# Parallel transport invariance

1. May 4, 2010

### bueller11

I'm trying to show that $$\frac{d}{dt}\; g_{\mu \nu} u^{\mu} v^{\nu} = 0$$ in the context of parallel transport (or maybe not zero), and I'm rather insecure about the procedure. This is akin to problem 3.14 in Hobson's et al. book (General Relativity an introduction for physicists).

As a guess, I tried the take the time derivative:
$$\frac{d}{dt}\; g_{\mu \nu} u^{\mu} v^{\nu}+g_{\mu \nu} \dot{u^{\mu}} v^{\nu}+g_{\mu \nu} u^{\mu} \dot{v^{\nu}}=0$$

I was assuming a stationary metric, so the first part would be zero, leaving

$$g_{\mu \nu} \dot{u^{\mu}} v^{\nu}+g_{\mu \nu} u^{\mu} \dot{v^{\nu}}=0$$

From there I can substitute in for $$\dot{u^{\mu}}$$ and $$\dot{v^{\nu}}$$.

Is this the right path to take? It seems there's then some index trickery involved to solve this.

Thanks!

2. May 5, 2010

### bueller11

Did I post this in the wrong forum?

Some more info: vectors $$u$$ and $$v$$ are parallel transported along a geodesic, the point is to show that the dot product is invariant under parallel transport.

Also, this isn't homework, but it was a question asked on a midterm.

3. May 5, 2010

### zhentil

I'm a bit confused how general relativity treats this. It seems that this is the very definition of metric-compatible connection and parallel transport.

4. May 5, 2010

### bueller11

That was similar to my thought. I was thinking: well, we are only including the things that make this true so what's the point in showing it? As a result, I was thinking it was more of a mathematical question about topology and manifolds and I lacked understanding. In other words, it seemed that it's a mathematical argument not a physical one and I don't understand it. Overall, I'm really confused.

But what I was trying... Parallel transport says the covariant derivative is zero along a curve parameterized by, in this case, t. That means for a generic vector $$v$$,

$$\frac{dv^{a}}{du} = -\Gamma^{a}_{bc}v^{b} \frac{dx^{c}}{du}$$,

And I can rewrite the equations to be:

$$g_{\mu \nu} (-\Gamma^{\mu}_{\beta \sigma}u^{\beta} \dot{dx^{\sigma}}) v^{\nu}+g_{\mu \nu} u^{\mu} (-\Gamma^{\nu}_{\beta \sigma}v^{\beta} \dot{dx^{\sigma}})=0$$

Is that true? This doesn't seem like it's going anywhere.

5. May 5, 2010

### zhentil

The definition of a metric-compatible connection is that

$$\frac{d}{dt} <X,Y> = <D_tX,Y> + <X,D_tY>$$

By the definition of parallel transport, the right-hand side is zero. You're on the right track, but you don't have to assume a stationary metric.

6. May 6, 2010

### RedX

Do you mean:

$$D_t <X,Y> = <D_tX,Y> + <X,D_tY>$$ ?

Then this is just the Leibniz rule and there is no need for metric compatibility.

In all the standard physics books the proof goes something like this:

$$\partial_k (V^iU_i)=\partial_k (V^i) U_i+V^i\partial_k (U_i)$$

parallel transport implies:

$$D_kV^i=\partial_k V^i+\Gamma V=0$$
$$D_k U_i=\partial_k U_i-\Gamma U=0$$

where the indices on second term on the RHS have been suppressed.

If you solve for the partial derivative of the vector in terms of gamma, and then plug that into the above equation, you get it's equal to zero.

7. May 6, 2010

### bueller11

Ok, I got it. It's so much simpler than I was making it. For some reason it didn't connect to drop an index so I could get the signs to change in the covariant derivative then to the math.

Thank you so much zhentil and RedX!

8. May 6, 2010

### zhentil

That's the definition of metric compatibility. You can easily construct connections that don't obey this rule. I'm not sure why you've drawn a distinction between $$D_t<X,Y>$$ and $$\frac{d}{dt} <X,Y>$$. By definition, a covariant derivative acts like a normal derivative on functions.

9. May 6, 2010

### RedX

You're right. Your proof is much simpler than the one my instructor gave. In fact, the proof is practically tautological.

The reason I made a distinction is because usually little d means a coordinate derivative and big D means a covariant derivative, and the person wrote it with a little d. I forgot that when acting on a scalar, they are the same. So that is why I wasn't able to come up with your simple proof. Once you note that little d acting on a scalar is the same as big D acting on a scalar, then the proof is just automatic.

10. May 7, 2010

### Ben Niehoff

Little d/dt means total derivative; a total derivative with respect to a parameter along a path is generally assumed to mean the total covariant derivative.

The coordinate derivative is $\partial/\partial x$ where x is some coordinate (NOT a parameter along a path).

11. May 7, 2010

### RedX

Yeah. I got confused because of the overloading of the parameter t. So it'd be:

t=t
x,y,z=x(t),y(t),z(t)

t,x,y,z={t(s)=s},x(s),y(s),z(s)

Also, it is true that for parallel transport:

$$\partial_k <X,Y>=0$$

but I guess that's just a path with a direction only on the kth-coordinate basis.

So the total derivative along a path parametrized by s would be:

$$\frac{dx^k}{ds}D_k X$$ for a vector X and for a scalar you can replace D with a partial derivative.

I think I got it now.

btw, I have written in my notes:

<df,X>=Xf
<dxj,X>=$$\delta^{i}_j$$

where X is a basis vector (an operator). Is that a general rule, that you ignore the 'd' in the first term of the angled brackets, and let X operate on the term next to the 'd'? What would <f,X> mean (i.e., without a 'd' in front of the f)?