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Big test

February 18, 1999

Web URL: http://newton.ex.ac.uk/teaching/rj/

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Summary

Syllabus

Special relativity
Equivalence principle
Riemannian spaces
Curvature
General tensor analysis
Einstein's theory
Principle of covariance
Affine connection
Riemannian curvature
Precession of perihelion of mercury
Schwarzschild metric
Bending of light by sun
Black holes
Cosmology: Roberson-Walker metric
Friedmann models
Big-bang theory
Helium production

Special Relativity
Event vector

$\begin{displaymath}x^\mu = (ct, x^1, x^2, x^3) \end{displaymath}$

In inertial reference frame, space-time interval:

$\begin{displaymath}ds^2 = \eta_{\mu \nu} dx^\mu dx^\nu \end{displaymath}$

$\begin{displaymath}ds^2 = c^2 d \tau^2 = c^2 dt^2 - \{ dx^{1^2} +d x^{2^2} +d x^{3^2}\} \end{displaymath}$
Inertial reference frame is one in which free particles move with constant speed along straight lines.
Assumptions in special relativity:
Speed of light is independent of frame;
Free particles always move with constant speed on straight line;
Empty space is homogeneous and isotropic.
Lorentz transformation: For prime frame moving along x-axis with speed v.
$dx^{0^\prime} = \gamma \{ dx^0 - {v\over c} dx^1\}$
$dx^{1^\prime} = \gamma \{ d x^1 - {v \over c} dx^0 \}$
$dx^{2^\prime} = dx^2$
$dx^{3^\prime} = dx^3$
$\gamma =\frac{1}{\surd{1- v^2/c^2}}$

$\begin{displaymath}dx^{\prime \mu} = L^\mu_\nu(v) dx^\nu \end{displaymath}$
It can be shown: $dx^{\mu} = L^\mu_\nu(-v) dx^{\prime \nu}$ and hence $L^\mu_\nu(-v) L^\nu_\kappa(v) = \delta^\mu_{~\kappa}$
4-velocity $u^\mu = \frac{d x^\mu}{d \tau}$
4-momentum $p^\mu = m u^\mu$ , m is proper mass.
Transformation of $u^\mu$ :

$\begin{displaymath}u^{\prime \mu} = L^\mu_\nu u^\nu \end{displaymath}$
Equation of motion of a free particle:

$\begin{displaymath}\dot{p}^\mu = m \frac{ d^2 x^\mu}{d\tau^2} = 0 \end{displaymath}$
Composition of velocities:
If a particle has 3-velocity (w_x, w_y, w_z) in frame S, then velocity in $S^\prime$ is

$\begin{eqnarray*} w_x^\prime = \frac{ w_x-v}{1- v w_x/c^2} \\ w_y^\prime = \fra... ... v w_x/c^2)}\\ w_z^\prime = \frac{ w_z}{ \gamma( 1- v w_x/c^2)} \end{eqnarray*}$
$\begin{eqnarray*} { F_i} = \frac{d { p_i}}{dt} = m \frac{d \gamma {v_i}}{dt} \\ \end{eqnarray*}$

Rate of working:

$\begin{eqnarray*} \frac{d E}{dt} = { F_i} {v_i} = m \gamma^3 {v_i}.\frac{d { v_i... ...dt} = m c^2 \frac{d \gamma}{dt} \\ and ~~E= mc^2 \gamma = c p^0 \end{eqnarray*}$
Law of conservation of energy-momentum: $\Delta p^\mu =0$

General Relativity
Newton's Gravitation law

$\begin{displaymath}{ F_j} = m_g { g_j}, ~~~ {g_j} = - GM {r_j}/r^3, ~~~ { F_j} = m_i { \ddot{r_j}} \end{displaymath}$
For particle moving vertically in a lift accelerating upwards with acceleration a, then

$\begin{displaymath}m_i \ddot y = -m_g g -m_i a \end{displaymath}$

where y is local coordinate. Note g can be eliminated by choosing a = -g m_g/m_i.
Weak equivalence principle: m_i =m_g
Strong equivalence principle: gravity can be eliminated locally by choosing a = -g ie. in free -fall.
Consequences:
a) Bending of light by gravitational field;
b) Gravitational red-shift.
An accelerating reference frame transforms coordinates in LIF from $\zeta^\mu$ to $x^\mu$ whereupon

