# Help with McMahon's Relativity

1. Jul 5, 2008

### mikah

1) Lamda tensor metric conversion between local non-holonomic basis and global holonomic basis, how can you just "read off" the diagonal from the line element say in spherical coordinates and when do you choose the inverse metric (just take the inverse of each component of the diagonal?) when converting with either one-forms or vectors? pp 100 to 103.

2) Why did they choose to make the components of the tangent vector the partial diff operator making the vector an operator?

3) When calculating the Cartan coordinate free geodesic acceleration between two geodesics when the Lie bracket is zero (flat) how do you get the second total derivative with respect to tau as the product of tangent vector grad (tangent vector grad displacement vector). Do you take double Lie derivative on operators? pp 135

2. Jul 7, 2008

### cristo

Staff Emeritus
More people will be able to help you if you put your questions in context; for example, quoting the relevant part of the text. Otherwise, you may be waiting a while for someone with that book to come along.

3. Jul 7, 2008

### Fredrik

Staff Emeritus
Think about how to calculate the rate of change of a function $f:\mathbb R^n\rightarrow\mathbb R$ along a curve in $\mathbb R^n$. Suppose that $C:(a,b)\rightarrow\mathbb R^n$. Then the rate of change of f at C(t) is $(f\circ C)'(t)$. The chain rule now tells us that

$$(f\circ C)'(t)=f,_\mu(C(t)) C'^\mu(t)=C^\mu'(t) D_\mu|_{C(t)} f$$

Note that $C^\mu'(t)$ are the components of the tangent vector of the curve.

Now suppose that $C:(a,b)\rightarrow M$ and $f:M\rightarrow\mathbb R$. The rate of change of f at C(t) is still $(f\circ C)'(t)$, but now the first equality above doesn't make sense as it stands. We can however make sense of it using a coordinate system x:

$$(f\circ C)'(t)=(f\circ x^{-1}\circ x\circ C)=(x\circ C)^\mu'(t)\frac{\partial}{\partial x^\mu}\bigg|_{C(t)}f$$

Note the similarities between this and the previous result. In both cases, the rate of change of f at C(t) came out equal to a linear combination of derivative operators acting on f, and in both cases the coefficients are equal to the components of a tangent vector of a curve in $\mathbb R^n$. Also note that if we think of the identity map I (defined by I(x)=x for all x) on $\mathbb R^n$ as a coordinate system on $\mathbb R^n$, then we can put the first calculation into exactly the same form as the second (with I instead of x).