Confusion about vector and scalar field transformation laws

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.