Tensors for Physics

392 Appendix: Exercises... The result is Bμν(α)= ⎛ ⎝ cos 2α−sin 2α 0 sin 2α cos 2α 0 000 ⎞ ⎠. (A.8) Consequently. one has Bμν(α) ...

Appendix: Exercises... 393 3.5 Velocity of a Particle Moving on a Screw Curve(p.46) Hint: Useα=ωt for the parameter occurring in ...

394 Appendix: Exercises... (i) Tetrahedron: the position vectors of the corners are u^1 =ex+ey+ez, u^2 =−ex−ey+ez, u^3 =ex−ey−ez ...

Appendix: Exercises... 395 Due toc^3 =−(a^3 + 3 a^2 b+ 3 ab^2 +b^3 ), the diagonal elements are equal toa^3 +a^2 b+ ab^2 =a(a^2 ...

396 Appendix: Exercises... Here one has∇νvμ=εμκνwκ, and consequently ∇·v= 0 ,(∇×v)λ=ελνμεμκνwκ= 2 wλ, ∇νvμ = 0. 7.2 Test Solutio ...

Appendix: Exercises... 397 is made. The Maxwell equation−∂Bμ/∂t=εμνλ∇νEλleads to cB(μ^0 )=εμνλ̂kνE(λ^0 ). Thus the magnetic fiel ...

398 Appendix: Exercises... Differentiation of the second of these equations with respect tot, insertion of the time change of th ...

Appendix: Exercises... 399 which is the relation (7.83), and LμLν−LνLμ=rν∇μ−rμ∇ν. The right hand side of the last equation can b ...

400 Appendix: Exercises... (i)Homogeneous Field, wherev=e=const., witheparallel to thex-axis. The expectation is ∮ v·dr=I=0 sinc ...

Appendix: Exercises... 401 is made, with a scalar coefficientc 1. Multiplication of this equation byuμv̂νleads to c 1 =uμ̂vνSμν= ...

402 Appendix: Exercises... I≡ ∮ v·dr=R^2 w ∫ 2 π 0 dφ= 2 πR^2 w. The equalityS=Iis in accord with the Stokes law. 8.4 Moment of ...

Appendix: Exercises... 403 Exercise Chapter 9 9.1 Verify the Required Properties of the Third and Fourth Rank Irreducible Tensor ...

404 Appendix: Exercises... Due torκ∇κ=r∂/∂randXμ 1 μ 2 ···μ∼r−(+^1 ), the second term in the preceding equation is equal to − ...

Appendix: Exercises... 405 polar vector is the ansatzvκ=c(r)εκλνΩλXν, with a coefficientc. For the present problem, the creeping ...

406 Appendix: Exercises... Δ()μ 1 μ 2 ···μ− 1 λ,μ′ 1 μ′ 2 ···μ′− 1 λ = 2 + 1 2 − 1 Δ(μ−^1 ) 1 μ 2 ···μ− 1 ,μ′ 1 μ′ 2 ···μ ...

Appendix: Exercises... 407 Δ(μνλ,μ^3 ) ′ν′λ′= 1 6 [ (δμμ′δνν′+δμν′δνμ′)δλλ′+(δμμ′δνλ′+δμλ′δνμ′)δλν′ +(δμλ′δνν′+δμν′δνλ′)δλμ′ ] − ...

408 Appendix: Exercises... where the integrand is positive, furthermoreν 0 >0, and thus d dt sa≥ 0 holds true. 12.3 Second Or ...

Appendix: Exercises... 409 where the prime indicates the differentiation with respect tor. This corresponds to g+=−γτrg′eq, see ...

410 Appendix: Exercises... For circular polarized light with its electric field vector parallel toe=ex±iey, the vector polarizat ...

Appendix: Exercises... 411 13.2 Verify a Relation Peculiar for Spin 1 /2 (p.243) Fo r s p i n s= 1 /2,the peculiar relation sμsν ...