Unfortunately, it seems that you have some fundamental misunderstandings of how Einstein's summation convention works and what you are allowed and not allowed to do with it.
Spoonszz said:
Hi
@Orodruin ! :)
My understanding is that in 1.151 we replaced ##\mu## with ##\rho## and ##\nu## with ##\sigma## which we called our dummy indices.
A "dummy index" is not something that we spontaneously decide to call an index. In the Einstein summation convention, a repeated index should be summed over and it therefore does not matter what that index is called. Hence the nomenclature "dummy index". While you can rename your summation indices at will, you
cannot rename them to be the same as another index that already exists in your expression. Whenever you have more than two of the same index, your expression is ambiguous at best and you risk going very very wrong from there. This is one of the most common mistakes among students that have not familiarised themselves enough with the notation and a very essential thing to be aware of. In 1.151, ##\mu## and ##\nu## were repeated, and therefore summation indices, and could be replaced with ##\sigma## and ##\rho##, respectively. Note that we could not,
absolutely not, replace ##\nu## with ##\mu## as this would result in an expression with four ##\mu##.
I understand that in the line above 1.153 we consider the case where ##\delta _{\rho }^{\mu }## is equal to 1 which happens when ##\delta _{\rho}^{\mu }## is ##\delta _{\mu }^{\mu }##,
It is not clear what you mean here. You cannot just replace ##\rho## with ##\mu##. The Kronecker delta is one when both indices are the same, leading to the general relation ##\delta^\rho_\mu A_\rho = A_\mu##. Note that there is no summation index ##\rho## on the right. However, ##\mu## only appears once and is therefore a free index. You cannot rename free indices at will - unless you do so on both sides of the equation. Writing ##\delta^\mu_\mu## is confusing, according to the summation notation, this should be summed over ##\mu## and therefore be equal to 4 (in the cases of 4-dimensional space-time).
My issue now comes in getting ##\eta^{\mu \sigma }\partial _{\sigma }\phi + \eta^{\rho \mu }\partial _{\rho }\phi = 2\eta^{\mu \nu}\partial _{\nu }\phi ##. I replace the ##\sigma## and ##\rho## in 1.153 with ##\nu## and ##\mu## as they were dummy indices
yet I end it ##\eta^{\mu \nu }\partial _{\nu }\phi + \eta^{\mu \mu }\partial _{\mu }\phi## which is clearly not equal to ##2\eta^{\mu \nu}\partial _{\nu }\phi ##
While you are allowed to replace ##\sigma## with ##\nu## in the first term (there was no ##\nu## in the first term before and ##\sigma## is a summation index), you are not allowed to replace ##\rho## with ##\mu## as the free index in the equation is called ##\mu##. Your result therefore becomes nonsense with a term with three of the same index. Instead, you can replace ##\rho## with literally any letter
except ##\mu##. I will leave it up to you to choose this letter wisely at your own discretion.
! This also answers my question on why ##\eta^{\mu \nu }\partial _{\nu }\phi + \eta^{\mu \mu }\partial _{\mu }\phi## was incorrect to start with as I was replacing ##\rho## with an index that was already available in my expression. With that being said I should be replacing ##\rho## with ##\nu## which I then get ##\eta^{\mu \nu }\partial _{\nu }\phi + \eta^{\nu \mu }\partial _{\nu }\phi## and since my minkowski metric is symmetric I am allowed to add ##\eta^{\mu \nu }\partial _{\nu }\phi + \eta^{\nu\mu }\partial _{\nu }\phi## getting ##2\eta^{\mu \nu }\partial _{\nu }\phi##. 
