Recent content by Kunhee

It's been solved. Thanks again.

Hi Simon Thanks for the help. I think I am stuck at the expansion of K rel / K cl = (y-1) (m0 c^2) / (1/2)(m0 c^2) Is there a website link or formula that can assist me with the expansion of the above equation in the power of (v/c)^2? *and would it be okay for me to assume Kcl is (1/2)( m0...

Homework Statement [/B] An electron e- and positron e+ annihilate to produce two photons. a_ Why are two photons produced rather than one? b_ Assume that the e- and e+ are at rest just before they annihilate. In their rest frame, what are the energies and momenta of the photons? Define the +x...

Homework Statement [/B] By expanding Krel / Kcl in powers of (v/c)^2, estimate the value of v/c for which Krel differs from Kcl by 10%. Homework Equations Kcl = classical Kinetic Energy = 1/2 m0 v^2 Krel = relative Kinetic Energy = (y-1) (m0 c)^2 The Attempt at a Solution I did a binomial...

u' = [u - (-u)] / [1 - (u)(-u)/c^2] = 2u / [1+(u/c)^2 Plugging the u' into Lorentz factor for the Energy equation: E = y m0 c^2 = m0 c^2 / [1 - ( [2u / [1+(u/c)^2] / c )^2] ^ -1/2 And with further simplification I get m0 c^2 / (1/3) = 3 Got it. Thanks!

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I see. Thank you!

Does that mean I need to first do velocity transformation u = (u' + v) / (1 + u'v/c^2) And use the resulting velocity to plug into E = y m0 c^2 ?

Oh I see. Could you help me set up the equation in the particle's frame of reference? Before I can plug into E = y m0 c^2 I need to find the m0 when the Lorentz factor is applied first because we are observing it from the frame of the particle?

When (v/c)^2 = .5 then the equation is just that E = (1.414)(m0c^2). Where did the 3 come from?

Homework Statement [/B] Two particles of rest mass m0 approach each other with equal and opposite velocity v, in a laboratory frame. What is the total energy of one particle as measured in the rest frame of the other? But the question gives a clue which reads "if (v/c)^2 = .5, then E =...

Thanks for the responses. It was a quiz question for a Sp. Relativity class. I guess the answer is simply that the moving clock's time is dilated within the S frame by ((1-(u/c)^2)^(-1/2).

S' is moving with speed v relative to S. (x' direction) And a clock is moving with speed u in the S frame. (x-direction). If time t elapsed in the S frame, how much time elapsed for the moving clock in the S frame? t' = t / y or is it just t? The question didn't specify anything other than...

A Moving Clock runs slow. But, If time t has elapsed in the S frame, does SR apply to a clock moving with speed u in the x-direction in the S frame, relative to the S frame? Or does SR apply only when the clock is in another reference frame S' moving in the x'-prime direction, given that...

I see, thank you!