# Confusion about vector and scalar field transformation laws

I'm a little bit confused about the transformations for scalars fields and vector fields in classical field theory. I've learned that a scalar field is a smooth function

$$\phi : M \longrightarrow \mathbb{R} \tag1$$

where $$M$$ is the Minkowski space

Under a change of coordinates (actually a boost), this field satisfies

$$\phi'(x') = \phi( \Lambda^{-1} \Lambda x) = \phi(x) \tag2$$.

However, we still have smooth vector fields in Minkowski space wich are maps

$$A: M \longrightarrow M \tag3$$

If I choose some basis $$\{e_\mu\}$$, a vector field $$A$$ can be written as

$$A(x) = A^\mu(x) e_\mu(x)\tag4$$

I'm not sure about it, but I think that the components $$A^\mu(x)$$ are scalar fields

$$A^\mu : M \longrightarrow \mathbb{R} \tag5$$ such that scalar field transformation (2) should be correct. However, I know that the vector field components transform under a Boost as

$$A'^\mu(x') = {\Lambda}^\mu _{\ \nu}A^\nu(x)$$

Can someone explain my conceptual mistake?

• A scalar field is invariant under coordinate transformations. That's a necessary condition, no matter what else you've seen presented as a definition.
– J.G.
Jul 1 at 20:39

No, the components of $$A^\mu$$ are not scalars. For example, $$A^0$$ is not a scalar. Indeed,$$A^{\prime0}=\Lambda^0_\nu A^\nu=\Lambda^0_0A^0+\Lambda^0_iA^i\ne A^0.$$If you fix a vector $$V_\mu$$, $$A^\mu V_\mu$$ is a scalar. But $$A^0$$ isn't of this form. You might think you can just take $$V_\mu=\delta_\mu^0$$, but the RHS doesn't transform as a vector. Indeed, if the components of $$V_\mu$$ satisfy that equation in one coordinate system, they won't in a general alternative.

• Every (classical) scalar field is a smooth function from $M$ to $\mathbb{R}$. Not every smooth function from $M$ to $\mathbb{R}$ is a scalar field. Jul 1 at 19:22
• I don't know if there's much to add because it's just a definition. I.e. "being a scalar field" implies "being a smooth function" but not vice versa. If someone told you the implication goes both ways, he's wrong. Jul 1 at 20:52
• @Lil'Gravity That's 100% wrong! It is likely that he was simplifying for lay audiences or stating it for simplicity with the goal of clarifying later. If neither of these is true, you should stop listening to those lectures. Jul 2 at 11:38
• @PraharMitra If he does, it's in a later lecture, rather than later in lecture 2. The trouble starts here. Unfortunately, Orborne cites both the Higgs boson (a true scalar field) and charge densities (the $J^0$ of a vector field $J^\mu$) as examples of "scalar fields", and that may be why a viewer would expect e.g. $A^0$ to be "a scalar".
– J.G.
Jul 2 at 12:10
• @J.G. I'm trying my best to be helpful for the youtube guy here. Could he have meant scalar w.r.t. spatial rotations? After all, to use the term "scalar" you also need to say which symmetry group you are referring to. $A^0$, $J^0$ are scalars w.r.t. $SO(3)$ but not w.r.t. $SO(1,3)$. On second thought, I'm trying too hard here. I really can't see this being right in any way! Jul 2 at 12:13

I think that the confusion originates from a collision in naming conventions.

A scalar field is a real or complex function defined on a manifold. In this case, the Minkowski space.

Unfortunately, in linear algebra, when a vector space on a field $$K$$ is defined, the elements of $$K$$ are also called scalars. The introduction of a basis implies the possibility of expressing every vector as a linear combination (with scalar coefficients belonging to $$K$$) of the basis vectors.

Such a different meaning of the word scalar in two close concepts explains the possibility of confusion, in particular when there is not enough care in the choice of the words and the different meaning is not emphasized enough.

• It's also a collision in two meanings of vector.
– J.G.
Jul 2 at 5:56
• @J.G. I do not see problems with the vectors. Where there would be a collision of meainig? Jul 2 at 6:22
• "Member of vector space" vs "something that transforms as $A^{\prime\mu^\prime}=\dfrac{\partial x^{\prime\mu^{\prime}}}{\partial x^\mu}A^\mu$".
– J.G.
Jul 2 at 6:25
• @J.G. well, that's not something I would call a different meaning. It is a special case of the definition of vector when additional differential structures are added to the purely algebraic concept of vector (vector fields on a manifold). Jul 2 at 6:30

My favourite example of a vector field is the four-velocity. A four-velocity at a point $$x\in M$$ is the tangent vector of a smooth world line passing through $$M$$ and parametrized with proper time. For example in Sean Carroll's book on GR the proper time of a world line $$\lambda\mapsto x(\lambda)$$ is defined as $$\Delta\tau=\int\sqrt{ -\eta_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}\,d\lambda}\,,~~~~\eta_{\mu\nu}={\rm diag}(-1,1,1,1)\,.$$ Clearly, $$\frac{d\tau}{d\lambda}=\sqrt{-\Big(\frac{dx^0}{d\lambda}\Big)^2+\Big(\frac{dx^1}{d\lambda}\Big)^2+\Big(\frac{dx^2}{d\lambda}\Big)^2+\Big(\frac{dx^3}{d\lambda}\Big)^2}$$ so that the four velocity is $$U^\mu=\frac{dx^\mu}{d\tau}=\frac{dx^\mu}{d\lambda}\frac{d\lambda}{d\tau}\,.$$ So far we considered an observer whose coordinates are $$(x^0,x^1,x^2,x^3)$$. Another observer moving relative to the first observer has coordinates $$(x^{0'},x^{1'},x^{2'},x^{3'})$$. It is clear that $$x^{\mu}\not=x^{\mu'}$$ so the components of $$U^\mu$$ cannot be scalars. The four velocity transforms as $$U^{\mu'}={\Lambda^{\mu'}}_\nu U^\nu$$ and $$U^\mu U_\mu$$ is an invariant because it equals one in every Lorentz frame.