# Tensor notation issue

1. May 27, 2011

### dingo_d

1. The problem statement, all variables and given/known data

I'm looking at the wikipedia article about four-momentum and I can't seem to get things right. It says

Calculating the Minkowski norm of the four-momentum gives a Lorentz invariant quantity equal (up to factors of the ''c'') to the square of the particle's proper mass:

$$-||\mathbf{P}||^2 = - P^\mu P_\mu = - \eta_{\mu\nu} P^\mu P^\nu = {E^2 \over c^2} - |\vec p|^2 = m^2c^2$$

where we use the convention that

$$\eta^{\mu\nu} = \begin{pmatrix} -1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}$$

is the reciprocal of the metric tensor of special relativity.

So I get that I can do that straightforward by taking the dot product $$P^\mu P_\mu$$ (I don't know why there is a minus sign, but that's wikipedia after all). But how to calculate it from:

$$\eta_{\mu\nu} P^\mu P^\nu$$?

Should I contract the given $$\eta^{\mu\nu}$$ using $$g^{\mu\nu}$$? ($$eta^{\mu\nu}=g^{\mu \alpha}g^{\nu \beta}\eta_{\alpha \beta}$$)?

And how does that act on $$P^\mu P^\nu$$? Since the indices should stand for certain $$\eta$$ and $$\nu$$ in those tensors, right?

I'm kinda confused as to how did they manage to get the result using that notation :\

2. May 27, 2011

### n1person

My (probably crappy/incorrect) explanation of this is that $$P^\mu$$ is well defined but $$P_\mu$$ is not. In order to work with the later, we need to raise it's index, which is fairly trivial in minkowski space (but in curved spacetimes the metric will be more significant).

As far as how to calculate it, just work with the indices, notice that $$\eta_{\mu \nu}$$ is only non-zero when $$\mu = \nu$$ and is 1 except when $$\mu = \nu = 0$$ where it will be -1. From that I think you get the result.

Last edited: May 27, 2011
3. May 27, 2011

### dingo_d

Isn't a difference in covariant and contravariant 4 - vectors in just - signs for covariant space coordinates?

$$P^\mu=\begin{pmatrix} E\\ p_1\\ p_2\\ p_3\end{pmatrix}$$ and then

$$P_\mu=\begin{pmatrix} E\\ -p_1\\ -p_2\\ -p_3\end{pmatrix}$$? Because then the dot product is straightforward:

$$P^\mu P_\mu=E^2-p_1^2-p_2^2-p_3^2$$.

That part is clear, what is not is how to deal with the $$\eta_{\mu\nu} P^\mu P^\nu$$ part...

4. May 27, 2011

### n1person

The whole changing from covariant to contravariant is like that because the metric is simple. In some crazy metric it won't be that simple.

$$\eta_{\mu \nu} P^{\mu}=P_{\nu}$$

$$\eta_{\mu\nu} P^{\mu} = \begin{pmatrix} -1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} E\\ p_1\\ p_2\\ p_3\end{pmatrix}= - \begin{pmatrix} E\\ -p_1\\ -p_2\\ -p_3\end{pmatrix}$$

5. May 28, 2011

### dingo_d

Yeah, I assume that I work in standard basis (flat Minkowski space, right?).

I think I got it. Since

$$-\eta_{\mu\nu} P^{\mu} =- \begin{pmatrix} -1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} E\\ p_1\\ p_2\\ p_3\end{pmatrix}= \begin{pmatrix} E\\ -p_1\\ -p_2\\ -p_3\end{pmatrix}$$

Then $$-\eta_{\mu\nu} P^\mu P^\nu$$ is just

$$\begin{pmatrix} E\\ -p_1\\ -p_2\\ -p_3\end{pmatrix}\begin{pmatrix} E\\ p_1\\ p_2\\ p_3\end{pmatrix}=E^2-p_1^2-p_2^2-p_3^2$$

right?