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Friday, March 12, 2010

Some technical calculations of Klein-Gordon field

by Octavian

Problem

Classical electromagnetism (with no sources) follows from the action

\displaystyle S = \int d^4 x \, \Big(-\frac{1}{4} F_{\mu \nu} F^{\mu \nu} \Big) \ \ \ \ \ (1)

where {F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu}.

  1. Derive Maxwell’s equations as the Euler-Lagrange equations of this action, treating the components {A_\mu(x)} as the dynamical variables. Write the equations in standard form by identifying {E^i = -F^{0i}} and {\epsilon^{ijk} B^k = -F^{ij}}.
  2. Construct the energy-momentum tensor for this theory. Note that the usual procedure does not result in a symmetric tensor. To remedy that, we can add to {T^{\mu \nu}} a term of the form {\partial_\lambda K^{\lambda \mu \nu}}, where {K^{\lambda \mu \nu}} is antisymmetric in its first two indices. Such an object is automatically divergenceless, so

    \displaystyle \hat{T}^{\mu \nu} = T^{\mu \nu} + \partial_\lambda K^{\lambda \mu \nu} \ \ \ \ \ (2)

    is an equally good energy-momentum tensor with the same globally conserved energy and momentum. Show that this construction, with

    \displaystyle K^{\lambda \mu \nu} = F^{\mu \nu} A^\nu \ \ \ \ \ (3)

    leads to the an energy-momentum tensor {\hat{T}} that is symmetric and yields the standard formula for the electromagnetic energy and momentum densities:

    \displaystyle \varepsilon = \frac{1}{2} (E^2 + B^2), \qquad \textbf{S} = \textbf{E} \times \textbf{B} \ \ \ \ \ (4)

Solution

  1. The Lagrangian for this no-source classical electromagnetism is

    \displaystyle \L = -\frac{1}{4} F_{\mu \nu} F^{\mu \nu}, \qquad F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu \ \ \ \ \ (5)

    We can make the variation of its action:

    \displaystyle \begin{array}{rcl} \delta S &=& -\frac{1}{4} \int d^4 x \, 2 F_{\mu \nu} \delta F^{\mu \nu} \\ &=& -\frac{1}{2} \int d^4 x \, F_{\mu \nu} (\partial^\mu \delta A^\nu - \partial^\nu \delta A^\mu) \\ &=& -\frac{1}{2} \int d^4 x \, F^{\mu \nu} (\partial_\mu \delta A_\nu - \partial_\nu \delta A_\mu) \\ &=& -\frac{1}{2} \int d^4 x \, \Big( \partial_\mu (F^{\mu \nu} \delta A_\nu) - \partial_\mu F^{\mu \nu} \delta A_\nu \\ && - \partial_\nu (F^{\mu \nu} \delta A_\mu) + \partial_\nu F^{\mu \nu} \delta A_\mu \Big) \\ &=& \frac{1}{2} \int d^4 x \, 2 \partial_\mu F^{\mu \nu} \delta A_\nu \\ 0 &=& \int d^4 x \, \partial_\mu F^{\mu \nu} \delta A_\nu \end{array}

    Hence we have

    \displaystyle \partial_\mu F^{\mu \nu} = 0

    By noting that

    \displaystyle F^{0i} = -E^i, \qquad F^{ij} = - \epsilon^{ijk} B^k

    then we have for {\nu = 0},

    \displaystyle \begin{array}{rcl} \partial_0 F^{00} + \partial_i F^{i0} &=& 0 \\ \nabla \cdot \textbf{E} &=& 0 \end{array}

    and for {\nu = j},

    \displaystyle \begin{array}{rcl} \partial_0 F^{0j} + \partial_i F^{ij} &=& 0 \\ -\frac{\partial E^j}{\partial t} - \partial_i (\epsilon^{ijk} B^k) &=& 0 \\ \epsilon^{jik} \partial_i B_k &=& \frac{\partial E^j}{\partial t} \\ \nabla \times \textbf{B} &=& \frac{\partial \textbf{E}}{\partial t} \end{array}

  2. From the previous calculation we can write the variation of action as

    \displaystyle \delta S = -\int d^4 x \, \partial_\mu (F^{\mu \nu} \delta A_\nu) \ \ \ \ \ (6)

    Since the variation of Lagrangian is {\delta \L = a^\mu \partial_\mu \L}, then we have

    \displaystyle \begin{array}{rcl} 0 &=& \int d^4 x \, \Big( -\partial_\mu (F^{\mu \nu} a^\rho \partial_\rho A_\nu) - \partial_\mu (a^\rho \delta^\mu_\rho \L) \Big) \\ 0 &=& \int d^4 x \, a^\rho \partial_\mu \Big( -(F^{\mu \nu} \partial_\rho A_\nu) - \delta^\mu_\rho \L \Big) \end{array}

    such that we can write the energy-momentum tensor as

    \displaystyle T^{\mu \nu} = -F^{\mu \rho} \partial^\nu A_\rho - \delta^{\mu \nu} \L \ \ \ \ \ (7)

    We can write the new (symmetric) energy-momentum tensor by adding the new term to the old one:

    \displaystyle \begin{array}{rcl} \hat{T}^{\mu \nu} &=& T^{\mu \nu} + \partial_\rho (F^{\mu \rho} A^\nu) \\ &=& T^{\mu \nu} + \partial_\rho F^{\mu \rho} A^\nu + F^{\mu \rho} \partial_\rho A^\nu \\ &=& F^{\mu \rho} (\partial_\rho A^\nu - \partial^\nu A_\rho) - \delta^{\mu \nu} \L \\ &=& F^{\mu \rho} {F_\rho}^\nu - \delta^{\mu \nu} \L \end{array}

    Obviously {\hat{T}^{\mu \nu}} is symmetric. We can find the value of the components of this energy-momentum tensor. The energy density of the electromagnetic field is

    \displaystyle \begin{array}{rcl} \hat{T}^{00} &=& F^{0\rho} {F_\rho}^0 - \delta^{00} \L \\ &=& -\sum_i F^{0i} F^{i0} + \frac{1}{4} F_{\alpha \beta} F^{\alpha \beta} \\ &=& \sum_i (F^{0i})^2 + \frac{1}{4} (F_{0\beta} F^{0\beta} + F_{i\beta} F^{i\beta}) \\ &=& \sum_i \Big( (E^i)^2 + \frac{1}{4} ( - F^{0i} F^{0i} - F^{i0} F^{i0} + \sum_j F^{ij} F^{ij}) \Big) \\ &=& \sum_i \frac{1}{2} (E^i)^2 + \frac{1}{4} \sum_{i,j,k,l} \epsilon^{ijk} \epsilon^{ijl} B^k B^l \\ &=& \frac{1}{2} \sum_i \Big( (E^i)^2 + (B^i)^2 \Big) \end{array}

    And the Poynting vector is given by

    \displaystyle \begin{array}{rcl} \hat{T}^{0i} &=& F^{0\rho} {F_\rho}^i - \delta^{0i} \L \\ &=& F^{oj} {F_j}^i \\ &=& -\sum_j F^{0j} F^{ji} \\ &=& \sum_j F^{0j} F^{ij} \\ &=& \sum_{j,k} (-E^j) (-\epsilon^{ijk} B^k) \\ &=& \sum_{j,k} \epsilon^{ijk} E^j B^k \\ &=& (\textbf{E} \times \textbf{B})^i \end{array}

Problem

Consider the field theory of a complex-valued scalar field obeying the Klein-Gordon equation. The action of this theory is

\displaystyle S = \int d^4x \, (\partial_\mu \phi^\ast \partial^\mu \phi - m^2 \phi^\ast \phi ) \ \ \ \ \ (8)

It is easiest to analyze this theory by considering {\phi(x)} and {\phi^\ast(x)}, rather than the real and imaginary parts of {\phi(x)}, as the basic dynamical variables.

