#### 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):

gab(dxa/dτ)(dxb/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) = vxΦ where vx is the x-component of velocity

(dx2/dτ) = (dy/dτ) = vyΦ

(dx3/dτ) = (dz/dτ) = vzΦ

Therefore, the expression gab(dxa/dτ)(dxb/dτ) evaluates to:

2 + ( vx)2Φ2 + ( vy)2Φ2 + ( vz)2Φ2

This turns into:

2 + v2Φ2

which can be factored into:

(-1 + v22

The 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?

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#### Orodruin

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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

Mentor
Therefore, the expression gab(dxa/dτ)(dxb/dτ) evaluates to:

2 + ( vx)2Φ2 + ( vy)2Φ2 + ( vz)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

Mentor
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 $-1$.

Last edited:

#### 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 gab(dxa/dτ)(dxb/dτ)

If c is taken to be 1, then this would correspond with the whole gab(dxa/dτ)(dxb/dτ) = -1 rule.

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

gab(dxa/dτ)(dxb/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.

Mentor

#### vanhees71

Gold Member
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.

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