$\begin{displaymath}ds^2 = g_{\mu \nu}(x) dx^\mu dx^\nu \end{displaymath}$
2-dimensional surfaces:

ds² = g_{1 1} du² + 2 g_{1 2} du dv + g_{2 2} dv²

Let $du = A d \zeta_1 + B d \zeta_2$ , $dv = C d \zeta_1 + D d \zeta_2$ Then can choose A, B, C, D and 6 indept. derivatives $\frac{\partial A }{\partial \zeta_i}, \frac{\partial B }{\partial \zeta_i}, ...$ (ie 10 quantities) such that

$\begin{eqnarray*} ds^2 = L_1 d \zeta_1^2 + L_2 d \zeta^2 + 2 L_{1 2} d \zeta_1 ... ...ial \zeta_1} = \frac{\partial L_{1 2} }{\partial \zeta_2} =0 \\ \end{eqnarray*}$

ie 9 equations (last is orientation of $\zeta_1, ~ \zeta_2$ axes).
In terms of $\zeta$ , surface is locally Euclidean. Axes are initially parallel to geodesics.
$\zeta_1, \zeta_2$ are Riemannian coordinates. We have transformed back from accelerating ref. frame to LIF. $\zeta_i$ are arc-lengths along respective axes.
Useful for maps.
For sphere, two great circles meeting at pole can be taken as Riemannian coordinates. Polar angles are not Riemannian.
Extrinsic curvature: need to embed surface in higher dimensions.
Curvature of a simple curve in 2-dimensions is defined as rate of change of tangent at point- bring circle of varying radius up to curve until it osculates, ie has common tangent..
Radius of curvature = $\frac{\partial s }{\partial \psi}$
Curvature of surface: choose a point P and determine its tangent plane to the surface. Then select direction in this plane, determined by $\theta$ say, and find curvature of curve in surface whose tangent at P is parallel to this direction. Let curvature be $K(\theta)$ . Then rotate the direction, changing $\theta$ , and determine maximum and minimum values of $K(\theta)$ . These are extrinsic values of curvature.
Intrinsic curvature: measured by property of space. eg
squash test': difference in circumference of circular segment (divided by diameter) and $\pi$ .
Gaussian curvature: if surface defined by z= z(u,v), z axis normal to surface at (u, v) =0. Then near origin:

$\begin{displaymath} z= {1 \over 2}(\frac{\partial ^2 z }{\partial u^2} u^2 + 2 ... ... u \partial v} uv + \frac{\partial ^2 z }{\partial v^2} v^2 ) \end{displaymath}$

Diagonalise the Hessian by introducing new-coordinates $\zeta_1, \zeta_2$ , then

$\begin{displaymath}z= {1 \over 2} ( K_1 \zeta_1^2 + K_2 \zeta_2^2) \end{displaymath}$

Principal radii of curvatures are ${1\over K_1}, {1 \over K_2}$ and Gaussian curvature is K = K₁ K₂.
Gaussian curvature is intrinsic property: Simplified proof. Suppose

$\begin{displaymath}ds^2 = g_{r r} d r^2 + r^2 d \theta^2 \end{displaymath}$

then we need to show K determined by g_{r, r}.
On line of constant $\theta, ~~~K_1 = \frac{\partial \psi }{\partial s}$ as this is curvature of curve.
On line of constant r, curvature is 1/r, but this lies at angle $90- \psi$ to surface normal. Hence $K_2= {sin~\psi \over r}$ and $K = \frac{sin~\psi}{r} \frac{\partial \psi }{\partial s}$
Use $cos~\psi = \frac{\partial r }{\partial s} = {1 \over {\surd g_{r r}}}$ . Thus
$K= \frac{1}{2 r g^2_{ r r}} \frac{\partial g_{r r } }{\partial r}$ .
Demonstrates that Gaussian curvature determined by metric.
Geodesics : shortest distance between 2 points.
On sphere, 2 geodesics (which meet at pole) move apart so that angular distance

$\begin{displaymath}\eta= R \phi ~sin~\theta = R \phi~ sin({s \over R}) \end{displaymath}$

or

$\begin{displaymath}\frac{d^2 \eta}{ds^2} = - \eta K \end{displaymath}$

General Tensor Analysis
If frame $S^\prime$ is accelerating with respect to S, then event $x^\mu$ transformed into $x^{\prime \mu}$ which is non-linear function of x⁰, x¹, x², x³.

