Special Relativity Muon problem

Diracobama2181 · Mar 30, 2022

i) The muon reaches the ground

ii)
To a ground observer, the decay time is dilated
$$\Delta t_d=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\Delta\tau_d>\Delta \tau_d$$

The time for the muon to reach the ground is

$$\Delta t_g=\frac{10 km}{0.999c}< \Delta t_d$$

which is why it reaches the ground.

iii)
From the muon's frame, the ground comes toward it with a velocity v=0.999c[\itex], and hence the distance to the ground is length contracted by $$\Delta l'=\sqrt{1-\frac{v^2}{c^2}}10 km$$ so $$\frac{\Delta l'}{0.999c}=\frac{\sqrt{1-\frac{v^2}{c^2}}10 km}{0.999c}<2\times 10^{-6} s $$ So the muon would reach the ground. Does this sufficiently answer the above questions? Any feedback would be greatly appreciated. Thanks.

PeroK · Mar 31, 2022

I would like to see the actual numbers to support the conclusions.

Diracobama2181 · Mar 31, 2022

Diracobama2181 said:

Homework Statement:: The height of the Earth's atmosphere from top to ground is 10 km. Due to collisions of cosmic rays with the Earth's atmosphere, muons are produced at the top of the atmosphere
at a velocity of 0.999c where the speed of light is c=3.00*10^8 m/s[\tex]. At rest a muon decays in 2*10^-6[\tex] sec. i) Does the muon reach the ground? If it does reach the ground ii) How would an observer on the ground explain it? iii) How would an observer moving with the muon explain it? Relevant Equations:: $$\Delta t=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\Delta\tau$$ $$\Delta l&#039;=\sqrt{1-\frac{v^2}{c^2}}\Delta l$$ i) The muon reaches the ground ii) To a ground observer, the decay time is dilated $$\Delta t_d=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\Delta\tau_d&gt;\Delta \tau_d$$ The time for the muon to reach the ground is $$\Delta t_g=\frac{10 km}{0.999c}&lt; \Delta t_d$$ which is why it reaches the ground. iii) From the muon&#039;s frame, the ground comes toward it with a velocity v=0.999c[\itex], and hence the distance to the ground is length contracted&lt;br /&gt; by&lt;br /&gt; $$\Delta l&amp;#039;=\sqrt{1-\frac{v^2}{c^2}}10 km$$&lt;br /&gt; &lt;br /&gt; so&lt;br /&gt; $$\frac{\Delta l&amp;#039;}{0.999c}=\frac{\sqrt{1-\frac{v^2}{c^2}}10 km}{0.999c}&amp;lt;2\times 10^{-6} s $$&lt;br /&gt; &lt;br /&gt; So the muon would reach the ground.&lt;br /&gt; &lt;br /&gt; Does this sufficiently answer the above questions? Any feedback would be greatly appreciated. Thanks.&lt;br /&gt;

&lt;br /&gt; &lt;br /&gt; &lt;blockquote data-attributes=&quot;member: 493650&quot; data-quote=&quot;PeroK&quot; data-source=&quot;post: 6616605&quot; class=&quot;bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch&quot;&gt; &lt;div class=&quot;bbCodeBlock-title&quot;&gt; PeroK said: &lt;/div&gt; &lt;div class=&quot;bbCodeBlock-content&quot;&gt; &lt;div class=&quot;bbCodeBlock-expandContent js-expandContent &quot;&gt; I would like to see the actual numbers to support the conclusions. &lt;/div&gt; &lt;/div&gt; &lt;/blockquote&gt;i) The muon reaches the ground&lt;br /&gt; &lt;br /&gt; ii)&lt;br /&gt; To a ground observer, the decay time is dilated&lt;br /&gt; $$\Delta t_d=\frac{1}{\sqrt{1-\frac{0.999c^2}{c^2}}}\Delta\tau_d=22.4 \tau_d=4.5 *10^{-5}s&amp;gt;\Delta \tau_d$$&lt;br /&gt; &lt;br /&gt; The time for the muon to reach the ground is&lt;br /&gt; &lt;br /&gt; $$\Delta t_g=\frac{10 km}{0.999c}=3.3*10^{-5} s&amp;lt; \Delta t_d$$&lt;br /&gt; &lt;br /&gt; which is why it reaches the ground.&lt;br /&gt; &lt;br /&gt; iii)&lt;br /&gt; From the muon&amp;#039;s frame, the ground comes toward it with a velocity v=0.999c[\itex], and hence the distance to the ground is length contracted&amp;lt;br /&amp;gt; by&amp;lt;br /&amp;gt; $$\Delta l&amp;amp;#039;=\sqrt{1-\frac{v^2}{c^2}}10 km=447.1 m$$&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; so&amp;lt;br /&amp;gt; $$\frac{\Delta l&amp;amp;#039;}{0.999c}=\frac{\sqrt{1-\frac{v^2}{c^2}}10 km}{0.999c}=1.5*10^{-6}&amp;amp;lt;2\times 10^{-6} s $$&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; These should be the approximate values, but I am actually more concerned whether the logic of my argument actually works (ie, if the above relations actually hold, then I have shown that the muon does actually reach the ground) Thanks.

PeroK · Mar 31, 2022

Diracobama2181 said:

These should be the approximate values, but I am actually more concerned whether the logic of my argument actually works (ie, if the above relations actually hold, then I have shown that the muon does actually reach the ground) Thanks.

There's no logic to say whether the muon reaches the ground or not. It depends entirely on whether the muon is traveling fast enough. Your calculations show that ##0.999c## is fast enough.

PeroK · Mar 31, 2022

PS It's interesting that SR is taught in this way that emphasises the difference between reference frames. The physics is the same, so there must be a common test for whether the muon reaches the ground.

If we let the depth of the atmosphere be ##D## (in the Earth frame) and the proper lifetime of the muon to be ##\tau##, then:

In the ground frame the muon exists for a time ##\Delta t = \gamma \tau## and travels a distance ##d = v\Delta t = \gamma v \tau## in that time. The test for reaching the ground is ##d > D##. I.e.
$$\gamma v > \frac D \tau$$In the muon frame, we have a distance to Earth of ##D' = \frac D \gamma##, hence a time of ##\Delta t' = \frac {D'} v = \frac {D}{\gamma v}##. The test to hit the ground is ##\Delta t' < \tau## hence:$$\gamma v > \frac D \tau$$ So, from that point of view it's the same test/criteria in both frames. That seems to me more logical and more in keeping with the principle of relativity.

Also, if anyone else anywhere in the universe is given the same data, then they can do the same calculations and come to the same conclusion.

Diracobama2181 · Mar 31, 2022

By the logic of my argument, I simply mean that is showing the time of of travel less than the time of decay sufficient to show that the muon would have to reach the ground. I have recently contacted my professor however, and they said this form of argument dosen't work, and I would have to give the distance that the muon traveled before decaying. His answer confused me though and I suppose I am still confused as to why my solution would not sufficiently answer the above question.

PeroK · Mar 31, 2022

Diracobama2181 said:

By the logic of my argument, I simply mean that is showing the time of of travel less than the time of decay sufficient to show that the muon would have to reach the ground. I have recently contacted my professor however, and they said this form of argument dosen't work, and I would have to give the distance that the muon traveled before decaying. His answer confused me though and I suppose I am still confused as to why my solution would not sufficiently answer the above question.

I don't see it makes any difference. Either the muon hits the ground before it decays; or, the muon would travel further than the distance to the Earth's surface before it decays; it's all the same.

It's a shame if you have done the problem correctly and yet you're still confused.

Special Relativity Muon problem

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