Covarient Formulation Lecture 8

Transcription

1 Covarient Formulation Lecture 8 1 Covarient Notation We use a 4-D space represented by the Cartesian coordinates, x 0 (orx 4 ), x 1, x 2, x 3. The components describe a vector (tensor of rank 1) in this space. These components are contravarient, and have a superscript label. They transform as ; A α = x α x β Aβ Remember that covarient components transform as ; A α = xβ x α A β This space has a metric defined by; g ij = Note that the metric can be used to raise or lower an index, ie it changes the tensor type between covarient and contravarient. Thus since g αβ g βγ = δ γ α x α = g αβ x β x β = g βα x α For special relativity we let x 0 (orx 4 ) = ct. Obviously we use a metric which interchanges the 1 as needed. The contraction of the 4-vector ( x, ct) gives the length; ds 2 = (ct) 2 x x A scalar product is obtained by contracting covarient with contravarient components. Consider the covarient notation for the gradient; α = x α ( x 0, ) The contravarient representation is; 1

2 α = g αβ β ( x 0, ) 2 Contraction If we have the product of components of similar type, then the metric must be used to raise (or lower) and index to get the contraction. Later we use vector/matrix notation. Thus to obtain a contraction of two vectors represented by rows or columns of components, the algebra multiplies a row by a column vector, ie a vector by its conjugate. For example, the vector with components (a, d) has a squared length; ( ) a (a, b) = a b 2 + b 2 Also we obtain; α α = 2 x Note that this is a scalar operator. Matrix multiplication follows in the same way. ( ) a b (f, g) = (fa + gc, fb + gd) c d while; ( a c b d ) ( f g ) = ( af + cg bf + gd ) We require the maticies to be the transpose of each other. A common technique which does not take advantage of tensor algebra, is to multiply the time component of a 4-vector by ic where i 2 = 1. Then usual vector algebra is applied. 3 4-Vectors 3.1 Length Using the 4 components of a spatial 4-vector as defined above write the Lorentz transformation for a boost in the 1 direction as ; x 0 = γ(x 0 βx 1 ) 2

3 x 1 = γ(x 1 βx 0 ) x 2 = x 2 x 3 = x 3 The transformation in an arbitrary boost direction is ; x 0 = γ(x 0 β x) x = x + γ 1 β 2 ( β x) β γ βx Energy/Momentum The momentum and energy are written; Pc = βγm 0 c 2 E = γm 0 c 2 These are transformed using the velocity transformation equations. Use the expression; γ = 1 1 β 2 0 = γ 0 γ (1 + V 0 V z /c2 ) The above is obtained by transforming (1 β 2 ) to a new coordinate frame boosted by V 0. Using this result one finds; p z c = γ(p z c βe ) p x c = p x c p y c = p y c E = γ(e βp zc) The above is the general form for a 4-vector transformation. Now contract this 4-vector with itself. p i p i = (pc) 2 E 2 3

4 This result is a scalar and is invariant under a transformation. In the rest frame p = 0 and E = m 0 c 2. Thus; E 2 = (pc) 2 + (m 0 c 2 ) Velocity Next consider velocity transformations, and find that they can be written as; γu z = γ 0 (γ u z + β 0γ c) γu x = γ U x γu y = γ U y γc = γ 0 (γ c + β 0 γ U z) The velocity transforms as the 4-vector (γc, γ U) 3.4 Current density Consider the equation of continuity; J + ρ = 0 Suppose we operate on a form j = (cρ, J) with the 4-gradient α. If j α is a 4-vector, we recover the equation of continuity is an invarient, which is expected. 4 Gaussian Units To remain consistent with the text we now change from the MKS system to Gaussian units. This makes the fields and equations more symmetric in a relativistic formulation, at the expense of using units which must be converted when comparing calculated values to measured quantities. In gaussian units Maxwell s equations have the form; E = 4πρ ; B = 0 E = (1/c) B 4

