A proper lifetime of a certain particle is 100.0ns. a) How long does it live in the lab frame if it moves at

V=0.960c? b) How far does it travel in the lab during that time? c) What is the distance travelled in the
lab frame according to an observer moving with this particle.

What does c mean?

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To solve these problems, we need to use the concept of time dilation and length contraction in special relativity.

a) The proper lifetime of a particle is the time it would measure for itself if it were at rest. In this case, the proper lifetime is given as 100.0ns.

The time dilation formula relates the proper time (τ) to the observed time (t) in the lab frame:

t = τ / √(1 - (v^2 / c^2))

where v is the velocity of the particle and c is the speed of light.

Using the given velocity v = 0.960c and substituting it into the formula, we have:

t = 100.0ns / √(1 - (0.960c)^2 / c^2)
t = 100.0ns / √(1 - 0.9216)
t = 100.0ns / √(0.0784)
t = 100.0ns / 0.280

Calculating this, we find that the particle lives approximately 357.1ns in the lab frame.

b) To find the distance traveled in the lab frame, we can use the formula for velocity:

v = d / t

Since we know the velocity v and the time t, we can rearrange the formula to solve for distance traveled (d):

d = v * t

Substituting the values:

d = 0.960c * 357.1ns
d ≈ 0.345c * ns

Therefore, the particle travels approximately 0.345 times the speed of light multiplied by the time interval in the lab frame.

c) According to the observer moving with the particle, the distance traveled is zero. This is because, in the particle's frame of reference, it is at rest, and therefore it does not traverse any distance in its own frame.

So, in summary:
a) The particle lives approximately 357.1ns in the lab frame.
b) The particle travels approximately 0.345 times the speed of light multiplied by the time interval in the lab frame.
c) According to the observer moving with the particle, the distance traveled is zero.