  1. Find the conjugate momenta to {\phi(x)} and {\phi^\ast(x)} and the canonical commutation relations. Show that the Hamiltonian is

    \displaystyle H = \int d^3x \, (\pi^\ast \pi + \nabla \phi^\ast \cdot \nabla \phi + m^2 \phi^\ast \phi) \ \ \ \ \ (9)

    Compute the Heisenberg equation of motion for {\phi(x)} and show that it is indeed the Klein-Gordon equation.

  2. Diagonalize {H} by introducing creation and annihilation operators. Show that the theory contains two sets of particles of mass {m}.
  3. Rewrite the conserved charge

    \displaystyle Q = \int d^3x \, i (\phi^\ast \pi^\ast - \pi \phi) \ \ \ \ \ (10)

    in terms of creation and annihilation operators, and evaluate the charge of the particles of each type.

  4. Consider the case of two complex Klein-Gordon fields with the same mass. Label the fields as {\phi_a(x)}, where {a = 1,2}. Show that there are now four conserved charges, one given by the generalization of part 3, and the other three given by

    \displaystyle Q^i = \int d^3x \, \frac{i}{2} (\phi^\ast_a (\sigma^i)_{ab} \pi^\ast_b - \pi_a (\sigma^i)_{ab} \phi_b) \ \ \ \ \ (11)

    where {\sigma^i} are the Pauli sigma matrices. Show that these three charges have the commutation relations of angular momentum {(SU(2))}.

Solution

  1. We start with our Lagrangian density

    \displaystyle \L = \partial_\mu \phi^\ast \partial^\mu \phi - m^2 \phi^\ast \phi \ \ \ \ \ (12)

    then we can find the conjugate momenta, i.e.

    \displaystyle \begin{array}{rcl} \pi &=& \frac{\partial \L}{\partial \dot{\phi}} = \dot{\phi}^\ast \\ \pi^\ast &=& \frac{\partial \L}{\partial \dot{\phi}^\ast} = \dot{\phi} \end{array}

    We can introduce the operators {a} and {b} such that

    \displaystyle \begin{array}{rcl} \phi &=& \int \frac{d^3p}{(2\pi)^3} \, \frac{1}{\sqrt{2E_p}} (a_p e^{-ip \cdot x} + b_p^\dagger e^{ip \cdot x}) \\ \phi^\ast &=& \int \frac{d^3p}{(2\pi)^3} \, \frac{1}{\sqrt{2E_p}} (b_p e^{-ip \cdot x} + a_p^\dagger e^{ip \cdot x}) \end{array}

    If {a = b}, then the field is real. We also can find the expression for the conjugate canonical momenta

    \displaystyle \begin{array}{rcl} \pi &=& \int \frac{d^3p}{(2\pi)^3} \, (-i) \sqrt{\frac{E_p}{2}} (b_p e^{-ip \cdot x} - a_p^\dagger e^{ip \cdot x}) \\ \pi^\ast &=& \int \frac{d^3p}{(2\pi)^3} \, (-i) \sqrt{\frac{E_p}{2}} (a_p e^{-ip \cdot x} - b_p^\dagger e^{ip \cdot x}) \end{array}

    Let’s check the commutation relation between {\phi} and {\pi}.

    \displaystyle \begin{array}{rcl} [\phi(x), \pi(y)] &=& \int \frac{d^3p \, d^3q}{(2\pi)^6} \, \frac{(-i)}{2} \sqrt{\frac{E_q}{E_p}} \Big( [a_p, b_q] e^{-i(p \cdot x + q \cdot y)} \\ && - [a_p, a_q^\dagger] e^{-i(p \cdot x - q \cdot y)} \\ && + [b_p^\dagger, b_q] e^{i(p \cdot x - q \cdot y)} - [b_p^\dagger, a_q^\dagger] e^{i(p \cdot x + q \cdot y)} \Big) \\ &=& \int \frac{d^3p}{(2\pi)^3} \, \frac{(-i)}{2} \Big( -e^{-ip \cdot (x-y)} - e^{ip \cdot (x-y)} \Big) \\ &=& i \delta^{(3)}(x-y) \end{array}

    It shows us that the measurement of position and momentum of this particle in two distinct points of spacetime is independent each other; in every point of spacetime, we cannot measure these two simultaneously. The same calculation will work for {\phi^\ast} and {\pi^\ast}, i.e.  [\phi^\ast, \pi^\ast]=i \delta^{(3)}(x-y).  But it is not the case for {\phi} and {\phi^\ast}.

    \displaystyle \begin{array}{rcl} [\phi(x), \phi^\ast(y)] &=& \int \frac{d^3p \, d^3q}{(2\pi)^6} \, \frac{1}{2\sqrt{E_p E_q}} \Big( [a_p, b_q] e^{-i(p \cdot x + q \cdot y)} \\ && + [a_p, a_q^\dagger] e^{-i(p \cdot x - q \cdot y)} \\ &&+ [b_p^\dagger, b_q] e^{i(p \cdot x - q \cdot y)} + [b_p^\dagger, a_q^\dagger] e^{i(p \cdot x + q \cdot y)} \Big) \\ &=& \int \frac{d^3p}{(2\pi)^3} \, \frac{1}{2E_p} \Big( e^{-ip \cdot (x-y)} - e^{ip \cdot (x-y)} \Big) \\ &=& D(x-y) - D(y-x) \end{array}

    And the last relation maybe is the most simple, \displaystyle [\phi(x), \pi^\ast (y)] = 0 \ \ \ \ \ .

  2. The Hamiltonian density {\mathcal{H}} is expressed as

    \displaystyle \begin{array}{rcl} \mathcal{H} &=& \pi \dot{\phi} + \pi^\ast \dot{\phi}^\ast - \L \\ &=& \pi \pi^\ast + \pi^\ast \pi - \pi \pi^\ast + \nabla \phi^\ast \cdot \nabla \phi + m^2 \phi^\ast \phi \\ &=& \pi^\ast \pi + \nabla \phi^\ast \cdot \nabla \phi + m^2 \phi^\ast \phi \end{array}

