# Four velocity in GR

1. Jul 12, 2012

### GarageDweller

In SR, Four velocity was the proper time derivative of a world line, do we have to take the covariant derivative instead in GR?

2. Jul 12, 2012

### Muphrid

The covariant derivative is a generalization of the vector derivative. How and why exactly would this connect to the derivative with respect to a scalar parameter?

(They do connect, but I think you need to think this through to understand the question you're asking first.)

3. Jul 12, 2012

### GarageDweller

Through an absolute derivative? But that wouldn't make sense as the expression would involve the four velocity itself.

hmmm

4. Jul 12, 2012

### Muphrid

Yes, relationship $d/d\tau = v \cdot D$ holds, and it's useful for when you know something as a function of position and don't want to go through to the dependence of the path on proper time.

That said, this is obviously not useful to calculate the proper time itself from $x(\tau)$, because you need to know the four-velocity first.

5. Jul 12, 2012

### Bill_K

As pointed out in the previous thread, the covariant generalization of the proper time derivative is called the absolute derivative. For example the 4-acceleration is simply

Aμ = DUμ/Dτ

If the object being differentiated is a field (i.e. defined everywhere) you can express it as the covariant derivative projected along the world line,

DWμ/Dt = Wμ Uν

However it's more general than the covariant derivative, and can be applied to quantities that are defined only on the world line. In which case you write it instead explicitly using a Christoffel symbol, as was done in Wikipedia.

Last edited: Jul 12, 2012
6. Jul 12, 2012

### Boston_Guy

An event in spacetime is described by X = (ct, x, y, z). A displacement in spacetime is
dX = (cdt, dx, dy, dz). The 4-velocity is a displacement in spacetime divided by the propertime it took for the displacement to occur, i.e. U = dX/d$\tau$.

In terms of spacetime coordinates this reads Uα = dxα/d$\tau$. If one had to find the 4-acceleration then you'd take the total derivative of the 4-velocity, i.e. A = dU/dτ. In spacetime coordinates this reads Aα = DUα/Dτ.

D/Dτ is not the covariant derivative, its called the derivative along the curve. Its related to the covariant derivative by

DAα/Dτ = Aμ dxβ/dτ

or

DAα/Dτ = dAμ/dτ + $\Gamma^{μ}_{αβ}$ Aα dxβ/dτ

7. Jul 12, 2012

### stevendaryl

Staff Emeritus
I think nobody has really answered what I thought was the original question.

If $Q$ is any 4-vector defined along a path $P(\tau)$, then

$\dfrac{DQ^{\mu}}{d\tau} = \dfrac{dQ^{\mu}}{d\tau} + \Gamma^{\mu}_{\nu \lambda} Q^{\nu} U^{\lambda}$

where $U^{\mu}$ is the velocity 4-vector.

The way I understood the original question was: what about the case $Q^{\mu} = X^{\mu}$? (Where $X^{\mu}$ is the position 4-vector). Then it's definitely not the case that

$\dfrac{DX^{\mu}}{d\tau} = \dfrac{dX^{\mu}}{d\tau} + \Gamma^{\mu}_{\nu \lambda} X^{\nu} U^{\lambda}$

But why not? The real fact is that position is not a 4-vector in curved spacetime. It's really just a 4-tuple of scalars.

8. Jul 12, 2012

### Boston_Guy

I don't understand. That's what I just posted.

A point in spacetime not a 4-vector but a spacetime displacemen is a 4-vector.

9. Jul 12, 2012

### stevendaryl

Staff Emeritus
Well, I don't see how your post explained why

$\dfrac{DQ^{\mu}}{d \tau} = \dfrac{dQ^{\mu}}{d \tau} + \Gamma^{\mu}_{\nu \lambda} Q^{\nu} U^{\lambda}$

doesn't apply in the case $Q^{\mu} = X^{\mu}$. It doesn't apply because $X^{\mu}$ is not a 4-vector.

10. Jul 13, 2012

### Staff: Mentor

It certainly applies when using curvilinear coordinates in flat space. Try it with cylindrical coordinates and see what you get.

11. Jul 13, 2012

### DrGreg

No. If you were right then you would be saying that the 4-velocity U is defined by the equation
$$U^\mu= \dfrac{dX^\mu}{d\tau} + \Gamma^\mu_{\nu \lambda} X^\nu U^\lambda$$
That's a circular definition because U appears on both sides.

12. Jul 13, 2012

### stevendaryl

Staff Emeritus
What do you mean "it certainly applies"? The definition of $U^\mu$ is just

$U^\mu = \dfrac{dX^\mu}{d \tau}$,

not

$U^\mu = \dfrac{dX^\mu}{d \tau} + \Gamma^{\mu}_{\nu}{\lambda} U^\nu X^\lambda$

The latter is not a correct equation, and it's a circular definition of $U^{\mu}$, since that appears on both sides.

