Question

23. Consider a 2D spacetime whose metric is: ds^2 = −dt^2 + [f(q)]^2 dq^2 (3) where f(q) is any function of the spatial coordinate q. a) Show that the t component of the geodesic equation implies that dt dτ is contant for a geodesic. b) The q component of the geodesic equation is hard to integrate, but show that requiring u · u = −1 implies that f dq dτ is a constant for a geodesic. c) Argue then that the trajectory q(t) of a free particle in this spacetime is such that dq dt = constant f . d) Imagine that we transform to a new coordinate system with coordinates t and x where x = F(q) and F(q) is the antiderivative of f(q). Show that the metric in the new coordinate system is the metric for flat spacetime, so the spacetime described by equation 3 is simply flat spacetime in disguise. We know geodesics in flat spacetime obey dx dτ = (f dq) dτ = constant, so the result in part b) is not surprising.

Answer 1

a) For the metric

The Lagrangian is

or

where dot represents derivative with respect to tau.The geodesic is just the Euler Lagrange equation of motion for this Lagrangian

For t, the geodesic is

where dot represents derivative with respect to tau.

Since L is not an explicit function of t, we can write

HEnce

Hence

or

or

Hence

is constant for a geodesic.

......(1)

where k is a constant.

b)

or

or

or

....... (2)

which is a constant since k is a constant.

c) Divide equation (1) by equation (2)

which gives

which is a constant since k is a constant

d) given

F(q) is an antiderivative of f(q) which means

or

or

Therefore

The metric becomes

which is the flat spacetime