  3. We just do some straightforward calculations in this subproblem,

    \displaystyle \begin{array}{rcl} H &=& \int d^3x \int \frac{d^3p \, d^3q}{(2\pi)^6} \, \Bigg( -\frac{\sqrt{E_q E_p}}{2} \Big( a_p b_q e^{-i(p+q) \cdot x} - a_p a_q^\dagger e^{-i(p-q) \cdot x} \\ &&- b_p^\dagger b_q e^{i(p-q) \cdot x} + b_p^\dagger a_q^\dagger e^{i(p+q) \cdot x} \Big) \\ &&+ \frac{-\textbf{p} \cdot \textbf{q}}{2 \sqrt{E_p E_q}} \Big( b_p a_q e^{-i(p+q) \cdot x} - b_p b_q^\dagger e^{-i(p-q) \cdot x} \\ && - a_p^\dagger a_q e^{i(p-q) \cdot x} + a_p^\dagger b_q^\dagger e^{i(p+q) \cdot x} \Big) \\ &&+ \frac{m^2}{2 \sqrt{E_p E_q}} \Big( b_p a_q e^{-i(p+q) \cdot x} + b_p b_q^\dagger e^{-i(p-q) \cdot x} + a_p^\dagger a_q e^{i(p-q) \cdot x} \\ && + a_p^\dagger b_q^\dagger e^{i(p+q) \cdot x} \Big) \Bigg) \\ &=& \int \frac{d^3p}{(2\pi)^3} \, \Big( -\frac{E_p}{2} (a_p b_{-p} - a_p a_p^\dagger - b_p^\dagger b_p + b_p^\dagger a_{-p}^\dagger) \\ &&+ \frac{|\textbf{p}|^2 + m^2}{2E_p} (b_p a_{-p} + b_p b_p^\dagger + a_p^\dagger a_p + a_p^\dagger b_{-p}^\dagger) \Big) \\ &=& \int \frac{d^3p}{(2\pi)^3} \, \frac{E_p}{2} (-a_p b_{-p} + a_p a_p^\dagger + b_p^\dagger b_p - b_p^\dagger a_{-p}^\dagger \\ && + b_p a_{-p} + b_p b_p^\dagger + a_p^\dagger a_p + a_p^\dagger b_{-p}^\dagger) \\ &=& \int \frac{d^3p}{(2\pi)^3} \, \frac{E_p}{2} (-a_{-p} b_{p} + a_p a_p^\dagger + b_p^\dagger b_p - b_{-p}^\dagger a_{p}^\dagger \\ && + b_p a_{-p} + b_p b_p^\dagger + a_p^\dagger a_p + a_p^\dagger b_{-p}^\dagger) \\ &=& \int \frac{d^3p}{(2\pi)^3} \, \frac{E_p}{2} (a_p^\dagger a_p + a_p a_p^\dagger + b_p^\dagger b_p + b_p b_p^\dagger) \\ &=& \int \frac{d^3p}{(2\pi)^3} \, E_p (a_p^\dagger a_p + b_p^\dagger b_p + \frac{1}{2} ([a_p, a_p^\dagger] + [b_p, b_p^\dagger])) \\ &=& \int \frac{d^3p}{(2\pi)^3} \, E_p (a_p^\dagger a_p + b_p^\dagger b_p) \end{array}

  4. The charge operator is defined as

    \displaystyle Q = \int d^3x \, i (\phi^\ast \pi^\ast - \pi \phi) \ \ \ \ \ (13)

    Then we can do this calculation

    \displaystyle \begin{array}{rcl} Q &=& \int d^3x \int \frac{d^3p \, d^3q}{(2\pi)^6} \, i \Bigg( \frac{(-i)}{2} \sqrt{\frac{E_p}{E_q}} \Big( b_p a_q e^{-i(p+q) \cdot x} - b_p b_q^\dagger e^{-i(p-q) \cdot x} \\ &&+ a_p^\dagger a_q e^{i(p-q) \cdot x} - a_p^\dagger b_q^\dagger e^{i(p+q) \cdot x} \Big) \\ &&- \frac{(-i)}{2} \sqrt{\frac{E_p}{E_q}} \Big( b_p a_q e^{-i(p+q) \cdot x} + b_p b_q^\dagger e^{-i(p-q) \cdot x} \\ && - a_p^\dagger a_q e^{i(p-q) \cdot x} - a_p^\dagger b_q^\dagger e^{i(p+q) \cdot x} \Big) \Bigg) \\ &=& \int \frac{d^3p}{(2\pi)^3} \, \frac{1}{2} (b_p a_{-p} - b_p b_p^\dagger + a_p^\dagger a_p - a_p^\dagger b_{-p}^\dagger - b_p a_{-p} \\ && - b_p b_p^\dagger + a_p^\dagger a_p + a_p^\dagger b_{-p}^\dagger) \\ &=& \int \frac{d^3p}{(2\pi)^3} \, (a_p^\dagger a_p - b_p b_p^\dagger) \\ &=& \int \frac{d^3p}{(2\pi)^3} \, (a_p^\dagger a_p - b_p^\dagger b_p + [b_p^\dagger , b_p]) \\ &=& \int \frac{d^3p}{(2\pi)^3} \, (a_p^\dagger a_p - b_p^\dagger b_p) \end{array}

  5. Now we have two complex-valued Klein-Gordon fields {\phi_1} and {\phi_2}. Define

    \displaystyle \phi = \phi_1 + \phi_2 \ \ \ \ \ (14)

    then we can form the Lagrangian for this fields

    \displaystyle \begin{array}{rcl} \mathcal{L} &=& \partial_\mu \phi^\ast \partial^\mu \phi - m^2 \phi^\ast \phi \\ &=& \partial_\mu (\phi_1^\ast + \phi_2^\ast) \partial^\mu (\phi_1 + \phi_2) - m^2 (\phi_1^\ast + \phi_2^\ast) (\phi_1 + \phi_2) \\ &=& \partial_\mu \phi_1^\ast \partial^\mu \phi_1 + \partial_\mu \phi_1^\ast \partial^\mu \phi_2 + \partial_\mu \phi_2^\ast \partial^\mu \phi_1 + \partial_\mu \phi_2^\ast \partial^\mu \phi_2 \\ &&- m^2 \phi_1^\ast \phi_1 - m^2 \phi_1^\ast \phi_2 - m^2 \phi_2^\ast \phi_1 - m^2 \phi_2^\ast \phi_2 \end{array}

    We can find the current density of this Lagrangian under this invariant transformations

    \displaystyle \begin{array}{rcl} \delta \phi_1 = -i \alpha_1 \phi_1, \qquad \delta \phi_1^\ast = i\alpha \phi_1^\ast \\ \delta \phi_2 = -i \alpha_2 \phi_2, \qquad \delta \phi_2^\ast = i\alpha \phi_2^\ast \end{array}

    such that

    \displaystyle \begin{array}{rcl} \int d^4x \, \delta \L &=& \int d^4x \, \Big( \partial_\mu \delta \phi_1^\ast \partial^\mu \phi_1 + \partial_\mu \phi_1^\ast \partial^\mu \delta \phi_1 \\ && + \partial_\mu \delta \phi_1^\ast \partial^\mu \phi_2 + \partial_\mu \phi_1^\ast \partial^\mu \delta \phi_2 \\ &&+ \partial_\mu \delta \phi_2^\ast \partial^\mu \phi_1 + \partial_\mu \phi_2^\ast \partial^\mu \delta \phi_1 + \partial_\mu \delta \phi_2^\ast \partial^\mu \phi_2 + \partial_\mu \phi_2^\ast \partial^\mu \delta \phi_2 \\ &&- m^2 \delta \phi_1^\ast \phi_1 - m^2 \phi_1^\ast \delta \phi_1 - m^2 \delta \phi_1^\ast \phi_2 - m^2 \phi_1^\ast \delta \phi_2 \\ &&- m^2 \delta \phi_2^\ast \phi_1 - m^2 \phi_2^\ast \delta \phi_1 - m^2 \delta \phi_2^\ast \phi_2 - m^2 \phi_2^\ast \delta \phi_2 \Big) \\ 0 &=& \int d^4x \, \partial_\mu \Big( \delta \phi_1^\ast \partial^\mu \phi_1 + \partial^\mu \phi_1^\ast \delta \phi_1 + \delta \phi_1^\ast \partial^\mu \phi_2 + \partial^\mu \phi_1^\ast \delta \phi_2 \\ &&+ \delta \phi_2^\ast \partial^\mu \phi_1 + \partial^\mu \phi_2^\ast \delta \phi_1 + \delta \phi_2^\ast \partial^\mu \phi_2 + \partial^\mu \phi_2^\ast \delta \phi_2 \Big) \end{array}

