Why Is the Accelerated Friedmann Equation Giving Incorrect Signs?

[Cancer]

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 these, we get:
-R-4\Lambda = \frac{T}{M^2_p} = \frac{-\rho+3p}{M^2_p}
Where R=6[ \frac{\ddot{a}}{a}+H^2].
Using this, I get:
\frac{\ddot{a}}{a} = -\Lambda - \frac{1}{6}\frac{\rho+3p}{M^2_p}

The problem is that I shouldn't get a minus sing in the \Lambda and divided by 3, but I just CAN'T see what I'm doing wrong, as you can see here:
http://ned.ipac.caltech.edu/level5/Carroll2/Carroll1_2.html
I've tried to look for a book that does this derivation but I haven't found any...

If you could help me I would really appreciate it!
Thanks

 

[Chalnoth]

Why do you have the trace of the \Lambda as -4\Lambda? I don't think that's right.

 

[George Jones]

Why do you have the trace of the \Lambda as -4\Lambda? I don't think that's right.

The trace of the $\Lambda g_{\mu \nu}$ term is $+4 \Lambda$, which then gives a $+ \Lambda/3$ in the expression for $\ddot{a}/a$.

 

[Cancer]

The trace of the $\Lambda g_{\mu \nu}$ term is $+4 \Lambda$, which then gives a $+ \Lambda/3$ in the expression for $\ddot{a}/a$.

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!

 

[bapowell]

Please don't kill your professor

 

[ChrisVer]

Are you sure about your result? It doesn't seem like only a matter of sign, but also a matter of your factors (like 1/6 in front of the density/momentum)

I find it easier to take the FE and take the time derivative of it. Doing so (and using the fluid continuity equation) the answer would be like:

\frac{\ddot{a}}{a} = \frac{\Lambda}{3} - \frac{1}{2} \frac{\rho +3p}{2M_{Pl}^2}

This coincides also with the bibliography of 8 \pi G /3 since in your case you have written that as \frac{1}{2 M_{Pl}^2}. If you prefer $G$, it would be:

\frac{\ddot{a}}{a} = \frac{\Lambda}{3} - \frac{4 \pi G }{3} (\rho +3p)

 

[Cancer]

Please don't kill your professor

Just kidding ;)

Are you sure about your result? It doesn't seem like only a matter of sign, but also a matter of your factors (like 1/6 in front of the density/momentum)

I find it easier to take the FE and take the time derivative of it. Doing so (and using the fluid continuity equation) the answer would be like:

\frac{\ddot{a}}{a} = \frac{\Lambda}{3} - \frac{1}{2} \frac{\rho +3p}{2M_{Pl}^2}

This coincides also with the bibliography of 8 \pi G /3 since in your case you have written that as \frac{1}{2 M_{Pl}^2}. If you prefer $G$, it would be:

\frac{\ddot{a}}{a} = \frac{\Lambda}{3} - \frac{4 \pi G }{3} (\rho +3p)

As I said, it was a problem of the sign and also the factor 3.
I've already solved my problem, thanks a lot! ^^

 

[Chronos]

There is an ancient tradition for murder that inflicts all the victims responsibilities - including servicing their spouse.