Suppose ϕ ( x ) a scalar field, v μ a 4-vector. According to my...

Ciara Rose

Ciara Rose



Suppose ϕ ( x ) a scalar field, v μ a 4-vector. According to my notes a quantity of form v μ μ ϕ ( x ) will not be Lorentz invariant.
But explicitly doing the active transformation the quantity becomes
Λ ν μ v ν ( Λ 1 ) μ ρ ρ ϕ ( y ) = v ν ν ϕ ( y )
where y = Λ 1 x and the partial differentiation is w.r.t. y. This seems to suggest that the quantity is a Lorentz scalar, so could be used to construct a Lorentz invariant first order equation of motion.
I'm clearly making a mistake here. But I don't see what I've done wrong. Am I wrong to think that v transforms nontrivially under the active transformation?

Answer & Explanation




2022-07-18Added 23 answers

The function v a a ϕ is a scalar field. Nonetheless, an equation like this is ugly because v a points in some "preferred" direction.
Here's another point of view, and I think this gets at what you were saying about "no first order equations". Suppose that v a is not a vector but is instead just a collection of four fixed real numbers. Suppose we consider the equation μ v μ μ ϕ = 0now. This equation is not Lorentz invariant anymore since the numbers v μ don't change.
Another approach: think of v a as a new spacetime-dependent vector field. Then, v a a ϕ = 0 is Lorentz invariant equation, but it involves two fields. This is nicer than choosing a preferred direction.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get your answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?