    By inserting the variations of each fields, we will have the current and charge densities,

    \displaystyle \begin{array}{rcl} j^\mu &=& \delta \phi_1^\ast \partial^\mu \phi_1 + \partial^\mu \phi_1^\ast \delta \phi_1 + \delta \phi_1^\ast \partial^\mu \phi_2 + \partial^\mu \phi_1^\ast \delta \phi_2 \\ &&+ \delta \phi_2^\ast \partial^\mu \phi_1 + \partial^\mu \phi_2^\ast \delta \phi_1 + \delta \phi_2^\ast \partial^\mu \phi_2 + \partial^\mu \phi_2^\ast \delta \phi_2 \\ j^0 &=& i \Big( \alpha_1 (\phi_1^\ast \pi_1^\ast - \pi_1 \phi_1) + \alpha_1 (\phi_1^\ast \pi_2^\ast - \pi_2 \phi_1) \\ &&+ \alpha_2 (\phi_2^\ast \pi_1^\ast - \pi_1 \phi_2) + \alpha_2 (\phi_2^\ast \pi_2^\ast - \pi_2 \phi_2) \Big) \end{array}

    Next we make a new convention for the symbols {\phi} and {\pi}, by overriding the previous one, i.e.

    \displaystyle \begin{array}{rcl} \phi = \begin{pmatrix} \phi_1 \\ \phi_2 \end{pmatrix}, \qquad \phi^\ast = \begin{pmatrix} \phi_1^\ast & \phi_2^\ast \end{pmatrix} \\ \pi = \begin{pmatrix} \pi_1 & \pi_2 \end{pmatrix}, \qquad \pi^\ast = \begin{pmatrix} \pi_1^\ast \\ \pi_2^\ast \end{pmatrix} \end{array}

    By tuning the values of {\alpha_1} and {\alpha_2}, we have these charges

    \displaystyle \begin{array}{rcl} Q^0 &=& \int d^3x \, i \Big( \phi^\ast \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \pi^\ast - \pi \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \phi \Big) \\ &=& \int d^3x \, i (\phi^\ast \pi^\ast - \pi \phi) \\ Q^1 &=& \int d^3x \, i \Big( \phi^\ast \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \pi^\ast - \pi \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \phi \Big) \\ &=& \int d^3x \, i (\phi^\ast \sigma^1 \pi^\ast - \pi \sigma^1 \phi) \\ Q^2 &=& \int d^3x \, i \Big( \phi^\ast \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \pi^\ast - \pi \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \phi \Big) \\ &=& \int d^3x \, i (\phi^\ast \sigma^2 \pi^\ast - \pi \sigma^2 \phi) \\ Q^3 &=& \int d^3x \, i \Big( \phi^\ast \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \pi^\ast - \pi \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \phi \Big) \\ &=& \int d^3x \, i (\phi^\ast \sigma^3 \pi^\ast - \pi \sigma^3 \phi) \end{array}

    with {\sigma^i} is the Pauli sigma matrices. We can conclude that

    \displaystyle Q^i = \int d^3x \, i (\phi^\ast \sigma^i \pi^\ast - \pi \sigma^i \phi) \ \ \ \ \ (15)

    where {i = 1, 2, 3}. Hence we can see easily that {Q^i} obeys the commutation relations of {SU(2)} group. Namely,

    \displaystyle \begin{array}{rcl} [Q^i, Q^j] &=& \int d^3x \, i (\phi^\ast [\sigma^i, \sigma^j] \pi^\ast - \pi [\sigma^i, \sigma^j] \phi) \\ &=& i \epsilon_{ijk} \int d^3x \, i (\phi^\ast \sigma^k \pi^\ast - \pi \sigma^k \phi) \\ &=& i \epsilon_{ijk} Q^k \end{array}

4 Comments
Thursday, March 11, 2010

[Useless] theory for logical and lovical behaviors

by Octavian

Few weeks ago my friend and I made a theory that describes our logical and lovical behaviors. It’s disgusting.

We, the human, or more specifically, the nerd human being, have generally two states that represent our mind and heart states. The fields who represent them are called \Psi_g, the state of logic, and \Psi_v, the state of love. If we are known to be in the logic (love) state, then at that time our mind (heart) works maximally without being interfered by the others. Since ideally in our daily life we cannot be in one of these states perfectly (suppose you are in a full concentration to read a paper or book in campus park, but suddenly there is a beautiful girl walks beside you), then our state is the superposition of both of them:

\Psi = a_g \Psi_g + a_v \Psi_v + \epsilon

where the state \Psi also contains the state \epsilon which comes from the fact that we cannot always categorize our behaviors as logical or lovical, but we hope that \epsilon \ll. After doing some experiments, it is the result: |a_v|^2 \to 0, |a_g|^2 \to 1, and \epsilon \gg. Well, at least we must keep |a_v|^2 not to be zero identically. However, what is the meaning of the large value of \epsilon? It’s simple. Suppose you are in a full concentration to read a paper or book in campus park, but suddenly there is a beautiful girl walks beside you, and indeed she comes to you, also brings many papers and books which are about her homework, and asks you with the dangerous eyes to work that problem. And we cannot say any words while solving it out. And finally the story is ended with a thank you and she walks away, leaves us alone in a park, and we continue to read the damn paper/book.

Well, perhaps we should throw any calculation about this stuff, before everything goes to useless.

Useless

When you try your best but you don't succeed...

2 Comments
Wednesday, March 10, 2010

The very brief review of Lagrangian and Hamiltonian field theory

by Octavian

— 1. Lagrangian and Hamiltonian Theory —

— 1.1. Lagrangian Field Theory —

Suppose we have a system with the Lagrangian {L}. Then the actual process that can happen is the one where the action {S = \int{L dt}} is minimum, i.e. for

\displaystyle S = \int{L dt} = \int{{\mathcal L}(\phi, \partial_\mu \phi) d^4x} \ \ \ \ \ (1)

then we have

\displaystyle 0 = dS \nonumber = \int{d^4x \Bigg( \frac{\partial \mathcal{L}}{\partial \phi} \delta \phi - \partial_\mu \Bigg( \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} \Bigg) \delta \phi + \partial_\mu \Bigg( \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} \delta \phi \Bigg) \Bigg)} \ \ \ \ \ (2)

such that we have the equation of motion

\displaystyle \partial_\mu \Bigg( \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} \Bigg) - \frac{\partial \mathcal{L}}{\partial \phi} = 0 \ \ \ \ \ (3)

As an example, if we have the Lagrangian

\displaystyle \mathcal{L} = \frac{1}{2} \partial_\mu \phi \partial^\mu \phi - \frac{1}{2} m^2 \phi^2 \ \ \ \ \ (4)

then the equation of motion resulted from the Lagrangian above is the Klein-Gordon equation

\displaystyle \Big( \partial_\mu \partial^\mu + m^2 \Big) \phi = 0 \ \ \ \ \ (5)

Note that the field {\phi} in this equation is not the quantum-mechanical wavefunction in nonrelativistic quantum mechanics, but the classical field instead.

