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?

Question

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&#039;m clearly making a mistake here. But I don&#039;t see what I&#039;ve done wrong. Am I wrong to think that   v transforms nontrivially under the active transformation?

Kitamiliseakekw · Accepted Answer

The function       v    a        ∂    a    ϕ is a scalar field. Nonetheless, an equation like this is ugly because       v    a   points in some &quot;preferred&quot; direction.Here&#039;s another point of view, and I think this gets at what you were saying about &quot;no first order equations&quot;. 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&#039;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.