Lagrangian for a single scalar field

[Spoonszz]

Homework Statement

Hello all !
Over the past few day's, I've been trying to understand how Sean Carroll comes to the conclusion that he does on equation 1.153. I've tried to look for various resources online but I still have trouble understanding how he is able to add both partial derivatives in 1.153.

Homework Equations

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The Attempt at a Solution

I tried to replace the dummy indices with our initial indices on 1.151 yet that seems to not bring me to the conclusion that one is supposed to get on 1.153. I would appreciate any guidance or help thank you !

 

[Orodruin]

Could you please write down what you do get? Exactly what is the problem in doing the last step?

Note that you can use LaTeX to write equations, see LaTeX Primer.

 

[Spoonszz]

Hi @Orodruin ! :)

My understanding is that in 1.151 we replaced $\mu$ with $\rho$ and $\nu$ with $\sigma$ which we called our dummy indices. 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 $\delta _{\mu }^{\mu }$, which also turns our metric $\eta^{\rho \sigma }$ into $\eta^{\mu \sigma }$ and we also follow the exact procedure on the second term above 1.153. 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 $2\eta^{\mu \nu}\partial _{\nu }\phi$

 

[Mike Stefan]

If Two identical indices are present inside a product then you must sum over
this (repeated indice) .In the equation above I can tell that
you must sum over the sigma indices (within the range of the sigma indices)

in the first term(on the left side of the equation) and obtain a result(regarding the first term). Then on
the the second term you must sum over the (repeated indice ro ) and obtain a result(regarding the second term). You add both results for the left
side of the equation. On the right side of the equation you must sum over the repeateded indice(niu) and obtain a result.
It seem on the left side of the equation
you have a contraction of a second degree contravariant tensor with a first degree covariant tensor(vector). Same happen
for the right side of the equation.
Now in a contraction (sum over repeated indices) between a second
degree contravariant tensor and a first
degree covariant tensor you will obtain
a contravariant first degree tensor(vector).

 

[Orodruin]

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.

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.

 

[Mike Stefan]

Again I said make a sum over the repeated indice (whatever notation you
want to chose) In the First term (on the
left side of the equation) and obtain a result. Then make a sum over the repeated indices (whatever notation you want to use)inside the Second Term (present on the left side of our equation) and obtain another result.
Then you add both results.You Will obtain a contravariant tensor of rank
One(a 4-Vector).
Then Look at the right side of our equation and sum over the repeated indices (whatever notation you want to
use) and obtain a result which should should be again a contravariant tensor of rank One(4- Vector).
Now If you compare the result from the
left side of the equation with the Other result from the right side of the equation you should obtain an equality.

 

[Orodruin]

Again I said make a sum over the repeated indice (whatever notation you
want to chose) In the First term (on the
left side of the equation) and obtain a result. Then make a sum over the repeated indices (whatever notation you want to use)inside the Second Term (present on the left side of our equation) and obtain another result.
Then you add both results.You Will obtain a contravariant tensor of rank
One(a 4-Vector).
Then Look at the right side of our equation and sum over the repeated indices (whatever notation you want to
use) and obtain a result which should should be again a contravariant tensor of rank One(4- Vector).
Now If you compare the result from the
left side of the equation with the Other result from the right side of the equation you should obtain an equality.

Why are you replying again? The OP has not been back since your last post and I have no problems getting this right. I also think your posts are off the mark for the OP. His problem is understanding what he is allowed and not allowed to do when using the summation convention, in this case how you are allowed to rename summation indices. In particular, how you are not allowed to give summation indices the same name as an already existing index.

Also, please do not use line breaks and capitalisation randomly in your posts. It makes them essentially impossible to read.

 

[Spoonszz]

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 $\mu$.

Glad to see that you were able to explain to me the workings behind the "dummy index" ! 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 $2\eta^{\mu \nu }\partial _{\nu }\phi$.

@Mike Stefan Thanks for you advice as well !