# Line element and derivation of lagrange equation

1. Jan 17, 2016

### Alain De Vos

With coordinates q en basis e ,textbooks use as line element :
ds=∑ ei*dqi But ei is a function of place, as one can see in deriving formulas for covariant derivative. Why don't they use as line element the correct:
ds=∑ (ei*dqi+dei*qi)
Same question in deriving covariant derivative,

2. Jan 18, 2016

### Staff: Mentor

Can you give an example of a textbook that uses this form of the line element? I see at least two obvious problems with it:

(1) It sums the coordinate differentials, not their squares. For example, in 2-dimensional Cartesian coordinates, the line element should be $ds^2 = dx^2 + dy^2$, but your form would give (assuming that you don't really mean "basis" when you say "basis"--see problem #2 below) $ds = dx + dy$, which is obviously wrong.

(2) The basis is a set of vectors, but the line element is a scalar. Strictly speaking, for Cartesian coordinates, your form of the line element would give $ds = \hat{e}_x dx + \hat{e}_y dy$, which would make $ds$ a vector, not a scalar.

3. Jan 31, 2016

### Alain De Vos

I was reading a book by John Dirk Walecka, but his notation confused me, I switched to Lambourne.
I have a comparable question in the formula for geodesics.
Why is the formula for geodesics a covariant derivative of a standard derivative (being the tangent)?
Why is it not a covariant derivative of a covariant derivative?
Is it because in the first derivative the connection coefficients are zero ?

4. Jan 31, 2016

### Staff: Mentor

What do you mean by "standard derivative"? If you mean "partial derivative", a tangent vector is not a partial derivative.

I recommend looking at chapters 2 and 3 of Carroll's lecture notes on GR:

http://arxiv.org/abs/gr-qc/9712019

These chapters discuss the issues you are talking about, and I find them much easier to follow than many textbook treatments of the subject.

5. Jan 31, 2016

### stevendaryl

Staff Emeritus
Here's what I'm guessing you're talking about.
1. You start (or can start) with a parametrized path $\mathcal{P}(s)$, which can be described in a particular coordinate system by 4 functions $x^\mu(s)$.
2. You take the first derivative of $x^\mu$ to get a 4-velocity, $U^\mu$
3. You take the second derivative of $x^\mu$ to get a 4-acceleration, $A^\mu$
4. For a geodesic, $A^\mu = 0$
So, perhaps what you're talking about is the fact that
• $U^\mu = \frac{d}{ds} x^\mu$
• $A^\mu = \frac{d}{ds} U^\mu + \Gamma^\mu_{\nu \lambda} U^\nu U^\lambda$
So you're asking why the connection coefficients $\Gamma^\mu_{\nu \lambda}$ appear in the second equation, but not in the first.

A technical reason, which may not be satisfying to you, is that $x^\mu$ is NOT a vector, it's a collection of 4 scalar functions. You have to use a covariant derivative when you take the derivative of a vector, but not when you take the derivative of a scalar.

6. Jan 31, 2016

### stevendaryl

Staff Emeritus
Here's what I'm guessing he's talking about. If you describe a particle's path by giving its coordinates $x^\mu(s)$ as a function of proper time, $s$, then:
• The 4-velocity, $U^\mu$ is defined by: $U^\mu = \frac{d}{ds} x^\mu$
• The 4-acceleration, or proper acceleration, is defined by $A^\mu = \frac{d}{ds} U^\mu + \Gamma^\mu_{\nu \lambda} U^\nu U^\lambda$
So I think he's asking why $\Gamma^\mu_{\nu \lambda}$ appears in the second equation, but not the first.

7. Feb 1, 2016

### Alain De Vos

Exactly Peter, this was my question, i'm currently reading Schutz, afterwards i'll raid Carroll.

8. Feb 22, 2016

### Alain De Vos

Currently reading Carroll spacetime and geometry page 16 and 17.
Still unclear for me why tangent vector to a curve does not include change of basis-vectors as lambda changes.
I.e. the connection coefficients.
Anyone who can shed a light on this very basic issue ?

9. Feb 22, 2016

### Orodruin

Staff Emeritus
There is no vector basis involved here, just coordinates. The key point here is that there is no such thing as a position vector in a general manifold and I think this is something which has passed you by. By definition, a vector is defined as a directional derivative of a scalar field. A curve, i.e., giving the coordinates as a function of the curve parameter, naturally identifies such a derivative with components $dx^\mu/ds$, where $s$ is the curve parameter.

10. Feb 22, 2016

### Orodruin

Staff Emeritus
I think this reads wrong out of context. A tangent vector is a partial derivative operator (in some coordinate system - a linear combination of them in a general coordinate system).

11. Feb 23, 2016

### stevendaryl

Staff Emeritus
There is a big distinction between a location (described in coordinates as $x^\mu$) and a tangent vector, and that is that a location is not a vector. It doesn't make any sense to add two locations together. What does it mean to add the location of New York City to the location of Paris? Locations don't transform as vectors. It doesn't make any sense to write a location as a linear combination of basis locations. So throwing in connection coefficients in computing $\frac{dx^\mu}{ds}$ would not really make any sense.

12. Feb 23, 2016

### Alain De Vos

So all points in curved spacetime form a manifold but not a vectorspace.
It is the collection of all tangents to all curves going through a specific point on this manifold which form a vectorspace.
(I must have been thinking 3-D where points and vectors can be interchanged)

13. Feb 23, 2016

### Alain De Vos

The basis in this tangent space is then : e(μ)=∂P/∂xμ

14. Feb 23, 2016

### Orodruin

Staff Emeritus
No, the basis of the tangent space is $e_\mu = \partial/\partial x^\mu$ (no P). A vector is a directional derivative.