Relativity problem, calculating invariant P2u*Pu2

[lebashad]

Hello

good veiled I have a problem in relativity ,how to show the following relation disintegration is considered one particle a of mass Ma was at rest desintegre in two particles 1 and 2 of mass M1 and M2

convention:

Pij: I in index and J in top
P2u=Pau-P1u and P1u=Pau-P2u
I have of problems to calculate l'invaraint P2u*Pu2

in P2u:2 and U in index ,I makes: P2u*Pu2=(Pau-P1u) * (- P1u+Pau)
I developpe and nothing with the result

thank you for the blow of inch!

 

[Jhero]

Hello,

Thank you for your question. In order to show the relation for the disintegration of one particle into two particles, we can use the conservation of momentum and energy in relativity.

First, let's define the four-momentum for the initial particle as Pau = (Ea/c, Pa), where Ea is the energy and Pa is the momentum. Similarly, for the two final particles, we can define their four-momenta as P1u = (E1/c, P1) and P2u = (E2/c, P2).

Using the conservation of momentum, we know that the initial momentum Pa must be equal to the sum of the final momenta P1 and P2, so we can write the following equation:

Pa = P1 + P2

Similarly, using the conservation of energy, we know that the initial energy Ea must be equal to the sum of the final energies E1 and E2, so we can write the following equation:

Ea = E1 + E2

Now, using the convention you mentioned (Pij), we can rewrite these equations as:

Pau = P1u + P2u
Ea/c = E1/c + E2/c

Next, we can use the relation P2u = Pau - P1u to substitute for Pau in the first equation:

P2u = P1u + P2u - P1u

Simplifying, we get:

P2u = P2u

Therefore, we can see that the relation P2u = Pau - P1u is satisfied.

To calculate the invariant P2u*Pu2, we can use the definition of the invariant in relativity, which is given by:

P2u*Pu2 = (P2u)^2 - (P2)^2

Substituting in our expressions for P2u and Pu2, we get:

P2u*Pu2 = (Pau - P1u)^2 - (P1)^2

Expanding, we get:

P2u*Pu2 = (Pau)^2 - 2Pau*P1u + (P1u)^2 - (P1)^2

Using the conservation equations from earlier, we know that (Pau)^2 = (P1u)^2 = (P1)^2,