-2

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?

Charles
  • 43
  • No, in general that won't be an invariant. Are you mixing it up with the norm, which is an invariant? If not can you give us some idea of the background to your question? – John Rennie Nov 17 '17 at 10:58
  • Actually there is no background to the question. I just failed it on an exam and I do not understand what I shall have done. Also I'm new on this website. – Charles Nov 17 '17 at 11:01
  • No, but $u^\alpha v_\alpha$ would have been an invariant. – Photon Nov 17 '17 at 11:06
  • Can you explain me why ? – Charles Nov 17 '17 at 11:10
  • Depending on the index position the components transform differently. If you don't want to dive into transformation properties of lower index components, you can also use that $v_\alpha=\eta_{\alpha\beta}v^\beta$ and $\eta_{\alpha\beta}=\eta_{\gamma\delta}\Lambda^\gamma_\alpha\Lambda^\delta_\beta$ – Photon Nov 17 '17 at 11:15
  • I am sorry but I don't know what Λ is supposed to be. – Charles Nov 17 '17 at 11:24
  • The Lorentz transformation matrix (and $\eta$ is the Minkowski metric). – Photon Nov 17 '17 at 11:26

1 Answers1

4

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

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

eranreches
  • 4,199