1

if $v_{\gamma,p}$ is the velocity along a curve at point $p$ on the manifold of space time, and $g$ is the metric tensor, then $g(v_{\gamma,p},v_{\gamma,p})^{1/2}$, calculated in tensor notation by $(g_{ij}v^iv^j)^{1/2}$ gives the speed.

I don't understand why the speed isn't given by $g(v_{\gamma,p},\vec 0)$ instead. For example, in the Euclidean metric in 2D, $g(v,w)=\sqrt{(v_1-w_1)^2+(v_2-w_2)^2}$, and then if $v=w$, $g$ simply evaluates to $0.$

Shouldn't the speed along a curve at a point be given by the distance between the velocity tangent vector and the "standing-still" velocity vector, rather than the distance between of the velocity and itself?

user56834
  • 1,928

2 Answers2

1

That is not the Euclidean metric. The Euclidean metric in 2 dimensions would look like $g(v,w) = v_1w_1 + v_2w_2$, and the speed would similarly be given by $g(v,v)^{1/2}$.

We define the length of a (timelike) parametrized path $\gamma: [a,b] \to M$, where M is our spacetime, as $$ L(\gamma) = \int_\gamma d\tau \sqrt{|g_{ij}\dot{x}^i\dot{x}^j|}, $$ where the overdot notation indicates differentiation with respect to the path parameter (this length is independent of parametrization). Then the "distance" between two points $x$ and $y$ with timelike separation is defined as $$ d(x,y) = \mathrm{Sup}\{L(\gamma): \text{ $\gamma$ is a path connecting $x$ and $y$}\}. $$ Since geodesics maximize the length locally this equates to integration along some geodesic.

In the Euclidean case the distance is instead defined as $\mathrm{Inf}\{L(\gamma)\}$ (this is because of the signature difference), which picks out the straight line, and integration yields the usual distance formula.

As a final note, one does often use the term "metric" to refer to the distance function, while the metric tensor induces an inner product. So the confusion is understandable.

Erik Jörgenfelt
  • 1,573
0

The motion of a material particle, under the action only of inertia and gravitation, is described by the equation

$$ \frac{d^2x_{\mu}}{ds^2} + \Gamma^{\mu}_{\alpha \beta} \frac{dx_{\alpha}}{ds} \frac{dx_{\beta}}{ds} = 0$$

Which comes from the fact that the parallel displacement of a vector along a curve is given by :

$$ \Delta A^\mu = \oint \delta A^\mu $$ $$ \Delta A^\mu = - \oint \Gamma^{\mu}_{\alpha \beta}A^\alpha dx_\beta $$

If $A^\mu$ is a 4-vector velocity then this is how you compute it's displacement from a point $P_1$ on the curve to point $P_2$.

Mihai B.
  • 582