— 1.2. Hamiltonian Field Theory —

For the discrete system, the canonical coordinate is denoted as {q} and its canonical momentum is {p = {\partial L}/{\partial \dot{q}}}, such that the Hamiltonian of this system is

\displaystyle H = \sum{p\dot{q} - L} \ \ \ \ \ (6)

But if we work in the continuous system where we must work with the field {\phi(\textbf{x})}, then its canonical momentum is

\displaystyle \pi(\textbf{x}) = \frac{\partial \mathcal{L}}{\partial \dot{\phi}} \ \ \ \ \ (7)

such that the Hamiltonian {H} of this system can be expressed as

\displaystyle H = \int{d^3\textbf{x} \Big( \pi(\textbf{x}) \dot{\phi}(\textbf{x}) - \mathcal{L} \Big)} = \int{d^3\textbf{x} \, \mathcal{H}} \ \ \ \ \ (8)

For the Lagrangian {\mathcal{L} = \frac{1}{2} \partial_\mu \phi \partial^\mu \phi - \frac{1}{2} m^2 \phi^2}, we will have the Hamiltonian

\displaystyle H = \int{d^3\textbf{x} \Big( \frac{1}{2} \pi^2 + \frac{1}{2} (\nabla \phi)^2 + \frac{1}{2} m^2 \phi^2 \Big)} \ \ \ \ \ (9)

— 1.3. Noether Theorem —

There is a close link between the concept of symmetry of the Lagrangian and the conservation law that is observed in nature. If we change the field {\phi} such that it will have the form

\displaystyle \phi(\textbf{x}) \rightarrow \phi'(\textbf{x}) = \phi(\textbf{x}) + \alpha \Delta \phi(\textbf{x}) \ \ \ \ \ (10)

such that the Lagrangian will be of the form

\displaystyle \mathcal{L}(\textbf{x}) \rightarrow \mathcal{L}(\textbf{x}) + \alpha \partial_\mu \mathcal{T}^\mu (\textbf{x}) \ \ \ \ \ (11)

The relation between {\mathcal{T}^\mu(\textbf{x})} and {\Delta \phi(\textbf{x})}, then, is

\displaystyle \alpha \partial_\mu \mathcal{T}^\mu = \alpha \Delta \mathcal{L} \ \ \ \ \ (12)

\displaystyle \alpha \partial_\mu \mathcal{T}^\mu = \alpha \Bigg( \frac{\partial \mathcal{L}}{\partial \phi} \Delta \phi + \alpha \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} \partial_\mu (\Delta \phi) \Bigg) \ \ \ \ \ (13)

\displaystyle \alpha \partial_\mu \mathcal{T}^\mu = \alpha \partial_\mu \Bigg( \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} \Delta \phi \Bigg) + \alpha \Bigg( \frac{\partial \mathcal{L}}{\partial \phi} - \partial_\mu \Bigg( \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} \Bigg) \Bigg) \Delta \phi \ \ \ \ \ (14)

\displaystyle \alpha \partial_\mu \mathcal{T}^\mu = \alpha \partial_\mu \Bigg( \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} \Delta \phi \Bigg) \ \ \ \ \ (15)

such that we can conclude that this equation holds

\displaystyle \partial_\mu j^\mu(\textbf{x}) = 0, \qquad j^\mu (\textbf{x}) \equiv \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} \Delta \phi - \mathcal{T}^\mu \ \ \ \ \ (16)

An important conserved variable for this current {j^\mu} is the charge,

\displaystyle Q = \int{d^3\textbf{x} j^0} \ \ \ \ \ (17)

such that

\displaystyle \frac{dQ}{dt} = \int{d^3\textbf{x} \frac{\partial j^0}{\partial t}} = - \int{d^3\textbf{x} \partial_i j^i} = - \int{d\textbf{S} \cdot \textbf{j}} = 0 \ \ \ \ \ (18)

The infinitesimal change of field {\phi} can also be viewed as the transformation of the coordinate in the spacetime. As an example, the coordinate transformation

\displaystyle x^\mu \rightarrow x^\mu - a^\mu \ \ \ \ \ (19)

corresponds with the change of field

\displaystyle \phi(\textbf{x}) \rightarrow \phi'(x) = \phi(x+a) = \phi(x) + a^\mu \partial_\mu \phi(x) \ \ \ \ \ (20)

such that the Lagrangian {\mathcal{L}} also changes like {\phi},

\displaystyle \mathcal{L}(x) \rightarrow \mathcal{L}(x) + a^\mu \partial_\mu \mathcal{L}(x) = \mathcal{L}(x) + a^\nu \partial_\mu (\mathcal{L} \delta^\mu_\nu) \ \ \ \ \ (21)

and hence we can write

\displaystyle j^\mu = \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)} a^\mu \partial_\mu \phi - a^\nu {\delta^\mu}_\nu \mathcal{L} \ \ \ \ \ (22)

\displaystyle a^\nu {T^\mu}_\nu = a^\nu \Bigg(\frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)} \partial_\nu \phi - {\delta^\mu}_\nu \mathcal{L} \Bigg) \ \ \ \ \ (23)

\displaystyle {T^\mu}_\nu = \Bigg(\frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)} \partial_\nu \phi - {\delta^\mu}_\nu \mathcal{L} \Bigg) \ \ \ \ \ (24)

where the tensor {{T^\mu}_\nu} is called the energy-momentum tensor.

The Hamiltonian density then can be described as the zero-zero component of this energy-momentum tensor, i.e.

\displaystyle H = \int{d^3\textbf{x} \, T^{00}} = \int{d^3\textbf{x} \Big( \frac{1}{2} \pi^2 + \frac{1}{2} (\nabla \phi)^2 + \frac{1}{2} m^2 \phi^2 \Big)} = \int{d^3\textbf{x} \, \mathcal{H}} \ \ \ \ \ (25)

and the momentum is

\displaystyle P^i = \int{d^3\textbf{x} \, T^{0i}} = - \int{d^3\textbf{x} \, \pi \partial_i \phi} \ \ \ \ \ (26)

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Thursday, March 4, 2010

Riemannian Manifolds – Part 2

by Octavian

— 1.4. Torsion and Curvature Tensors —

We have talked so far about the properties which differ the curved manifold with the flat one, such as whether the connection coefficient vanishes or not, but we still don’t have the intrinsic mechanicsm which allows us to determine the curvature of manifold, in the sense that this mechanism must not rely on the coordinate system heavily. The nonvanishing connection coefficient cannot provide these requirements for us for two reasons. First, because it depends on how we make the system of coordinate basis vectors over the manifold and it ends up with how we choose the coordinate system. And second, because it cannot tell us how much the curvature of some region of a manifold. In this subsection we will discuss about the properties of torsion tensor and some curvature tensors. We also discuss the torsion tensor here because its geometrical meaning is as important as the curvature tensors. Here, we will talk about torsion tensor first before attacking the curvature tensors because of the former is simpler than the latter.

