Muon Decay: Special Relativity WS#1 Homework

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Homework Statement


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1.) A muon is created by a cosmic ray interaction at an altitude of 60km. Imagine that after its creation, the muon hurtles downward at a speed of 0.998, as measure by a ground-based observer. After the muon’s “internal clock” registers 2.0μs , the muon decays?

a.) If the muon’s internal clock were to measure the same time between its birth and death as clock on the ground do (i.e. if special relativity is not true and time is absolute), about how far would this muon have traveled before it decayed?

b.) How far will this muon really travel before it decays?

Homework Equations


300m = 300m(1s/3x10^8m) = 10^-6s = 1μs
∆Sab = √∆t^2ab - ∆x^2ab (?)

The Attempt at a Solution


60km = 200μs

a) If time is absolute:

Muon decays at 600m, or 2μs away from the starting point of 60km, or 200μs (?)

b) 300m = 1μs , thus 2μs = 600m
600m/0.998 = 601.2 m (?)

Anyone have any advice? I'm not looking for the whole solution to be given to me, but I'm at a standstill in terms of my understanding of how the problem works...
 

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Orodruin said:
Do you know what time dilation is?

I only recently learned what time dilation is, so I'm still very unfamiliar with how the concept works within the math aspect of special relativity. What confuses me most is that the first part of this problem asks for an absolute time version, while the second part asks for something different...
 
The first question asks you what would happen if there was not time dilation due to special relativity, so of course it will be different.

The second question asks you what happens when special relativity holds, i.e., when there is time dilation. Note that the muon will only decay when 2 microseconds have passed according to its "clock", not the clock in the Earth rest frame.
 
Orodruin said:
The first question asks you what would happen if there was not time dilation due to special relativity, so of course it will be different.

The second question asks you what happens when special relativity holds, i.e., when there is time dilation. Note that the muon will only decay when 2 microseconds have passed according to its "clock", not the clock in the Earth rest frame.
I'm still unsure about how to approach the first part of the problem, but with the second part:

The muon decays when two microseconds have passed by its clock. It's traveling downwards at a speed of .998, so almost the speed of light. Thus:
2 microseconds = 600 meters
600m/0.998 = 601.2 m

In this case, I'm assuming that dividing by the speed will give me the actual distance that the muon has traveled. Yet something still feels off to me.

Alternatively, should I attempt to plugin the values into the formula ∆Sab = √∆t^2ab - ∆x^2ab?
 
Orodruin said:
No, you are computing it with the time elapsed on an Earth clock. This is not what the question asks for. The muon survives until 2 microseconds has passed in its own rest frame.
Orodruin,

I assume then that I just solved for problem a), which assumes that the muon's clock measures the same as one on the ground.

As for problem b, I have no clue where to start. I feel like I may be missing a formula or equation, or maybe it's one I have not learned yet.

I do appreciate the assistance, I'm just not sure where to go next within this problem.