Special Relativity- Muons in a Storage Ring

[Vickitty]

Another question, in which I believe I've gotten the same wrong answer two different ways now.

Muons have a mass m = 105 MeV/c^2. They are accelerated to a kinetic energy of 2 TeV in a storage ring with radius r = 2 km. A student speculates that since muons have a lifetime of only T = 2x10^-6 s, they can only go at most cT = 3x10^8m/s*2x10^-6s = 600 m, which means they can't even make a single loop around the storage ring. Is the student right? Calculate the number of loops that the muon can actually make, because of time dilation, before you calculate the total distance the muon can complete.

I know the student is wrong, because of the special relativity theories. I keep getting stuck at trying to figure out the velocity, though. I've been using the formula Ekin = Etotal - Erest, where Etotal = (mc^2)/(1 - v^2/c^2)^1/2, and Erest = mc^2, and then solving for v, but I keep getting 195.16 m/s, which can't be right.

Thank you for any help!

 

[turin]

E_kinetic = E_total - E_rest = γ E_rest - E_rest = ( γ - 1 ) E_rest = ( γ - 1 ) m c²

=>

γ = 1 + { E_kinetic / ( m c² ) } = 1 / √{ 1 - β² }

=>

β = √{ 1 - ( 1 - ( E_kinetic / ( m c² ) )² ) }

=>

v = √{ 1 - ( 1 - ( E_kinetic / ( m c² ) )² ) } c ~ c.

You should calculate the time dilation from the γ in the first step.

 

[M98Ranger]

The student's reasoning is incorrect because they are not taking into account the effects of time dilation on the muons. According to special relativity, as an object approaches the speed of light, time slows down for that object as observed by an outside observer. This means that the muon's lifetime will also appear longer to the observer in the storage ring due to its high speed.

To calculate the number of loops the muon can make, we need to first find its velocity. Using the formula you mentioned, we can rearrange it to solve for v:

v = c * √(1 - (Erest/Etotal)^2)

Plugging in the values given, we get:

v = c * √(1 - (105 MeV/c^2)/(2 TeV)^2) = 0.99999999998c

This means that the muon is traveling at a speed very close to the speed of light, and its time dilation factor is:

γ = 1/√(1 - v^2/c^2) = 223.60

This means that for every second that passes for the observer in the storage ring, 223.60 seconds will pass for the muon. So, in its 2x10^-6 s lifetime, the muon will actually experience:

T' = T * γ = (2x10^-6 s) * 223.60 = 0.0004472 s

This is the time that the muon will experience, and during this time it will travel a distance of:

d = vt' = (0.99999999998c)(0.0004472 s) = 447.20 m

This means that the muon can actually make multiple loops around the storage ring before decaying. To find the number of loops, we can divide the total distance (2 km) by the distance traveled in one loop (447.20 m):

N = (2 km)/(447.20 m) = 4463.8 loops

So, the muon can actually make over 4000 loops around the storage ring before decaying due to the effects of time dilation. This shows how special relativity plays a crucial role in understanding the behavior of particles at high speeds.