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!
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?
Homework Statement
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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...