Geodesics - Reparameterization I am reading Wald's textbook Chapter 3. I am struggling with Section 3.3 and problem 5. 
It states that any curve that satisfies the weaker condition $T^{a}\nabla_{b}T^{b} = \alpha T^{a}$ is $eq.(3.3.2)$ where $\alpha$ is an arbitrary function on the curve can be reparameterized such that eq.(3.3.1), 
namely $T^{a}\nabla_{b}T^{b}=0$ is satisfied. 
How could I show this ? I am struggling to formulate an answer, hints are welcomed. 
EDIT:
Following user Avantgarde comment and I am sorry for what is probably a silly question and yes the expression is different the book (sorry for that),what is the difference between $T^{b}\nabla_{b}T^{a}=0$ and $T^{a}\nabla_{b}T^b=0$ ?
Thank you ! All answers and comments were helpful
 A: Remember that $T^a$ represents not just some vector field, but the tangent to a curve (the geodesic) with a given parameterization.  You should be able to figure this out for yourself from his description of tangent vectors, but Wald happens to say explicitly just above Eq. (3.3.9) that if $t$ is the original parameter, and you define a new parameterization $s = s(t)$, then the new tangent will be $S^a = (dt/ds) T^a$.  Assuming well-behaved parameters, this also implies $T^a = (ds/dt) S^a$.  Plug that into $T^a \nabla_a T^b = \alpha T^b$, find an expression involving $s(t)$, and you've got it.
A: Note that your equation does not have the right index contractions. The correct version is below.
Take the geodesic equation in the canonical form:$$T^b \nabla_b T^a = 0$$
and explicitly expand it out
$$\frac{d^2 x^a}{d \lambda^2} + \Gamma^a_{b c} \frac{dx^b}{d \lambda} \frac{dx^c}{d \lambda} = 0$$
where $\lambda$ is the affine parameter that makes the RHS of geodesics equation zero. Now express this equation in terms of a new parameter $\sigma$ which is any general parameter that parameterizes the path. We wish to find the relationship between $\lambda$ and $\sigma$. To do so, use the chain rule
$$\frac{dx^a}{d \lambda} = \frac{dx^a}{d\sigma} \frac{d\sigma}{d \lambda}$$
The geodesic equation parameterized by $\sigma$ then becomes:
$$\frac{d^2 x^a}{d \sigma^2} + \Gamma^a_{b c} \frac{dx^b}{d \sigma} \frac{dx^c}{d \sigma} = -\frac{dx^a}{d \sigma} \left( \frac{d^2 \sigma}{d \lambda^2}\right) \left( \frac{d \sigma}{d \lambda}\right)^{-2} = \alpha T^a$$
which has the solution
$$\frac{d \lambda}{d \sigma} = \exp\left[ {\int^\sigma dx \ \alpha(x)} \right]$$
So, one can always go from $\sigma$ to $\lambda$ to parameterize geodesics so that its equation has RHS $=0$.
A: Maybe using vector notation can help. $T^b \nabla_b T^a$ is equivalent to $(\vec{T} \cdot \nabla) \vec{T}$, while $T^a \nabla_b T^b$ is $\vec{T} (\nabla \cdot \vec{T})$.
The first is the directional derivative of $\vec{T}$ in its own direction. It being equal to zero is the condition for parallel transport, where the vector is kept as parallel as possible while it advances. The second expression is completely different: it's the product of $\vec{T}$ (a vector) with its divergence (a scalar). It is zero when either of its factors is zero, and it doesn't really have a particular geometric interpretation.