The torsion tensor {T : \chi M \times \chi M \rightarrow \chi M} is defined as

\displaystyle T(X,Y) = \nabla_X Y - \nabla_Y X - [X,Y] \ \ \ \ \ (1)

for the vector fields {X, Y \in \chi M}. Componentwise, it is described as

\displaystyle {T^\lambda}_{\mu \nu} e_\lambda = T(e_\mu, e_\nu) = ({\Gamma^\lambda}_{\mu \nu} - {\Gamma^\lambda}_{\nu \mu}) e_\lambda \ \ \ \ \ (2)

such that we have

\displaystyle {T^\lambda}_{\mu \nu} = {\Gamma^\lambda}_{\mu \nu} - {\Gamma^\lambda}_{\nu \mu} \ \ \ \ \ (3)

Therefore, torsion tensor measures how much we differ from the symmetric property of connection coefficient in the Levi-Civita case. If {{T^\lambda}_{\mu \nu} = 0}, then the manifold is called torsionless manifold.

If we have three points {p, q, r \in M} with their coordinates {\{x^\mu\}}, {\{x^\mu + \delta^\mu\}} and {\{x^\mu + \epsilon^\mu\}} respectively, with {\delta^\mu, \epsilon^\mu \ll} for all {\mu}. If we construct the vectors {X = \delta^\mu e_\mu} and {Y = \epsilon^\mu e_\mu} such that {X} ({Y}) is the vector which connects the point {p} to {q} ({r}). If we parallel transport the vector {X} ({Y}) along the infinitesimal line {pr} ({pq}), then the component of vector {X} ({Y}) becomes {\delta^\mu - \delta^\lambda \epsilon^\nu {\Gamma^\mu}_{\nu \lambda}} ({\epsilon^\mu - \epsilon^\nu \delta^\lambda {\Gamma^\mu}_{\lambda \nu}}). This transported vector of {X} ({Y}) connects a point {r} ({q}) to a new point {s_1} ({s_2}), such that we have the vector which connects points {p} and {s_1} ({p} and {s_2}) is {pr + rs_1} ({pq + qs_2}), i.e.

\displaystyle pr + rs_1 = \epsilon^\mu + \delta^\mu - \delta^\lambda \epsilon^\nu {\Gamma^\mu}_{\nu \lambda} \ \ \ \ \ (4)

\displaystyle pq + qs_2 = \delta^\mu + \epsilon^\mu - \epsilon^\nu \delta^\lambda {\Gamma^\mu}_{\lambda \nu} \ \ \ \ \ (5)

The difference between the vectors {ps_1} and {ps_2} is

\displaystyle \delta^\lambda \epsilon^\nu ({\Gamma^\mu}_{\lambda \nu} - {\Gamma^\mu}_{\nu \lambda}) \ \ \ \ \ (6)

where we notice that the term inside the parenthesis is the torsion tensor {{T^\mu}_{\lambda \nu}}. Hence, if the torsion tensor vanishes, then the parallelogram made of small displacement vectors and their parallel transport is closed.

The first curvature tensor I want to discuss is the Riemann curvature tensor {R : \chi M \times \chi M \times \chi M \rightarrow \chi M}. It is defined as

\displaystyle R(X, Y) Z = \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X,Y]} Z \ \ \ \ \ (7)

for {X, Y, Z \in \chi M}. The component of Riemann tensor is described as

\displaystyle {R^\lambda}_{\mu \nu \sigma} e_\lambda = (\partial_\nu {\Gamma^\lambda}_{\sigma \mu} - \partial_\sigma {\Gamma^\lambda}_{\nu \mu} + {\Gamma^\omega}_{\sigma \mu} {\Gamma^\lambda}_{\nu \omega} - {\Gamma^\omega}_{\nu \mu} {\Gamma^\lambda}_{\sigma \omega}) e_\lambda \ \ \ \ \ (8)

such that we have

\displaystyle {R^\lambda}_{\mu \nu \sigma} = \partial_\nu {\Gamma^\lambda}_{\sigma \mu} - \partial_\sigma {\Gamma^\lambda}_{\nu \mu} + {\Gamma^\omega}_{\sigma \mu} {\Gamma^\lambda}_{\nu \omega} - {\Gamma^\omega}_{\nu \mu} {\Gamma^\lambda}_{\sigma \omega} \ \ \ \ \ (9)

The geometrical meaning of Riemann tensor is about the difference of two vectors after they are parallel transported from the same vector along two different paths. Suppose we have four points {p, q, r, s} which have coordinates {\{x^\mu\}}, {\{x^\mu + \delta^\mu\}}, {\{x^\mu + \epsilon^\mu\}} and {\{x^\mu + \delta^\mu + \epsilon^\mu\}} respectively. If we have a vector {V \in T_pM}, then after parallel transport to point {s} along the path {pqs}, this vector has the form

\displaystyle V^\mu - V^\kappa \delta^\nu {\Gamma^\mu}_{\nu \kappa} - V^\kappa \epsilon^\nu {\Gamma^\mu}_{\nu \kappa} - V^\kappa \epsilon^\lambda \delta^\nu (\partial_\nu {\Gamma^\mu}_{\lambda \kappa} - {\Gamma^\omega}_{\nu \kappa} {\Gamma^\mu}_{\lambda \omega}) \ \ \ \ \ (10)

and similarly for the path {prs},

\displaystyle V^\mu - V^\kappa \epsilon^\nu {\Gamma^\mu}_{\nu \kappa} - V^\kappa \delta^\nu {\Gamma^\mu}_{\nu \kappa} - V^\kappa \epsilon^\lambda \delta^\nu (\partial_\lambda {\Gamma^\mu}_{\nu \kappa} - {\Gamma^\omega}_{\lambda \kappa} {\Gamma^\mu}_{\nu \omega}) \ \ \ \ \ (11)

where we omit the terms larger than second order in {\delta} and {\epsilon}. The difference between these two vectors is

\displaystyle V^\kappa \epsilon^\lambda \delta^\nu (\partial_\lambda {\Gamma^\mu}_{\nu \kappa} - \partial_\nu {\Gamma^\mu}_{\lambda \kappa} + {\Gamma^\omega}_{\nu \kappa} {\Gamma^\mu}_{\lambda \omega} - {\Gamma^\omega}_{\lambda \kappa} {\Gamma^\mu}_{\lambda \omega}) \ \ \ \ \ (12)

and again, we can identify the term in parenthesis as the Riemann tensor {{R^\mu}_{\kappa \lambda \nu}}.

The Ricci curvature tensor is defined as

\displaystyle Ric (X,Y) \equiv \langle dx^\mu, R(e_\mu, Y) X \rangle \ \ \ \ \ (13)

and its component is

\displaystyle R_{\mu \nu} = Ric(e_\mu, e_\nu) = {R^\lambda}_{\mu \lambda \nu} \ \ \ \ \ (14)

such that we can form the Ricci scalar curvature, defined as the contraction of Ricci curvature tensor with the metric,

\displaystyle R = g^{\mu \nu} R_{\mu \nu} \ \ \ \ \ (15)

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Friday, February 26, 2010

Feynman plays bongo

by Octavian

I know that Feynman’s hobby is to play bongo (and I also see his photo in his Feynman Lectures book), but this video tells me about his skill.

