GR0177 #33
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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). | ![Alternate Solution - Unverified](/img/altsol-unverified.gif) | 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 ,
![\triangle t'=\sqrt{909/c^2}=sqrt{101}\times 10^{-8} s](/cgi-bin/mimetex.cgi?\triangle t'=\sqrt{909/c^2}=sqrt{101}\times 10^{-8} s)
Thus, .
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.
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| ![Alternate Solution - Unverified](/img/altsol-unverified.gif) |
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Comments |
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). | ![Alternate Solution - Unverified](/img/altsol-unverified.gif) | 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 ,
![\triangle t'=\sqrt{909/c^2}=sqrt{101}\times 10^{-8} s](/cgi-bin/mimetex.cgi?\triangle t'=\sqrt{909/c^2}=sqrt{101}\times 10^{-8} s)
Thus, .
hanin 2009-10-04 22:13:57 |
Why ?
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kroner 2009-10-05 11:36:41 |
The pion's own rest frame is the frame where it doesn't move so .
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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.
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| ![Alternate Solution - Unverified](/img/altsol-unverified.gif) | 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?
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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) 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.
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| ![NEC](/img/nec.gif) | 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
=![\frac{L^2 c^2}{t^2 c^2+L^2}](/cgi-bin/mimetex.cgi?\frac{L^2 c^2}{t^2 c^2+L^2})
now plug and chug...
=![\frac{(30m)^2 ((3)(10^8)(\frac{m}{s}))^2}{(10^{-8}s)^2 ((3)(10^8)(\frac{m}{s}))^2+(30m)^2}](/cgi-bin/mimetex.cgi?\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})](/cgi-bin/mimetex.cgi?(10^{-8})(10^{8}))
in the first term in the denominator, leaving only
![3m^2+30m^2](/cgi-bin/mimetex.cgi?3m^2+30m^2)
in the denominator,
but ![3m^2\ll30m^2](/cgi-bin/mimetex.cgi?3m^2\ll30m^2)
so we simplify
<![\frac{(30m)^2 ((3)(10^8)(\frac{m}{s}))^2}{(30m)^2}](/cgi-bin/mimetex.cgi?\frac{(30m)^2 ((3)(10^8)(\frac{m}{s}))^2}{(30m)^2})
after canceling we see that
v<![(3)(10^8)\frac{m}{s}](/cgi-bin/mimetex.cgi?(3)(10^8)\frac{m}{s})
but only by a very small amount thus we have D | ![NEC](/img/nec.gif) | 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
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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.
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VanishingHitchwriter 2006-12-01 14:07:02 |
Surround your expressions with dollar signs for latex equations. Here's the guy's comment back again...
where , . So:
![30 = v\gammat_0 = v*(10^{-8})/(\sqrt{1-v^2/c^2})](/cgi-bin/mimetex.cgi?30 = v\gammat_0 = v*(10^{-8})/(\sqrt{1-v^2/c^2}))
![30*\sqrt{1-v^2/c^2} = v*(10^{-8})](/cgi-bin/mimetex.cgi?30*\sqrt{1-v^2/c^2} = v*(10^{-8}))
![900*(1-v^2/c^2) = v^2(10^{-16})](/cgi-bin/mimetex.cgi?900*(1-v^2/c^2) = v^2(10^{-16}))
![900 = v^2*10^{-16} + v^2*900/c^2](/cgi-bin/mimetex.cgi?900 = v^2*10^{-16} + v^2*900/c^2)
==================> ![900/c^2=(900/9)*10^{-16}=10^{-14}](/cgi-bin/mimetex.cgi?900/c^2=(900/9)*10^{-16}=10^{-14})
![900 = v^2(10^{-16} + 10^{-14})](/cgi-bin/mimetex.cgi?900 = v^2(10^{-16} + 10^{-14}))
![\sqrt(900/(10^-16+10^-14)) = v](/cgi-bin/mimetex.cgi?\sqrt(900/(10^-16+10^-14)) = v)
![30/\sqrt(1.01*10^-14) = v](/cgi-bin/mimetex.cgi?30/\sqrt(1.01*10^-14) = v)
![30/1.005*10^-7 = v](/cgi-bin/mimetex.cgi?30/1.005*10^-7 = v)
![v ~ 2.98*10^8](/cgi-bin/mimetex.cgi?v ~ 2.98*10^8)
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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?
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| ![Answered Question!](/img/answered.gif) | 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](/img/nec.gif) |
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