5 B = 4π/c J + (1/c) E 4.1 Wave Eqn Previously we noted that the contraction; α α = (1/c 2 ) This must be a scalar quantity. Using Maxwell s equations above we have the wave equation for the vector and scalar potentials α α A δ = (4π/c)j δ where j δ is the current density 4-vector previously obtained. The vector and scalar potentials form a 4-vector, and the Lorentz condition is; α A α = (1/c) φ + A = 0 5 Dual Tensors As discussed in a previous lecture, all anti(skew)-symmetric tensors have a dual representation. Of present interest, use a rank 3 Levi-Civita tensor to contract an antisymmetric tensor of rank 2 as follows; A α = ǫ αβγ B βγ Then look at the tensor transformations ; B βγ = x β x δ x γ x λ Cδλ ǫ αβγ = xδ x α xλ x β xτ x γ ǫ δλτ Combining above differntials, we are able to make cancellations to show at only a form x α x β remaims so that this represents the transformation of a vector as expected. This vector is an axial vector as it has the reflection properties of a tensor of rank 2 instead of a tensor of rank 1 (a true vector) 5

6 6 Field Tensors Above we observed that the vector and scalar potentials form components of a 4-vector. The E and B fields are obtained from these potential forms, (φ, A), using the equations (Gaussian units); E x = (1/c) A x φ x = ( 0 A 1 1 A 0 ) B x = ( 2 A 3 3 A 2 ) and so forth. By inspection it then appears that the fields are components of an antisymmetric tensor of rank 2. The tensor has the form; β 0 E x E y E z α F αβ = E x 0 B z B y E y B z 0 B x v E z B y B x 0 Therefore in covarient form Maxwell s equations are; α F αβ = (4π/c) j β This equation produces the 2 equations; E = 4πρ B = (4π/c) J + (1/c) E The other 2 Maxwell equations are satisfied by requiring that the fields are obtained from the potentials. Now the field strength tensor is anti-symmetric so it has a dual representation. This obtained from; F αβ = (1/2)ǫ αβγδ F γδ We require F αβ = 0. This produces the Maxwell equations; B = 0 E = (1/c) B The dual field strength tensor has the form; 6

7 F αβ = β 0 B x B y B z B x 0 E z E y B y E z 0 E x B z E y E x 0 Note that it is obtained by E B and B E. 7 Lorentz Group α v Suppose we look at a matrix representation of a Lorentz transformation. Use x 0 = ct and put the boost in the 1 direction. Define β = tanh(ξ), γ = cosh(ξ) and γβ = sinh(ξ). The parameter ξ is called the rapidity, x 0 x 1 x 2 x 3 = cosh(ξ) sinh(ξ) 0 0 sinhh(ξ) cosh(ξ) Using matrix multiplication this has the form; x = A x Now A has an inverse and the matrix multiplication operation looks like a rotation which preserves length. Thus we have a unitary, linear transformation which generates a continuous group. This transformation matrix has 4 4 = 16 elements, but the matrix has only 6 independent parameters (count them). Therefore we can generate this matrix by choosing 3 of these parameters to be the Euler rotation angles and 3 the boost along the 3 spatial dimensions. Any unitary transformation can be represented by the operator A = e L so that the inverse transform is A = e L. As the transformation represents a continuous group, we need only consider infinitesmal rotations ie the Lie algebra of the group. Thus for infinitesmal angle rotations; A I + ω S Here ω are the 3 Euler angles and S is a matrix which would result in a rotation about one of the axes. For example note that as ω 0 cos(ω) 1 and sin(ω) ω. Therefore a rotation about the x axis would occur by applying the matrix; cos(ω) sin(ω) 0 sin(ω) cos(ω) ω 0 ω 1 Then introduce rotations defined by the boosts using sinh and cosh. The operator L is x 0 x 1 x 2 x 3 7

8 defined by; L = ω s ξ κ The matricies s and κ are defined in the text. These matricies are constructed to satisfy the Lie angebra of the Lorentz group; [s i, s j ] = ǫ ijk s k [s i, κ j ] = ǫ ijk κ k [κ i, κ j ] = ǫ ijk s k 8 Field Transformation Now in matrix form, the Lorentz transformation is; x = A x x = x A Because we expect x F αβ x = x F αβ x Substitute using the Lorentz transformation matrix and its transpose to obtain; F αβ = AF αβ A Multiply the above matricies to obtain a new field tensor and compare the components. This gives the transformation properties of the fields. For a boost in the 1 direction E 1 = E 1 ; B 1 = B 1 E 2 = γ(e 2 βb 3 ) ; B 2 = γ(b 2 + βe 3 ) E 3 = γ(e 3 + βb 2 ) ; B 3 = γ(b 3 βe 2 ) If there is ONLY a magnetic field in the unprimed frame, then; E = β B 8