$\begin{displaymath}dx^{\prime \mu} = \frac{\partial x^{\prime \mu}}{\partial x^{\nu}} dx^\nu\end{displaymath}$

Note:

$\begin{eqnarray*} \frac{\partial x^\alpha }{\partial x^{\prime \mu}} \frac{\partial x^{\prime \mu} }{\partial x^\beta} = \delta^\alpha_{~\beta} \end{eqnarray*}$
Contravariant vector $p^\mu$ satisfies transformation:

$\begin{displaymath}p^{\prime \mu} = \frac{\partial x^{{\prime \mu}}}{\partial x^{\nu}} p^\nu\end{displaymath}$

example: $d x^\mu$ .
In LIF:

$\begin{displaymath} ds^2 = \eta_{\alpha \beta} d \zeta^\alpha d \zeta^\beta \\ \end{displaymath}$

In accelerating frame where event is $x^\mu$

$\begin{displaymath}ds^2= \eta_{\alpha \beta} \frac{\partial \zeta^\alpha }{\pa... ...\partial x^\nu} d x^\mu d x^\nu = g_{\mu \nu} d x^\mu d x^\nu \end{displaymath}$

Note Riemannian metric tensor $g_{\mu \nu}$ depends on event $x^\mu$ .
Can show (problem) in different accelerating frame S $^\prime$ :

$\begin{eqnarray*} ds^2 = g^{\prime}_{{\prime \mu}{\prime \nu}} d x^{\prime \mu}d... ...ac{\partial x^{\prime \nu} }{\partial x^\beta} g_{\alpha \beta} \end{eqnarray*}$

If the relation

$\begin{eqnarray*} F^\prime_{\mu \nu} = \frac{\partial x^{\prime \mu} }{\partial ... ...{\partial x^{\prime \nu} }{\partial x^\beta} F_{\alpha \beta} , \end{eqnarray*}$

we say that $F_{\alpha \beta}$ is a second rank covariant tensor.
Define unit mixed tensor, $\delta_\mu^{~\nu}$ , which is unity when indices equal and zero otherwise.
Covariant vector satisfies

$\begin{displaymath}p^\prime_\mu = \frac{\partial x^\nu }{\partial x^{{\prime \mu}}} p_\nu\end{displaymath}$

example (problem) $\frac{\partial \phi}{\partial x^\mu}$
$g_{\mu \nu}$ is a covariant tensor: Show:

$\begin{eqnarray*} g^{\prime}_{\mu \nu} = \frac{\partial x^\alpha }{\partial x^{\... ...frac{\partial x^\beta }{\partial x^{\prime\nu}} g_{\alpha \beta} \end{eqnarray*}$
Generally can define covariant vector from contravariant one by

$\begin{displaymath}p_\mu = g_{\mu \nu} p^\nu \end{displaymath}$
Define contravariant metric tensor: $g^{\mu \nu}$ is $\mu \nu$ element of g^-1, ie

$\begin{displaymath}g_{\mu \nu} g^{\nu \kappa} = \delta_\mu^{~\kappa} \end{displaymath}$
General tensors: $F_{\alpha \beta \gamma }^{~~~~\mu \nu \kappa }$ is a mixed tensor of rank equal to the number of components.
Contraction of tensors: if $A_\mu^{~\nu}$ and $B_\kappa^{~\alpha}$ are tensors, then the contraction

