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  GR0177 #33
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Special Relativity}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.

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
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).Alternate Solution - Unverified
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.
ramparts
2009-10-07 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.
Alternate Solution - Unverified
Comments
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).Alternate Solution - Unverified
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.
hanin
2009-10-04 22:13:57		 Why \Delta x=0?
kroner
2009-10-05 11:36:41		 The pion's own rest frame is the frame where it doesn't move so \Delta x = 0.
ramparts
2009-10-07 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.
Alternate Solution - Unverified
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.
noether
2009-11-05 15:28:05		 If 2.99*10^8 were listed as an answer, would you have chosen that?
gravity
2010-11-10 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.
NEC
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		NEC
boundforthefloor
2006-11-26 05:25:22		Can anyone clarify this? I'm befuddled and can't find much that helps.
johnyp03
2006-11-29 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
boundforthefloor
2006-12-01 11:40:08		 Thanks johnyp. The actual equation setup was killing me, now that I've seen the calcualtions it makes much more sense.
VanishingHitchwriter
2006-12-01 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
matweiss
2010-09-29 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?
Answered Question!
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.NEC

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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.

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