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I cannot for the life of me figure out what I am doing wrong with this problem:

If a high speed particle travels a distance of 39 Km with a half life of 1 micro sec, what must its speed be to travel said distance? I keep getting Erroneous calculations.

$$\Delta t' = \Delta t\gamma = \frac{\Delta x}{v}\gamma$$

therefore, the speed observed by the neutrino or whatever is

$$\begin{gather}\Delta t' = 1 \times 10^{-6} s = \frac{39 \times 10^3 m}{v\sqrt{1 -\frac{v^2}{c^2}}}\\ \sqrt{v^2 - \frac{v^4}{c^2}} =39\times 10^9 \frac{m}{s}\\ - c^{-2} v^4 + v^2 - (39)^2\times 10^{18} = 0\end{gather}$$

So I solve this quartic equation and I get really strange results

$$v^2 = -\frac{1}{2} \pm \frac{\sqrt{1 - 4c^{-2}\times(39)^2\times 10^{18}}}{2}$$

Which always yields imaginary values for $v$. what did I do wrong? Should my $\Delta x$ be negative? I don't see why it would matter since I can define positive in any direction.

Below the problem is stated in its entirety: "The particle is unstable. It decays into other particles with a half-life of only t = 1 microseconds. That means that, even if it were travelling at the speed of light, it would only be able to travel 300 meters

before disappearing. However," continues Professor Smith, "we find plenty of these Bluto particles at the ground level. We believe that they are created in the upper atmosphere at an altitude of H = 39 km above the ground. How in the world can they reach the ground before decaying?"

"Ah", answers Professor Jones, "have you taken into account the time dilation factor? If the Bluto particles are created with a high enough initial speed, they could reach the ground during their brief lifetime."

How fast must the particles be moving to reach the ground? Express your answer as a fraction of the speed of light. You must be correct to the sixth digit in order to be marked correct.

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Are you sure you've given the exact statement of the problem? Because the way you've written it, it's not solvable. The half-life is related to the average lifetime of a large number of these particles, but it doesn't tell you anything about an individual particle. –  David Z Sep 24 '12 at 23:50
    
@David Zaslavasky: Doesn't it tell you the time taken for a particle's probability of decaying to reach 1/2?. –  Dimensio1n0 Jun 3 '13 at 13:43
    
@dimension10 yes, but probability can only be related to reality by observing a large number of particles. If you only observe one unstable particle of unknown type, you cannot determine anything about its half-life. –  David Z Jun 3 '13 at 18:48
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2 Answers

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Be careful to distinguish correctly between lab frame and particle frame. In the particle frame $$\text{$\Delta $t}=10^{-6}\sec$$ In the lab frame $$\text{$\Delta $t}_L=\frac{\text{$\Delta $x}}{v}$$ Finally the relation between the time lapse in the particle frame and the lab frame is $$\text{$\Delta $t}_L=\text{$\gamma \Delta $t}=\frac{\text{$\Delta $t}}{\sqrt{1-v^2/c^2}}=\frac{\text{$\Delta $x}}{v}$$ Solving for v yields $$v=\frac{c \text{$\Delta $x}}{\sqrt{c^2 \text{$\Delta $t}^2+\text{$\Delta $x}^2}}$$

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I think I see, so you multiply gamma to the "Proper Time" ? Correct? –  Cactus BAMF Sep 25 '12 at 0:44
    
@CactusBAMF That's correct. The time that elapses measured by the lab observer's clock is longer. The lab observer sees the particle observer's clock (that is measuring the proper time) running slower than his own. –  Robert Miller Sep 25 '12 at 1:01
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Whenever I see questions about special relativity I always advise students to write down the relevant spacetime points in one frame than apply the Lorentz transformations to get the corresponding co-ordinates in the boosted frame. If you do this carefully the answer drops out (fairly!) easily. Looking at Robert's answer this is what he's done, and I end up with the same final equation as him, but I thought it would be worth making the calculation explicit.

Start in the rest frame of the muon. We have two points: the start point $(0, 0)$ and the point where the muon decays $(t, 0)$, where $t$ is the time the muon takes to decay. I'm guessing we're supposed to take $t$ to be 1$\mu$sec.

Now transform these points to the Earth frame. The Earth frame is moving at velocity $v$ relative to the muon rest frame, and for convenience we'll define $(0, 0)$ in the Earth frame to co-incide with $(0, 0)$ in the muon frame. Then we just have to transform $(t, 0)$. We simply plug time $t$ and distance $0$ into the Lorentz transformations to get:

$$ t' = \gamma \left( t - \frac{vx}{c^2} \right) = \gamma t $$

$$ x' = \gamma \left( x - vt \right) = -\gamma v t$$

So in our frame the muon starts at $(0, 0)$ and decays at $(\gamma t, -\gamma v t)$.

The rest is easy. We're told that the muon travels 39km in our frame: let's call this $d$. Then (dropping the minus sign for convenience since the direction of $d$ doesn't matter):

$$ d = \gamma v t = \frac{vt}{\sqrt{1 - v^2/c^2}} $$

and some quick algebra rearranges this to get:

$$ v = \frac{cd}{\sqrt{c^2t^2 + d^2}} $$

Which is of course exactly what Robert concluded a lot more quickly than I've done! The point of all the above rigmarole is to try to make it obvious where this equation comes from.

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