$\begin{displaymath}C_\mu^{~\alpha} = A_\mu^{~\nu} B_\nu ^{~\alpha} \end{displaymath}$

is also a tensor.
Quotient theorem: If A is an object and the contraction of A with a tensor B with arbitrary components produces a tensor C, then A must be a tensor.
Principle of Covariance: Laws of SR are valid in LIF; extend to accelerating frames by transforming from $\zeta^\mu \rightarrow x^\mu$ . This is simpler if the laws are written using tensors.
or, the principle of covariance states that an equation holds in a gravitational field if two conditions are met: a) The equation holds in the absence of gravitation when $g_{\mu \nu}$ becomes $\eta_{\mu \nu}$ and the affine connection (see later) vanishes, ie at the origin of the LIF, b) the equation preserves its form under a general coordinate transformation $x \rightarrow x^{\prime}$ .
Equivalence principle requires that same equations hold in gravitational fields.
In LIF:

$\begin{displaymath}ds^2 = -c^2 d\tau^2 = \eta_{\alpha \beta} d\zeta^\alpha d\zeta^\beta \end{displaymath}$

and equation of motion of free particle:

$\begin{displaymath}\frac{d^2 \zeta^\alpha}{d \tau^2} = \ddot{\zeta^\alpha} =0 \end{displaymath}$
In accelerating frame, equation of motion becomes:

$\begin{eqnarray*} \frac{d }{d \tau}( \frac{\partial \zeta^\alpha }{\partial x^\... ...pha }{\partial x^\mu \partial x^\nu} \dot{x}^\mu \dot{x}^\nu = 0 \end{eqnarray*}$

or

$\begin{eqnarray*} \ddot{x}^\lambda + \Gamma^\lambda_{~\mu \nu} \dot{x}^\mu \dot{x}^\nu = 0 \\ \end{eqnarray*}$

where the affine connection is

$\begin{eqnarray*} \Gamma^\lambda_{~\mu \nu} = \frac{\partial x^\lambda }{\parti... ...} \frac{\partial^2 \zeta^\alpha }{\partial x^\mu \partial x^\nu} \end{eqnarray*}$

Note $\Gamma^\lambda_{~\mu \nu} = \Gamma^\lambda_{~\nu \mu}$ .
We now show that affine connection is determined by metric.

$\begin{eqnarray*} g_{\mu \nu} = \frac{\partial \zeta^\alpha }{\partial x^\mu} \frac{\partial \zeta^\beta }{\partial x^\nu} \eta_{\alpha \beta} \end{eqnarray*}$

$\begin{eqnarray*} \frac{\partial g_{\mu \nu} }{\partial x^\lambda} = \frac{\part... ...^\beta }{\partial x^\nu \partial x^\lambda} \eta_{\alpha \beta} \end{eqnarray*}$

Recall

$\begin{eqnarray*} \Gamma^\lambda_{\mu \nu} = \frac{\partial x^\lambda }{\partial... ...frac{\partial^2 \zeta^\alpha }{\partial x^\mu \partial x^\nu}\\ \end{eqnarray*}$

Mult. by $\frac{\partial \zeta^\alpha }{\partial x^\lambda}$

$\begin{eqnarray*} \Gamma^\lambda_{\mu \nu} \frac{\partial \zeta^\alpha }{\partia... ... \frac{\partial^2 \zeta^\alpha }{\partial x^\mu \partial x^\nu} \end{eqnarray*}$

Hence

$\begin{eqnarray*} \frac{\partial g_{\mu \nu} }{\partial x^\lambda} = \Gamma^\rho... ...+ \Gamma^\rho_{\lambda \nu} g_{\rho \mu} = g_{\mu \nu, \lambda} \end{eqnarray*}$
We can invert these equations to give

$\begin{eqnarray*} 2 g_{\kappa \nu} \Gamma^\kappa_{\lambda \mu} = g_{\mu \nu, \lambda}+ g_{\lambda \nu, \mu} - g_{\mu \lambda, \nu} \end{eqnarray*}$

Proof

$\begin{eqnarray*} g_{\mu \nu, \lambda} = \Gamma^\rho_{\lambda \mu} g_{\rho \nu} + \Gamma^\rho_{\lambda \nu} g_{\rho \mu} ~~~~~.... A \end{eqnarray*}$

