# Zero order Tensor

One of the definition of tensor is that it should be invariant under transformation. Thus a tensor of order zero is a scalar should be invariant under transformation. For a tensor component $$v^{\mu}$$ the inner product $$v^{\mu}v_{\mu}$$ is a tensor of rank zero and should be invariant under Lorentz transformation.But, $$v'^{\mu}v'_{\mu}={\Lambda_{\mu}}^{\alpha}{\Lambda^{\mu}}_{\beta}v^{\beta}v_{\alpha}$$
$${\Lambda_{\mu}}^{\alpha}{\Lambda^{\mu}}_{\beta}$$ not equal to $${\delta_{\beta}}^{\alpha}$$, which implies $$v'^{\mu}v'_{\mu}$$ not equal to $$v^{\mu}v_{\mu}$$. This shows that $$v^{\mu}v_{\mu}$$ is not invariant under transformation(not a tensor). Can anybody suggest where am I wrong?

• I'm sorry. Its typing mistake. – walber97 Feb 26 at 17:54
• I want to know wheather the inner product of two tensors which contract to a zero order tensor is invariant under transformation. – walber97 Feb 26 at 17:57
• Yes, it is. To see how, first you have to be careful about index placement. You cannot just write $\Lambda^\alpha_\mu$ because ${\Lambda^\alpha}_\mu$ is not the same thing as ${\Lambda_\mu}^\alpha$. Second, you need to write down the defining relation for $\Lambda$ to be a Lorentz transformation. – G. Smith Feb 26 at 18:02
• OK. Do you know what relation $\Lambda$ has to satisfy to be a Lorentz transformation? – G. Smith Feb 27 at 3:00
• Right. Work that out in index notation and you should find that ${\Lambda_{\mu}}^{\alpha}{\Lambda^{\mu}}_{\beta}$ *is* equal to ${\delta_{\beta}}^{\alpha}$ because the mixed-index ${\eta^\mu}_\nu$ is ${\delta^\mu}_\nu$. – G. Smith Feb 27 at 3:12