thanks - but length and time has dimension of -1 , so how can we say Mass dimension is the only physical dimension left?Orodruin said: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).
got it - thanksOrodruin said: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##.
Yes, what is your point?dx said:(∂φ)2 has mass dimension 4.
Yes, length and time have mass dimension -1. They do not have independent dimensions as in SI units.zaman786 said:thanks - but length and time has dimension of -1 , so how can we say Mass dimension is the only physical dimension left?
Literally what I already said …apostolosdt said:Recall that action, ##\int d^4 x\,\cal{L}##, should be dimensionless.
Orodruin said:integrate to a dimensionless action
got it- thanksOrodruin said:Yes, length and time have mass dimension -1. They do not have independent dimensions as in SI units.
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.apostolosdt said:Sorry, I was referring to the OP. You indeed mentioned it first.
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 said: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 said: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.
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 said: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.
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 said: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 ?
thanks alotOrodruin said: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.