Expressions accompanying a Christoffel Symbol (a notation question)

[help1please]

Just been reading on Christoffel Symbols and I am having a notational mind block.

Say we have:

\Gamma^{k}_{ij} v^{i} v^{j}

The velocity depends on this expression, but I don't read this term v^{i} v^{j} as a velocity squared do I? It's just the one velocity, are the superscripts here denoting different ''points''?

 

[Muphrid]

No, it's not. It literally just means v^i v^j = (v \cdot e^i)(v \cdot e^j).

I'm not a fan of Christoffel symbols because they're dependent on coordinates and resist geometric interpretation as a result, but at least the other vectors involved can be interpreted in a coordinate-free way.

 

[help1please]

So what's it actually doing here?

How do you interpret (v.e^i)(v.e^j)?

 

[help1please]

For instance, and this is just a guess, are these basis vectors calculating the curve for the velocity?

 

[Muphrid]

They're just basis vectors in whatever coordinate system you've chosen. They don't calculate anything. They just extract the components needed to match up with the ones the Christoffel symbol wants.

Other than that, I'm not sure what else to say to illuminate the situation. Is this just a general confusion about index notation and summation convention, or?

 

[help1please]

Just a confusion of the interpretation of the indices.

So, they are the basis vectors ''linking up'' one might say (?) to the coordinates you have chosen. What would be the difference between coordinates e^i and those which match e^j?

Why would you want two coordinates? What role will it play?In my example above, we have the christoffel symbol dependent on the velocity. How does velocity comes into play with two separate coordinates?

 

[phyzguy]

Try writing out one component of the 4-acceleration, for example:
a^x = \frac{du^x}{d\tau} = \Gamma^x_{ij} v^i v^j
a^x = \Gamma^x_{xx}(v^x)^2 + \Gamma^x_{xy}v^x v^y + \Gamma^x_{xz}v^x v^z + ...

 

[Muphrid]

I think what you need an understanding of is why the four-velocity is there twice?

The four-acceleration is given by a = (v \cdot \mathcal D)v. It may seem weird that the four-velocity appears twice, but this is the correct equation. When you calculate the components, the Christoffel connection helps spit out the components of the acceleration based on combinations of the components of velocity. The i and j labels are just placeholders to tell you which Christoffel symbol goes with which components of velocity.

 

[help1please]

Try writing out one component of the 4-acceleration, for example:
a^x = \frac{du^x}{d\tau} = \Gamma^x_{ij} v^i v^j
a^x = \Gamma^x_{xx}(v^x)^2 + \Gamma^x_{xy}v^x v^y + \Gamma^x_{xz}v^x v^z + ...

Ok, so

\Gamma_{xx}^{x} (v^x)^2

Is a velocity squared, but

\Gamma^x_{xy}v^x v^y

Isn't... right?

 

[help1please]

I think what you need an understanding of is why the four-velocity is there twice?

The four-acceleration is given by a = (v \cdot \mathcal D)v. It may seem weird that the four-velocity appears twice, but this is the correct equation. When you calculate the components, the Christoffel connection helps spit out the components of the acceleration based on combinations of the components of velocity. The i and j labels are just placeholders to tell you which Christoffel symbol goes with which components of velocity.

Can you show me why the four velocity appears twice?

 

[phyzguy]

Ok, so

\Gamma_{xx}^{x} (v^x)^2

Is a velocity squared, but

\Gamma^x_{xy}v^x v^y

Isn't... right?

They both have units of velocity squared. The first term is the x-component of velocity times itself, and the second term is x-component of velocity times y-component of velocity. Since, as we siad earlier, the Christoffel symbols have units 1/length, then the RHS has units of 1/length * (length/time)^2 = length/time^2, which is an acceleration. I don't see where your confusion is.

 

[phyzguy]

Can you show me why the four velocity appears twice?

The easiest way to see why the 4-velocity appears twice is to study the derivation of a geodesic as a curve of extremal length. Here's a good starting point.

 

[Chestermiller]

Hi help1please,

I have attached a simple analysis in Microsoft Word to help explain in detail where the products of the velocity components come from in the expression for the acceleration of a particle. I have a feeling that my friend Muphrid will not be totally thrilled with the interpretation I present, but I think the development will work for you.

Chet