# Proper Acceleration and Christoffel Symbols

1. Aug 2, 2012

### TheEtherWind

I don't know exactly what I'm looking for in this question so I'll ask it in a vague way. What is the connection between a particle's proper acceleration and the christoffel symbol of the second kind (single contravariant and double covariant) ? Is this correct?

$\frac{∂^2x^\alpha}{∂\tau^2}=-\Gamma^\alpha{}{}_\beta\gamma\frac{∂x^\beta}{∂\tau}\frac{∂x^\gamma}{∂\tau}$

What are the physical meanings behind these quantities?

2. Aug 3, 2012

### Muphrid

Do you understand what the Christoffel symbols mean in general?

You're trying to get toward the geodesic equation. Christoffel symbols arise in general from trying to take derivatives of vectors. A coordinate-free version can be written like this:

$$(v \cdot D) v = 0$$

In other words, the covariant derivative of the four-velocity along the direction of the four-velocity is zero. This encapsulates the basic idea behind there being no acceleration. When this equation is expanded out in coordinates, you get

$$(v \cdot D) v = v^i \partial_i v^j + v^i \Gamma_{ik}^j v^k$$

The $v^i \partial_i v^j$ term is identified as $d^2 x/d\tau^2$, in direct analogy to the flat space case. The connection is involved because of the use of the covariant derivative instead of the ordinary vector derivative, and in general because we can replace derivatives with respect to $\tau$ with an appropriate directional derivative in spacetime.

3. Aug 3, 2012

### Bill_K

Sorry, I have to disagree with the notation being used here. A partial derivative ∂μ or ∂/∂xμ, or also a covariant derivative, can be applied only to a field. That is, to a quantity which is a function of all four coordinates and which is defined everywhere, at least in a certain region of spacetime, and therefore we can meaningfully speak of its having partial derivatives in all four directions.

But when we talk about the properties of a particle: velocity, acceleration, etc, we refer to properties that are defined only along a single curve. For example the 4-velocity vμ(s) is a function of a single parameter, usually the proper time, or in the case of a null curve an affine parameter. In this situation you cannot write ∂vμ/∂xν, and without that you can't write vν∂vμ/∂xν either. The distinction is vital.

What you can write is an ordinary derivative d/ds along the curve. The covariant generalization of the ordinary derivative is called the absolute derivative, written δ/δs. The relationship between the two involves Christoffel symbols, similar to the relationship between the covariant and partial derivatives of a field.

For example for covariant and contravariant vectors,

δWμ/δs = dWμ/ds + ΓμνσWνvσ

δWμ/δs = dWμ/ds - ΓνμσWνvσ

In these terms, particle dynamics becomes quite simple. E.g. the 4-acceleration is simply aμ = δvμ/δs

4. Aug 3, 2012

### Mentz114

$$\frac{∂^2x^\alpha}{∂\tau^2}=-\Gamma^\alpha{}{}_\beta\gamma\frac{∂x^\beta}{∂\tau }\frac{∂x^\gamma}{∂\tau}$$
This is called the geodesic equation. If it is satisfied then the worldline $x(\tau)$ has no proper acceleration. The proper acceleration is given by ∇σuμuσ, where ∇σuμ is the covariant derivative, which uses the Christoffel symbols. uμ is $\frac{∂x^\mu}{∂\tau}$

5. Aug 3, 2012

### Bill_K

No, for the reasons I just stated.

6. Aug 3, 2012

### Muphrid

It may not make sense to talk about a velocity field for a single particle, but a single particle is just a limiting case of a general matter distribution (single particle -> a delta distribution), is it not? In that light, I think it's reasonable to consider $v \equiv v(x(\tau))$, and the chain rule would apply.

7. Aug 3, 2012

### Bill_K

One would never try to solve a problem in Newtonian mechanics by replacing all the particles by little tubes of fluid, and there is no reason to do it that way in general relativity.

Surely you would not try to find the deflection of light by considering a tube filled with an electromagnetic field. This is just making an easy problem difficult.

It's preferable to just use notation that is a) simpler and b) makes physical sense.

8. Aug 3, 2012

### Mentz114

Have I been calculating the wrong thing all this time ? I don't understand what you're saying.

In Stephani's book, he writes $\dot{u}=u_{i;n}u^n=D\dot{u}/D\tau$ (page 175).

Last edited: Aug 3, 2012
9. Aug 3, 2012

### Muphrid

Maybe one wouldn't be so explicit as to talk about "tubes of fluid," but even in the case of just a single particle, $v \cdot D$ is a well-defined derivative operator everywhere on the particle's worldline. Suggesting that we need another derivative operator that is superficially similar to the covariant derivative yet different on a technical level just doesn't strike me as useful--or even meaningful. It introduces an unnecessary distinction when the two concepts are really the same.

10. Aug 3, 2012

### Bill_K

v·D = δ/δs is well defined, but D by itself is not -- the notation is misleading. More importantly, it's also misleading to think that general relativity deals only with continuous matter, and that any problem which is discrete must be represented by a delta function. These are test particles we're talking about. They're not made of anything smaller, they just trace out the geometry.

