Interpretation for negative energy of curves Let $(M,g)$ be a Lorentz manifold and $\gamma :[a,b] \to M$ a differentiable curve. I understand we define the energy of $\gamma $ as: $$E[\gamma] = \frac{1}{2} \int_a^b g_{\gamma(t)}(\gamma'(t),\gamma'(t))\,{\rm d}t.$$
For timelike curves we can have negative energy. Can someone explain what this means? Thanks.
(I am a math student who would like to gain some physics intuition on this, in case that matters)
 A: If you computed arc length you'd chop your curve into pieces, find the distance between the end points of each piece, add them up, and then take the limit as the pieces get individually smaller and hence more numerous.
Now image that instead you took the square of the distance between the end points of each piece, that wouldn't work because you would get zero in the limit. 
But the energy is very similar to the method of the above paragraph. The number you get by integration can be approached in the following way (for curves that are pretty ordinary). Start with your time interval T. Then for each n: break your time interval T into n pieces, compute the squared distance between the endpoints, add them all up, then multiply by n/2T. Do that for each n. Then, in thee limit as n goes to infinity, you get the same number as the integral you mentioned. 
There are at least a couple reasons to consider the energy. First, mathematicians often like to consider quadratic forms. Second, in relativity the squared length could be positive or negative and then you don't have to worry about taking the square root of a negative number if you compute the energy rather than the length of the curve.
So now let's address that it can be negative. Whether timeline curves have positive energy or negative energy is purely a convention, I just checked 36 relativity books and 11 out of 36 used the convention that timelike has negative energy, 1 used the one convention in one edition and the opposite convention in another edition and 1 used both and the other 26 used the convention that the energy was positive for time like curves.
In particle physics settings I think everyone measures timelike to be positive. Then you can say simple things like "rest mass is the length of the energy momentum vector on shell" but some in relativity like to make it so energy is negative perhaps just to have (some) equations look (more) like euclidean geometry.
The real fact that isn't just convention is that the energy of timelike and spacelike have opposite sign so if one is positive the other is negative. And that is just because the squared length is positive for one and negative for the other so energy inherits that from squared length, which obviously is not the square of a length, but a separate thing, the dot product of a vector with itself when the dot product comes from a nondegenerate bilinear form with signature from (1,3) or from (3,1).
Sometimes mathematicians steal a word just because the math is the same as a situation where the word meant something. But there are lots if situations in math where you want to define an energy just to minimize it and it is often a quadratic form, so that might be all that is going on here.
Physical energy in relativity is a different beast, it is frame dependent so not geometrical (no more geometrical than the x component of a vector when there is no obvious reason to choose any direction to be the x direction), and it is conserved in that it flows from here to there rather than being locally created or locally destroyed.
As doetoe mentions there is another reason to consider energy. If you consider the Energy function as a function of dynamic functions (as functional, taking dynamic functions and giving a scalar) then you can do something like taking the derivative in the direction of a function and for certain dynamical functions those directional derivatives could be zero. I'll go into more detail.
This often happens in Physics that you have a functional that takes dynamical functional and gives scalars and for a special critical dynamical function, the directional derivatives are zeros, and if so the special dynamical function that does that is physical. But obviously it is only physical if you selected the right functional.  Since the energy functional gives the geodesics as those critical dynamical functions then there is a sense where is is a correct kind of functional.  Of course multiplying the expression by 2 or -1 or something won't change which curves are special, so there isn't anything special about the sign of the energy functional, and as I mentioned before relative sign can potentially tell us something because it can distinguish between a curve that is always timelike and a curve that is always spacelike.
So now on to the more detail about how to take a derivative of a curve in the direction of another curve.  Consider a curve $\gamma$ and then consider another curve $f$ that is close to $\gamma$ in the sense that $f(a)=\gamma(a)$ and $f(b)=\gamma(b)$ and that the interval $[a,b]$ can be broken into pieces $[a,a_1],$ $[a_1,a_2],$ $[a_2,a_3],$ $\dots,$ $[a_{n-2},a_{n-1}],$ $[a_{n-1},a_n],$ $[a_n,b],$ such that for each interval (e.g. $[a_k,a_{k+1}]$) there is a coordinate chart where both $\gamma(t)$,  $f(t)$ and the whole coordinate line segment between them in is the chart for every $t$ in the interval (e.g. $t\in[a_k,a_{k+1}]$).  Then you can define the function $g(t)=f(t)-\gamma (t)$ as the coordinate different between the two. And then for any $h$ in $[0,1]$ you can define the function $H(h)=\gamma+hg$  where $H(h)$ is a function $\gamma+hg$, and the function $\gamma+hg$ sends $t$ to $(\gamma+hg)(t)=\gamma(t)+h*g(t)$. When $h$ is zero you get $\gamma$ and when $h$ is one you get $f$, so $H(0)=\gamma$ and $H(1)=f$. Then, if you consider the function that sends $h$ to $E(H(h))$ then it is a simple function from scalars to scalars. So the derivative could be zero at $h=0$ (at $H(0)=\gamma$). If this happens for any such $f$, then the function $\gamma$ is critical because any parametrized small change deviates less than order $h$ in the parametrization of deviation.  For the energy functional you give, doetoe pointed out that it is the geodesics that are critical.
