# Nature of Geodesic: Determine Without Knowing Metric?

• shubham agn
In summary, the conversation discusses the ability to determine if a geodesic is spacelike, timelike, or null without knowing the metric. It is concluded that the 4-velocity is not dependent on the curvature of the geodesic, but the metric is still needed to find the covariant components and determine the magnitude of the velocity. Thus, it is not possible to determine the type of geodesic without knowing the metric.
shubham agn
Hello!
Please help: A world line is given to us. It is known that it is a geodesic. The metric, however, is not known. Since we don't know the metric, it should not be possible to tell whether the geodesic is spacelike/timelike/null (Right?)
But since the geodesic is known (x,y,z,t), we can find out ∂x/∂t, ∂y/∂t, ∂z/∂t. That is, we can find out the speed of the object which will travel that geodesic. Depending on whether it is greater than, equal to or less than 'c' (light speed), we can say that the geodesic is spacelike, null or timelike. Thus, it seems we can tell this without knowing the metric. What is wrong with this?

Thank you!

shubham agn said:
Hello!
Please help: A world line is given to us. It is known that it is a geodesic. The metric, however, is not known. Since we don't know the metric, it should not be possible to tell whether the geodesic is spacelike/timelike/null (Right?)
But since the geodesic is known (x,y,z,t), we can find out ∂x/∂t, ∂y/∂t, ∂z/∂t. That is, we can find out the speed of the object which will travel that geodesic. Depending on whether it is greater than, equal to or less than 'c' (light speed), we can say that the geodesic is spacelike, null or timelike. Thus, it seems we can tell this without knowing the metric. What is wrong with this?
I don't think being a geodesic or not is relevant to whether the 4-velocity is timelike, spacelike or null because ##u^\mu u_\mu## depends only on ##u## and not on its curvature.

So, if you are given the contravariant components you need the metric to find the covariant components and ##u^\mu u_\mu##.

shubham agn said:
Hello!
Please help: A world line is given to us. It is known that it is a geodesic. The metric, however, is not known. Since we don't know the metric, it should not be possible to tell whether the geodesic is spacelike/timelike/null (Right?)
But since the geodesic is known (x,y,z,t), we can find out ∂x/∂t, ∂y/∂t, ∂z/∂t. That is, we can find out the speed of the object which will travel that geodesic. Depending on whether it is greater than, equal to or less than 'c' (light speed), we can say that the geodesic is spacelike, null or timelike. Thus, it seems we can tell this without knowing the metric. What is wrong with this?

The conclusion that $|\frac{\partial x}{\partial t}| < c$ for a timelike path is only necessarily true in Cartesian coordinates, and you need to know the metric in order to know whether your coordinates are Cartesian. For example, in polar coordinates $r, \theta, \phi$, $\frac{\partial \theta}{\partial t}$, the "speed" in direction $\theta$, can be greater than $c$.[/QUOTE]

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Mentz114
stevendaryl said:
...
Just to let you know that your post needs an edit ( at the time of writing this).

stevendaryl
Mentz114 said:
I don't think being a geodesic or not is relevant to whether the 4-velocity is timelike, spacelike or null because ##u^\mu u_\mu## depends only on ##u## and not on its curvature.

So, if you are given the contravariant components you need the metric to find the covariant components and ##u^\mu u_\mu##.

I was just thinking of the 3-velocity. Its components are clearly ∂x/∂t, ∂y/∂t, ∂z/∂t. I can find these since the (x,y,z,t) is given to me and I don't need the metric for that. Thus I can find the magnitude of the 3-velocity (without knowing the metric) and compare it with 'c' (the speed of light).

shubham agn said:
I was just thinking of the 3-velocity. Its components are clearly ∂x/∂t, ∂y/∂t, ∂z/∂t. I can find these since the (x,y,z,t) is given to me and I don't need the metric for that. Thus I can find the magnitude of the 3-velocity (without knowing the metric) and compare it with 'c' (the speed of light).

You can't find the magnitude of a vector without using the metric.

