That was extremely helpful, thanks. Oh and that was meant to be a negative sign in front of the c^2... just mistakenly left it out of the post.
I'm not sure if I followed you to the letter, but I read through your post and then ran with the idea (perhaps deviating a little).
Since my...
Yup, fabulous. After doing all that and solving for my A(r) and B(r), as well as setting a constant equal to c^2 to accommodate the minkowski metric as r approaches infinity, I end up with:
d\tau^2 = (1+\frac{1}{Kr})^{-1}dr^2 + r^2(d\theta^2 + \sin^2\theta d\phi^2) + c^2 (1+\frac{1}{Kr})dt^2...
I thought you meant consecutively descending. Ok, so the way I thought about it was this:
You can add the number of ways you can get a descending order where your highest card (call it M) is 4 (since M can't be less than 4), plus the number of possibilities when M=5, M=6, all the way to M=36...
Actually, your estimate is much larger than the actual number of possibilities. Try to think of it this way:
How many possibilities are there for the first card? (Ans: Obviously 36)
How many possibilities are there for the second card? (Hint: Not 36)
How many possibilities are there for...
It's actually fairly simple. First you can easily determine how many different sets of 4 cards can you get that are in descending order, and you can probably figure out how many total hands there are. And well, yea... you know what to do from there.
Well how many different sets of 4 cards can you get that are in descending order? (i.e. {36, 35, 34, 33} is one, {35, 34, 33, 32} is another... how many are there?)
I see. That would explain a lot then because I was summing the entire thing because of that extra repeated index 'p' (not just summing the terms with p).
Wait, since nothing is fixing p, don't you mean
R^1{}_{111} = \Gamma^1{}_{11,1}-\Gamma^1{}_{11,1} +
(\Gamma^1{}_{11} \Gamma^1{}_{11}...
Okay, Maxima seems correct as it would end up giving you the correct Schwarzschild metric. So I've come up with a system of checkpoints to see where I'm making my mistake. If you could possibly flag the checkpoint that is wrong, that would really help me try to fix my error.
Maxima agrees...
That looks fantastic, thanks. I'm not sure why my way kept getting a strange result, but now that I have this to go by I can do some tests.
Thanks a lot Pervect. This program looks neat :smile:
I agree completely, but just as a nit-picky note, I believe you mean Colombia (It's more of a pet-peeve than anything else I guess... it's just that being Colombian and seeing your country constantly spelled like a US/Canadian city get's on that pet-peeve list real fast).
Pervect, this is what you got using my form of the metric, right? Do you also use the same Christoffel symbols I've been using?
\Gamma^1{}_{1 1} = \frac{A'}{2A}
\Gamma^2{}_{2 1} = \Gamma^3{}_{3 1} = \Gamma^2{}_{1 2} = \Gamma^3{}_{1 3} = \frac{1}{r}
\Gamma^3{}_{3 2} = \Gamma^3{}_{2 3} =...
Whilst redoing it on my own, and trying to simplify, I actually did A(r)*R_tt - B(r)*R_rr = 0. (Only to then come back to the forum and realize that is what you suggest). However, I end up with:
\frac{6}{r^2} + \frac{(B')^2}{4B}(7 + \frac{1}{B}) + \frac{1}{r}(\frac{B'}{B} + \frac{A'}{A}) -...
So the two Christoffel symbols I had incorrect didn't end up being present in the equations for R_{1 1} and R_{4 4} anyway so no progress there. Nevertheless, I am going back and doing the whole thing over to see if I can figure something out.
I'm working this out for myself (https://www.physicsforums.com/showthread.php?t=80565) and I ended up with all the same Christoffel symbols except for one:
\Gamma^1{}_{4 4} = \frac{-A'}{2B} (Note: If you're looking at my thread, I defined my A and B in reverse. So your A is my B and vice...