3 by 𝑟, 𝑠, 𝑡 in ∆ Page | 15  𝑎𝑟1 𝑎𝑟2 𝑎𝑟3 ∆ = |𝑎𝑠1 ′ 𝑎𝑠2 𝑎𝑠3 | = 𝜖𝑖𝑗𝑘 𝑎𝑟𝑖 𝑎𝑠𝑗 𝑎𝑡𝑘 = 𝜖𝑖𝑗𝑘 𝑎𝑖𝑟 𝑎𝑗𝑠 𝑎𝑘𝑡 (18) 𝑎𝑡1 𝑎𝑡2 𝑎𝑡3 then ∆′ = 𝜖𝑟𝑠𝑡 ∆ ∆′ = ∆ 𝑒𝑣𝑒𝑛 𝑝𝑒𝑟𝑚. ∆′ = −∆ 𝑜𝑑𝑑 𝑝𝑒𝑟𝑚. where ∆′ = 0 𝑛𝑜 𝑝𝑒𝑟𝑚. (𝑡𝑤𝑜 𝑟𝑜𝑤𝑠 𝑎𝑟𝑒 𝑒𝑞𝑢𝑎𝑙) 𝑖. 𝑒. 𝜖𝑟𝑠𝑡 ∆ = 𝜖𝑖𝑗𝑘 𝑎𝑟𝑖 𝑎𝑠𝑗 𝑎𝑡𝑘 = 𝜖𝑖𝑗𝑘 𝑎𝑖𝑟 𝑎𝑗𝑠 𝑎𝑘𝑡 Example: Prove that 𝑢 ⋅ (𝑣 × 𝑤) = 𝜖𝑖𝑗𝑘 𝑢𝑖 𝑣𝑗 𝑤𝑘 Solution: 𝑢 ⋅ (𝑣 × 𝑤) = 𝑢𝑖 𝑒 𝑖 ⋅ (𝑣𝑗 𝑤𝑘 𝜖𝑗𝑘𝑟 𝑒 𝑟 ) = 𝑢𝑖 𝑣𝑗 𝑤𝑘 𝜖𝑗𝑘𝑟 (𝑒 𝑖 ⋅ 𝑒 𝑟 ) = 𝑢𝑖 𝑣𝑗 𝑤𝑘 𝜖𝑗𝑘𝑟 𝛿𝑖𝑟 = 𝑢𝑖 𝑣𝑗 𝑤𝑘 𝜖𝑗𝑘𝑖 = 𝑢𝑖 𝑣𝑗 𝑤𝑘 𝜖𝑖𝑗𝑘 Example: Suppose that we have the determinant 𝛿𝑟1 𝛿𝑟2 𝛿𝑟3 | 𝛿𝑠1 𝛿𝑠2 𝛿𝑠3 | 𝛿𝑡1 𝛿𝑡2 𝛿𝑡3 then find a form for 𝜖𝑖𝑗𝑘 𝜖𝑟𝑠𝑡 and prove that (𝑖) 𝜖𝑖𝑗𝑘 𝜖𝑟𝑠𝑘 = 𝛿𝑖𝑟 𝛿𝑗𝑠 − 𝛿𝑖𝑠 𝛿𝑗𝑟 (𝑖𝑖) 𝜖𝑖𝑗𝑘 𝜖𝑟𝑗𝑘 = 2 𝛿𝑖𝑟 (𝑖𝑖𝑖) 𝜖𝑖𝑗𝑘 𝜖𝑖𝑗𝑘 = 6 Solution: From (17) 𝛿𝑟1 𝛿𝑟2 𝛿𝑟3 | 𝛿𝑠1 𝛿𝑠2 𝛿𝑠3 | = 𝜖𝑖𝑗𝑘 𝛿𝑖𝑟 𝛿𝑗𝑠 𝛿𝑘𝑡 = 𝜖𝑟𝑠𝑡 𝛿𝑡1 𝛿𝑡2 𝛿𝑡3 𝛿𝑟1 𝛿𝑟2 𝛿𝑟3 𝛿𝑟𝑖 𝛿𝑟𝑗 𝛿𝑟𝑘 𝜖𝑖𝑗𝑘 𝜖𝑟𝑠𝑡 = 𝜖𝑖𝑗𝑘 | 𝛿𝑠1 𝛿𝑠2 𝛿𝑠3 | = | 𝛿𝑠𝑖 𝛿𝑠𝑗 𝛿𝑠𝑘 | 𝛿𝑡1 𝛿𝑡2 𝛿𝑡3 𝛿𝑡𝑖 𝛿𝑡𝑗 𝛿𝑡𝑘 Page | 16 because multiplying by 𝜖𝑖𝑗𝑘 replacing 1,2,3 by 𝑖, 𝑗, 𝑘 (𝑖) 𝑙𝑒𝑡 𝑡 = 𝑘 𝛿𝑟𝑖 𝛿𝑟𝑗 𝛿𝑟𝑘 ∴ 𝜖𝑖𝑗𝑘 𝜖𝑟𝑠𝑘 = | 𝛿𝑠𝑖 𝛿𝑠𝑗 𝛿𝑠𝑘 | 𝛿𝑘𝑖 𝛿𝑘𝑗 𝛿𝑘𝑘 = 𝛿𝑟𝑖 ( 𝛿𝑠𝑗 𝛿𝑘𝑘 − 𝛿𝑠𝑘 𝛿𝑘𝑗 ) − 𝛿𝑟𝑗 ( 𝛿𝑠𝑖 𝛿𝑘𝑘 − 𝛿𝑠𝑘 𝛿𝑘𝑖 ) + 𝛿𝑟𝑘 ( 𝛿𝑠𝑖 𝛿𝑘𝑗 − 𝛿𝑠𝑗 𝛿𝑘𝑖 ) = 3 𝛿𝑟𝑖 𝛿𝑠𝑗 − 𝛿𝑟𝑖 𝛿𝑠𝑗 − 3 𝛿𝑟𝑗 𝛿𝑠𝑖 + 𝛿𝑟𝑗 𝛿𝑠𝑖 + 𝛿𝑠𝑖 𝛿𝑟𝑗 − 𝛿𝑠𝑗 𝛿𝑟𝑖 = 2 𝛿𝑟𝑖 𝛿𝑠𝑗 − 2 𝛿𝑟𝑗 𝛿𝑠𝑖 + 𝛿𝑟𝑗 𝛿𝑠𝑖 − 𝛿𝑠𝑗 𝛿𝑟𝑖 = 𝛿𝑟𝑖 𝛿𝑠𝑗 − 𝛿𝑟𝑗 𝛿𝑠𝑖 (𝑖𝑖) if 