I Equivalent statements of the weak energy condition

[hideelo]

Im reading Carroll's Spacetime and Geometry section 4.2 where he claims the following as the weak energy condition:

Given a energy momentum tensor T and a timelike vector t then T_μν t^μ t^ν ≥ 0.

He claims that for a perfect fluid this is equivalent to the statement that ρ ≥ 0 and (ρ+P) ≥ 0. Where ρ is the density and P the pressure

He shows this by first showing that if the timelike vector t was the velocity of the fluid (i.e. we were in the rest frame of the fluid) then the weak energy condition reduces to ρ ≥ 0. So far so good. Then he claims that for a lightlike vector l, T_μν l^μ l^ν ≥ 0 implies that (ρ+P) ≥ 0. I see that. He then says that I should convince myself that these conditions on ρ and P therefore hold for any timelike vectors as a consequnce, I don't see how, can anyone explain it to me?

 

[PeterDonis]

Consider the following proposition: given any timelike vector $t^\mu$, we can express any other timelike vector $v^\mu$ as a linear combination of $t^\mu$ and some null vector $l^\mu$ (which might be a different null vector for different $v^\mu$). Is it true? And if it is, does it help in answering your question?

 

[hideelo]

Consider the following proposition: given any timelike vector $t^\mu$, we can express any other timelike vector $v^\mu$ as a linear combination of $t^\mu$ and some null vector $l^\mu$ (which might be a different null vector for different $v^\mu$). Is it true? And if it is, does it help in answering your question?

That's what I thought as well. However, If you think about it it must be wrong. If given some timelike vector t^μ any other timelike vector v^μ can be written as a sum of t^μ and some lightlike vector l^μ, flip that statement around and you get that the difference between any two timelike vectors is a lightlike vector, and that's not true

 

[PeterDonis]

If you think about it it must be wrong.

I stated it sloppily. The correct statement is that, given any future-directed timelike vector $t$, any other future-directed timelike vector $v$ can be expressed as a linear combination, with positive coefficients (positive provided that $v \neq t$), of $t$ and some future-directed lightlike vector $l$. The "future directed" part is crucial.

flip that statement around and you get that the difference between any two timelike vectors is a lightlike vector

No, because the second timelike vector has a positive coefficient in front of it, and both timelike vectors are future directed. So all you really get by flipping the statement around is that, by subtracting some particular coefficient times a timelike vector $v$ from another timelike vector $t$ (both being future-directed), you can get a lightlike vector. But given $t$ and $u$, only one particular value for the coefficient will work. So it's much more specific than the statement that the difference between any two timelike vectors is lightlike (which is indeed false); that would amount to claiming that you can choose the coefficient arbitrarily, which is not the case.

To state things concretely: we have

$$
v = a t + b l
$$

where $t$ and $v$ are timelike and $l$ is null, $a > 0$ and $b > 0$. Then we have

$$
\frac{1}{b} v - \frac{a}{b} t = l
$$

Furthermore, to obtain specific expressions for $a$ and $b$, we can always choose a local inertial coordinate chart in which $t = (u, 0, 0, 0)$, where $u > 0$, and in which $v = (w, x, 0, 0)$, where $w > x > 0$. (This amounts to choosing a chart in which $t$ corresponds to the 4-momentum of an observer at rest, and the first spatial direction points along the direction of motion of $v$ with respect to $t$.) Then, if we write $l = (k, k, 0, 0)$, where $k > 0$ and $|k|$ can be chosen arbitrarily, we obviously have

$$
v = \frac{w - x}{u} t + \frac{x}{k} l
$$

This can be inverted to express $l$ in terms of $t$ and $v$, but only with one particular set of coefficients.