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.
In classical mechanics, the dimension of the Lagrangian, which is typically represented as L, is the same as the dimension of energy. Therefore, it has the dimension of mass times length squared per time squared (ML²/T²).
To determine the dimension of a term in the Lagrangian, you need to ensure that each term has the same dimension as energy (ML²/T²). This involves analyzing the dimensions of each variable and constant in the term, such as mass (M), length (L), and time (T), and combining them appropriately.
All terms in the Lagrangian must have the same dimension to ensure the consistency and physical meaningfulness of the equations of motion derived from it. This requirement stems from the principle of dimensional homogeneity, which states that equations must be dimensionally consistent.
In field theory, the Lagrangian density, often denoted as ℒ, has the dimension of energy per unit volume. Therefore, its dimension is ML⁻¹T⁻², which is energy (ML²/T²) divided by volume (L³).
The dimensions of fields play a crucial role in determining the dimension of the Lagrangian. For instance, in a scalar field theory, the field itself must have a specific dimension to ensure that the Lagrangian density has the correct dimension. Typically, the dimension of the field is chosen such that when combined with derivatives (which have dimensions of L⁻¹) and other constants, the resulting term has the dimension of energy density (ML⁻¹T⁻²).