Interchange $\lambda, \mu$

$\begin{eqnarray*} g_{\lambda \nu, \mu} = \Gamma^\rho _{\mu \lambda } g_{\rho \nu} + \Gamma^\rho_{\mu \nu} g_{\rho \lambda} ~~~~~ .... B \end{eqnarray*}$

Interchange $\nu, \lambda,$ in A

$\begin{eqnarray*} g_{\mu \lambda, \nu} = \Gamma^\rho _{\nu \mu } g_{\rho \lambda} + \Gamma^\rho_{\nu \lambda} g_{\rho \mu} ~~~~~ .... C \end{eqnarray*}$

A + B - C gives result
$\begin{eqnarray*} \Gamma^\kappa_{~\lambda \mu} = {1 \over 2} g^{\kappa \nu} \{ g... ...u \nu, \lambda}+ g_{\lambda \nu, \mu} - g_{\mu \lambda, \nu} \} \end{eqnarray*}$
Note that $\Gamma$ vanishes in LIF.
Equation for free particle is also the equation for a geodesic (see problem).
Transformation of the affine connection

$\begin{eqnarray*} \Gamma^\lambda_{\mu \nu} = \frac{\partial x^\lambda }{\partial... ...frac{\partial^2 \zeta^\alpha }{\partial x^\mu \partial x^\nu}\\ \end{eqnarray*}$

$\begin{eqnarray*} \Gamma^{\prime \lambda}_{\mu \nu } = \frac{\partial x^{\prime ... ...ime \nu}} \frac{\partial \zeta^\alpha }{\partial x^\sigma}\Bigr) \end{eqnarray*}$

$\begin{eqnarray*} = \frac{\partial x^{\prime \lambda} }{\partial x^\rho} \frac{... ... }{\partial x^{\prime \mu} \partial x^{\prime \nu}} \\ ...... D \end{eqnarray*}$

The second term destroys tensor character for $\Gamma$
Same for derivatives of vectors:

$\begin{eqnarray*} V^{\prime \mu} = \frac{\partial x^{\prime \mu} }{\partial x^\nu} V^\nu \end{eqnarray*}$

$\begin{eqnarray*} \frac{\partial V^{\prime \mu} }{\partial x^{\prime \lambda}} =... ...\partial x^\rho }{\partial x^{\prime \lambda}} V^\nu ~~~~......B \end{eqnarray*}$

-- they are not tensors.
Conclusion: differentiation is NOT a good idea!
Need covariant differentiation - which generates a tensor.
First show:

$\begin{eqnarray*} \Gamma^{\prime \lambda}_{~\mu \nu} = \frac{\partial x^{{\prime... ... \lambda} }{\partial x^\rho \partial x^\sigma}\\ ~~~~~~.... A \end{eqnarray*}$

Proof

$\begin{eqnarray*} \frac{\partial x^{\prime \lambda} }{\partial x^\rho} \frac{\partial x^\rho }{\partial x^{\prime \nu}} = \delta^\lambda_{~ \nu} \end{eqnarray*}$

Differentiate wrt $~x^{\prime \mu}$

$\begin{eqnarray*} \frac{\partial x^{\prime \lambda} }{\partial x^\rho} \frac{\pa... ...\frac{\partial x^\sigma }{\partial x^{{\prime \mu}}} ~~~~.....C \end{eqnarray*}$

Hence, from C and D,

$\begin{eqnarray*} \Gamma^{{\prime \lambda}}_{~\mu \nu} = \frac{\partial x^{\pri... ...artial^2 x^{\prime \lambda} }{\partial x^\rho \partial x^\sigma} \end{eqnarray*}$

Then from A

$\begin{eqnarray*} \Gamma^{\prime \mu}_{~\lambda \kappa} V^{\prime \kappa}= \{ \f... ...frac{\partial x^\rho }{\partial x^{\prime \lambda}} V^\sigma \\ \end{eqnarray*}$

add to B

$\begin{eqnarray*} \frac{\partial V^{\prime \mu} }{\partial x^{\prime \lambda}} ... ...}{\partial x^\rho} + \Gamma^\nu_{~\rho \sigma } V^\sigma \Bigr) \end{eqnarray*}$

