1. Aug 23, 2011

### off-diagonal

Hi, everyone

I'm now reading A Relativist's Toolkit by Eric Poisson.

In chapter 3, he say something about stationary and static spacetime.

For static spacetime, it's just like stationary (admits timelike Killing vector $t^{\alpha}$) with addition that metric should be invariant under time reversal $t \rightarrow -t$. Which implies in some specific coordinate $g_{t\mu} = 0$.

He also say that, this also implies that $t_{\alpha} = g_{tt}\partial_{\alpha}t$ and he concludes that "a spacetime is static if the timelike Killing vector field is hypersurface orthogonal".

I'd like to ask about these two implications. I have no idea how these two become relevant with static spacetime. Because, as far as I know, hypersurface orthogonal is the congruence of geodesics which have no rotating part.

2. Aug 23, 2011

### WannabeNewton

Well, hypersurface orthogonality basically means that the vector field in question is everywhere orthogonal to a family of hypersurfaces. If you take a single hypersurface from that family, you know that the normal vector $n_a = \frac{\partial f}{\partial x^{a}}$ will be orthogonal to that selected hypersurface at some point P ($n_{a}dx^{a} = 0$ in the neighborhood of P). So a vector field is hypersurface orthogonal if it is proportional to the normal vector everywhere i.e. $t^{a} = \sigma (x)n^a$. One condition that comes out of this is (and you can show this yourself from killing's equation $\triangledown _{\beta }t_{\alpha } + \triangledown _{\alpha }t_{\beta } = 0$) that $t_{a} = t^{2}\frac{\partial f}{\partial x^{a}}$ for an arbitrary scalar field f. If you consider this condition specifically for a time - like killing field then $t_{a} = \frac{\partial }{\partial t} = \delta ^{a}_{0}$ so $t_{a} = g_{ab}t^{a} = g_{0a}$ and similarly, $t^{2} = g_{00}$. Setting a = 0 in $g_{0a} = g_{00}\frac{\partial f}{\partial x^{a}}$ gives $f = x^{0} + h(x^{i})$. Now, make the coordinate transformation $x^{'0} = x^{0} + h(x^{i}), x^{'i} = x^{i}$ and you will find that $g_{'0i} = 0$. I am not sure about the second statement you made regarding $t_{\alpha }$; are you sure it was written that way?

3. Aug 23, 2011

### bobc2

O.K., WannabeNewton, are you really a junior in high school?

4. Aug 23, 2011

### WannabeNewton

Yeah, I go to Bronx High School of Science, NYC. Why o.0?

5. Aug 23, 2011

### Staff: Mentor

Wow, I thought that was a joke. That is impressive.

Did you teach yourself tensors or is it part of your curriculum?

6. Aug 23, 2011

### WannabeNewton

Used Wald's text and Caroll's text to self - study the basics then used Bishop's Tensor Analysis on Manifolds. My school curriculum only goes up to AP Calculus BC.

7. Aug 23, 2011

### off-diagonal

Wow, thank you WannabeNewton for the proof of the first statement that helps me a lot.

About the second statement, I am sure because I just copy+paste from the original textbook.

8. Aug 23, 2011

### WannabeNewton

Oh that is just the statement that the killing field will be proportional to the gradient because then the killing field is orthogonal to the family of hypersurfaces everywhere since it is proportional, at each point on the family of hypersurfaces, to the gradient (which will be orthogonal to a hypersurface at a point). The metric is the proportionality factor. I just don't get where he got the $g_{tt}$ from in that statement because I always thought it was $t^{a} = g^{ab}\partial _{b}t$ if you are going to use the metric as the proportionality.

9. Aug 23, 2011

### bobc2

Hey, boys. I think we have a boy genius here. (I didn't start studying this stuff until a physics student in grad school). From now on I learn at your feet, WannabeNewton (I think you are already Newton).

10. Aug 23, 2011

### WannabeNewton

I wish hehe. I'm not smart at all, I just like learning these things is all. Cheers my good man.