- 4.5.1 Properties of the Riemann tensor 145. - 5.5 Transformation of the densities 184. - This is the GR analogue of the Coulomb field from E&M,. - From there, we will move on to the Hamiltonian formulation of the same problem.. - The Euler–Lagrange equations come from the extremization, in the variational calculus sense, of the action:. - constant of the motion. - The Euler–Lagrange equations come from extremization of the action. - T by appropriate choice of the coefficients { α j. - Then the kinetic term in the Lagrangian will be made up of the following derivatives:. - Finally, the explicit form of the metric can be recovered from the “line element”. - 2 There are a few general properties of the metric that we can assume for all metrics considered here. - take a ˙ x α derivative of the kinetic term:. - (a) For the metric g µν in spherical coordinates, with x 1 = r, x 2 = θ, x 3 = φ, find the r component of the equation of motion (i.e. - (b) In terms of the decomposition, T µν = S µν + A µν , evaluate the sums:. - The point of that introduction to the “covariant formulation” (meaning coordinate-independent) of the equations of motion will become clear as we pro- ceed. - the full dynamics of the system. - Putting θ = π 2 reduces the dimensionality of the problem. - The importance of the inverse is to recover the tautology:. - We turn now to Hamilton’s formulation of the equations of motion. - (1.110) Notice that we have performed the transformation on only one of the two variables in the Hamiltonian. - Remember the point of the Hamiltonian approach – we want to treat p α and x α as independent entities. - corresponding directly to the transformation of the original equations of motion: ∂p ∂H. - Then we will develop constants of the motion for Euclidean space written in spherical coordinates. - The advantage of the generator ˆ K(x, p), and the transformation it generates. - r The quantity J is a constant of the motion ˙ J = −[H, J. - The generator J = a p is obviously a constant of the motion (since p itself is for uniform motion).. - x, p), which was ¯ the point of the Poisson brackets in the first place.. - and from the form of the Hamiltonian in (1.171):. - Because of the functional dependence here (the first term in parentheses depends only on φ, the second only on θ), we have a separation constant. - The motivation there came from the second piece of the [H, J. - so suppose we take the first of the three transformations, then:. - If we took the Cartesian form of the transformation. - so that’s four constants of the motion.. - Let’s think about what we can get out of the above with no work. - r = 0, which we associate with turning points of the motion. - M p corresponding to the direction (counterclockwise or clockwise) of the orbit.. - Notice the role played by the metric in the determination of the orbits (both for L and H. - g µν ) that arise in general relativity in terms of the particle orbits (geodesics) they produce. - Note that the semi-major axis of the orbit is just:. - don’t worry about the transformation of the metric itself).. - where the matrix in the middle defines the covariant form of the metric.. - This provides another parametrization for the Lorentz transformation, in terms of η, the “rapidity”.. - To get the form of the relativistic Lagrangian, we can motivate from classical mechanics. - If we think of this in terms of the infinitesimal motion of a particle along a curve, we can write:. - (a) Write ¯ x µ (t), the four-vector (just the c t ¯ and ¯ x entries, for this problem) location of the clock in ¯ O (its rest frame), and x µ , the four-vector location of the clock in O (the “lab”).. - (c) Make a Minkowski diagram of the clock’s motion in O. - We have the relativistic Lagrangian, or at least one form of the relativistic Lagrangian. - Turning to the generator of the above transformation, we have:. - The problem, if one can call it that, is in the form of the Lagrangian. - The Legendre transform of the Lagrangian is:. - and then the rest of the equations read:. - The full form of the relativistic solution is not particularly enlightening – we provide a plot in Figure 2.8 for comparison, using the same t. - The magnitude of the velocity of a particle is:. - In the rest frame of the test mass m (which we will call O. - Figure 2.13 In the stationary frame of the test mass m (here called O), the lab frame is moving to the left with speed ¯ u. - compare ds 2 in the lab and in the rest frame of the particle). - The point is, given the motion of a particle in the lab, we can find the proper time of the particle (the time in its instantaneous rest frame).. - (b) Express your result in terms of the rapidities η, ψ and α defined by:. - When we write an equation like (3.1), we are referring to the components of the tensor.. - identical to the scalar case, but with the nontrivial involvement of the basis vectors. - and using this, we can write ¯ f in terms of the original basis vectors. - so that d x ¯ α = T α β dx β and the matrix T is the inverse of the matrix in (3.26).. - This type of (second-rank) tensor transformation is not obvious under rotations since rotations leave the metric itself invariant ( R T g R = g with g the matrix representation of the metric), and we don’t notice the transformation of the metric.. - From the figure, the covariant component ¯ v 1 corresponds to the perpendicular projection of the vector onto the ˆ¯ x-axis. - The whole point of the definition is that the transformation for the components should “undo” the transformation for the basis vectors – then:. - in terms of the contravariant components before interpreting as a metric. - δx γ f α ,γ + O(δx and we would say that the (non-tensorial) change of the vector is given by:. - For now, we will define the “covariant” derivative in terms of the unspecified factor δf α. - This difference is called the “torsion” of the geometry. - f = f ˜ θ (θ, φ) ˆe θ + f ˜ φ (θ, φ) ˆe φ (3.80) in the natural orthonormal basis defined on the surface of the sphere. - What is the conserved quantity J and the interpretation of the infinitesimal transformation if you set f µν = g µν. - where C βγ is the cofactor matrix (and is independent of the element h αβ. - (a) Using this definition, find the derivatives of the determinant w.r.t. - “torsion” of the manifold. - The metric transforms in the usual way, and we can write the original metric in terms of the new one via:. - g µν , (4.26) and using the definition of the coordinates (4.24) gives:. - If we calculate the transformation of the metric:. - What is this vector in terms of the derivatives of x α w.r.t. - Indeed its magnitude is special, it is called the “curvature” of the curve. - (4.58) Collecting our results, we have two nonzero components of the Riemann tensor:. - find the curvature of the ellipse as a function of φ – plot for φ = 0. - 4.5.1 Properties of the Riemann tensor. - it in terms of more derivatives of the metric. - This led us to the definition of the Riemann tensor in (4.13):. - Suppose we know the geodesics of the space x α (τ. - Figure 4.4 Top and side views of our torus – the radius R is the distance from the center to the middle of the “tube”, and a is the radius of the tube.. - So we consider the classical and relativistic form of the “geodesic deviation equations”.. - (4.108) where the last line follows from the definition of the Riemann tensor. - What can we say about the derivatives of the Ricci tensor?. - The first thing we can do is simplify the form of the metric based on physical arguments (symmetries, for example). - Using this factor, find the mass of the sun in meters. - T µν − 1 2 g µν T (4.124) for T ≡ T α α = g αβ T αβ , the trace of the stress tensor. - We want to find the time-dependence of the motion of the two masses: x 1 (t) and x −1 (t).. - Transforming the Lagrangian is simple – for each of the terms in the potential, we have (ignoring endpoints):. - 0, but what physical description should we give to the stiffness of the rod?. - x, representing our continuum view of the function φ (x, t). - This is the Lagrangian for the continuum form of the ball and spring problem.. - An action like (5.26) depends on the derivatives of the field φ(x, t). - but what happens to the field equations that come from a Lagrange density of the form L (φ, φ ,µ. - ∂φ ,µ which is a natural generalization of the Euler–Lagrange equations of motion from classical mechanics.