@ecterrab 

Substitue is what I was looking for, but there are still some issues...

1. In my manual method, I still have a exp term floating around. This is because I did not explicitly make a substitution for the metric term will all upper indices. If I do that it works perfectly, and I get the correct expression but you need to make sure that you have a negative exponent for the upper index metric. 

2. If I use Substitue as you propose, albeit with a rather cheeky modification, I was expecting it to take care of it and understand that for the upstairs index metric you need the negative exponential, but it sadly did not.

3. Taking your file explicitly and ammending to look at the Christoffel of the second kind, you can also see that it fails to understand that the upstairs metric needs to have the exponential being negative; yielding the exp(4phi) term. 

I can, in theory, do everything with the Christoffel symbolds of first kind and then "manually" raise one index, I was just curious if Maple could do it easily. I am starting to believe that it can't do it nicely with the one upstairs index on the Christoffel symbol. 

Could I be correct? 

Thanks! 

@ecterrab 

Thank you for this, but I wanted something a little bit different, if Maple can do that. I should also note, I should have written Phi(X), not just Phi -- it needs to be coordinate dependant. Yours will work great for known metric, but I am trying to do it with arbitray metrics, and thus want the index expressions explicitly, and how the original vs. conformal expressions are related. Example: 

I will attach a screenshot of my worksheet and re-upload and hopefully you will be able to access it. 

MyConformal.mw

So I guess my question is -- based on my worksheet/screenshot if you can see it this time -- can I take my equation (3) and apply the transformation g_[] = exp(2 Phi(X)) g_[] easily, without needing to input each index pair, and arrive at my equation (5)? 

Thank you! 

Yes you can solve it. 

Do you have a file with an attempt at it? I can help you with it. 

@segfault 

Can you post the metric you are trying to work with? I promise I wont scoop any ideas, I have far to many projects of my own to worry about. 

I am sorry if I am thinking about it too simply. 

If you have a metric that you want to use, why not just manually enter it at the start and use the Setup command? Instead of initializing a metric then trying to change it? 

It would be easiest if we could see what you have attempted. Can you upload your worksheet? 

@segfault 

In the help page search  Physics,d_ or on your worksheet just do ?Physics,d_ 

Here is the link to the online version Physics,d_

@ecterrab 

Thank you for the response. Some of those commands I was aware of but some are new to me, I will use them in the future! 

Can you upload a worksheet with attempted solution? 

To @acer and @mmcdara

I am sorry for my very late response. I posted this question and then I had to step away from the problem for a while and now I am back into it. 

Thank you both for the replies, I was unaware that I could get it to run that quickly, must not have been thinking hard enough. Both of your responses make sense, but if something else comes up I will be sure to ask. 

Thank you again. 

@Scot Gould 

..

@dharr 

Thank you. Yes something like that but I don't think it fully solves my problem. I will reformulate. 

However, you are correct E2 is a mess, and that is a construct of how I want to solve the problem. I can elaborate in more detail and that might make it more clear what I was hoping for, I will edit my question afterwords. The differential equation I want to solve is: 

Sum((-1)^n*n*2^(n - 1)*a[n]*diff(Phi(r), r)^(-1 + 2*n), n = 1 .. N)=Q/r

I want the solution for Phi(r) to be of the form

diff(Phi(r), r) = Sum(b[i]/r^i, i = 1 .. infinity)

Now say I truncate the first sum to N=2 and I want to get the first few coefficients b[i] which solve this equation, obviously this is an infinite sum but if we consider just the first few terms for simplicity

We have that b[1]=-Q,b[2]=0,b[3]=4a[2]Q^3 so on and so on. Is there a command that can do this while considering what the previous found coefficients are (ie set b[1]=Q,b[2]=0)? 

For some reason it doesn't like using f[4](0). If you change to f[4] it works fine for me. 

@Paras31 

I will continue to look at this. 

@Paras31 

I can plot yes. If those are the plots you are looking for let me dive a little deeper into it and get back to you ASAP.