# Affine parameter

1. Feb 15, 2010

### Fredrik

Staff Emeritus
Wald (p. 41) defines a geodesic as a curve whose tangent vector satisfies

$$T^a\nabla_aT^b=0$$ . . . . . (3.3.1)

Then he says that we could have defined it by requiring

$$T^a\nabla_aT^b=\alpha T^b$$ . . . . . (3.3.2)

instead, where $\alpha$ is "an arbitrary function on the curve", but we choose the former because a reparametrization can turn the second equation into the first anyway. He calls a parameter such that the first equation is satisified an "affine parameter".

I don't see how it's possible to get the first equation from the second. In fact it looks impossible, so I'm assuming that I've made a mistake somewhere. I prefer a coordinate indpendent notation, so I would write the second equation as

$$(\nabla_TT)_{\gamma(t)}=\alpha(t)T_{\gamma(t)}$$

where $\gamma:I\rightarrow M$ is the curve, and T is an extension of the velocity vector field $\dot\gamma(t):I\rightarrow TM$ to a neighborhood of the curve. If we choose a frame $\{E_i\}$ such that $\nabla_TE_i$=0 (a "parallel frame"), we get

$$\nabla_TT=T^i\nabla_{E_i}(V^jE_j)=T^iE_iT^jE_j=TT^jE_j$$

so

$$(\nabla_TT)_{\gamma(t)}=T_{\gamma(t)}T^jE_j|_{\gamma(t)}=\dot\gamma(t)T^jE_j|_{\gamma(t)}=(T^j\circ\gamma)'(t)E_j|_{\gamma(t)}$$

and

$$\alpha(t)T_{\gamma(t)}=\alpha(t)T^j(\gamma(t))E_j|{\gamma(t)}$$

If we define $$x:I\rightarrow\mathbb R^n[/itex] by $x^i(t)=V^i\circ\gamma(t)$, the equation we started with turns into [tex]x'(t)=\alpha(t)x(t)$$

To reparametrize $\gamma$ is to replace it with $\gamma\circ s$ where s is a smooth strictly increasing function on I=[a,b] that preserves the endpoints of the interval. But we have

$$y'(t)=x'(s(t))s'(t)=\alpha(s(t))x(s(t))s'(t)=\alpha(s(t))y(t)s'(t)\neq 0$$

I don't think it was my choice to use a parallel frame that messed something up. It just removed some extra terms. So what am I doing wrong?

Last edited: Feb 15, 2010
2. Feb 15, 2010

### dx

Simply set $$\alpha = 0$$.

The two equations are not equivalent to each other, so if you were trying to show that, you won't succeed. The class of curves that satisfy definition I ('affinely parametrized geodesics') is a subclass of the curves that satisfy definition II ('arbitrarity parametrized geodesics'). The difference is essentially illustrated by the following example: in the plane E2 with the usual coordinate system, the curve γ : R → R2, defined by γ(λ) = (a + mλ, b + m'λ) is a straight line, and the parameter λ is affine, because it agrees with the affine structure of R2. This satisfies definition I, and also definition II with α = 0 . The curve γ': R → R2 defined by γ'(λ') = (a + m(λ')2, b + m'(λ')2) is also a straight line, but the parameter λ' is not affine. This curve γ' satisfies definition II, with a suitable function $\alpha$, but not definition I.

Last edited: Feb 15, 2010
3. Feb 15, 2010

### Fredrik

Staff Emeritus
The exact quote from Wald (p. 41) is "However, it's easy to show that given a curve that satisfies equation (3.3.2) we can always reparameterize it so that it satisfies equation (3.3.1) (see problem 5)". It's clear that he's not talking about setting $\alpha=0$.

Your second curve, I'll call it $\sigma$ because I want to use the prime symbol for the derivative, satisfies $\sigma(t)=(a+mt^2,b+m't^2)$. A curve in $\mathbb R^2$ satisfies (3.3.1) if and only if the derivative of its components are constant. We have $\sigma'(t)=(2mt,2m't)$, so it doesn't satisfy (3.3.1), and $(2mt,2mt')$ isn't $=\alpha(t)(a+mt^2,b+m't^2)$ for any $\alpha$, so it doesn't satisfy (3.3.2) either.

