# Relativity: Computing 4-momentum components in a local inertial frame

So I have a massive particle with some velocity $$\vec V$$. In the global frame I assume the 4-momentum is given by $$\mathbf p = ( m\gamma_{\vec v}, m\gamma_{\vec v} \vec V)$$.

Now instead of using the global frame, I want to calculate the 4-momentum components as observed in the local inertial frame of an observer moving with speed $$v$$. The orthonormal basis of this frame is given by:

$$\mathbf e_0 = (\gamma, \gamma v, 0, 0) ,\;\;\; \mathbf e_1 = (\gamma v, \gamma, 0, 0),\;\;\; \mathbf e_2 = (0,0,1,0),\;\;\;\mathbf e_3 = (0,0,0,1)$$

We are told to calculate the dot product of $$p^{0'}= -\mathbf p\;\cdot\;\mathbf e_0$$ to get the 4-momentum time component . And so my first instinct is to straight up just take the dot product of those two vectors.

However, earlier in my class we were told that by definition, the dot product between two four vectors, $$\mathbf a$$ and $$\mathbf b$$, is given by $$\mathbf a \cdot \mathbf b = \eta_{\alpha\beta}a^{\alpha}b^{\beta}$$

I'm confused as to whether or not I should include the $$\eta$$ term and why the time component has a negative in front of the dot product. Does the negative come from already multiplying through with $$\eta$$ or am I missing something?

• There is no frame that you can consider “the global frame”. Commented Mar 26, 2020 at 23:25
• Perhaps not realistically, but is that not how most special relativity classes begin - using global coordinate systems ? Commented Mar 26, 2020 at 23:36
• I didn’t say that there are no global frames in SR. There are. There are an infinite number of them, and none is preferred as the global frame. Commented Mar 26, 2020 at 23:38

Assume you have $$\mathbf p = p^{\alpha'} \mathbf e_\alpha$$. Then $$\mathbf p\cdot\mathbf e_\beta = p^{\alpha'} \mathbf e_\alpha \cdot\mathbf e_\beta = p^{\alpha'}\eta_{\alpha\beta}$$ so $$p^{\alpha'} = \eta^{\alpha\beta}\mathbf p\cdot\mathbf e_\beta$$ In particular $$p^{0'} = \eta^{00} \mathbf p\cdot\mathbf e_0 = -\mathbf p\cdot\mathbf e_0$$
And yes, you should use $$\eta_{\alpha\beta}$$ to calculate scalar product.
• So $p^{0'} = \eta^{00}( \eta_{\alpha\beta} p^{\alpha}e_0^{\beta})$ ? Any intuitive reason as to why we have two $\eta$ terms? Commented Mar 26, 2020 at 23:34
• @ZacharyC look at the general formula $$p^{\alpha'} = \eta^{\alpha\beta}e_\beta^\gamma\eta_{\gamma\delta}p^\delta$$ You can see you need two tensors $\eta$ to put the indices in the right positions. Commented Mar 27, 2020 at 9:14
• @ZacharyC You can also notice that the result should be independent on whether you use (1,3) or (3,1) signature for $\eta$, that is changing $\eta\rightarrow-\eta$ shouldn't affect the result. So you have to have an even number of $\eta$. Commented Mar 27, 2020 at 9:16
• @G.Smith I follow the convention introduced by the OP. As I understand, $p^\alpha$ are the components of $\mathbf p$ in the standard base, and $p^{\alpha'}$ are the components of $\mathbf p$ in base $\mathbf e_\alpha$. Not my choice of notation, but I stick to it. Commented Mar 27, 2020 at 9:17