Relativistic Energy

lylos

1. The problem statement, all variables and given/known data
An Omega- particle has rest energy 1672 MeV and mean lifetime 8.2X10-11 s. It is created and decays in a particle track detector and leaves a track 24mm long. What is the total energy of the Omega- particle?


2. Relevant equations
E=E0/Sqrt(1-v^2/c^2)
E0=1672 MeV
Lorrentz Equations


3. The attempt at a solution
I took the distance and divided by time and found the V would be .967c. Once that was used to then find E. The answer was off. The correct answer was 2330 MeV. I don't know how they got that. My answer was 7678 MeV.

meopemuk

Did you take into account that a moving particle lives longer due to Einstein's time dilation?

Eugene.

lylos

Please ignore.

lylos

I thought that to be the case. Considering the life of the particle is given as 8.2X10^-11 s, I figured that would be the time the particle sees. The 24mm would be the distance it travels that the observer sees. Now connecting these two is where I had issues. Should I assume that the 8.2x10^-11 s is the life of the particle that we observe?

meopemuk

Normally, "mean lifetime" is supposed to be in the particle's rest frame. You were asked to find particle's energy with respect to the observer. The track length is also measured in the observer's frame. Therefore, you should also use the (dilated) lifetime in the same (observer's) frame of reference.

Eugene.

dynamicsolo

Something doesn't seem right about one of the numbers in your problem statement. The light-travel time for 24 mm is 8.01x10^-11 sec. If the particle appears time-dilated in the lab frame, shouldn't the light-travel time for its track length be longer than the rest lifetime of the particle? [The lifetime you quote for the omega-minus appears correct; are you sure about the track length?]

meopemuk

I don't see any contradiction here. You, basically, need to solve a system of two equations. Denoting d=24mm the traveled distance, [itex]t_0 [/itex]= 8.01x10^-11 sec the lifetime at rest, and

[tex] t = t_0 (1 - v^2/c^2)^{-1/2} [/tex]

the lifetime of the moving particle, we can find the velocity from equation

[tex] d = vt [/tex]

This velocity should be lower than the speed of light, of course.

Eugene.

dynamicsolo

Isn't [tex]t_0[/tex] the rest lifetime of the particle? It's the observed time in the lab-frame that should be longer ([tex]\gamma[/tex] > 1). The difficulty is that, in the expression [tex] t = t_0 (1 - v^2/c^2)^{-1/2} [/tex], it is [tex]t_0[/tex] that equals 82 picoseconds, while the lab-frame time is about 80 picoseconds.

The muon illustration is representative. Its rest lifetime is 2.2 microseconds; if it were to travel at the speed of light, it would decay in a distance of 0.66 km. The fact that muons from high-altitude cosmic ray showers are detected on the ground is taken to indicate the effect of time dilation, in which [tex]\gamma[/tex] is in the range of at least 50 to 100, giving it an Earth-frame lifetime sufficient to travel some tens of kilometers, in accordance with d ~ ct. In the rest frame of the muon, it is the distance from its point of creation to the ground that appears contracted by a factor of at least 50 to 100, so it only needs to cover a fraction of a kilometer to the apparently approaching ground, again in accordance with d ~ ct.

If the omega were observed in the lab frame to have a lifetime of 82 picoseconds and covers 24 mm in that time, then its lab-frame velocity would be 0.9763 c, giving
[tex]\gamma[/tex] = 4.617 . But isn't it the moving omega which has the "dilated" lifetime? We would then find a rest lifetime of 82/4.617 = 17.8 picoseconds. If it's the rest lifetime that is "dilated", we should see a lifetime of (4.617)·82 = 379 picoseconds for the lab-frame lifetime and the track would be
(379 picoseconds)(0.9763)(2.998x10^10 cm/sec) = 11.1 cm.

I think there is something in the problem statement that is not right. (The lifetime issue would be fine if this were the charmed-omega, but the rest energy is wrong.)


(After some further thought...) I finally decided to break down and "reverse engineer" the given answer. When you find gamma from the ratio total energy/rest energy, solve for v, time-dilate the rest lifetime, and now calculate the track length using

d = v · ( [tex]\gamma[/tex][tex]t_0[/tex] ) , you get 23.87 mm.

SO, the issue appears to be that the precision of the track length is not given at a sufficiently high level to resolve the needed values accurately. The observed track length is too close to d = c[tex]t_0[/tex] to get a clear result at low precision.

To sum up, [tex]t_0[/tex] in the dilation equation is the rest lifetime and the method meopemuk describes is otherwise correct in principle. However, the measurement given for the track length needs to be given to at least three significant figures (four would be better) in order to find v with enough precision to solve the problem.

meopemuk

Hi dynamicsolo,

If I solve my two equations (with two unknowns v and t) with respect to v, I obtain

[tex] v = \frac{d}{\sqrt{t_0^2 + d^2/c^2}} [/tex]

Upon substitution of numerical values, I got [itex] v \approx 2 \cdot 10^{8} [/itex] m/s which looks reasonable.

Eugene.

dynamicsolo

... presumably with a square root in the denominator. I get 0.6967 c for the velocity from the analysis I made, so I agree with your velocity (which I get, with the formula correction indicated).

I think I realize the other mistake that was being made from the beginning. The track length in the lab frame is d = [tex]\beta\gamma[/tex]·c·[tex]t_0[/tex]. [This is the observed velocity in the lab frame, u = (beta)·c , times the observed lifetime in the lab frame, which is the time-dilated lifetime (gamma)·t0.] This gives

[tex]\beta\gamma[/tex] = 2.4 cm / (82 picoseconds)(2.998x10^10 cm/sec)

= 2.4 / 2.458 = 0.9763 .

So, lylos, the 0.9763 we saw back at the start of this thread is not (beta) = v/c , but [tex]\beta\gamma[/tex] , which must be solved for v. (I think particle physicists have a name for this quantity, but I forget what that is.)

I still hold to my earlier posts. The track length must be given to enough precision to determine [tex]\beta\gamma[/tex] accurately.

meopemuk

Yes, of course. I corrected my post. Thanks.

Eugene.