Spacetime: Einstein's Theory Of Space Time

General relativity is Einstein’s theory of space, time and gravitation. It is the most beautiful physical theory ever invented. Nevertheless, it has a reputation of being extremely difficult, primarily for two reasons: tensors are everywhere, and space time is curved. But at heart it is a very simple subject. The essential idea is perfectly straightforward: Spacetime is a curved pseudo-Riemannian manifold with a metric signature of (-+++) and the relationship between matter and the curvature of spacetime is contained in the equation

R_μν- 1/2 Rg_μν=8πGT_μν (1)

This is simply an equation between 4x4 matrices, and the subscripts label elements of each matrix. The expression on the left hand side is a measure of the curvature of spacetime,

Spacetime is a four dimensional set, with elements labeled by three dimensions of space and one time. An individual point in spacetime is called an event. The path of a particle is a curve through spacetime, a parameterized one dimensional set of events, called the world line. The stage on which SR is played out is a specific four dimensional manifold, known as Minkowski spacetime.

Spacetime indices are always in Greek; occasionally we will use Latin indices if we mean only the spatial components, e.g. I = 1, 2, 3

Vectors in spacetime are four dimensional, and are often referred to as four vectors and written in components as V^μ.

It is also convenient to write the spacetime interval in a more compact form. We therefore introduce the 4x4 matrix, the metric. The metric gives us a way of taking the norm of a vector, or the dot product of two vectors. η_μν=(█(-1 0 0 0@ 0 1 0 0@ 0 0 1 0@ 0 0 0 1 ))

Then the dot product of two vectors is defined to be[5]

A∙B ≡ η_μν A^μ B^ν=-A^0 B^0+A^1 B^1+A^2 B^2+A^3 B^3 (2)

This is especially useful for taking the infinitesimal interval, or line element: ds^2=η_μν dx^μ dx^ν

Find the equation of the geodesic between (0,0) and (π/2 ,1)

Solution:

Let γ(t)=(x(t),y(t),z(t)), With γ(0)=(1,0,0) and γ(t_0 )=(0,1,1)

Then the length L of the curve is given by:

L|γ|=∫_(t_0)^(t_1)▒‖γ^' (t)‖ dt

Expanding the integrand H:= ‖γ^' (t)‖

H=√(〖 (dx/dt)^2〗^ +(dy/dt)^2+(dz/dt)^2 )=√((-sin⁡θ(dθ/dt) )^2+(cos⁡θ⁡(dθ/dt) )^2+(dz/dt)^2 )=√(〖(〖cos〗^2 θ+〖sin〗^2 θ)〗^2 (dθ/dt)^2+(dz/dt)^2 )=√((dθ/dt)^2+(dz/dt)^2 )

Re parameterizing with the arc length s. the arc-length of a curve γ at any point t can be written as s(t)=∫_0^t▒‖γ^' (τ)‖dτ⟹ds/dt=‖γ^' (t)‖

Let t=t(s)so that γ(t(s))=γ ̅(s), then

‖γ ̅^' (s)‖=‖d/ds γ ̅(s)‖=‖d/ds γ(t(s))‖=‖γ^' (t)dt/ds‖=‖(γ^' (t))/‖γ^' (t)‖ ‖=‖γ^' (t)‖/‖γ^' (t)‖

Reparametrise γ using arc length gives:

L|γ|=∫_( 0)^L▒‖γ^' (s)‖ ds=∫_0^L▒√((dθ/ds)^2+(dz/ds)^2 ) ds=∫_0^L▒√(〖(θ')〗^2+〖(z')〗^2 ) ds=∫_0^L▒1ds

Thus √(〖(θ')〗^2+〖(z')〗^2 )=1, and applying the Euler Lagrange equation we yield following system of equations: d/ds z'/√(〖(θ')〗^2+〖(z')〗^2 )=d/ds z'/1=0⟹z^'=c_1 d/ds θ'/√(〖(θ')〗^2+〖(z')〗^2 )=d/ds