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g

_{ab}(dx

^{a}/dτ)(dx

^{b}/dτ) = -1

Well, I tested this formula out with Minkowski space (- + + + signature):

If I understand correctly, then in Minkowsi space:

(dx

^{0}/dτ) = (dt/dτ) = Φ where Φ is the relativity factor 1/sqrt(1 - v

^{2}/c

^{2})

(dx

^{1}/dτ) = (dx/dτ) = v

_{x}Φ where v

_{x}is the x-component of velocity

(dx

^{2}/dτ) = (dy/dτ) = v

_{y}Φ

(dx

^{3}/dτ) = (dz/dτ) = v

_{z}Φ

Therefore, the expression g

_{ab}(dx

^{a}/dτ)(dx

^{b}/dτ) evaluates to:

-Φ

^{2}+ ( v

_{x})

^{2}Φ

^{2}+ ( v

_{y})

^{2}Φ

^{2}+ ( v

_{z})

^{2}Φ

^{2}

This turns into:

-Φ

^{2}+ v

^{2}Φ

^{2}

which can be factored into:

(-1 + v

^{2})Φ

^{2}

The above expression does not always = -1.

In fact the above expression is:

(-1 + v

^{2}) / (1 - v

^{2}/c

^{2})

This expression would only = -1 if either v = 0 or c = 1 (and the latter case is a matter of convention, I usually keep c = 3 * 10^8 m/s).

Having said this, where am I going wrong in my understanding (or my math in case I made any simple mistakes)? Apparently, this conservation law can be taken advantage of in order to solve some geodesic equations, but the fact that this does not = -1 for all v is throwing me off when it comes to the law.

I have one hypothesis as to the answer of my own question:

Geodesics describe the world lines of objects that are in a state of "free fall". They have no forces acting on them aside from gravity (which is not a force, but spacetime curvature in the sense of relativity). Is it possible then that this law expects v = 0 to be the case?