Scalar invariance under Lorentz-transformation

Question

With $u^\alpha v^\beta$ the components of two vector fields, is

$$u^\alpha v^\alpha=u^1v^1+u^2v^2+u^3v^3+u^4v^4$$

a scalar invariant under a Lorentz-transformation? And why?

Answer 1

First note that your expression of $u^{\alpha}v^{\alpha}$ is wrong. Einstein summation convention tell you that you sum if an index appears twice - once up and once down. Thus, in fact

$$u^{\alpha}v_{\alpha}=u^1v_1+u^2v_2+u^3v_3+u^4v_4$$

The quantity $u^\alpha v^\beta$, on the other hand, is a tensor of rank $(2,0)$, which can be represented by a $4\times 4$ matrix

$$\begin{pmatrix}u^1v^1&u^1v^2&u^1v^3&u^1v^4\\u^2v^1&u^2v^2&u^2v^3&u^2v^4\\u^3v^1&u^3v^2&u^3v^3&u^3v^4\\u^4v^1&u^4v^2&u^4v^3&u^4v^4\end{pmatrix}$$

so by writing $u^\alpha v^\alpha$ you refer to its diagonal elements.

Second, you are confusing upper and lower indices. The location of the index determines its transformation properties. Take for example a vector $v^\alpha$, it transform in the following manner

$$v^{\alpha^\prime}=\frac{\partial x^{\alpha^\prime}}{\partial x^\alpha}v^\alpha$$

On the other hand, a co-vector $u_{\alpha}$ transforms differently

$$u_{\alpha^\prime}=\frac{\partial x^{\alpha}}{\partial x^{\alpha^\prime}}u_{\alpha}$$

In your case, the quantity $u^\alpha v^\beta$ transform like this

$$u^{\alpha^\prime} v^{\beta^\prime}=\frac{\partial x^{\alpha^\prime}}{\partial x^\alpha}\frac{\partial x^{\beta^\prime}}{\partial x^\beta}u^\alpha v^\beta$$

and in particular, the diagonal elements

$$u^{\alpha^\prime} v^{\alpha^\prime}=\frac{\partial x^{\alpha^\prime}}{\partial x^\alpha}\frac{\partial x^{\alpha^\prime}}{\partial x^\beta}u^\alpha v^\beta$$

In general, it is not invariant under the Lorentz transformations. However, the quantity $u^\alpha v_\alpha$ has no free indices since you sum over $\alpha$, and thus it remains invariant under transformations. You can also see it by transforming $u^\alpha$ and $v_\alpha$ separately

$$u^{\alpha^\prime}v_{\alpha^\prime}=\frac{\partial x^{\alpha^\prime}}{\partial x^\beta}u^\beta\frac{\partial x^{\gamma}}{\partial x^{\alpha^\prime}}v_\gamma=\frac{\partial x^{\gamma}}{\partial x^\beta}u^\beta v_\gamma=\delta^{\gamma}_{\beta}u^\beta v_\gamma=u^\beta v_\beta$$