Learn About Conservation Law & Geodesic Equation

[space-time]

In my quest to learn how to solve the geodesic equation, I came across this law which apparently holds true for all metrics (according to what I read):

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:

(dx⁰/dτ) = (dt/dτ) = Φ where Φ is the relativity factor 1/sqrt(1 - v²/c²)

(dx¹/dτ) = (dx/dτ) = v_xΦ where v_x is the x-component of velocity

(dx²/dτ) = (dy/dτ) = v_yΦ

(dx³/dτ) = (dz/dτ) = v_zΦTherefore, the expression g_ab(dx^a/dτ)(dx^b/dτ) evaluates to:

-Φ² + ( v_x)²Φ² + ( v_y)²Φ² + ( v_z)²Φ²

This turns into:

-Φ² + v²Φ²

which can be factored into:

(-1 + v²)Φ²The above expression does not always = -1.

In fact the above expression is:

(-1 + v²) / (1 - v²/c²)

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?

 

[Orodruin]

You are mixing up expressions from texts using c=1 and texts not using such units. As a result, your results are only valid in units where c=1 and there is nothing strange about that.

In general units where c is not equal to one the 4-velocity is normalised such that $V^2 = -c^2$ and the metric has factors of c in it that you missed.

 

[space-time]

You are mixing up expressions from texts using c=1 and texts not using such units. As a result, your results are only valid in units where c=1 and there is nothing strange about that.

In general units where c is not equal to one the 4-velocity is normalised such that $V^2 = -c^2$ and the metric has factors of c in it that you missed.

Thank you for this post. I have just one more question. You said that I am missing factors of c in the case where c is not 1. Where am I missing those factors of c?

The form of the line element that I used when evaluating the metric tensor contracted over the derivatives is:

ds² = -c²dt² + dx² + dy² + dz²

The c term is there. Where then am I missing factors of c? Thank you for your assistance.

 

[Nugatory]

Therefore, the expression g_ab(dx^a/dτ)(dx^b/dτ) evaluates to:

-Φ² + ( v_x)²Φ² + ( v_y)²Φ² + ( v_z)²Φ²

What value are you using for $g_{00}$ here?

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?

If an object is in free fall (no proper acceleration, no force acting on it) then there always exists a local inertial frame in which the $v=0$ - but once you restore your lost $c$ you'll find that the invariant works even for non-zero $v$.

 

[space-time]

What value are you using for $g_{00}$ here?
If an object is in free fall (no proper acceleration, no force acting on it) then there always exists a local inertial frame in which the $v=0$ - but once you restore your lost $c$ you'll find that the invariant works even for non-zero $v$.

I'm using -1 for g₀₀.

Also, what missing c?

I'm using

ds² = -c²dt² + dx² + dy² + dz²

 

[Nugatory]

I'm using -1 for g₀₀.
...
ds² = -c²dt² + dx² + dy² + dz²

You may be using -1 for $g_{00}$, but the line element you just wrote down says that it is $-c^2$ not $-1$.

 

[Ibix]

Where then am I missing factors of c?

When $c$ has been left out, you can put it back in by dimensional analysis. All terms in the expression $g_{ab}U^aU^b=1$ must have the same units, but note that $dt/d\tau$ has different units to $dx/d\tau$ if $c\neq 1$, and the terms on the left hand side largely have dimensions of velocity squared.

 

[space-time]

You may be using -1 for $g_{00}$, but the line element you just wrote down says that it is $-c^2$ not $-1$.

When $c$ has been left out, you can put it back in by dimensional analysis. All terms in the expression $g_{ab}U^aU^b=1$ must have the same units, but note that $dt/d\tau$ has different units to $dx/d\tau$ if $c\neq 1$, and the terms on the left hand side largely have dimensions of velocity squared.

Thanks guys I got it. When I did the calculation again using -c² as g₀₀ , I got -c² as my answer to g_ab(dx^a/dτ)(dx^b/dτ)

If c is taken to be 1, then this would correspond with the whole g_ab(dx^a/dτ)(dx^b/dτ) = -1 rule.

In that case, I guess the general rule for any units (not just c = 1 units) would be:

g_ab(dx^a/dτ)(dx^b/dτ) = -c²

for the ( - + + +) signature. Is this correct?

 

[Ibix]

Looks right. I must say I've taken to dropping the factors of $c$ and just working in units like light seconds and seconds. Translating back to SI units is trivial.

 

[Nugatory]

Is this correct?

Looks right, and it's what @Orodruin said back in post #2

 

[vanhees71]

It's simply $x^0=c t$. Then $g_{\mu \nu}=\mathrm{diag}(-1,1,1,1)$. Just set $c=1$, which are the natural units, and you don't stumble over these annoying factors of $c$ anymore ;-))). SCNR.