Perhaps the reason you think δ/δs is "another" derivative that's "superficially similar" is that you're less familiar with it. In fact it's more basic than the covariant derivative, and should be introduced first.

One should use the tool that's appropriate to the problem at hand, and δ/δs is what's appropriate for any situation where test particles are involved. In addition to geodesic motion this includes such things as parallel transport, Fermi-Walker transport, geodesic deviation, etc. It would be quite cumbersome to do these in terms of continuous matter distributions.

11. Aug 3, 2012

### TrickyDicky

Bill, I think your distinction is superfluous in the case the covariant derivative is the Levi-Civita connection of a certain metric, then the geodesics for the connection are precise
ly the geodesics of the metric that are parametrised by arc length(proper time).
The geodesic equation equates the covariant derivative of the metric tensor field to the d/ds derivative along a curve.
Your distinction is valid in the general case though.

12. Aug 3, 2012

### Muphrid

I don't think I agree with the notion that $D$ by itself can't be defined. This is like saying, given $a \cdot \nabla$, the directional derivative in flat space, you can't define $\nabla$ by itself. We know that's not true; you just do $e^i e_i \cdot \nabla = \nabla$ and you're done. The same can be done with $D$.

The notion for $D$ comes from basic ideas about coordinate transformations and rotations. How does the absolute derivative operator come about in a more fundamental way?

13. Aug 3, 2012

### Bill_K

Ok, I'll say it again, Muphrid, but this is the last time. D can only be applied to a field, a function of four variables φ(x, y, z, t). It cannot be applied to a function which is defined only along a curve, such as vμ(τ), and that is what we're dealing with. If you still don't see the distinction, sorry. But here's an example, the 4-velocity for a particle in a circular orbit:

vμ = (γa cos ωτ, γa sin ωτ, 0, γa)

If you really think that Dvμ has meaning, please calculate it for this curve.

14. Aug 3, 2012

### Muphrid

I didn't mean to imply that $D$ by itself makes sense in this particular situation; I realize now that was what you must've meant, so my apologies in that regard.

15. Aug 3, 2012

### Staff: Mentor

See Attached Word Document to thread Expressions accompanying a Christoffel Symbol (a notation question)

16. Aug 4, 2012

### TrickyDicky

I understood the OP to be in the context of GR and the geodesic equation that he wanted to relate proper acceleration(a 3-vector, not a 4-acceleration) to the Christoffel symbols.
In that context with vanishing proper acceleration, that is geodesic motion, ((v.D)v=0 in Muphrid notation) one can use indistinctly the covariant derivative of the metric field and the absolute derivative along the geodesic curve.
In other circumstances like calculating any other non-vanishing 4-acceleration this is not the case of course, but I (and i guess Muphrid and Mentz) interpreted the OP to refer to the one situation where this can be done so he could relate easily " the connection between a particle's proper acceleration and the christoffel symbol" and since this is the relativity forum in the context of GR, and the EFE are anyway only exactly solvable for test particles in vacuum.
It is actually useful to take into account also that this doesn't hold in the general case, so Bill's point is pertinent.

17. Aug 4, 2012

### Staff: Mentor

18. Aug 4, 2012

### stevendaryl

Staff Emeritus
The way I prefer to think about the covariant derivative, and the geodesic equation, and proper acceleration and all that is to start with VECTOR equations (or more generally, tensor equations) and then look at specific components, rather than starting off with components.

So if you have a particle that is at a postion P(s) as a function of a path parameter s (proper time, typically), then there is a corresponding velocity 4-vector U(s) defined by

$U = \dfrac{d}{ds} P$

Then there is a corresponding acceleration 4-vector A(s) defined by:

$A = \dfrac{d}{ds} U$

It's completely straight-forward. The complexity comes in when you start looking at the components of the acceleration 4-vector. To do this, we write:

$U = \sum U^{\alpha} e_{\alpha} = U^{\alpha} e_{\alpha}$ (sum is implicit).

where $e_{\alpha}$ is a set of basis vectors, and the $U^{\alpha}$ are the components of $U$ in that basis. Then we use the product rule to compute $\dfrac{d}{ds}U$:

$\dfrac{d}{ds}U = (\dfrac{d}{ds}U^{\alpha}) e_\alpha + U^{\alpha} (\dfrac{d}{ds}e_{\alpha})$

Now we apply the chain rule: $\dfrac{d}{ds} f(g(s)) = \dfrac{df}{dg} \dfrac{dg}{ds}$

$\dfrac{d}{ds}e_{\alpha} = \dfrac{\partial e_{\alpha}}{\partial x^{\beta}} \dfrac{d x^{\beta}}{ds} = \dfrac{\partial e_{\alpha}}{\partial x^{\beta}} U^{\beta}$

So what is the meaning of the expression $\dfrac{\partial e_{\alpha}}{\partial x^{\beta}}$? This expression shows how the basis vector $e_{\alpha}$ changes as one moves in the direction of increasing $x^{\beta}$. For example, think about polar coordinates for the 2D plane. The two basis vectors are $e_r$ which points in the direction of increasing radial distance $r$ and $e_\theta$, which points in the direction of increasing angle $\theta$. These two vectors are not constant; when you are to the left of the origin ($\theta = \pi$), $e_r$ points to the left, and when you are to the right of the origin ($\theta = 0$), $e_r$ points to the right. So changing $\theta$ changes the basis vector $e_{r}$ (and $e_\theta$ as well).

