Oh my god, thank you, I was using
R_{\mu \nu} -\frac{1}{2}R g_{\mu \nu} -\Lambda g_{\mu \nu} = 8\pi G T_{\mu \nu}
Because that's what the problem said, but I see now that it is a typo in the sign... I'm going to kill my professor he has done 3 typos in this problem already T_T
Thanks a lot people!
Hi!
I'm struggling to derive the accelerated Friedmann equation (shame on me!)...
I'll tell you what I'm doing and maybe you can find where the mistake is.
First of all, we know that:
H^2 = \frac{\rho}{2M^2_p}+\frac{\Lambda}{3}
Now, using the Einstein Equations and doing the trace of...
Yeah, but why that doesn't happen in the first case? :(
In addition, if we take \epsilon =1 \rightarrow a(t) = t, then:
dt = \pm t dx \rightarrow \ln (t)=\pm x \rightarrow t = \exp (\pm x)
That case is also reaaally weird, as you have line cossing in the future and in the past...
Thanks for...
At the beginning I thought that the second case could be because a(0)=0 (even though t=0 is not defined...), so every observer is like conected at t=0... but then I wouldn't understand the first case! Because both are expansions, they should look like similar.
Hi everyone!
I'm solving some GR problems and I have a question.
The problem is that we have a metric like the FRW metric but in 1+1 dimensions, i.e.:
ds^2 = -dt \otimes dt + a(t)^2 dx \otimes dt
Where a(t)=t^{1/\epsilon} (for t>0) and a(t)=(-t)^{1/\epsilon} (for t<0 ).
We take a vector V^\mu...
Hi people!
Today I was doing some QFT homework and in one of them they ask me to calculate the Harmonic Oscillator propagator, which, as you may know is:
W(q_2,t_2 ; q_1,t_1) = \sqrt{\frac{m\omega}{2\pi i \hbar \sin \omega (t_2-t_1)}} \times \exp \left(\frac{im\omega}{2\hbar \sin \omega...
I think I did it in #6, is anything wrong in that calculation?...
I think you might be right, but still I can't see it.
Why do I have to emit the photon from x=-\frac{-c^2}{c}?
The observer at the begining (x^0 = 0) is at a positive ##x##, and another observer sends the beam from ## x = 0##...
But, those two conditions are in fact just one:
x\geq \omega
As I understand, this x will tell me when will the rest observer will reach that point.
The observer will never pass through a point x which is less than \omega because is accelerating to the other direction (to x positive), that's why...
Yes. In my view, if I replace that x by ct (i.e. a photon sent from x=0) then the t I find will tell me when are they going to coincide... And I am always able to find a solution, which is:
t=\frac{g\omega^2}{2c^3}+\frac{\omega}{c}
Am I doing something stupid maybe? xD
I understand the theory...
Hi everybody,
I know that there are a lot of threads in this forum about Rindler coordinates but none of them have helped me :confused:
I'll explain you my problem. First of all, my coordinates (x^0,x) (Cartesian coord., where x^0=ct) are related to the Rindler coordinates (\omega ^0,\omega)...