𝑠 = 𝑗 in (𝑖) then 𝜖𝑖𝑗𝑘 𝜖𝑟𝑗𝑘 = 𝛿𝑟𝑖 𝛿𝑗𝑗 − 𝛿𝑟𝑗 𝛿𝑗𝑖 = 3 𝛿𝑟𝑖 − 𝛿𝑟𝑖 = 2 𝛿𝑟𝑖 (𝑖𝑖𝑖) if 𝑖 = 𝑟 in (𝑖𝑖) then 𝜖𝑖𝑗𝑘 𝜖𝑖𝑗𝑘 = 2 𝛿𝑖𝑖 = 2 × 3 = 6 The axial tensor: Suppose that [ℓ𝑖𝑗 ] represents the rotation matrix of a normal transformation ∴ ℓ𝑟𝑖 ℓ𝑠𝑗 ℓ𝑡𝑘 𝜖𝑖𝑗𝑘 = 𝜖𝑟𝑠𝑡 ∆ (1) where ∆ = det ℓ𝑖𝑗. Equation (1) can be written in the form 𝜖𝑟𝑠𝑡 = ∆ ℓ𝑟𝑖 ℓ𝑠𝑗 ℓ𝑡𝑘 𝜖𝑖𝑗𝑘 i.e. 𝜖𝑖𝑗𝑘 satisfy the transformation law of order three since there is a transform from right-hand system to left-hand system then ∆ must be there so it has the value -1. If we have such transform, we call 𝜖𝑖𝑗𝑘 here the axial tenor. If ∆ = 1 then this tensor ′ is called proper tensor. For a 𝐶𝑇(2) 𝑇𝑝𝑞 = ℓ𝑝𝑖 ℓ𝑞𝑗 𝑇𝑖𝑗 and if the transformation law ∗ of the form 𝑇𝑝𝑞 = ∆ ℓ𝑝𝑖 ℓ𝑞𝑗 𝑇𝑖𝑗 where ∆ = ±1 then 𝑇𝑖𝑗 is called the axial tensor of order two. In general, the transformation law of order n of an axial tensor is ∗ 𝑇𝑝𝑞𝑟… = ∆ ℓ𝑝𝑖 ℓ𝑞𝑗 ℓ𝑟𝑘 … 𝑇𝑖𝑗𝑘… Page | 17 Note that: - Addition or subtraction of two axial tensors of the same order is an axial tensor of the same order. - Product of an axial tensor with a proper tensor is an axial tensor. - Product of two axial tensors is a proper tensor. - The contraction of an axial tensor of order n produce