Dimension of terms in Lagrangian

[zaman786]

TL;DR

what do we mean by dimensions of terms in lagrangian?

hi, when we say in SM , we can add terms having dimension 4 or less than that- in this to what dimension we are refering ? kindly help how do you measure the dimension of terms in Lagrangian. thanks

 

[Orodruin]

Mass dimension, the only physical dimension left once you introduce natural units. And we are talking only about the dimension of the field content - excluding any dimension of constants.

To find the mass dimension of a field, look at its kinetic term, which contains derivatives and that field only. The total mass dimension of the Lagrange density should be 4 (or it would not integrate to a dimensionless action) and derivatives have mass dimension 1 (as length and time have mass dimension -1).

 

[Orodruin]

Example: The kinetic term for a real scalar field has the content $(\partial \phi)^2$. If $\phi$ has mass dimension $k$, then this term has mass dimension $2(k+1)$ where the 2 comes from the square and the 1 from the derivative. This needs to be equal to 4 so $k =1$.

 

[zaman786]

Mass dimension, the only physical dimension left once you introduce natural units. And we are talking only about the dimension of the field content - excluding any dimension of constants.

To find the mass dimension of a field, look at its kinetic term, which contains derivatives and that field only. The total mass dimension of the Lagrange density should be 4 (or it would not integrate to a dimensionless action) and derivatives have mass dimension 1 (as length and time have mass dimension -1).

thanks - but length and time has dimension of -1 , so how can we say Mass dimension is the only physical dimension left?

 

[zaman786]

Example: The kinetic term for a real scalar field has the content $(\partial \phi)^2$. If $\phi$ has mass dimension $k$, then this term has mass dimension $2(k+1)$ where the 2 comes from the square and the 1 from the derivative. This needs to be equal to 4 so $k =1$.

got it - thanks

 

[dx]

(∂φ)² has mass dimension 4.

 

[Orodruin]

(∂φ)² has mass dimension 4.

Yes, what is your point?

The thing is that you know it needs to be 4 in order to have the right mass dimension of the Lagrange density. The question of the OP was how you deduce the mass dimension of the fields. If the field has mass dimension k, then that term has mass dimension 2(k+1) = 4 from which it can be concluded that k = 1.

 

[Orodruin]

thanks - but length and time has dimension of -1 , so how can we say Mass dimension is the only physical dimension left?

Yes, length and time have mass dimension -1. They do not have independent dimensions as in SI units.

 

[apostolosdt]

Recall that action, $\int d^4 x\,\cal{L}$, should be dimensionless.

 

[Orodruin]

Recall that action, $\int d^4 x\,\cal{L}$, should be dimensionless.

Literally what I already said …

integrate to a dimensionless action

 

[apostolosdt]

Sorry, I was referring to the OP. You indeed mentioned it first.

 

[zaman786]

Yes, length and time have mass dimension -1. They do not have independent dimensions as in SI units.

got it- thanks

 

[zaman786]

Sorry, I was referring to the OP. You indeed mentioned it first.

now, in SM Lagrangian terms has dimension 4 - than how do we allow dimension 5 operator in SM , i know it is some what related as perturbation to SM.

 

[Orodruin]

now, in SM Lagrangian terms has dimension 4 - than how do we allow dimension 5 operator in SM , i know it is some what related as perturbation to SM.

The SM as such has only d=4 operators. If you do SM effective field theory you will typically add operators of higher dimension. In the Lagrangian these will be accompanied by a dimensional prefactor, typically assumed to be the mass scale of new physics to an appropriate power.

 

[zaman786]

now, in SM Lagrangian terms has dimension 4 - than how do we allow dimension 5 operator in SM , i know it is some what related as perturbation to SM.

 

[zaman786]

The SM as such has only d=4 operators. If you do SM effective field theory you will typically add operators of higher dimension. In the Lagrangian these will be accompanied by a dimensional prefactor, typically assumed to be the mass scale of new physics to an appropriate power.

thanks - but as say action should be dimensionless - so if we introduce dimension 5 operator , than do we have to introduce dimensional prefactor - so that overall dimension remains to be 4 ?

 

[Orodruin]

thanks - but as say action should be dimensionless - so if we introduce dimension 5 operator , than do we have to introduce dimensional prefactor - so that overall dimension remains to be 4 ?

Yes, that is what I said. This prefactor is typically assumed to be a power of the scale of the new physics related to the effective operator.

 

[zaman786]

Yes, that is what I said. This prefactor is typically assumed to be a power of the scale of the new physics related to the effective operator.

thanks a lot