But since you suggested working with cylindrical coordinates, let me work out the latter. Let's have coordinates $r$ and $\theta$, which are defined in terms of cartesian coordinates $x$ and $y$ via

$x = r cos(\theta)$
$y = r sin(\theta)$

The connection coefficients $\Gamma^{\mu}_{\nu \lambda}$ turn out to be (derivation skipped)

$\Gamma^{r}_{r r} = \Gamma^{r}_{r \theta} = \Gamma^{r}_{\theta r} = 0$
$\Gamma^{r}_{\theta \theta} = -r$
$\Gamma^{\theta}_{r r} = \Gamma^{\theta}_{\theta \theta} = 0$
$\Gamma^{\theta}_{r \theta} = \Gamma^{\theta}_{\theta r} = \dfrac{1}{r}$

So the path-derivative, or whatever it is called is:

$\dfrac{DQ^r}{dt} = \dfrac{dQ^r}{dt} - r Q^{\theta} U^{\theta}$

$\dfrac{DQ^\theta}{dt} = \dfrac{dQ^\theta}{dt} + \dfrac{1}{r} Q^{\theta} U^{r} + \dfrac{1}{r} Q^{r} U^{\theta}$

In the particular case $Q^\mu = X^{\mu}$, with $X^r = r$ and $X^\theta = \theta$, this becomes:

$\dfrac{Dr}{dt} = \dfrac{dr}{dt} - r \theta U^{\theta}$

$\dfrac{D \theta}{dt} = \dfrac{d \theta}{dt} + \dfrac{1}{r} \theta U^{r} + \dfrac{1}{r} r U^{\theta}$
$= \dfrac{d \theta}{dt} + \dfrac{1}{r} \theta U^{r} + U^{\theta}$

I don't know what those quantities are supposed to mean. Typically, the quantity $X^{\mu}$ is not a vector, and it doesn't make sense to take its covariant derivative, or to parallel-transport it.

13. Jul 14, 2012

### GarageDweller

Stevens first post cleared it up for me, i dont see my why there need be this debate. So essentially dX is a vector, and X is not, hence dX/dt is really just the vector dX divided by dt.

Last edited: Jul 14, 2012
14. Jul 14, 2012

### Staff: Mentor

In cylindrical coordinates, the component of the position vector in the θ direction is zero, not Xθ. See what happens when you make this correction in your equations. You will find that the covariant derivative of the 4 position vector actually is the 4 velocity vector, particularly when you revert to the original form of the right hand side of the equation, where there are partial derivatives of the position vector components with respect to tau, rather than the components of the 4 velocity vector. We already know that the position vector in flat space is a 4 vector, since it transforms according to the Lorentz Transformation. In fact, that is how the Lorentz Transformation is usually presented for flat space.

15. Jul 14, 2012

### stevendaryl

Staff Emeritus
This might be a terminology issue about what is the "position" of an object. You are interpreting it to mean the displacement vector from the origin to the location of the object. A displacement vector certainly is a vector (well, in curved spacetime, that's only true in the limit as the displacement is small).

However, the whole point of a coordinate system is to be able to identify points in space by an n-tuple of coordinate values. In order to specify the location of an object in polar coordinates, you have to give two numbers: the distance $r$ from the origin, and the angle $\theta$ that one must pass through on a circle of radius $r$ centered at the origin until one gets to the object. So the location of the object is specified by the pair $(r,\theta)$. This pair is NOT a vector. However, the velocity vector $U^{\mu}$ created by differentiating the pair with respect to $t$ IS a vector: $U^r = \dfrac{dr}{dt}$, $U^{\theta} = \dfrac{d \theta}{dt}$

As I said, it's a matter of terminology, I suppose, but I think it's pretty useless to describe the "position" of an object by saying "It's at a distance r from the origin along the $\widehat{r}$ direction. It's much more useful to say that the location is the n-tuple of coordinates needed to locate the object.

16. Jul 14, 2012

### Muphrid

Then what you're doing is not choosing a coordinate basis but an orthogonal (or orthonormal) basis. I think the mathematics (and the picture built on that) is a little different from the common choice of a coordinate basis.

17. Jul 15, 2012

### Staff: Mentor

Yes. I finally see what you have been trying to say regarding the notational ambiguity involved here. It is something I've known all along, but was not able to make the connection with your explanation. Here is my own spin on the subject, in terms of notation that I am used to:

Let s be a position vector drawn from an arbitrary origin to a point in flat space. If curvilinear coordinates are being used, then s can be expressed in component form in terms of coordinate basis vectors by:

s = si ai

where ai are the coordinate basis vectors.

But, in curvilinear coordinates, the si are not equal to the coordinates themselves:

si ≠ Xi

They are functions of the coordinates:

si = si (X0, ... , Xn)

Unlike the spatial position vector s, the differential position vector between two neighboring points ds can be expressed linearly in terms of the differentials of the spatial coordinates according to:

ds = dXi ai

From this, it follows immediately that the 4 velocity is given by:

U = $\frac{ds}{dτ}$ = $\frac{dXi}{dτ}$ ai

Yiiiii! I HATE LATEX

It is also possible to obtain this same result using the covariant differentiation formula, starting from the equations:

s = si ai

and si = si (X0, ... , Xn)

specifically for the particular coordinate system involved. We already showed this in our earlier posts for the case of cylindrical coordinates.

I hope this gives us some closure on the subject.

Chet