an axial tensor of order (𝑛 − 2). Example: 1- If 𝑇𝑖𝑗𝑘 , 𝑇𝑝𝑞 are two axial tensors of order 3 and 2 respectively, then ∗ ∗ 𝑇𝑟𝑠𝑡 𝑇𝑚𝑛 = (∆ ℓ𝑟𝑖 ℓ𝑠𝑗 ℓ𝑡𝑘 𝑇𝑖𝑗𝑘 )(∆ ℓ𝑚𝑝 ℓ𝑛𝑞 𝑇𝑝𝑞 ) = ℓ𝑟𝑖 ℓ𝑠𝑗 ℓ𝑡𝑘 ℓ𝑚𝑝 ℓ𝑛𝑞 𝑇𝑖𝑗𝑘 𝑇𝑝𝑞 which is a proper tensor of order 5. 2- (𝑐𝑢𝑟𝑙 𝑢 ⃗Ԧ)𝑖 = 𝜖𝑖𝑗𝑘 𝑢𝑘,𝑗 , The curl is an axial vector. The characteristic equation of a 𝑪𝑻(𝟐) det(𝑇 − 𝑡 𝐼) = 0 𝑇11 − 𝑡 𝑇12 𝑇13 | 𝑇21 𝑇22 − 𝑡 𝑇23 | = 0 𝑇31 𝑇32 𝑇33 − 𝑡 Which is a cubic equation in t. clearly 𝑡 3 − 𝐼1 𝑡 2 + 𝐼2 𝑡 − 𝐼3 = 0 Where 𝐼1 , 𝐼2 , 𝐼3 are principal invariant of 𝑇 such that 𝐼1 = 𝑡𝑟 (𝑇) = 𝑇𝑖𝑖 = 𝑇11 + 𝑇22 + 𝑇33 1 𝐼2 = {𝐼12 − 𝑡𝑟(𝑇 2 )} 2 𝑇 11 𝑇 12 𝑇 13 𝐼3 = det 𝑇 = |𝑇 21 𝑇 22 𝑇 23 | 𝑇 31 𝑇 32 𝑇 33 Page | 18 Cayley – Hamilton theorem Every Cartesian tensor of order two 𝐶𝑇(2) satisfy the characteristic equation. 𝑖. 𝑒. 𝑇 3 − 𝐼1 𝑇 2 + 𝐼2 𝑇 − 𝐼3 𝐼 = 0 Calculus of tensors: The partial derivative according to the coordinates 𝑥𝑖 is denoted by , 𝑖 such that 𝜕𝜙 𝜕𝑢𝑖 𝜕2𝜙 𝜙,𝑖 = , 𝑢𝑖,𝑗 = , 𝜙,𝑖𝑗 = 𝜕𝑥𝑖 𝜕𝑥𝑗 𝜕𝑥𝑖 𝜕𝑥𝑗 𝜕 2 𝑢𝑖 𝜕2𝜙 𝜕2𝜙 𝜕2𝜙 𝑢𝑖,𝑗𝑗 = , 𝛻2𝜙 = + + = ϕ,𝑖𝑖 𝜕𝑥𝑗 𝜕𝑥𝑗 𝜕𝑥12 𝜕𝑥22 𝜕𝑥32 If we have a vector 𝐴 whose components are 𝐴𝑖 , then 𝐴 = 𝐴𝑖 𝑒 𝑖 and 𝜕 𝜕𝐴𝑗 𝜕𝐴𝑗 𝜕𝐴𝑖 𝑑𝑖𝑣 𝐴Ԧ = (𝑒 𝑖 ) ∙ ( 𝐴𝑗 𝑒 𝑗 ) = (𝑒 𝑖 ∙ 𝑒 𝑗 ) = 𝛿𝑖𝑗 = = 𝐴𝑖,𝑖 𝜕𝑥𝑖 𝜕𝑥𝑖 𝜕𝑥𝑖 