So in general, $\dfrac{\partial e_{\alpha}}{\partial x^{\beta}}$ will be non-zero. Since for a specific choice of coordinates, and for specific $\alpha$ and $\beta$, $\dfrac{\partial e_{\alpha}}{\partial x^{\beta}}$ is the derivative of a vector $e_\alpha$ with respect to a scalar, $x^{/beta}$, the result will be another vector.

Since we can write any vector as a linear combination of basis vectors, we can write:
$\dfrac{\partial e_{\alpha}}{\partial x^{\beta}} = \Gamma^{\lambda}_{\alpha \beta} e_{\lambda}$

So the coefficients $\Gamma^{\lambda}_{\alpha \beta}$ are just the components of the vector $\dfrac{\partial e_{\alpha}}{\partial x^{\beta}}$. They just describe how the basis vectors $e_\alpha$ vary from place to place.

Putting it all together, we have:

$\dfrac{d}{ds} U = (\dfrac{d}{ds}U^{\alpha}) e_{\alpha} + (\Gamma^{\lambda}_{\alpha \beta} \ e_{\lambda} \ U^{\beta}) U^{\alpha}$

The first term on the right side of the = takes into account how $U^\alpha$ varies as $s$ increases, and the second term takes into account how $e_{\alpha}$ varies. Since $\lambda$ and $\alpha$ are dummy indices, we can switch them without changing anything:
$(\Gamma^{\lambda}_{\alpha \beta} \ e_{\lambda} \ U^{\beta}) U^{\alpha}= (\Gamma^{\alpha}_{\lambda \beta} \ e_{\alpha} \ U^{\beta}) U^{\lambda}$. So we can factor out $e_\alpha$ and write:

$\dfrac{d}{ds} U = (\dfrac{d}{ds}U^{\alpha} + \Gamma^{\alpha}_{\lambda\beta} \ U^{\beta} U^{\lambda}) e_{\alpha}$

So $\dfrac{d}{ds} U$ is a 4-vector $A$ with components

$A^{\alpha} = \dfrac{d}{ds}U^{\alpha} + \Gamma^{\alpha}_{\lambda\beta} \ U^{\beta} U^{\lambda}$

The notation
$A^{\alpha} = \dfrac{D U^{\alpha}}{ds}$ is sometimes used, but it should be remembered that this is an operation on the entire 4-vector $U$, not just on the component $U^{\alpha}$. A more accurate notation would be
$A^{\alpha} = (\dfrac{d U}{ds})^\alpha$. It is the $\alpha$ component of $\dfrac{d U}{ds}$, not any kind of derivative of $U^{\alpha}$.

19. Aug 10, 2012

### TheEtherWind

Ok, so let me see if I got this straight. I was mistaking $\frac{∂^2x^\alpha}{∂\tau^2}$ with $\dfrac{d}{ds}U^{\alpha}$

So...

$$A^{\alpha} = \dfrac{d}{ds}U^{\alpha} + \Gamma^{\alpha}_{\lambda\beta} \ U^{\beta} U^{\lambda}$$

And when $\dfrac{d}{ds}U^{\alpha}=0$ (when there's no proper acceleration) This is the geodesic equation:

$$A^{\alpha} = \Gamma^{\alpha}_{\lambda\beta} \ U^{\beta} U^{\lambda}$$

or

$$\frac{∂^2x^\alpha}{∂\tau^2}=-\Gamma^\alpha{}{}_\beta\gamma\frac{∂x^\beta}{∂\tau }\frac{∂x^\gamma}{∂\tau}$$

Why's there a negative sign difference?

Also, what's significant about the geodesic equation? i.e. why does it have a name?

Also, if I interpreted stevendaryl's post correctly...$\dfrac{\partial e_{\alpha}}{\partial x^{\beta}}$ is essentially the Christoffel symbol, but how? It does make sense however, that a rate of change of a basis vector would be a christoffel symbol considering the christoffel symbol is comprised entirely of the metric (and derivatives of) and both are therefore geometric, or am I thinking about that wrong?

Thanks for all the help

20. Aug 10, 2012

### Staff: Mentor

I think stevendaryl did a masterful job of explaining this, but you still don't seem to get it. (It is basically what I was trying to do in my response, but he did a much better and thorough job). I suggest you go through stevendaryl's development again, and study each an every step.

In curved spacetime, if all the spatial components of 4 velocity are equal to zero (as reckoned from a given frame of reference), this does not necessarily mean that the 4 acceleration components are zero, since the time component of the 4 velocity is non-zero. This means that in such a frame of reference, even if the objects are at rest relative to the frame of reference, the objects are still experiencing acceleration, and they must have forces acting on them. This is how we perceive gravity.