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GR0177 #33
Problem
 GREPhysics.NET Official Solution Alternate Solutions
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Special Relativity}Frames

The distance the pion must travel is in the lab frame, and thus .

The decay time is given in its proper time, and thus . Solving for v, one finds that choice (D) works.  Alternate Solutions
danielkwalsh
2010-09-05 18:25:18
A quick trick: the equation 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 , since . This simplifies the math a lot more in solving for . This gives , which is closest to choice (D). Herminso
2009-09-21 13:42:50
Use . In the own rest frame of the pion , and thus:

.

But in the Lab frame,

.

Now, using the invariant ,

Thus, .
 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 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 , since . This simplifies the math a lot more in solving for . This gives , which is closest to choice (D). Herminso
2009-09-21 13:42:50
Use . In the own rest frame of the pion , and thus:

.

But in the Lab frame,

.

Now, using the invariant ,

Thus, .
 hanin2009-10-04 22:13:57 Why ?
 kroner2009-10-05 11:36:41 The pion's own rest frame is the frame where it doesn't move so .
 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) 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 t
L =v t
some simplification steps
=
now plug and chug...
=
Note now that we have a

in the first term in the denominator, leaving only

in the denominator,
but
so we simplify
<
after canceling we see that
v<
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... where , . So: ==================>
 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$produces . type this... to get...$\int_0^\infty\partial\Rightarrow\ddot{x},\dot{x}\sqrt{z}\langle my \rangle\left( abacadabra \right)_{me}\vec{E}\frac{a}{b}\$