𝜕𝑥𝑖 The curl of 𝑢 where 𝑢 = 𝑢𝑖 𝑒 𝑖 can be written as 𝑒1 𝑒2 𝑒3 𝜕 𝜕 𝜕 𝜕𝑢𝑘 𝑐𝑢𝑟𝑙 𝑢 = (∇ × 𝑢) = | | = 𝜖𝑖𝑗𝑘 𝑒 𝑖 𝜕𝑥1 𝜕𝑥2 𝜕𝑥3 𝜕𝑥𝑗 𝑢1 𝑢2 𝑢3 Note that 𝑒 𝑖 × 𝑒 𝑗 = 𝜖𝑖𝑗𝑘 𝑒 𝑘 , hence ∴ 𝑎 × 𝑏 = 𝑎𝑖 𝑒 𝑖 × 𝑏𝑗 𝑒 𝑗 = 𝑎𝑖 𝑏𝑗 (𝑒 𝑖 × 𝑒 𝑗 ) = 𝑎𝑖 𝑏𝑗 𝜖𝑖𝑗𝑘 𝑒 𝑘 𝜕 𝜕𝑢𝑗 𝑐𝑢𝑟𝑙 𝑢 = (∇ × 𝑢) = (𝑒 𝑖 ) × (𝑢𝑗 𝑒 𝑗 ) = (𝑒 × 𝑒 𝑗 ) 𝜕𝑥𝑖 𝜕𝑥𝑖 𝑖 𝜕𝑢𝑗 ∴ 𝑐𝑢𝑟𝑙 𝑢 = 𝜖𝑖𝑗𝑘 𝑒 = 𝜖𝑖𝑗𝑘 𝑢𝑗,𝑖 𝑒 𝑘 𝜕𝑥𝑖 𝑘 suppose that 𝜙 , 𝑢 , 𝑇 are scalar function, vector function and tensor function respectively of the position vector 𝑥. The gradient operator denoted by ∇ of the basis {𝑒 𝑖 } has the following effects on 𝜙 , 𝑢 , 𝑇 Page | 19  𝜕𝜙 𝑔𝑟𝑎𝑑 𝜙 ≡ ∇ 𝜙 = 𝑒 = 𝜙,𝑖 𝑒 𝑖 𝜕𝑥𝑖 𝑖 𝜕 𝜕𝑢𝑃 𝑔𝑟𝑎𝑑 𝑢 ≡ ∇ ⊗ 𝑢 = ( ) (𝑢𝑝 𝑒 𝑝 ) ⊗ 𝑒 𝑞 = (𝑒 ⊗ 𝑒 𝑞 ) = 𝑢𝑝,𝑞 (𝑒 𝑝 ⊗ 𝑒 𝑞 ) 𝜕𝑥𝑞 𝜕𝑥𝑞 𝑝 𝜕 𝜕 𝑔𝑟𝑎𝑑 𝑇 ≡ ∇ ⊗ 𝑇 = (𝑇) = (𝑇 𝑒 ⊗ 𝑒𝑞 ⊗ 𝑒𝑟 ⊗ …) ⊗ 𝑒𝑖 𝜕𝑥𝑖 𝜕𝑥𝑖 𝑝𝑞𝑟… 𝑝 𝜕𝑇𝑃𝑞𝑟… = (𝑒 𝑃 ⊗ 𝑒 𝑞 ⊗ 𝑒 𝑟 … ⊗ 𝑒 𝑖 ) = 𝑇𝑝𝑞𝑟…,𝑖 (𝑒 𝑃 ⊗ 𝑒 𝑞 ⊗ 𝑒 𝑟 … ⊗ 𝑒 𝑖 ) 𝜕𝑥𝑖 The contraction of (∇ ⊗ 𝑢) given by (∇ ⊗ 𝑢)𝑝𝑝 = ∇ ⋅ 𝑢 = 𝑢𝑝,𝑝. We can do the contraction of ∇ ⊗ 𝑇 several times, so if we define 𝑑𝑖𝑣 𝑇 to be the contraction resulting from 𝑃 ↔ 𝑖 , then 𝑑𝑖𝑣 𝑇 = 𝑇𝑖𝑞𝑟…,𝑖 (𝑒 𝑖 ⊗ 𝑒 𝑞 ⊗ 𝑒 𝑟 … ⊗ 𝑒 𝑖 ) and if 𝑇 is a 𝐶𝑇(2) then (∇ ⊗ 𝑇) = 𝑇𝑝𝑞,𝑖 (𝑒 𝑃 ⊗ 𝑒 𝑞 ⊗ 𝑒 𝑖 ) After contraction, it will be 𝜕𝑇𝑖𝑞 𝑑𝑖𝑣 𝑇 = 𝑒 = 𝑇𝑖𝑞,𝑖 𝑒 𝑞 𝜕𝑥𝑖 𝑞 The above formulae are true if the base {𝑒 𝑖 } is a cartesian base, i.e. 