Question

If we intercept an electron having total energy 1533 MeV that came from Vega, which is 26 ly from us, how far in light-years was the trip in the rest frame of the electron?

Answer 1

You need to work backward from the relativistic momentum-energy formula to find the velocity of the electron in our reference frame, then use that with the Lorentz-Fitzgerald contraction to find the ratio between the distance in the electron's frame and in ours.

Answer 2

Here's the first part:

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/releng.html

Answer 3

From the special theory of relativity \[t' =t/\sqrt{1-v ^{2}/c ^{2}}\]where t is the time in the electron's reference frame and t' is the time in our reference frame and v is the velocity we observe . Also from the special theory we have\[E=mc ^{2}=m _{0}c ^{2}/\sqrt{1-v ^{2}/c ^{2}}\]where \[m _{0}\]is the rest mass of the object .
The rest mass of an electron is .511MeV

Solve for t in terms of v,E and mo.

Answer 4

mc^2=8.187*10^-14 J
(8.187*10^-14)/1533=5.34*10^-17
What do I do after that?

Answer 5

Because energy can be expressed in electron volts and mass can be expressed as energy (E=m0*c^2) you do not have to change to joules and kgms in this problem. You can if you want but you must work with the rest mass in kgm the velocity of light in m/sec and energy in joules.

what do we need to do first. We want the time in the moving reference frame of the electron,t. We have the equation that relates that time to the observer time on earth, t'. we know c. So we must find the velocity of the electron with an total energy of 1533 MeV. Actually we only need to find (v/c)^2. since that quantity appears in the two equations we are using.

Since we have the total energy of the electron (i.e. rest mass energy +KE) E=1533MeV we can find its velocity. with the equation\[E=mc ^{2}=m _{0}c ^{2}/\sqrt{1-(v/c)^{2}}\]
E=mc^2 =.1533MeV
solve for v/c and substitute that into the time equation to find t. remember m0c^2=0.511Mev and t'=27 years.

Answer 6

I get it now. Thanks.