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Friday, February 26, 2010

This is the last day, but the beginning of new adventure; Gell-Mann Conference – Day 3

by Octavian

Yesterday after the conference was finished, all participants went to Regent Hotel to attend the banquet. Well I’m not interested to tell something about the banquet, because it’s typical. Moreover, I just took photo of the front of NEC that evening, and took no photo in the banquet party, which means it’s not my enjoyment to come to this type of event.

NEC, right before banquet

The conference today is very exciting, especially in the debates from Dvali and others, which make the situation alive. This is the list of speakers, complete with their titles of presentations:

Stephen Adler, from IAS, Princeton University. Dark Matter Scattering and the Flyby Anomalies.

Stephen Adler

Georgi Dvali, from New York University. Microscopic Gravity and Particle Physics.

Georgi Dvali

Paul Frampton, from University of North Carolina. Primordial Black Holes as All Dark Matter.

Paul Frampton

Viatcheslav Mukhanov, from Ludwig-Maximilians University. Quantum Origin of the Universe Structure: Theory and Observations.

Viatcheslav Mukhanov

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Friday, February 26, 2010

Crazy!; Gell-Mann Conference – Day 2

by Octavian

This is my report about what happened yesterday. If you ask it about how it is, I’ll say it’s crazy.

The conference was started by the talk from Hagen Kleinert, Free University of Berlin, about Gauge Structures in Multivalued Fields. He had clear ways to explain something, and in fact, he also made a book about Path Integral and Multivalued Fields which are sold here in front of the Auditorium by the World Scientific. The next speaker is Rabindra Mohapatra from University of Maryland who talked about Neutrino Masses and Grand Unification of Flavor. The last speaker in this first session of the day before tea break is Serguey Petcov from SISSA, Italy, and his presentation titled Neutrino Mixing, Oscillations, Leptonic-CP Violation, the See-Saw Mechanism and Beyond.

Hagen Kleinert

Rabindra Mohapatra

Serguey Petcov

Session 2 after tea break was started by a presentation about Some Recent Progress in AdS/CFT by John Schwarz from Caltech. I think I enjoy this talk very much. Then, it is followed by the presentations by Ignatios Antoniadis from CERN, Dieter Lust from Ludwig-Maximilians University, Lars Brink from Chalmers University of Technology, Sweden, and Spenta Wadia from Tata Institute, India, with the topics respectively, Aspects of String Phenomenology, String Corrections to QCD, Maximal Supersymmetry and Exceptional Groups, and Gauge/Gravity Duality and Some Applications.

John Schwarz

Ignatios Antoniadis

Dieter Lust

Lars Brink

Spenta Wadia

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Wednesday, February 24, 2010

It is a really great day; Gell-Mann Conference – Day 1

by Octavian

And the first day of Gell-Mann conference finally finished tonight. There were so many interesting things happened today, and those all started with a kind of not-so-delicious-because-I’m-not-so-hungry breakfast this morning, yet because I had eaten too much that night.

In front of the Auditorium, NEC

Anyway, the conference was started at 8 a.m. this morning, begun by the opening talk of Su Guaning, the President of NTU. Following a speech from Bertil Andersson (provost of NTU), there were some sequential talks from Francis Yeoh from the National Research Foundation, Singapore, whose topic Strengthening Research. Growing Innovation and Enterprise, Teck Seng Low from Agency for Science, Technology and Research, Singapore, with a topic The Nexus of Science: Key to Spawning New Innovations and Discoveries, and Choy Heng Lai from NUS, where his talk is about Overview of Complexity and Quantum Information Research in Singapore. These talks are mainly about the research situation in Singapore, not much the topics I want to listen in this conference.

The next speaker is Harald Fritzch from Ludwig-Maximilians University. He talked about Murray Gell-Mann – A Scientific Biography, but actually it was not only about the biography; the Gell-Mann’s idea about quarks concept is also explained. After that the physicist from MIT, George Zweig, also gave a talk titled Memories of Murray and the Quark Model. Generally the Zweig’s talk is more detail than Harald’s about the concept of quark revealed by Gell-Mann, but Harald Fritzch’s talk is broader, in the sense that he also told us about the personal life of Gell-Mann; something that is quite rare in this first day of the conference.

Harald Fritzch

George Zweig

After this first session there was a tea break. Well, it’s the part that I cannot tell too much. The whole story is about eating and eating (and drinking, of course).

The situation in a tea break

Auditorium before the second session

After the tea break, the first speaker is the Nobel laureate, Chen Ning-Yang, from Tsinghua University (or Brookhaven?). He talked much about Some Problems in Cold Atom Research. Well, rumor (ya, this rumor is from my friends in NTU, they heard it from the Yang’s talk 2 years ago) said that when Yang was 72 years old, his wife was 27. The first time I heard it, besides counting the difference of these numbers, was thinking about the breaking of this permutability after the time goes by; well don’t take it too seriously. Next is Nicholas Samios from Brookhaven National Laboratory, who talked about Murray and \Omega^-. And the next two speakers are Marek Karliner from Tel-Aviv University, Israel, and Gabriel Karl from University of Guelph, who talked about From \Omega^- to \Omega_b and Early History of QCD and Quarks, respectively.

Chen Ning-Yang

Nicholas Samios

Marek Karliner

Grabriel Karl

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Tuesday, February 23, 2010

And finally I arrived in Singapore; Gell-Mann Conference – Day 0

by Octavian

Yak, here’s the thing. I just arrived in Nanyang Executive Center (NEC), one of convenient place here in Nanyang Technological University (NTU), where since the last 4 years I have visited this place twice; the first is when I took part in IPhO XXXVII, July 2006. And of course my arrival at this time has a strong reason. You can assume that I take a holiday in Singapore, but it’s actually not true, because if I wanna enjoy my holiday, I won’t go to NTU. :)

Identity card and banquet invitation letter

I am here to attend the Conference in Honour of Murray Gell-Mann’s 80th Birthday, with the topics are all about Quantum Mechanics, Elementary Particles, Quantum Cosmology and Complexity. I can say that this is a great conference held near my university, because it has many notable speakers in the main session. I see many physicists outside there but I still cannot recognize who they are. Just be patient until they present their talks tomorrow.

It's taken right before I write this post. Guess what book is that!

Above is one of corners in my room. Well, I try to write shortly about my route to arrive in this place. When I landed in Changi Airport, my position was in Terminal 1. After taking the free-of-charge Skytrain to Terminal 2, I continue walking to the MRT (Mass Rapid Transfer) station in the basement of Terminal 2. There, I bought a ticket to the Boon Lay MRT station, with a transfer in Tanah Merah. In Boon Lay, I walked to the nearest BUS station, in a direction for NTU and NIE. I took BUS 179 and paid it using coins, S$1.20. Then I stopped in B05 Canteen 2, a bus stop inside NTU. After 5-minutes walking finally I arrived here in NEC. Well, the real condition was not so straight and easy like I describe above. There were too many distractions, such as asking someone or losing direction, which surely made me pain and tired. But overall, this place is kinda easy to find.

Tomorrow will be a fantastic day, and of course it will be energy-consuming. Now I want to take a nap.