4. Feb 15, 2010

### Ben Niehoff

Hmm...your method is a little convoluted, so I'm not sure precisely where you went wrong. Here's how I would think of it. Remember that T is defined as $\dot \gamma(t)$, so we can write the geodesic equation as

$$\nabla_{\dot \gamma(t)} \dot \gamma(t) = \alpha(t) \dot \gamma(t)$$

Under reparametrizations, we have

$$\frac{d}{dt} \gamma(s(t)) = \gamma'(s(t))s'(t)$$

Now, $\nabla$ is Fu-linear in its first argument, and follows the Leibniz rule in its second argument, so we can write

$$(s'(t))^2 \nabla_{\gamma'(s)} \gamma'(s) + \gamma'(s)s'(t) \nabla_{\gamma'(s)} s'(t) = \alpha(t) \gamma'(s) s'(t)$$

$$s'(t) \nabla_{\gamma'(s)} \gamma'(s) = (\alpha(t) - \nabla_{\gamma'(s)} s'(t)) \; \gamma'(s)$$

On functions, $\nabla$ is just the ordinary derivative, so if we set the RHS to zero ($X[f]$ means the vector X acting on f as a derivation),

$$\alpha(t) - \nabla_{\gamma'(s)} s'(t) = 0$$

$$\alpha(t) - \gamma'(s(t)) [s'(t)] = 0$$

$$\alpha(t) s'(t) - \dot \gamma(t) [s'(t)] = 0$$

$$\alpha(t) s'(t) - s''(t) = 0$$

which obviously has a solution.

5. Feb 15, 2010

### dx

How is it clear he's not talking about setting alpha = 0?

You want to show that $$\nabla_{\sigma'}\sigma' = \alpha \sigma'$$, not $$\sigma ' = \alpha \sigma$$; the latter doesn't actually make sense since σ is not a vector.

6. Feb 15, 2010

### dx

To make it completely clear, here's an example on the line E1. The curve γ(t) = t is an affinely parametrized geodesic, since γ' = ∂1, and ∇γ'γ' = 0. The curve σ(t) = t3 is also a geodesic, but it is not affinely parametrized. σ' = 3t21, and ∇σ'σ' = 6t∂1 = 2(1/t)3t21. So ∇σ'σ' = α(t)σ', where α(t) is 2/t.

The second curve satisfies (3.3.2), with α(t) = 2/t, but it doesn't satisfy (3.3.1). The first curve satisfies (3.3.1), which is a special case of (3.3.2) for α(t) = 0.

Last edited: Feb 15, 2010
7. Feb 15, 2010

### Fredrik

Staff Emeritus
For starters, the stuff I quoted wouldn't make sense if it's that easy (and the problem wouldn't have anything to do with reparameterizations), and as I said in #1, he also said explicitly that alpha is arbitrary.

My mistake.

Most of the things I do are just to make sure that everything is well-defined at each step of the calculation. For example, the domain of $\gamma$ is the interval I, but $\nabla$ acts on vector fields whose domains are open subsets of the manifold. That's why I can't avoid talking about extensions.

I don't understand exactly what you're doing, since many of the things you write down look ill-defined to me, but I think I get the basic idea. I'll try to work it out in my notation. I think I confused myself by not thinking of a different symbol for the the extension of the reparameterized field.

I'll post the calculation when I'm done.

8. Feb 15, 2010

### dx

The problem does have to do with reparametrizations, and alpha is arbitray, but I don't see how that contradicts anything I wrote. Just to make sure, you're not interpreting alpha as the new parameter are you?

The arbitrary alpha is there is (3.3.2) to allow the parameter to vary along the geodesic in a way that need not respect the affine structure of the tangent spaces, i.e. essentially the parameter is arbitrary for curves that satisfy (3.3.2). If, however, you need the curve parameter to be affine, then that will happen if and only if alpha is equal to zero.

