-1
$\begingroup$

I'm following a special relativity course and I'm trying to understand how the invariant mass works. In particular I don't get how the following passages work.

We have a collision between two particles and I get that the invariant mass is the sum squared of the 4-momenta. I also get the second passage, but I don't understand how to go through the last passage, in particular where each piece comes from.

Could you please help me? Thanks! :)

$$M^2 = (p^{\mu}_1 + p^{\mu}_2)^2 = (p^{\mu}_1)^2 + (p^{\mu}_2)^2 + 2 p^{\mu}_1 p_{\mu 2} = m_1^2 + m_2^2 + 2(E_1E_2 - \vec{p}_1 \cdot \vec{p}_2)$$

$\endgroup$

closed as off-topic by Aaron Stevens, GiorgioP, ZeroTheHero, Jon Custer, Kyle Kanos Mar 17 at 11:37

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Aaron Stevens, GiorgioP, ZeroTheHero, Jon Custer, Kyle Kanos
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Do you know what are the components of the 4-momentum $p_\mu$? $\endgroup$ – md2perpe Mar 16 at 10:50
  • $\begingroup$ Yes they are (E/c, p_x, p_y, p_z) $\endgroup$ – Matte Mar 16 at 10:54
  • $\begingroup$ Ok i think i got it, it should be because p²_mu=E²-p²=m² and for the last term from matrix multiplication $\endgroup$ – Matte Mar 16 at 10:57
  • $\begingroup$ So when can we just use p²_mu=m² $\endgroup$ – Matte Mar 16 at 10:58
  • $\begingroup$ Please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead. $\endgroup$ – Kyle Kanos Mar 17 at 11:37
-1
$\begingroup$

In general, if you have two four vectors $A^\mu=(A^0,\vec{A})$, $B^\mu=(B^0,\vec{B})$ then their Minkowski four-product is $$ A^\mu B_\mu = A^0B^0-\vec{A}\cdot\vec{B}. $$

In your case particle 1 you have that $$ (p_1^\mu)^2 \equiv P_1^\mu P_{1\mu}=E_1^2-\vec{p}_1^2 =m_1^2 $$ (I used the shorthand $\vec{p}^2\equiv\vec{p}\cdot\vec{p}$). The same holds for particle 2 so that the first two terms should be ok. The last term is $$ p_1^\mu p_{2\mu} = E_1E_2-\vec{p}_1\cdot\vec{p}_2 $$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.