# I Lie Derivatives vs Parallel Transport

1. Nov 6, 2017

### Silviu

Hello! In my GR class we were introduced to the parallel transport as the way in which 2 tensors can be compared with each other at different points (and how one reaches the curvature tensor from here). I was wondering why can't one use Lie derivatives, instead of parallel transport. As far as I understand, both define the transport of a tensor along a vector field, so why is one used instead of the other i.e. why is Lie derivative not good to define directional derivatives on a manifold? Thank you!

2. Nov 6, 2017

### Staff: Mentor

You can, but you'll get a different answer than you would using parallel transport. (And those two possibilities are not the only ones: in general there are an infinite number of possible ways to transport vectors/tensors between points in a curved manifold.) So the question is, which answer is the one you want to use? And the answer to that question depends on what physical problem you are trying to solve.

The physical problem for which "parallel transport" is the right answer is the problem: which curves are geodesics? The answer to that is, they are curves that parallel transport their tangent vectors along themselves. And that turns out to be the "right" answer because those curves turn out to be the ones that describe the worldlines of freely falling objects, i.e., weightless objects, feeling no acceleration. Since objects moving solely under the influence of "gravity" are weightless, these weightless, freely falling worldlines are the ones that tell us about the geometry of spacetime, so GR gives them a special status and uses parallel transport for that reason.

However, there are other physical problems for which "Lie derivative" is the right answer. Here's one: suppose I have an observer "hovering" at a fixed altitude above a large gravitating body like a planet or star (and not in orbit, his angular coordinates are also fixed, so he is firing a rocket or otherwise under acceleration). This person wants to keep a telescope oriented directly radially outward, i.e., in the opposite direction from the planet/star. How will the spacelike vector that describes the direction the telescope is pointing be transported along the observer's worldline? The answer is, using the Lie derivative (the more usual name for this in GR is "Fermi-Walker transport").

In your GR class they probably were only considering the former type of problem, not the latter, which is why they gave "parallel transport" as the answer. But it's good to be aware that there are other problems you will eventually encounter where other types of transport are relevant.

3. Nov 6, 2017

### Silviu

I understand what you mean, but I am still confused about when you want to use one over the other. Like why is the geodesic not defined by a Lie derivative? Is just that in that case the observation wouldn't match the theory or is there a deeper reason? In the end the approach to a physics problem should give you the same answer, but in this case it certainly doesn't and I am kinda confused...

4. Nov 6, 2017

### Staff: Mentor

This is like asking "why does a triangle not have four sides"? A geodesic is defined as a curve that parallel transports its tangent vector along itself--not a curve that Lie transports its tangent vector along itself.

I don't know what you mean. Parallel transport and Lie transport are distinct, different things. Why would you expect both of them to give you the same answer?

5. Nov 7, 2017

### romsofia

This picture that my friend made should help you see the difference between the two derivatives. Don't worry if it takes you a while to understand the differences between the two derivatives, I've struggled, and still struggle sometimes, talking about the difference between the two. It just takes a while to get use to them.

https://imgur.com/a/D638e (not sure how to embed the picture)

6. Nov 7, 2017

### martinbn

One reason is that the Lie derivative is not a good directional derivative. Reason being that at a point it doesn't depend only on the direction you differentiate in i.e. $(\mathcal L_XY)_p$ doesn't depend only on $X_p$, but also on values of the field $X$ near the point $p$.

7. Nov 7, 2017

### martinbn

I am confused by this. The two are different. Even in this context they don't seem the same to me?

8. Nov 7, 2017

### Staff: Mentor

You're right, I was thinking of Killing vector fields, which involve the Lie derivative of the metric.