Covariant derivative

$\begin{eqnarray*} V^\mu_{~;~\lambda} \equiv \frac{\partial V^\mu }{\partial x^\lambda} + \Gamma^\mu_{~\lambda \kappa} V^\kappa \end{eqnarray*}$
Covariant derivative is tensor
Covariant differentiation along a curve:

$\begin{displaymath}A^{\prime \mu}= \frac{\partial x^{\prime \mu} }{\partial x^\nu} A^\nu \end{displaymath}$

If $A^\mu = A^\mu(\tau)$ then

$\begin{eqnarray*} \frac{d A^{\prime \mu}}{d \tau} = \frac{\partial x^{\prime \mu... ...al x^\nu \partial x^\lambda} \frac{d x^\lambda }{ d \tau} A^\nu \end{eqnarray*}$

Second term involves transformation of affine connection. Define

$\begin{eqnarray*} \frac{D A^\mu}{D \tau} \equiv \frac{d A^\mu}{d \tau} + \Gamma^\mu_{~\nu \lambda} \frac{d x^\lambda }{ d \tau} A^\nu \end{eqnarray*}$

If this is a tensor operator, then

$\begin{eqnarray*} \frac{D A^{\prime \mu}}{D \tau} =\frac{d A^{\prime \mu}}{d \ta... ...nu \lambda} \frac{d x^{\prime \lambda} }{ d \tau} A^{\prime \nu} \end{eqnarray*}$

$\begin{eqnarray*} = \frac{\partial x^{\prime \mu} }{\partial x^\nu} \frac{d A^\n... ...d \tau} \frac{\partial x^{\prime \nu} }{\partial x^\tau} A^\tau \end{eqnarray*}$

Sum over $\nu$ in term 3 gives $\tau= \kappa$ . Also

$\begin{displaymath}\frac{\partial x^\sigma }{\partial x^{\prime \lambda}} \frac{d x^{\prime \lambda} }{ d \tau} = \frac{d x^\sigma }{ d \tau}\end{displaymath}$

In last term, sum over $\nu$ gives $\delta^\sigma_{~\tau}$ while sum over $\lambda$ gives $\delta^\rho_{~\kappa}$ . Last term is then

$\begin{displaymath}- \frac{\partial^2 x^{\prime \mu} }{\partial x^\sigma \partial x^\rho} \frac{d x^\rho}{d \tau} A^\sigma \end{displaymath}$

This cancels second term leaving:

$\begin{eqnarray*} \frac{D A^{\prime \mu}}{D \tau} = \frac{\partial x^{\prime \mu... ...ma^\nu_{~\kappa \sigma } \frac{d x^\sigma }{ d \tau} A^\kappa \} \end{eqnarray*}$

Hence covariant derivative of a vector along a curve is a tensor.
For a particle in a LIF:

$\begin{eqnarray*} \frac{d p^\mu}{d \tau} =0 \\ \Gamma^\mu_{~\kappa \sigma } =0 \end{eqnarray*}$

Hence $\frac{D p^\mu}{D \tau} =0$
For particle in accelerating ref. frame
$\frac{D p^\mu}{D \tau} =0$ or

$\begin{eqnarray*} \frac{d p^\mu}{d \tau} = - \Gamma^\mu_{~\kappa \sigma } \frac{... ...bda = -\Gamma^\mu_{~\kappa \sigma }\dot{x}^\kappa \dot{x}^\sigma \end{eqnarray*}$
Given a vector at a point x, and a curve $x^\mu = x^\mu (\tau)$ , we can transport the vector along the curve so that it reaches the point $x^\mu (\tau)$ with value determined by requiring $\frac{D B^\mu}{D \tau} =0$ . Such a vector is said to be parallel transported.
Geodesic Deviation: two close-by particles satisfy:

$\begin{eqnarray*} \frac{d^2 x^\mu}{d \tau^2} + \Gamma^\mu_{~\nu \lambda }\frac{d x^\nu}{d \tau}\frac{d x^\lambda}{d \tau} =0 ~~~~ ....A \end{eqnarray*}$