In 2D spatial coordinates, let me call them $p$ and $q$, what is the magnitude of the "velocity vector" $\vec{v} = (\frac{dp}{dt}, \frac{dq}{dt})$? If $p$ and $q$ are cartesian coordinates, then

$|\vec{v}|^2 = \frac{dp}{dt}^2 + \frac{dq}{dt}^2$

On the other hand, if they are polar coordinates (with $q$ radial and $p$ angular), then

$|\vec{v}|^2 = \frac{dq}{dt}^2 + q^2 \frac{dp}{dt}^2$

In general,
$|\vec{v}|^2 = g_{qq} \frac{dq}{dt}^2 + g_{qp} \frac{dq}{dt}\frac{dp}{dt} + g_{pq} \frac{dp}{dt}\frac{dq}{dt} + g_{pp} \frac{dp}{dt}^2$

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shubham agn said:
I was just thinking of the 3-velocity. Its components are clearly ∂x/∂t, ∂y/∂t, ∂z/∂t. I can find these since the (x,y,z,t) is given to me and I don't need the metric for that. Thus I can find the magnitude of the 3-velocity (without knowing the metric) and compare it with 'c' (the speed of light).
You need all four components to get ##u^\mu u_\mu##. There are no shortcuts to a general covariant result.

stevendaryl said:
You can't find the magnitude of a vector without using the metric.

Ok. So, if the components of the 3-velocity are v(x), v(y) and v(z) then is it not true that the magnitude of the "speed" V is given by V^2 = v(x)^2 + v(y)^2 + v(z)^2 ?
I mean does the validity of the above equation depend on what the metric is?

shubham agn said:
Ok. So, if the components of the 3-velocity are v(x), v(y) and v(z) then is it not true that the magnitude of the "speed" V is given by V^2 = v(x)^2 + v(y)^2 + v(z)^2 ?
I mean does the validity of the above equation depend on what the metric is?

Think about the simple case of an object orbiting the Earth above the equator at a constant radius $r$ and a constant angular velocity $\frac{d\phi}{dt}$. What is the speed of this object? It is $v = r \frac{d\phi}{dt}$. It isn't simply $\sqrt{(\frac{dr}{dt})^2 + (\frac{d\phi}{dt})^2}$. (That wouldn't even make sense, in terms of units)

stevendaryl said:
Think about the simple case of an object orbiting the Earth above the equator at a constant radius $r$ and a constant angular velocity $\frac{d\phi}{dt}$. What is the speed of this object? It is $v = r \frac{d\phi}{dt}$. It isn't simply $\sqrt{(\frac{dr}{dt})^2 + (\frac{d\phi}{dt})^2}$. (That wouldn't even make sense, in terms of units)

Just to elaborate a little: The fact that we call the coordinates $x$ and $y$, as opposed to $r$ and $\phi$, doesn't mean anything. You can pick whatever names for your coordinates you like, and that doesn't change the mathematics. What makes the difference between Cartesian coordinates and polar coordinates is the metric.

shubham agn said:
Hello!
Please help: A world line is given to us. It is known that it is a geodesic. The metric, however, is not known. Since we don't know the metric, it should not be possible to tell whether the geodesic is spacelike/timelike/null (Right?)
But since the geodesic is known (x,y,z,t), we can find out ∂x/∂t, ∂y/∂t, ∂z/∂t. That is, we can find out the speed of the object which will travel that geodesic. Depending on whether it is greater than, equal to or less than 'c' (light speed), we can say that the geodesic is spacelike, null or timelike. Thus, it seems we can tell this without knowing the metric. What is wrong with this?

Thank you!
You know absolutely nothing about the metric? Not even its signature?
If it is a Lorentz-signature metric, you might not need to know the whole metric to answer your question.

Do you know one timelike vector at a point along the geodesic? (I guess you must if you are assuming that you can compute a 3-velocity.)
By the way, to find the magnitude of a 3-velocity, you need at least a spatial metric (as stevendaryl suggests).

shubham agn said:
Ok. So, if the components of the 3-velocity are v(x), v(y) and v(z) then is it not true that the magnitude of the "speed" V is given by V^2 = v(x)^2 + v(y)^2 + v(z)^2 ?
I mean does the validity of the above equation depend on what the metric is?

The metric is in that equation too - you just aren't seeing it because you've written that equation in Cartesian coordinates and in those coordinates the metric components are equal to one: ##v^2=g_{xx}v_x^2+g_{yy}v_y^2+g_{zz}v_z^2=v_x^2+v_y^2+v_z^2##. If you used some other coordinate system (polar coordinates, for example) some of the metric components would have values other than one, and they'd have to appear explicitly in the calculation.
(You'll also notice that this equation is careless about the distinction between upper and lower indices - that also only works for the special case of Cartesian coordinates).

