Recent content by 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!

Hope it is clear enough.

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...

Ah, I think I got it. We can use reverse Lorentz transsform $$ t = \gamma (t' + \frac {vx'} {c^2}) $$ where the prime is the Earth so t' = 0 and x' = 2.9*10^12

Two spaceships are heading towards each other on a collision course. The following facts are all as measured by an observer on Earth: spaceship 1 has speed 0.74c, spaceship 2 has speed 0.62c, spaceship 1 is 60 m in length. Event 1 is a measurement of the position of spaceship 1 and Event 2 is a...

Yes, it would take me a while to see it. It is more elegant no doubt. But nothing wrong with the way I did it right?

I did that, except that I used $$ E_{\mu} = \gamma m_{\mu} c^2 $$for the energy of the muon. With the way you mentioned, I was getting 3 unknowns (the moment of the photon and the muon and the gamma/speed of the muon). Getting stuck at this: $$ E_{\mu} = m_{\pi} - \gamma m_{\mu} v_{\mu} $$ But...

Yes I will re-work my steps to see if I can do it in the correct order

I think I figured it out. I was doing in the rest frame and the exercise asked in the frame of the original pion. I still got the answers in the opposite sequence they asked... I first c) then b).

This is rest of the same exercise. b) Determine the energy (in MeV) of the µ+ as determined in the rest frame of the original π+. You should assume that the µ+ has a mass of 106 MeV/c2 , the π+ has a mass of 140 MeV/c2 and the νµ has zero mass. c) What is the speed of the µ+ as determined in...

I got it now. We can only use that trick when the distance is in light-year (or this case light-second), correct? Once again thank you very much.

Yes, I used this trick (you mentioned it before on another thread). I am just annoyed I keep getting the exact same result. $$ v \gamma \tau = d $$ then $$ v \gamma = \frac d \tau $$ square both sides $$ \frac {v^2} {1-v^2} = \frac {d^2} {\tau^2} $$And so on to give me v = 0.993c I...

This is really stressful... I just can't see it. :)

Can you explain a little more? This should be really simple as far as I can see and I can't spot my mistake. From what I can see I am using the same equations... $$ d = vT, T = \gamma \tau $$ therefore $$ d = v \gamma \tau $$

I did: Which gives me that answer.PS: What should I use to write the equations the way you guys do?