Last edited: Feb 15, 2010
9. Feb 15, 2010

### dx

What Wald means is the following: Any curve γ that satisfies ∇γ'γ' = αγ', for arbitrary α, can be reparametrized to get a new curve σ, which satisfies ∇σ'σ' = 0. α is arbitrary for γ, but for σ it must be zero, since that's the definition of an affinely parametrized curve.

Showing formally that any curve that satisfies (3.3.2) can be reparametrized as claimed is of course not totally trivial. I didn't mean to say in my first post that setting α = 0 proves this. That was just a response to you saying "I don't see how it's possible to get the first equation from the second", which I think you intended in a different way than I assumed.

Last edited: Feb 15, 2010
10. Feb 15, 2010

### Fredrik

Staff Emeritus
It sure sounded like you did. OK, there's no disagreement then.

11. Feb 15, 2010

### Fredrik

Staff Emeritus
Here's my calculation. I found it easier to work with the "covariant derivative along a curve" operator than with a connection. (And I'm calling it D, not Dt, since I didn't see the point of that t subscript. When I write Dt, I mean the operator that takes a real-valued function of one real variable to its derivative at t).

Let $I=[a,b]$ and let $\gamma:I\rightarrow M$ be a smooth curve that satisfies

$$D\dot\gamma(t)=\alpha(t)\dot\gamma(t)$$ . . . . . (1)

We're going to show that there exists a smooth strictly increasing bijection $s:I\rightarrow I$ such that $\sigma=\gamma\circ s$ satisfies

$$D\dot\sigma(t)=0$$ . . . . . (2)

Let $x:U\rightarrow\mathbb R^n$ be a coordinate system such that $\gamma(t)\in U$.

$$\dot\gamma^i(t)x^i=\gamma_*D_tx^i=(x^i\circ\gamma)'(t)=(x^i\circ\sigma\circ s)'(t)=(x^i\circ\sigma)'(s(t))s'(t)$$

$$=s'(t)\sigma_*D_tx^i=s'(t)\dot\sigma(t)x^i$$

This implies that

$$\dot\gamma(t)=s'(t)\dot\sigma(t)$$

We use this to rewrite (1).

$$\alpha(t)\dot\gamma(t)=\alpha(t)s'(t)\dot\sigma(t)$$

$$D\dot\gamma(t)=D(s'\dot\sigma)(t)=s''(t)\dot\sigma(t)+s'(t)D\dot\sigma(t)$$

Solve for $D\dot\sigma(t)$.

$$D\dot\sigma(t)=\Big(\alpha(t)-\frac{s''(t)}{s'(t)}\Big)\dot\sigma(t)$$

(2) is satisfied if the parenthesis is =0, so we're looking for an s that satisfies

$$\alpha(t)=\frac{s''(t)}{s'(t)}=\frac{d}{dt}\log s'(t)$$

$$\log s'(t)=\alpha(t)+C$$

$$s'(t)=A\exp\left(\int_{a}^{t}\alpha(t_1)dt_1\right)$$

$$s(t)=A\int_{a}^{t}\exp\left(\int_{a}^{t_2}\alpha(t_1)dt_1\right)dt_2+B$$

The condition s(a)=a gives us B=a, and...I don't see how to get anything simple from s(b)=b. But since A=eC we can at least see that s'(t)>0 for all t, which guarantees that s is strictly increasing on I.

Thank you both for helping me with this. Ben's post pointed me in the right direction to solve the problem, and dx's posts also helped me see things a bit more clearly.

12. Feb 15, 2010

### Ben Niehoff

The answer to that is simple: Just restrict $\nabla$ to act on vector fields whose domains are open subsets of the image of $\gamma$ in M, valued in some vector bundle (which happens to be TM). This restriction is fairly natural.

I'm not sure what you think is ill-defined. I did write $\gamma(s)$ instead of $\gamma(s(t))$ as shorthand, but I thought that would be clear.

13. Feb 15, 2010

### Fredrik

Staff Emeritus
After thinking a bit about what I've read in the book I linked to, in particular that $(\nabla_XY)_p$ only depends on the value of X at p and the values of Y on an infinitesimal curve segment through p that has $X_p$ as tangent vector, I agree that it's natural. (But I don't think I would use the word "restriction", since the set of vector fields along the curve isn't a subset of the set of vector fields on the manifold).

