GREPhysics.NET: GR0177 #99

GR0177 #99

Problem

GREPhysics.NET Official Solution

This problem is still being typed.

Special Relativity $\Rightarrow$ }Conservation of Energy

The tricky part of this problem is to note that the momentum of the photon is shared equally between all three final particles. (This insight was supplied by Felipe Birk). Thus, $p_e = p/3$ , where p is the momentum of the photon and $p_e$ is the momentum of the final electron or photon.

The initial energy before the photon strikes the electron is $E_0=pc+m_e c^2$ , which is just the energy of the photon plus the rest energy of the electron.

The final energy after the collision is $E_f = 3 \sqrt{(pc/3)^2+(m_e c^2)}$ , which is the sum total energy of all three final particles, i.e., the positron and two electrons. (A positron is the electron's antiparticle, and thus they have the same mass.) Note that the momentum split relation mentioned above is used to equate the final particle momentum with the initial photon momentum.

Conjure up the good conservation of energy idea. Equating $E_0=E_f$ , one gets $(pc)^2+(m_ec^2)^2 + 2pcm_ec^2 = 9((pc/3)^2+(m_ec^2)^2)$ . Canceling the $(pc)^2$ terms on both side, then solving, one arrives at $pc=4m_ec^2$ , which is choice (D).

Alternate Solutions

opalqnka 2014-10-15 01:29:53	You all post nasty solutions. Here is an easy one: The 4-momentum of the photon in the lab frame and the 4-momentum of the stationary electron are: $P_{1}$ = $\left( E/c, E/c, 0, 0 \right)$ and $P_{2}$ = $\left( mc, 0, 0, 0 \right)$ After the interaction 3 particles of equal masses and equal energies and momenta are created with the same 4-momentum vector of $P_{3}$ . We do not care about the form of the $P_{3}$ momentum, you will see why below: Use conservation of 4-momenta: $P_{1}$ + $P_{2}$ = 3 $P_{3}$ Now square (using the dot product with [-, +, +, +] metric) and use the fact that for a photon P.P = 0 and in the rest frame of the electrons/protons $P_{2}$ . $P_{2}$ = $P_{3}$ . $P_{3}$ = - $m^2$ $c^2$ , ie we exploit the fact that P.P is invariant! => $P_{1}$ . $P_{1}$ + 2 $P_{1}$ . $P_{2}$ + $P_{2}$ . $P_{2}$ = 9 $P_{3}$ . $P_{3}$ or 0 - 2Em - $m^2$ $c^2$ = -9 $m^2$ $c^2$ or E = 4m $c^2$ as in answer (D). I hope you will find this approach useful. Thank you for hosting the site! Reply to this comment
ArslanUyghur 2014-09-25 03:50:08	No four vectors and any frames, you can use just Energy–momentum relation and Energy and momentum conservation law: $E^2=p^2+m^2$ (basic equation) for the photon we have: $E=p$ (remember its mass is zero) for the electron before collision $E = m$ (here p=0) so the total energy of the system before collision is: $E = p +m$ or $E^2=p^2+2pm+m^2$ Now let's see what is the energy of the system after collision, using Energy–momentum relation we get: $E^2=p^2+9m^2$ where p is the same momentum of the photon or total linear momentum of three particles(since momentum is conserved). Equating the two expressions for square of energy, we obtain the following: $p^2+2pm+m^2$ =p^2+9m^2$ The solution is: p=4m or in SI units: p=4mc and energy of the photon is 4mc^2. The answer is D. Reply to this comment
vadim 2012-11-08 12:52:15	Shorter solution using relativistic invariant: $E^2-p^2=inv$ Comparing this value in initial laboratory frame and in final frame of particles one can obtain: $(E+m)^2-p^2=(3m)^2$ As far as photon is concerned, $E=p$ , which brings us to $E=4m$ , or in SI $E=4mc^2$ Reply to this comment
insertphyspun 2011-06-26 12:13:41	I have not yet seen the simplest solution, although dryu and matonski came close. Yes, you should use four vectors, but recall that the square of the momentum four vector is invariant! So, jump into the CM reference frame when determining the momentum four vector after the collision to make life simpler! To be more explicit, here it goes: The momentum four vector before the collision, where $E_{\nu}$ is the energy of the incoming photon and $mc^{2}$ is the rest energy of the electron, is $p_{\mu,i}=\left ( \frac{E_{\nu}+mc^{2}}{c},p_{\nu},0,0 \right )$ $p_{\mu,i}p^{\mu,i}=\frac{E_{\nu}^{2}+2E_{\nu}mc^{2}+m^{2}c^{4}}{c^{2}}-p_{\nu}^{2}=\frac{2E_{\nu}mc^{2}+m^{2}c^{4}}{c^{2}}$ (Using $p_{\nu}^{2}=E_{\nu}^{2}/c^{2}$ should be second nature by now!) The momentum four vector after the collision, taken in the CM frame of the three particles, is awesomely simple, $p_{\mu,f}=\left ( \frac{3mc^{2}}{c},0,0,0 \right )$ $p_{\mu,f}p^{\mu,f}=\frac{9m^{2}c^{4}}{c^{2}}$ The squared four vectors are equal, so put them together and solve for $E_{\nu}$ ! $2E_{\nu}mc^{2}+m^{2}c^{4}=9m^{2}c^{4}$ $E_{\nu}=4mc^{2}$ Reply to this comment
dryu 2008-10-17 22:32:47	It's simplest to use four vectors. In natural units: The initial four-momentum is $(E+m,E,0,0)$ and the final four-momentum is $3*(\gamma m, \gamma m v)$ . The invariant mass, e.g. these four vectors squared, is conserved: $(E+m)^2-E^2=9(\gamma^2m^2-\gamma^2m^2v^2)$ $2mE+m^2=9 m^2$ $E=4m$ To convert this into SI units, just let $m\rightarrow mc^2$ . Reply to this comment

Comments

opalqnka
2014-10-15 01:29:53

You all post nasty solutions.
Here is an easy one:

The 4-momentum of the photon in the lab frame and the 4-momentum of the stationary electron are:

$P_{1}$ = $\left( E/c, E/c, 0, 0 \right)$ and $P_{2}$ = $\left( mc, 0, 0, 0 \right)$

After the interaction 3 particles of equal masses and equal energies and momenta are created with the same 4-momentum vector of $P_{3}$ . We do not care about the form of the $P_{3}$ momentum, you will see why below:

Use conservation of 4-momenta:

$P_{1}$ + $P_{2}$ = 3 $P_{3}$

Now square (using the dot product with [-, +, +, +] metric) and use the fact that for a photon P.P = 0 and in the rest frame of the electrons/protons $P_{2}$ . $P_{2}$ = $P_{3}$ . $P_{3}$ = - $m^2$ $c^2$ , ie we exploit the fact that P.P is invariant!

=>

$P_{1}$ . $P_{1}$ + 2 $P_{1}$ . $P_{2}$ + $P_{2}$ . $P_{2}$ = 9 $P_{3}$ . $P_{3}$

or

0 - 2Em - $m^2$ $c^2$ = -9 $m^2$ $c^2$

or E = 4m $c^2$ as in answer (D).

I hope you will find this approach useful. Thank you for hosting the site!

shka
2018-07-09 17:59:48

*reads Yohni Kahn once* *becomes Einstein*

shka
2018-07-09 18:02:05

nice typo btw

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