Invariant mass and energy balance

[R3ap3r42]

Homework Statement

This is a question from a Special Relativity course. Uses invariant mass, Laurentz Transformation

Relevant Equations

Invariant mass, Lorentz Transformation, Conservation of Energy, Conservation of Momentum

a) Two particles have energies E1 and E2, and momenta p1 and p2. Write down an expression for the invariant mass of this two-particle system. Leave your answer in terms of E1 and E2, and p1 and p2.

b) A typical photon (γ) in the Cosmic Microwave Background (CMB) has an energy of kBTCMB, where TCMB = 2.73 K and kB is the Boltzmann constant. Such a photon can collide with a high-energy proton via the reaction p + γ → ∆+, where the ∆+ particle has a mass of 1.23 GeV/c2 .

i) Calculate the energy of the CMB photon in eV. [2 marks]

ii) If the proton and photon collide head-on, show that their invariant mass, $$ M_{inv} $$, satisfies

$$ M_{inv}^2 c^4 = m_p^2 c^4 + 2k_B T_{CMB} (E + cp) $$

where E, p and mp are the proton’s energy, momentum and mass. [4 marks] Hence show that the proton energy can be written

$$ E= \frac {m_∆^2 c^4−m_p^2 c^4} {4k_BT_{CMB}}+δE $$ and determine δE in terms of the particle masses and the photon energy.

iii) Compute the numerical values of δE and E in eV. [2 marks] iv) How would your expression for the proton energy change if the photon and proton collided at right angles?I got all the way to b) ii but I could not get to the given expression for E.
Can anyone point give me any clues? I am pretty sure it's just algebraic work that I can seem to simplify.

I manage to get to this:

$$ E= \frac {m_∆^2 c^4−m_p^2 c^4} {2k_BT_{CMB}} - cp $$

This seems close but I can't get rid of the cp (also note that mine is divided by 2 not 4 as expected).

Thanks a lot.

 

[Orodruin]

Homework Statement:: This is a question from a Special Relativity course. Uses invariant mass, Laurentz Transformation
Relevant Equations:: Invariant mass, Lorentz Transformation, Conservation of Energy, Conservation of Momentum

I got all the way to b) ii but I could not get to the given expression for E.

Show your work please.

 

[R3ap3r42]

Hope it is clear enough.

 

[PeroK]

I manage to get to this:

$$ E= \frac {m_∆^2 c^4−m_p^2 c^4} {2k_BT_{CMB}} - cp $$

This seems close but I can't get rid of the cp (also note that mine is divided by 2 not 4 as expected).

Thanks a lot.

You're missing the same trick I showed you yesterday. You have:
$$E + pc = X \ \ (1)$$where$$X = \frac {m_∆^2 c^4−m_p^2 c^4} {2k_BT_{CMB}}$$Now$$m_p^2c^4 = E^2 - p^2c^2 = (E-pc)(E+pc) = (E-pc)X$$$$\Rightarrow \ E - pc = \frac{m_p^2c^4}{X} \ \ (2)$$Now, add equations (1) and (2).

 

[R3ap3r42]

Wow! This is unbelievable. I was blind and now I see. :)
I guess a need to do 100 more of these.

Thanks a lot!