Intro special relativity problem regarding time dilation

[RagedFountain25]

Homework Statement

The problem says that a muon with a proper lifetime of 2.2 microseconds is produced 100 km above the ground in the reference frame of Earth. We need to find the minimum speed the muon must travel that allows it to reach the ground in time before the end of its life.

Relevant Equations

time dilation equation: delta t from Earth frame of reference = (delta t from muon frame of reference) / Sqrt[1-(u/c)^2]

This example is worked out in the book, and at the beginning, they make the assumption that the muon is traveling at c, and then find the change in time from the Earth reference frame using delta t=100km/c. Then delta t is plugged into the time dilation equation on the left side and we solve for u. We then find that u is .999978c. My issue is I don't understand how we are able to assume that the muon was traveling at c from the beginning only to find that it was actually traveling at .999978c. Is this just an issue of approximation or is there something to the method I have misunderstood here?

This is my first post, so I'm sorry if I've done anything wrong in posting this question. I've also attached a screenshot of the problem for clarity.

 

[Halc]

Assuming light speed gives approximately the time it takes for the muon to reach Earth relative to the Earth frame. In this case it is close enough, with 5 digits of precision. Divide that time by 2.2 usec and you get a gamma (dilation factor) needed. 333.33usec/2.2 usec is essentially the same (only 2 digits of precision were given for the decay) as 333.34/2.2usec

Your point is noted. The approximation would not work for something moving significantly slower.

 

[anuttarasammyak]

In your idea time consumed for muon to arrive alive is
$$\frac{L}{v}=\frac{t_M}{\sqrt{1-\frac{v^2}{c^2}}}$$
where L=100km,$t_M$=2.2 $\mu$ sec. We get solution v
$$\frac{v}{c}=... (1)$$ I will leave it for your homework.

In the way you quote
$$\frac{L}{c}=\frac{t_M}{\sqrt{1-\frac{v^2}{c^2}}}$$
$$\frac{v}{c}=1-\frac{c^2t_M^2}{L^2}...(2)$$

Exact solution (1) is larger than the approximate one (2) in the order of $(\frac{c t_M}{L})^4=(6.6 \times 10^{-3})^4$.

 

[RagedFountain25]

Assuming light speed gives approximately the time it takes for the muon to reach Earth relative to the Earth frame. In this case it is close enough, with 5 digits of precision. Divide that time by 2.2 usec and you get a gamma (dilation factor) needed. 333.33usec/2.2 usec is essentially the same (only 2 digits of precision were given for the decay) as 333.34/2.2usec

Your point is noted. The approximation would not work for something moving significantly slower.

Ok I understand. So it is just about precision. And of course for something that is moving far slower than c, you couldn't start the problem with that assumption, (which was what threw me off from the beginning). Thank you very much.

 

[Ibix]

You can do it formally correctly - say the muon velocity is $v$ then $\Delta t=100\mathrm{km}/v$ and solve. But how precise to you reckon that 100km figure is? And the 2.2$\mu$s? How much difference do you reckon $100/c$ versus $100/v$ will make to (a) the answer, and (b) the maths?

You should, of course, check that your result is reasonable. Is the $v$ you get actually near $c$? You might, as an exercise, do the same with a distance of 500m and see what $v$ you get using the approximation. Is $v\approx c$ reasonable? Then try using the full method.

Edit: too slow, I see