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Monday, February 22, 2010

Riemannian Manifolds – Part 1

by Octavian

— 1. Riemannian Manifolds —

— 1.1. Metric Tensor —

Now it’s the time to introduce the concept of distance into our manifold. In the previous posts, we have defined the curved space and generalized some objects which initially only exist in the flat space such as vector and dual vector fields, and regard them as the quantities which have the representations in terms of coordinate basis and hence also have the components, relative to this coordinate. However, in daily life we must tackle the space which has a distance concept within it and the concept of length of vector. The latter has a great significant functionality because often in physics we need to know the magnitude of quantity defined as the vector and dual vector fields. The development of the topics discussed in this post is useful for the construction and formulation of general relativity.

For if we want to define a distance notion in a given manifold {M}, then we can simply take any two arbitrary points in {M} and define how much the distance between them. Of course, we can say that our manifold has a distance notion only after the distance of any pairs of its points has been defined. Notice that this distance is not affected by our choice of coordinate; the type of coordinate system and the way we use it won’t change the distance we have defined between pairs of points. Hence, we should construct a mechanism which allows us to pretend the distance of two fixed points although we change the coordinate system of our manifold.

If we have two nearby points {p, q \in M} which have the coordinate values {\{x^\mu\}} and {\{x^\mu + dx^\mu\}}, then the distance between them should be comparable to the difference between their coordinates, i.e. {dx^\mu}, for {1 \leq \mu \leq n}. Precisely, if we make a vector {\textbf{ds}} which connects {p} and {q}, then we can represent it in terms of basis vectors as

\displaystyle \textbf{ds} = dx^\mu e_\mu \ \ \ \ \ (1)

where {e_\mu = {{\partial}}/{{\partial} x^\mu}} is the basis vector, such that the distance {ds} between these points can be stated as the length of this vector, i.e.

\displaystyle ds^2 = \textbf{ds} \cdot \textbf{ds} = dx^\mu dx^\nu e_\mu \cdot e_\nu \ \ \ \ \ (2)

If we define

\displaystyle g_{\mu \nu} \equiv e_\mu \cdot e_\nu \ \ \ \ \ (3)

then we can have the convenient description for the distance of points {p} and {q} as

\displaystyle ds^2 = g_{\mu \nu} dx^\mu dx^\nu \ \ \ \ \ (4)

Since {g_{\mu \nu}} is the inner product of two basis vectors, we can interpret it as the component of a {(0,2)}-tensor {g \in \mathcal{T}^0_2 M} where for the next discussion will be called as the metric tensor, and we can write it in terms of basis of the vector space {\mathcal{T}^0_2 M},

\displaystyle g = g_{\mu \nu} dx^\mu \otimes dx^\nu \ \ \ \ \ (5)

where now the {dx^\mu} is not the difference between coordinates of {p} and {q} which most physicists will imagine it as the infinitesimal distance, but it should be considered as the basis of dual vectors in {M}. Being a {(0,2)}-tensor, it needs two vectors to produce a real number, and if we provide the vector {ds} for its argument we will get the equation of {ds^2 = g_{\mu \nu} dx^\mu dx^\nu} above. Since {dx^\mu} and {dx^\nu} can be interchanged, then we need the metric {g} to be symmetric in its indices,

\displaystyle g_{\mu \nu} = g_{\nu \mu} \ \ \ \ \ (6)

If we have two vectors {U, V \in T_pM}, then the inner product between them is

\displaystyle g(U,V) = g_{\mu \nu} U^\alpha V^\beta (dx^\mu \otimes dx^\nu) (e_\alpha, e_\beta) = g_{\mu \nu} U^\mu V^\nu \ \ \ \ \ (7)

such that if we set {V = U}, then we have the length of vector {U},

\displaystyle |U|^2 = g(U,U) = g_{\mu \nu} U^\mu U^\nu \ \ \ \ \ (8)

But it still doesn’t tell us the whole story. We can restrict the metric {g} to be positive-definite, in the sense that if we have a vector {U \in T_pM}, then the norm of this vector {|U|^2} must be nonnegative, or {|U|^2 = g(U,U) \geq 0}, where the equation holds only for the case of the nullity of {U}. If a manifold {M} is equipped with the metric {g} which is symmetric and positive-definite, then this manifold is called the Riemannian manifold, and its metric {g} is called the Riemannian metric. Since it is positive-definite, all eigenvalues of Riemannian metric are positive, and hence it has the inverse denoted as {g^{\mu \nu}}. Moreover, it also can be diagonalized by using certain procedure involving the orthogonal matrix, and rescaled such that all entries in the diagonal of diagonalized metric are the unity. This matrix is nothing other than the identity metric, which is also known as the metric tensor of the Euclidean space {{\mathbb R}^n}, denoted as {\delta_{\mu \nu}}. Therefore, we conclude that the Riemannian metric {g} of a Riemannian manifold {M} can be reduced to the identity metric {\delta} of the Euclidean space {{\mathbb R}^n}.

We can weaken our restriction about the positive-definiteness of metric to a more flexible condition: {g(U,U)} can be zero although {U} is not zero, and it is possible to have {g(U,U) < 0} for a certain {U \in T_pM}. If a manifold {M} is equipped with this kind of symmetric metric, then {M} is called pseudo-Riemannian manifold, and its metric {g} is called pseudo-Riemannian metric. We can principally reduce this metric into the diagonal matrix, and hence we will get some positive and negative eigenvalues, or after rescaling and reordering, we will have the diagonal matrix {\eta = diag(1, \ldots, 1, -1, \ldots, -1)}, in which there are {r} numbers of {1} and {s} numbers of {-1}. The signature of metric is defined as {r - s}, and it is same for all points in the manifold. For the specific case, if the number of positive eigenvalues is only one, then {M} is called the Lorentzian manifold, and its metric {g} is called the Lorentzian metric. The diagonal metric corresponding to the Lorentzian one is called the Minkowski metric, {\eta = diag(1, -1, \ldots, -1)}.

Let’s back to the case of Riemannian manifold {M} with a Riemannian metric {g}. If this metric is provided only with one vector {U = U^\mu {\partial}/{\partial x^\mu} \in T_pM}, i.e. {g(U, \cdot)}, then it will produce something like the dual vector, since this object needs one more vector to produce a real number. Therefore, the metric {g} can be a bridge between the set of vectors and dual vectors in the manifold. If we denote the dual vector corresponding to {g(U, \cdot)} as {U_\nu dx^\nu}, then

\displaystyle U_\nu dx^\nu = g(U, \cdot) \ \ \ \ \ (9)

such that componentwise it has the form

\displaystyle U_\nu = g_{\mu \nu} U^\mu \ \ \ \ \ (10)

If the vector {V \in T_pM} is taken into the empty slot of {g(U, \cdot)}, then we have

\displaystyle g_{\mu \nu} U^\mu V^\nu = U_\nu V^\nu \ \ \ \ \ (11)

where the operation of inner product now can be carried out easily by using this new tool.

Here is the clear ground for this matter. In the manifold where we don’t introduce the metric, the inner product takes place between a vector and a dual vector. And since there is no link between the set of tangential vectors {T_pM} and dual vectors {T_p^\ast M}, we can take a product between any vectors and any dual vectors without an interesting intuitive sense. If we are working in the (pseudo-)Riemannian manifold, then the inner product also takes place between a vector and a dual vector. But since there is a linking chain between them, so called as the metric, then we will have an interesting situation when we take an inner product between a vector and its corresponding dual, in which it will give us the norm or length of that vector (or, dual vector). Of course we can freely take an inner product between a vector with any dual vectors which don’t correspond to that vector, in which we will have the ordinary inner product between two different vectors.

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