The Oh-My-God particle was an ultra-high-energy cosmic ray (most likely an iron nucleus) detected on the evening of 15 October 1991 over Dugway Proving Ground, Utah, by the University of Utah's Fly's Eye Cosmic Ray Detector. Its observation was a shock to astrophysicists (hence the name), who estimated its energy to be approximately 3×1020 [eV]. a) If the OMG particle came to a stop inside of an apple, by how much would the apple's temperature rise?
b) If the Oh-My-God particle were subjected to a uniform magnetic field of B = 1 [T], what would be the radius of particle's trajectory?
If anyone could give me any hints on how to start this problem in the comments section that would be helpful.© BrainMass Inc. brainmass.com June 21, 2018, 6:33 am ad1c9bdddf
If the OMG particle is stopped in an apple, then in principle we have to calculate the change in kinetic energy of the apple sue to the transferred momentum, the remaining energy is then the change in internal energy of the apple, leading to the temperature change. Now, you can easily see that kinetic energy change of the apple is negligible compared to the total energy transfer, so the answer is just going to involve estimating the heat capacity of the apple and then calculating the temperature rise by taking the total energy dumped in the apple to be 3 10^(20) eV. But we need to demonstrate first that what we think is obvious, i.e. that the apple as a whole won't start to move, is indeed correct.
Let's then consider conservation of momentum in this case. The momentum of the OMG particle is:
P = gamma m v
the energy is:
E = gamma m c^2
Since E >> m c^2, we have that v = approximately c and gamma ...
I give a detailed answer that is accessible to someone with only a basic knowledge of special relativity.