Recent content by John_Doe

No, it just seemed to me at first, by eye, that the equation should simpllify, which is obviously wrong. It's not supposed to say much now - I'm just insterested in the fact that all the working is now correct. Except, one last thing: what if I have g_{\alpha\beta}g^{\gamma\delta} and...

G_{\mu\nu} =R_{\mu\nu}-\frac{1}{2} g_{\mu\nu}g^{\alpha\beta}R_{\alpha\beta} =R_{\mu\nu}-\frac{1}{8} g_{\mu\nu}g^{\sigma\tau}g_{\sigma\tau}g^{\alpha\beta}R _{\alpha\beta} So all the working here is now correct, even if it doesn't help in the slightest? That's a good thing. Thank you all...

I'm sorry, but you've lost me there. It's not equal to \delta_{\alpha}^{\beta}? I thought that g^{\mu\nu} was defined as the inverse of the g_{\mu\nu}. After all, g^{\mu\nu}=\frac{G(\mu,\nu)}{g} if G(\mu,\nu) denotes the cofactors of g_{\mu\nu} and g = |g_{\mu\nu}|. Edit: Taking into...

Thank you very much, except now I seem to have g_{\mu\alpha}g^{\mu\beta}=\delta_{\alpha}^{\beta}=\left\{\begin{array}{cc}1,&\mbox{ if } \alpha=\beta\\0, & \mbox{ if } \alpha\neq\beta\end{array}\right. but also g_{\mu\nu}g^{\mu\nu}=4 Marginal confusion there... and I also don't quite...

Yeah - there should be an = there so that it's G_{\mu\nu} =R_{\mu\nu}-\frac{1}{2} g_{\mu\nu}g^{\alpha\beta}R_{\alpha\beta} =R_{\mu\nu}-\frac{1}{2} g_{\mu\nu}g^{\mu\beta}g_{\mu\beta}g^{\alpha\beta}R _{\alpha\beta} Then the metric tensors contract, yielding the kronecker deltas. I'm not sure...

Dumb question, but... G_{\mu\nu}=R_{\mu\nu}-\frac{1}{2} g_{\mu\nu}R Since R=g^{\mu\nu}R_{\mu\nu} and g^{\mu\nu}g_{\mu\nu}=1 it would appear that G_{\mu\nu} =R_{\mu\nu}-\frac{1}{2} g_{\mu\nu}g^{\alpha\beta}R_{\alpha\beta} =R_{\mu\nu}-\frac{1}{2}...

Could someone please explain how gravitational waves are modeled within the theory? Is it some sort of time dependent metric, or is it simply an indirect consequence of the theory, etc.? Also, I am self-taught GR, and I learned almost everything I know about the theory straight out of...

I have obtained Kerr's paper, "Gravitational Field Of A Spinning Mass As An Example Of Algebraically Special Metrics", and was wondering if someone would be able to provide an explanation of the mathematics, or at least some direction in which I should investigate in order to learn the...

y'' = \frac{d^2 y}{dx^2}

\frac{\partial f}{\partial x} = \dot{y} \sqrt{1 + \dot{y}^2} + \frac{y \dot{y}}{\sqrt{1 + \dot{y}^2}} = \dot{y} \frac{1 + \dot{y}^2 + y}{\sqrt{1 + \dot{y}^2}}

Yes. Oops. \frac{d^2 y}{dx^2} = \frac{d \dot{y'}}{d \dot{x}} = \frac{dy'}{dx} = y''

f = y\sqrt{1 + \dot{y}^2} \frac{\partial{f}}{\partial{y}} = \sqrt{1 + \dot{y}^2} \frac{\partial{f}}{\partial{\dot{y}}} = \frac{y\dot{y}}{\sqrt{1 + \dot{y}^2}} \frac{d}{dx}(\frac{y\dot{y}}{\sqrt{1 + \dot{y}^2}}) = \frac{\dot{y}^2 + y\ddot{y} - \frac{y\dot{y}^2\ddot{y}}{\sqrt{1 + \dot{y}^2}}}{1...

How is y the gravitational potential?

I originally thought that the way to tackle the problem would be to set \int^b_{a}{\sqrt{1 + \dot{y}^2}dx} = c as a constraint equation, and minimise \int^b_{a}{ydx}. Can you explain geometrically \int^b_{a}{y \sqrt{1 + \dot{y}^2}dx?

\frac{\partial f}{\partial y} - \frac{d}{dx}(\frac{\partial f}{\partial \dot{y}}) + \lambda (\frac{\partial g}{\partial y}) = 0 The constaint equation is g(y,\dot{y}, x) = 0