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GR0177 #33
Problem
 GREPhysics.NET Official Solution Alternate Solutions
This problem is still being typed.
Special Relativity$\Rightarrow$}Frames

The distance the pion must travel is in the lab frame, and thus $L=30=vt$.

The decay time is given in its proper time, and thus $t=\gamma t_0 \Rightarrow 30=v\gamma t_0$. Solving for v, one finds that choice (D) works.

Alternate Solutions
danielkwalsh
2010-09-05 18:25:18
A quick trick: the equation $\frac{30m}{v}=\gamma t_0$ is correct, but for an extremely relativistic particle, as we clearly have here, the decay time in the lab frame can be well approximated as $\frac{30m}{c}$, since $v\approx c$. This simplifies the math a lot more in solving for $v$. This gives $\beta=\frac{3\sqrt{11}}{10}$, which is closest to choice (D).
Herminso
2009-09-21 13:42:50
Use $\triangle S^2=\triangle x^2-c^2\triangle t^2$. In the own rest frame of the pion $\triangle x=0$, and thus:

$\triangle S^2=-c^2\(10^{-8})^2=-9$.

But in the Lab frame,

$\triangle S'^2=(30)^2-c^2\triangle t'^2$ .

Now, using the invariant $\triangle S^2=\triangle S'^2$,

$(30)^2-c^2\triangle t'^2=-9$ $\Rightarrow$ $\triangle t'=\sqrt{909/c^2}=sqrt{101}\times 10^{-8} s$

Thus, $v=\frac{\triangle x'}{\triangle t'}=\frac{30 m}{\sqrt{101}\times 10^{-8} s}\simeq2.98\times 10^8 m/s$.
 ramparts2009-10-07 13:32:13 By far the best solution, because it's the most elegant and because it makes the algebra significantly easier without a calculator.
danielkwalsh
2010-09-05 18:25:18
A quick trick: the equation $\frac{30m}{v}=\gamma t_0$ is correct, but for an extremely relativistic particle, as we clearly have here, the decay time in the lab frame can be well approximated as $\frac{30m}{c}$, since $v\approx c$. This simplifies the math a lot more in solving for $v$. This gives $\beta=\frac{3\sqrt{11}}{10}$, which is closest to choice (D).
Herminso
2009-09-21 13:42:50
Use $\triangle S^2=\triangle x^2-c^2\triangle t^2$. In the own rest frame of the pion $\triangle x=0$, and thus:

$\triangle S^2=-c^2\(10^{-8})^2=-9$.

But in the Lab frame,

$\triangle S'^2=(30)^2-c^2\triangle t'^2$ .

Now, using the invariant $\triangle S^2=\triangle S'^2$,

$(30)^2-c^2\triangle t'^2=-9$ $\Rightarrow$ $\triangle t'=\sqrt{909/c^2}=sqrt{101}\times 10^{-8} s$

Thus, $v=\frac{\triangle x'}{\triangle t'}=\frac{30 m}{\sqrt{101}\times 10^{-8} s}\simeq2.98\times 10^8 m/s$.
 hanin2009-10-04 22:13:57 Why $\Delta x=0$?
 kroner2009-10-05 11:36:41 The pion's own rest frame is the frame where it doesn't move so $\Delta x = 0$.
 ramparts2009-10-07 13:32:13 By far the best solution, because it's the most elegant and because it makes the algebra significantly easier without a calculator.
Ning Bao
2008-02-01 06:26:28
Pions don't reach speed of light (incidentally, this eliminates E). It must be going very close, however, to contract 30m to a little less than 3 meters to not violate this. This means D.
 noether2009-11-05 15:28:05 If 2.99*10^8 were listed as an answer, would you have chosen that?
 gravity2010-11-10 00:37:22 Yeah. This one was tricky. I did most of what everybody else did and ended with 30 c/(909)$^{1/2}$ which I figured looked more like more like c than it did 2.98 x 10^8. Gah. I should have known better! At least it's not the real test.
StrangeQuark
2007-05-12 12:03:43
To make this faster (avoid the "messy" fraction)
Note to start that this is not a photon thus answer E is out.
now as above,
L=v $\gamma$ t
L $\sqrt{1-\frac{v^2 }{c^2}$=v t
some simplification steps
$v^2$=$\frac{L^2 c^2}{t^2 c^2+L^2}$
now plug and chug...
$v^2$=$\frac{(30m)^2 ((3)(10^8)(\frac{m}{s}))^2}{(10^{-8}s)^2 ((3)(10^8)(\frac{m}{s}))^2+(30m)^2}$
Note now that we have a
$(10^{-8})(10^{8})$
in the first term in the denominator, leaving only
$3m^2+30m^2$
in the denominator,
but $3m^2\ll30m^2$
so we simplify
$v^2$<$\frac{(30m)^2 ((3)(10^8)(\frac{m}{s}))^2}{(30m)^2}$
after canceling we see that
v<$(3)(10^8)\frac{m}{s}$
but only by a very small amount thus we have D
boundforthefloor
2006-11-26 05:25:22
Can anyone clarify this? I'm befuddled and can't find much that helps.
 johnyp032006-11-29 17:48:43 So, you know L=30=vt in the lab frame. But, in the pion's frame, there is time dilation. So, t=t'=(gamma)t0 where t0=3*10^-8, gamma=1/(sqrt(1-v^2/c^2)). So: 30 = v(gamma)t0 = v*(10^-8)/(sqrt(1-v^2/c^2)) 30*sqrt(1-v^2/c^2) = v*(10^-8) 900*(1-v^2/c^2) = v^2(10^-16) 900 = v^2*10^-16 + v^2*900/c^2 ==================> 900/c^2=(900/9)*10^-16=10^-14 900 = v^2(10^-16 + 10^-14) sqrt(900/(10^-16+10^-14)) = v 30/sqrt(1.01*10^-14) = v 30/1.005*10^-7 = v v ~ 2.98*10^8 Hope this helps
 boundforthefloor2006-12-01 11:40:08 Thanks johnyp. The actual equation setup was killing me, now that I've seen the calcualtions it makes much more sense.
 VanishingHitchwriter2006-12-01 14:07:02 Surround your expressions with dollar signs for latex equations. Here's the guy's comment back again... $t=t'=\gamma t_0$ where $t_0=3*10^{-8}$, $\gamma=1/(\sqrt{1-v^2/c^2})$. So: $30 = v\gammat_0 = v*(10^{-8})/(\sqrt{1-v^2/c^2})$ $30*\sqrt{1-v^2/c^2} = v*(10^{-8})$ $900*(1-v^2/c^2) = v^2(10^{-16})$ $900 = v^2*10^{-16} + v^2*900/c^2$ ==================> $900/c^2=(900/9)*10^{-16}=10^{-14}$ $900 = v^2(10^{-16} + 10^{-14})$ $\sqrt(900/(10^-16+10^-14)) = v$ $30/\sqrt(1.01*10^-14) = v$ $30/1.005*10^-7 = v$ $v ~ 2.98*10^8$
 matweiss2010-09-29 09:31:49 hey, I don't understand why t0 is 3 x 10^-8. it seems like I thought t in the frame of the particle was 10^-8 s and that (if you assume the particle to essentially be moving at light speed) then t in the lab frame equals 30/c= 1x 10^-7. any help?
senatez
2006-11-02 12:46:13
I ended up with a messy fraction 10c/10.1 which you estimate as .99c. This is about 2.98*10^8. I sure wish they would give easier arithmatic, it is waists a lot of time.

LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$