Recent content by vizart

I am reading a paper and the authors read the tt component of the metric from the line element ds^2=f(r)[g(r) dt^2+h(r) dr^2] as g_{tt}=g(r) instead of (what I expect to be) g_{tt}=f(r) g(r) Could somebody please explain to me why? Thank you.

Thanks for the explanations CPT, but due to opposite variables in the theta function, when you take the first derivative, a relative negative sign between exponential functions appears which means it's more like \delta'(\tau)(e^{i\omega\tau}-e^{-i\omega\tau}) which as I mentioned is proportional...

Alright, I see that the cosine term that is multiplied by a delta function could be taken as one, but the sine term is multiplied by the derivative of delta function and I can't see how you could set it equal to zero.

Thanks for your response. I knew about the derivative of the delta function, but the problem is when you take first and second derivatives of the theta function, you get one term with delta function and cos\left (\omega(t-t^\prime)\right ) (the other derivative is carried over the exp function)...

I am having bit of a problem proving Eq. (0.2): (0.1) \text{ } G_{\omega}(t-t^\prime)=\theta(t-t^\prime) \frac{e^{-i\omega(t-t^\prime)}}{2\omega}+\theta(t^\prime-t) \frac{e^{i\omega(t-t^\prime)}}{2\omega} (0.2) \text { }\left (-\partial^2_{t}-\omega^2 \right ) G_\omega...

I agree, but the "always" condition requires certain class of geometries, I think. For example assume a flat and finite universe with boundary, then you won't receive such signals (from a certain moment in the past) forever.

I see your point and apologize for misjudgment. But the question is still open to me: assuming a compact universe with no boundary, what will happen to the observed CMB in later times?

I beg to differ. The light cone is not attached to the event (emission from LSS), its origin coincides with the observer at all times. In this sense, as time goes by, whatever the event is, it will move down the past light cone (which is synonymous to "moving back in time"). I know that little...

So as long as we are dealing with photons, there will come a time that no CMB (in photonic sense of the term) would exist in the sky, is that correct?

I don't understand what you exactly mean by "our perception of CMB" but my question is essentially that if there is a temporal edge, then is it true that past a certain time we would see NO CMB in the sky?

Even if there is no spatial edge there is still a temporal one, that is, the time before decoupling.

This came up when I was trying to explain how we receive light from decoupling era: I know that the CMB we are getting is from some Last Scattering Surface (LSS) and this surface is moving back in time in some sense. But the question is: will there come a time that we are past the decouping era...

It inevitably follows from a theorem in Linear algebra that states every unitary matrix is diagonalizable. You can probably find it in some L(Alg) textbook or alternatively you can simply google that statement.

I must be hanged for not seeing the simple fact that when the metric has the given Lewis form (4.1) if you move to the rotating reference frame, then the near-horizon condition "RHS of Eqn. 4.2 approaches zero" will be obvious.

thank you for the tip but if you look at the Boyer-Lindquist form #(4.1), you see that it is quite general and i think their statement is somehow related to the fact that the horizon is a null hypersurface. yet i don't know the 'how' part :)