Yes, that's what I thought you meant (but when I see things I don't understand in other places, I start doubting even the things I do understand). When you wrote

$$\frac{d}{dt} \gamma(s(t)) = \gamma'(s(t))s'(t)$$

$\gamma'(s(t))$ really meant $(x\circ\gamma)'(s(t))$, right? (Compare it to my version). But that would make your $\gamma'(s)$ a member of $\mathbb R^n$, so what's it doing as the "downstairs" variable of the connection?

At the time I was also confused about the linearity properties of the "restricted" connection. For example, I didn't feel comfortable seeing a number where I expected a function (like when one of the objects that the connection is acting on is being multiplied by s'(t) rather than s' or maybe even $s\circ\gamma^{-1})$.

Last edited: Feb 15, 2010
14. Feb 15, 2010

### Ben Niehoff

Fair enough. It's not really a pullback, either, so I wasn't sure what to call it.

Ah, ok. The notation $\gamma'(s(t))$ means $(\gamma' \circ s)(t)$. I suppose to be completely proper I should introduce a new symbol $\sigma: J \rightarrow M$, where J is the codomain of s, such that $(\sigma \circ s)(t) = \gamma(t)$. Then the line reads

$$(\sigma \circ s)'(t) = (\sigma' \circ s)(t) \; s'(t)$$

Then replace all the $\gamma(s)$ with $\sigma(s)$ in the remainder of my post.

15. Feb 15, 2010

### Ben Niehoff

Or, to write the whole thing again with more clarity, we have the maps:

$$\gamma: I \rightarrow M$$

$$s: I \rightarrow J$$

$$\sigma: J \rightarrow M$$

And then we have

$$\nabla_{\gamma'(t)} \gamma'(t) = \alpha(t) \gamma'(t)$$

$$\nabla_{(\sigma \circ s)'(t)} (\sigma \circ s)'(t) = \alpha(t) \gamma'(t)$$

$$s'(t) \; \nabla_{\sigma'(s)} \left( \sigma'(s) s'(t) \right) = \alpha(t) \gamma'(t)$$

$$(s'(t))^2 \; \nabla_{\sigma'(s)} \sigma'(s) \; + \; \sigma'(s) s'(t) \nabla_{\sigma'(s)} s'(t) = \alpha(t) \gamma'(t)$$

$$(s'(t))^2 \; \nabla_{\sigma'(s)} \sigma'(s) = \gamma'(t) \left( \alpha(t) - \nabla_{\sigma'(s)} s'(t) \right)$$

where again, the RHS can be made to vanish if

$$\alpha(t) - \nabla_{\sigma'(s)} s'(t) = 0$$

$$\alpha(t) s'(t) - \nabla_{\gamma'(t)} s'(t) = 0$$

$$\alpha(t) s'(t) - \gamma'(t) [s'(t)] = 0$$

$$\alpha(t) s'(t) - s''(t) = 0$$

And finally, if $s'(t) \neq 0$, we get

$$\nabla_{\sigma'(s)} \sigma'(s) = 0$$

That was kinda fun. It's actually much simpler if you just write out where all your maps are going at the beginning.

16. Feb 15, 2010

### wofsy

If the curve has constant velocity then

$$\nabla_TT$$ is orthogonal to T (assuming that you have a Levi-Cevita connection which Wald certainly assumes). So

$$\alpha(t)$$

must be zero. Intuitively this normal vector should point off of the manifold and so not be observable in the tangent space. For instance, a great circle parametrized by arc length has acceleration normal to the sphere and thus has no component tangent to the sphere.

If it is not parametrized by arc length then there will be a tangential component. In this case you want it to point in the same direction as T for otherwise the curve would have a non zero normal component of acceleration and this violates the idea of a straight line. A straight line has the property that any object moving along it can only accelerate along it.

Write

T = $$\alpha(s)$$ N(s)

Where s is the arc length parameter, N is the unit tangent to the curve, and alpha is the scaling factor.

If you take the covariant derivative of T with respect to N you will see that the two conditions are equivalent.

Last edited: Feb 15, 2010