Suppose phi(x) a scalar field, v mu a 4-vector. According to my notes a quantity of form v mu ∂=d mu phi(x) will not be Lorentz invariant.

Ciara Rose

Ciara Rose

Answered question

2022-07-17

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

Kitamiliseakekw

Kitamiliseakekw

Beginner2022-07-18Added 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.

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