Muons are elementary particles that have a very short lifetime of about two millionth of a second (when at rest). They can be created in the upper atmosphere (at an altitude of say 10,000 meters) by cosmic rays, in which case they travel towards the surface of the Earth at very high speeds, say 99.9% of the speed of light. If we ignored special relativity, how far could they travel at this speed before they decay? You'll notice that even though they are very fast this distance is much less than 10,000 meters, and they would never reach the ground. This is not correct however, because we know that many of these particles do reach the ground, and experiments observe them all the time. How does special relativity resolve this apparent contradiction, and how far can they actually travel before the end of their life?
d = 300*10^6m/s * 2*10^-6s = 600m befor decaying.
To understand how special relativity resolves this apparent contradiction, let's first calculate the distance a muon can travel before it decays, ignoring special relativity.
Given:
- Muons have a lifetime of about two millionth of a second when at rest.
- Muons are created at an altitude of 10,000 meters and travel towards the Earth's surface at 99.9% of the speed of light.
Ignoring special relativity, we can calculate the distance a muon can travel as follows:
- Speed of light, c = 299,792,458 meters per second.
- Time dilation factor, γ = 1, since we are ignoring special relativity.
- Time taken for a muon to decay, t_decay = 2 microseconds = 2 x 10^(-6) seconds.
- Distance traveled by the muon before decay (ignoring special relativity), d = speed x time.
d = (0.999c) x t_decay
= (0.999 x 299,792,458 m/s) x (2 x 10^(-6) s)
≈ 599.58 meters.
According to this calculation, a muon would only be able to travel about 599.58 meters before decaying. This is much less than the 10,000-meter altitude from which it was created. Thus, if we only consider classical physics, most muons would never reach the ground.
However, special relativity provides a solution to this paradox by considering the effects of time dilation. According to special relativity, when objects move at relativistic speeds, time appears to slow down for them relative to a stationary observer. This means that the muon's perceived lifetime is longer from the stationary observer's perspective.
The time dilation factor, γ, can be calculated using the formula:
γ = 1 / √(1 - v^2/c^2)
where v is the velocity of the muon and c is the speed of light.
For v = 0.999c,
γ = 1 / √(1 - (0.999c)^2/(c^2))
= 1 / √(1 - 0.999^2)
≈ 7.09.
Now, considering time dilation, we need to calculate the distance the muon can travel before decaying in its own frame of reference and then transform it back to the stationary observer's frame.
In the muon's frame, the time taken for it to decay, t_muon, can be calculated by dividing the rest lifetime by the time dilation factor:
t_muon = t_decay / γ
= (2 x 10^(-6) s) / 7.09
≈ 2.82 x 10^(-7) seconds.
Using the muon's frame of reference, the distance it can travel before decay is:
d_muon = (0.999c) x t_muon
= (0.999 x 299,792,458 m/s) x (2.82 x 10^(-7) s)
≈ 84.72 meters.
From the stationary observer's perspective, the distance traveled by the muon is given by:
d_observer = γ x d_muon
≈ 7.09 x 84.72 meters
≈ 599.58 meters.
Interestingly, this calculation matches the previous result obtained by ignoring special relativity. However, the crucial difference is that the muon's perceived lifetime is longer due to time dilation. This allows the muon to travel a greater distance, as observed from the stationary observer's frame, before it decays.
So, special relativity resolves the apparent contradiction by taking into account the time dilation effect. It explains that, from the perspective of an observer on the ground, the relativistic muons can travel much farther than expected based on their short rest lifetime. This is why many muons created in the upper atmosphere by cosmic rays indeed reach the ground, as experiments have observed.