$\begin{eqnarray*} \frac{d^2( x^\mu + \delta x^\mu)}{d \tau^2} +\\ \Gamma^\mu_{... ...u) }{d \tau}\frac{d( x^\lambda + \delta x^\lambda) }{d \tau} =0 \end{eqnarray*}$

To first order in $\delta$

$\begin{eqnarray*} \frac{d^2 \delta x^\mu}{d \tau^2} + 2 \Gamma^\mu_{~\nu \lamb... ...{d \tau} \frac{d x^\lambda}{d \tau} ~ (term~4) \\ =0 .....~ B \end{eqnarray*}$

Write as covariant derivative

$\begin{eqnarray*} \frac{D }{D \tau} \Bigl( \frac{D \delta x^\mu}{D \tau} ) = \fr... ... \frac{d x^\kappa }{ d \tau} \frac{d \delta x^\sigma }{ d \tau} \end{eqnarray*}$

Use A to get rid of $\frac{d^2 x^\kappa }{ d \tau^2}$ . Third and last terms are equal.

$\begin{eqnarray*} \frac{D^2 \delta x^\mu}{D \tau^2} = \frac{d^2 \delta x^\mu}{d ... ...}{ d \tau} \frac{d x^\beta }{ d \tau} \delta x^\sigma ~(term~7) \end{eqnarray*}$

Use B to eliminate $\frac{d^2 \delta x^\mu }{ d \tau^2}+ 2 \Gamma^\mu_{~\nu \lambda} \dot{x} ^\nu \delta \dot{x}^\lambda$ .

$\begin{eqnarray*} \frac{D^2 \delta x^\mu}{D \tau^2} = - \Gamma^\mu_{~\nu \lambd... ...pha }{ d \tau} \frac{d x^\beta }{ d \tau} \delta x^\sigma ~ (7) \end{eqnarray*}$

Hence

$\begin{eqnarray*} \frac{D^2 \delta x^\mu}{D \tau^2} = \delta x^\sigma \frac{d x^... ...gma} - \Gamma^\mu_{~\rho \sigma}\Gamma^\rho_{~\alpha \kappa} \} \end{eqnarray*}$

$\begin{eqnarray*} \frac{D^2 \delta x^\mu}{D \tau^2}= R^\mu_{\alpha \sigma \kap... ...x^\sigma \frac{d x^\alpha }{ d \tau} \frac{d x^\kappa }{ d \tau} \end{eqnarray*}$

Here R is Riemann curvature tensor.
Analogous to equation for geodesic deviation

$\begin{displaymath}\frac{ d^2 \eta}{d s^2} = -K \eta \end{displaymath}$
It can be shown that R satisfies a number of identities including:

$\begin{displaymath}R^\mu_{\alpha \sigma \kappa} = R^\mu_{\kappa \sigma \alpha}\end{displaymath}$
Problem with principle of covariance: equation for a particle in flat space time:
$\begin{eqnarray*} \frac{D^2 \zeta^\mu}{D \tau^2} = 0 \end{eqnarray*}$

But in presence of gravitational field:

$\begin{eqnarray*} \frac{D^2 x^\mu}{D \tau^2} = R^\mu_{\alpha \beta \gamma } \fra... ...a }{d \tau} \frac{d x^\beta }{d \tau} \frac{d x^\gamma }{d \tau} \end{eqnarray*}$
Equation for world lines of photons. In LIF, where world line is $\zeta^\mu = \zeta^\mu(p)$

$\begin{eqnarray*} \frac{d^2 \zeta^\mu}{d p^2} =0 \\ \eta_{\mu \nu} \dot{ \zeta}^\mu \dot{\zeta}^\nu =0 \end{eqnarray*}$

That is, a null geodesic.
In accelerating reference frame:

$\begin{eqnarray*} x^\mu= x^\mu(p) \\ \ddot{x}^\mu + \Gamma^\mu_{\lambda \nu} \... ...lambda \dot{x}^\nu =0 \\ g_{\mu \nu} \dot{x}^\mu \dot{x}^\nu =0 \end{eqnarray*}$

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1999-02-18

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