# Learn About Conservation Law & Geodesic Equation

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• space-time
In summary, the conversation discusses the law gab(dxa/dτ)(dxb/dτ) = -1, which holds true for all metrics. The expression is evaluated in Minkowski space and is found to be (-1 + v2)Φ2, which can be factored into (-1 + v2) / (1 - v2/c2). It is pointed out that the results are only valid in units where c=1 and that there are factors of c that have been missed. The law can still be used to solve geodesic equations, even for non-zero v, as long as the units are correctly taken into account.
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) = vzΦTherefore, 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 + v22The 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?

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 and vanhees71
Orodruin said:
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.

space-time said:
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##.

Nugatory said:
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

space-time said:
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:
space-time said:
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.

Nugatory said:
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 said:
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?

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.

space-time said:
Is this correct?
Looks right, and it's what @Orodruin said back in post #2

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.

## 1. What is conservation law?

Conservation law is a fundamental concept in physics that states that certain physical quantities, such as energy, momentum, and angular momentum, remain constant in a closed system over time.

## 2. Why is conservation law important?

Conservation laws provide a framework for understanding and predicting the behavior of physical systems. They also allow us to make accurate calculations and predictions about the behavior of objects and systems.

## 3. What is the geodesic equation?

The geodesic equation is a fundamental equation in the theory of relativity that describes the path of a free particle in curved spacetime. It takes into account the curvature of spacetime caused by massive objects.

## 4. How does the geodesic equation relate to conservation law?

The geodesic equation is derived from the principle of least action, which states that the path a particle takes between two points is the one that minimizes the action (a measure of the energy) along that path. This principle is closely related to the conservation of energy and momentum.

## 5. How is conservation law applied in real-world situations?

Conservation laws are applied in various fields of science and engineering, such as mechanics, electromagnetism, and thermodynamics. They are used to analyze and predict the behavior of systems ranging from subatomic particles to celestial bodies, and to design and optimize technologies such as engines, turbines, and spacecraft.

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