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Ok. So I think I understand it now. To compute even the magnitude of the 3-velocity along the geodesic, I will need the metric.
Now just one more query : In cartesian coordinates in flat space-time, the minimum spatial between two neighbouring points is given by (dx)^2 +(dy)^2 +(dz)^2. But now if we are in curved space-time, the minimum spatial distance between two points is not given by above equation. So if I take two points in our 3D space (in presence of Earth's gravity), then the distance measured by placing a straight ruler between these two points is not the minimum possible distance. I am finding this hard to imagine. I mean how can you get a distance smaller than that?

shubham agn said:
Ok. So I think I understand it now. To compute even the magnitude of the 3-velocity along the geodesic, I will need the metric.
Now just one more query : In cartesian coordinates in flat space-time, the minimum spatial between two neighbouring points is given by (dx)^2 +(dy)^2 +(dz)^2. But now if we are in curved space-time, the minimum spatial distance between two points is not given by above equation. So if I take two points in our 3D space (in presence of Earth's gravity), then the distance measured by placing a straight ruler between these two points is not the minimum possible distance. I am finding this hard to imagine. I mean how can you get a distance smaller than that?

It's easier to see this if you start with a two-dimensional example. On a flat sheet of paper the shortest distance between two points is given by ##\Delta{s}^2=\Delta{x}^2+\Delta{y}^2##, just as you say. But that's not true on the curved surface of the earth. In fact, you can't even lay a rectangular x-y grid down on that curved surface; the closest you can come is lines of latitude and longitude.

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shubham agn said:
Ok. So I think I understand it now. To compute even the magnitude of the 3-velocity along the geodesic, I will need the metric.
Yes
Now just one more query : In cartesian coordinates in flat space-time, the minimum spatial between two neighbouring points is given by (dx)^2 +(dy)^2 +(dz)^2. But now if we are in curved space-time, the minimum spatial distance between two points is not given by above equation. So if I take two points in our 3D space (in presence of Earth's gravity), then the distance measured by placing a straight ruler between these two points is not the minimum possible distance. I am finding this hard to imagine. I mean how can you get a distance smaller than that?

I'm not sure how you draw this conclusion. The first point is that to define distance, you need to eliminate the time from space-time. There are a couple of ways to do this, the easiest way (which turns out to not always be possible) is to define hypersurfaces of constant time, then, with time no longer a factor, you can use your space-time metric to define the distances between points on this spatial hypersurface (this is usually called the induced metric). I won't go into the "hard way", but I'll mention that if you have a rotating frame, the "easy way" I described above won't work.

The next point is that on this spatial hypersurface (which is a volume element that we've extracted by eliminating time from space-time), the geodesic paths will be in GR (though not in some other theories, but I'll not go into details) the geodesics are the paths of shortest distance.

It may be helpful to consider the surface of the Earth as an example. The lines of shortest distances between two points are great circles, which are geodesics. Note that lines of constant lattitude on the Earth are not geodesics, they are not great circles and they aren't the shortest distance between two points.

Thanks Nugatory and pervect!

## 1. What is a geodesic?

A geodesic is a curve on a surface that represents the shortest distance between two points. It is the equivalent of a straight line in Euclidean geometry.

## 2. How is the nature of a geodesic determined?

The nature of a geodesic is determined by considering the surface it lies on and the metric of that surface. The metric is a mathematical measure of distance on the surface, which can be used to calculate the shortest path between two points.

## 3. Can the nature of a geodesic be determined without knowing the metric?

Yes, the nature of a geodesic can be determined without knowing the metric. This can be done by considering the properties of the surface, such as its curvature and topology, and using these to determine the behavior of geodesics on the surface.

## 4. What are some common surfaces where geodesics are studied?

Geodesics are commonly studied on surfaces such as spheres, hyperbolic planes, and other curved spaces. They are also studied in general relativity, where they represent the paths of free-falling particles in spacetime.

## 5. What practical applications does the study of geodesics have?

The study of geodesics has many practical applications, such as in navigation and mapping, as well as in fields like physics and engineering. It also has applications in computer graphics and animation, where geodesics are used to create realistic and efficient paths for moving objects.

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