𝑒 𝑖 do not depend on 𝑥. If we have a vector 𝐴 whose components are 𝐴𝑖 , then 𝐴 = 𝐴𝑖 𝑒 𝑖 and 𝑑𝑖𝑣 𝐴Ԧ = 𝐴𝑖,𝑖. The gauss’s theorem ∫ ∇ ⋅ 𝐴 𝑑𝑉 = ∫ 𝑛 ⋅ 𝐴 𝑑𝑠 𝑠 𝜕𝑠 can be written as ∫ 𝐴𝑖,𝑖 𝑑𝑉 = ∫ 𝑛𝑖 𝐴𝑖 𝑑𝑠 𝑠 𝜕𝑠 where 𝑛 = 𝑛𝑖 𝑒 𝑖 , 𝐴 = 𝐴𝑗 𝑒 𝑗 so Page | 20 𝑛 ⋅ 𝐴 = 𝑛𝑖 𝑒 𝑖 ⋅ 𝐴𝑗 𝑒 𝑗 = 𝑛𝑖 𝐴𝑗 𝑒 𝑖 ∙ 𝑒 𝑗 = 𝑛𝑖 𝐴𝑗 𝛿𝑖𝑗 = 𝑛𝑖 𝐴𝑖 Tensor field: If we define a unique tensor for every point 𝑃(𝑥1 , 𝑥2 , 𝑥3 ) in a region 𝑅 then we obtain a tensor field. If 𝜙(𝑥1 , 𝑥2 , 𝑥3 ) is a scalar function defined in the coordinates 𝑥𝑖 and defined in the coordinates 𝑥𝑖′ by 𝜙 ′ (𝑥1′ , 𝑥2′ , 𝑥3′ ) then 𝜙 ′ = 𝜙 𝜕𝜙 ′ 𝜕𝜙 𝜕𝑥𝑖 𝜕𝜙 = = ℓ 𝑟𝑖 𝜕𝑥𝑟′ 𝜕𝑥𝑖 𝜕𝑥𝑟′ 𝜕𝑥𝑖 𝜕𝑥𝑖 ∴ 𝜙,𝑟′ = 𝜙,𝑖 ℓ𝑟𝑖 = ℓ𝑟𝑖 𝜙,𝑖 where ℓ𝑟𝑖 = 𝜕𝑥𝑟′ 𝑔𝑟𝑎𝑑 𝜙 = (∇ 𝜙) is a proper vector. Similarly, we can prove that if 𝑢𝑖 is a vector field then 𝑢𝑖,𝑗 is a tensor field of order 2. ′ i.e. 𝑢𝑖,𝑗 = ℓ𝑖𝑟 ℓ𝑗𝑠 𝑢𝑟,𝑠 𝜕𝑢𝑖′ 𝜕𝑢𝑟 𝜕𝑢𝑟 𝜕𝑥𝑠 ∵ 𝑢𝑖′ = ℓ𝑖𝑟 𝑢𝑟 ∴ ′ 𝑢𝑖,𝑗 = ′ = ℓ𝑖𝑟 = ℓ 𝑖𝑟 = ℓ𝑖𝑟 ℓ𝑗𝑠 𝑢𝑟 𝜕𝑥𝑗 𝜕𝑥𝑗′ 𝜕𝑥𝑠 𝜕𝑥𝑗′ by contraction ′ 𝑢𝑖,𝑖 = ℓ𝑖𝑟 ℓ𝑖𝑠 𝑢𝑟,𝑠 = 𝛿𝑟𝑠 𝑢𝑟,𝑠 = 𝑢𝑟,𝑟 i.e. 𝑢𝑟,𝑟 is a scalar field. In the same way we can prove that if 𝑇𝑖𝑗 is a tensor field of order two then 𝑇𝑖𝑗,𝑘 is a tensor field of order three. Page | 21