Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Let's have metric $$ ds^2 = dt^2 - dx^2 - dy^2 - dz^2 - 2f(t - z, x, y)(dt - dz)^2. $$ I need to prove that it is an exact solution for Einstein equations in vacuum for $\partial_{x}^{2}f + \partial_{y}^{2}f = 0$.

The straightforward method is obvious, but does some method especially for this metric exist?

Edit. As I see, first is replacing the variables as $$ u = (t - z), \quad v = (t + z). $$

share|improve this question
    
Hi Andrew McAdams. If you haven't already done so, please take a minute to read the definition of when to use the homework tag, and the Phys.SE policy for homework-like problems. –  Qmechanic Nov 27 '13 at 12:37

1 Answer 1

up vote 3 down vote accepted

The metric is well-known, and can be shown to describe a gravitational wave. I will help prove the claim using a slightly different metric which is more convenient for the tetrad formalism. The metric:

$$ds^2=dt^2-dr^2+H(t-r,x^1,x^2)(dt-dr) - d(x^i)^2$$

where $i=1,2$ and $x^1,x^2$ are generic coordinates. We define an orthonormal basis,

$$\omega^t=dt+\frac{1}{2}H(...)(dt-dr) \quad \omega^r = dr +\frac{1}{2}H(...)(dt-dr)$$

and $\omega^i=dx^i$ such that $\underline{g}=\eta^{ab}\omega_a \omega_b$. Taking exterior derivatives of the basis yields,

$$d\omega^t=\frac{1}{2}H_{,i} dx^i \wedge (dt-dr)=-\frac{1}{2}H_{,i}(\omega^t -\omega^r)\wedge \omega^i$$

and similarly for the other basis; of course $d\omega^i = 0$. By Cartan's first equation,

$$d\omega^a=-\theta^a_b \wedge \omega^b$$

for the spin connection $\theta^a_b$. Applying the equation, we may deduce the non-zero components,

$$\theta^t_i=\frac{1}{2}H_{,i}(\omega^t-\omega^r)$$

and identically for the $\omega^r$ case; the rest vanish. Cartan's second equation dictates,

$$R^a_b = d\theta^a_b +\omega^a_c \wedge \omega^c_b$$

You should be able to take over the computation from here. As we are only interested in the intrinsically flat case, the Ricci scalar should vanish in any basis, hence there is no need to convert back to the coordinate basis. Eventually, you will find the condition,

$$\Delta H = 0$$

where $\Delta$ is the Laplace operator, the desired result. To ensure it is a gravitational wave, we demand $T_{\mu\nu}=0$, otherwise we can't, at least by inspection, immediately conclude the wave is gravitational, rather than for example electromagnetic. By the Einstein field equations, that implies $R=0$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.