I have to show that the following mapping of momenta is surjective. The mapping $\{p_i^{\mu},p_j^{\mu},p_k^{\mu}\}\rightarrow\{\tilde{p}_{ij}^{\mu},\tilde{p}_k^{\mu}\}$ is given by

$$ \tilde{p}_k^{\mu}=\frac{1}{1-y_{ij,k}}p_k^{\mu} $$

$$ \tilde{p}_{ij}^{\mu}=p_i^{\mu}+p_j^{\mu}-\frac{y_{ij,k}}{1-y_{ij,k}}p_k^{\mu}. $$

with $y_{ij,k}=\frac{p_ip_j}{p_ip_j+p_ip_k+p_jp_k}$.

In other words: Do we cover the whole phase space for the tilde quantities under the mapping from ordinary momenta to tilde momenta?

Furthermore momentum conservation and on-shellness is guaranteed.

$$ \tilde{p}_{ij}^{\mu}+\tilde{p}_k^{\mu}=p_i^{\mu}+p_j^{\mu}+p_k^{\mu} $$

$$ p_i^2=p_j^2=p_k^2=\tilde{p}_{ij}^2=\tilde{p}_k^2=0 $$

To me it seems to be a bit strange, since we have 3 momenta (9 degerees of freedom) that we map to 2 momenta (with 6 degrees of freedom).

In addition to that it was mentioned that the mapping is also injective. How is this possible?

I can provide more physical background if required.

  • 1
    $\begingroup$ Well, if you were only looking at Lorentz/rotational invariants, the system of 3 vectors has 3 magnitudes and 3 relative cosines, so 6 degrees of freedom, covering the structure... $\endgroup$ – Cosmas Zachos Dec 7 '18 at 1:56
  • $\begingroup$ @CosmasZachos Thank you for the answer! You mean, we have 6 degrees of freedom on both sides? And can you imagine why the mapping is one to one? $\endgroup$ – Schnarco Dec 7 '18 at 2:11
  • $\begingroup$ It looks that way; there is lots of freedom in picking your origins and orienting your solid (the space like part thereof...). Might take the x-axis to be along $p_k$, and $p_i$ to define the x-y plane, and only $p_j$ has a z component in this orientation... The mapping is 1-to-1 up to superfluous orientation of the system... $\endgroup$ – Cosmas Zachos Dec 7 '18 at 2:17
  • $\begingroup$ @CosmasZachos Could you be so kind and explain to me, why we have also 6 degrees of freedom on the left hand side? $\endgroup$ – Schnarco Dec 7 '18 at 2:21
  • $\begingroup$ I mean, actually it is 2 magnitudes and 2 angles on that side, isn‘t it? $\endgroup$ – Schnarco Dec 7 '18 at 2:22

OK, I am dumping a few illustrative notes to flesh out my comment that the number of parameters (d.o.f.) is 6 before and after the map. (I am not sure what you mean about the 2-vectors being parallel when the 3 vectors are on a plane. $\tilde p_k$ and $\tilde p_{ij}$ are not parallel--see illustration below.)

Given the arbitrariness of your basis choice and orientation, you may as well pick your axes to simplify them, the triangular pyramid found by the spacelike pieces of $p_k,p_i,p_j$, $$p_k=s(1,1,0,0); \qquad p_i= r(1,\cos \theta, \sin \theta, 0); \qquad p_j=u (1,\cos\phi \cos\chi, \sin \phi \cos\chi, \sin\chi). $$ To put all 3 space vectors on a plane, pick $\chi=0$, so the volume of the space pyramid collapses to zero.

Your total momentum is then $$ (s+r+u, s+r\cos\theta + u \cos\phi\cos\chi, r\sin\theta + u \cos\chi \sin \phi,u\sin \chi ), $$ so when all 3 spatial vectors are on a plane, the sum is on a plane.

The scale factors for $p_k$ are $$ \frac{y}{1-y} = \frac {ru(1-\cos\theta \cos\phi\cos\chi -\cos\chi \sin\phi \sin\theta )}{s(r+u-r\cos\theta -u\cos\phi \cos\chi ) }, $$ and $$ \frac{1}{1-y}= \\ ({sr (1-\cos \theta)+su(1-\cos \phi\cos\chi)+ru(1-\cos\theta \cos\phi \cos\chi -\cos\chi\sin\phi \sin\theta )}) \\ /~({s(r+u-r\cos\theta -u\cos\phi \cos\chi ) }). $$ For $\chi=0$ the "sum vector" tilts away from the "original" one , $\tilde p_k$, so I don't see the fear hinted at in your comment. I think you need not do any spherical trigonometry, but I am unfamiliar with the hyperbolic geometry involved.

In any case, you may directly verify the before and after sufficiency of the 6 parameters. Given a null tilded vector doublet, you might, in principle solve for these parameters, and plug them into 3 null untilded vectors... I am not seriously suggesting doing this (solving the 6 equations for the 6 parameters)... just a notional fantasy on your bijection. The extra 3 d.o.f. you might think they'd have, are the three 0 entries, above, and they are just a peculiarity of the coordinate axes' choices. (Failure of surjection is tantamount to failure to solve these 6 equations.)

To handle the strangeness, you might look at $\theta=\chi=\pi/2$, $s=r=u=1$, so the original triplet momenta space parts become the orthogonal axes $\hat x, \hat y, \hat z$, and the equivalent doublet of tilded output light-like vectors $\tilde p _k=\frac{3}{2}(1,\hat x)$, and $\tilde p _{ij}= (\frac {3}{2}, \hat y+\hat z-\hat x /2) $. Aren't they doing the same job?

In any case, if you wanted to hack around with hyperbolic geometry, given your total momentum $P=\tilde P$, distinctly non-null, you have $$ y=1-\frac{2P\cdot p_k}{P^2} =\frac{p_i\cdot p_j}{\tilde p_k\cdot \tilde p_{ij}} ~,\qquad 2P\cdot \tilde p_k = P^2 ,\\ \frac{1}{1-y}= \frac{P^2}{2P\cdot p_k}, \qquad \frac{y}{1-y}=\frac{P^2-2P\cdot p_k}{2P\cdot p_k} ~~. $$

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