Special relativity - Maximum mass

[Aleolomorfo]

Homework Statement

Finding the maximum mass $M_x$ which can be made from a collision of identical particles with mass $m$, in the laboratory frame, in which one particle is at rest and the other one has energy $E$. The reaction is the following: $a+b \rightarrow a+b+x$.

The Attempt at a Solution

I assume that the maximum mass is produced when the three resultant particles are at rest in the CM frame. So I use the invariant of the total momentum squared from the lab frame before the collision ($p$) and the CM frame after the collision ($k$):
$$p_1=(E,0,0,\sqrt{E^2-m^2}) \hspace{1cm} p_2=(m,0,0,0)$$
$$(p_1+p_2)^2 = 2m(E+m)$$
$$k_1=k_2=(m,0,0,0) \hspace{1cm} k_x=(M_x,0,0,0)$$
$$(k_1+k_2+k_x)^2=4m^2+4mM_x+m^2_x$$
I equal the invariants:
$$2m(E+m)=4m^2+4mM_x+M^2_x$$
After calculation I find a second order equation in $M_x$ whose solutions are:
$$-2m\pm\sqrt{2m^2+2mE}$$
I reject the solution with - because it is completely negative but also the solution with + is negative under a certain value of E. I do not know if this is because a mistake or it is the threshold energy of the reaction.

 

[PeroK]

Homework Statement

Finding the maximum mass $M_x$ which can be made from a collision of identical particles with mass $m$, in the laboratory frame, in which one particle is at rest and the other one has energy $E$. The reaction is the following: $a+b \rightarrow a+b+x$.

The Attempt at a Solution

I assume that the maximum mass is produced when the three resultant particles are at rest in the CM frame. So I use the invariant of the total momentum squared from the lab frame before the collision ($p$) and the CM frame after the collision ($k$):
$$p_1=(E,0,0,\sqrt{E^2-m^2}) \hspace{1cm} p_2=(m,0,0,0)$$
$$(p_1+p_2)^2 = 2m(E+m)$$
$$k_1=k_2=(m,0,0,0) \hspace{1cm} k_x=(M_x,0,0,0)$$
$$(k_1+k_2+k_x)^2=4m^2+4mM_x+m^2_x$$
I equal the invariants:
$$2m(E+m)=4m^2+4mM_x+M^2_x$$
After calculation I find a second order equation in $M_x$ whose solutions are:
$$-2m\pm\sqrt{2m^2+2mE}$$
I reject the solution with - because it is completely negative but also the solution with + is negative under a certain value of E. I do not know if this is because a mistake or it is the threshold energy of the reaction.

What values of $E$ are you worried about?

 

[vela]

After calculation I find a second order equation in $M_x$ whose solutions are:
$$-2m\pm\sqrt{2m^2+2mE}$$
I reject the solution with - because it is completely negative but also the solution with + is negative under a certain value of E. I do not know if this is because a mistake or it is the threshold energy of the reaction.

Remember $E$ is the energy of a particle of mass $m$.

 

[Aleolomorfo]

Yes, I get it. I've realized the stupidity of my question. $M_x$ is negative if $E < m$, but this is impossible. Thank you.