Relativistic Distance/Time Problem

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Homework Statement



A muon has a lifetime of 2.20 x10-6 s when at rest, after which time it decays into other particles.

a) Ignore any effects of relativity discussed in this section. If the muon was moving at 0.99c, how far would it travel before decaying into other particles, according to Newtonian mechanics?

b) How long would the muon last, according to an observer in the earth’s frame of reference who viewed the muon moving at 0.99c?

c) How far would the muon actually travel, when viewed moving at 0.99c?

d) Compare the two distances travelled. Explain why this type of evidence is excellent support for the theory of relativity.


Homework Equations


Δtm = Δts / √1- v2/c2


The Attempt at a Solution




a) With Newtonian mechanics, there would be no time dilation so:

0.99c(3.00x108)

= 2.97 x108m/s

(2.97 x108m/s) (2.20 x10-6 s)

= 653.4m
According to Newtonian mechanics, the muon would travel 653.4m.


b)

Δtm = Δts / √1- v2/c2

Δtm = 2.2 x10-6 / √1- 0.99c2 /c2

Δtm = 2.2 x10-6 / √1 - 0.9801

Δtm = 1.56 x10-5s

According to an observer on earth, the muon will last for 1.56 x10-5s.


c)
I am not sure how to calculate distance with respect to relativity, we've only learned equations for time dilation, mass and length contraction. Is there a specific formula for distance that I don't know about?

Also, I tried to calculate a with Newtonian mechanics but I'm not completely sure if that is what the question is asking for.

Any help on this is greatly appreciated, I know I must be missing something but just can't think of it; this question has been driving me crazy!





 

Answers and Replies

  • #2
PeroK
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What you're missing is the formula:

##s = vt##

This relates displacement, velocity and time. For constant velocity motion, of course.
 
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  • #3
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What you're missing is the formula:

##s = vt##

This relates displacement, velocity and time. For constant velocity motion, of course.
What you're missing is the formula:

##s = vt##

This relates displacement, velocity and time. For constant velocity motion, of course.
That is what I thought I used for a) :

s = vt

v = 0.99c(3.00x108)

v = 2.97 x108m/s

s = (2.97 x108m/s) (2.20 x10-6 s)

= 653.4m

Is that the same formula to be used for question c)? I assumed that it would be a different equation for the two questions?
 
  • #4
jbriggs444
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s that the same formula to be used for question c)? I assumed that it would be a different equation for the two questions?
In part c, one is applying the principles of special relativity. How long does a moving muon last according to your rest frame?
 
  • #5
PeroK
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That is what I thought I used for a) :

s = vt

v = 0.99c(3.00x108)

v = 2.97 x108m/s

s = (2.97 x108m/s) (2.20 x10-6 s)

= 653.4m

Is that the same formula to be used for question c)? I assumed that it would be a different equation for the two questions?
##s = vt## is the defining relationship between displacement, velocity and time. These quantities must be as measured in a single frame of reference.

How else would you define velocity, other than by a measured displacement in a measured time?

SR tells you that the elapsed time - e.g. for the lifetime of a muon - may be different in different reference frames. You cannot directly use a time interval measured in another reference frame for a calculation in your reference frame.

In Newtonian physics there is only one elapsed time for all reference frames. So, you don't need to worry about converting time intervals from one frame to another.
 

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