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GR0177 #33
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Alternate Solutions |
Herminso
2009-09-21 13:42:50 | Use  . In the own rest frame of the pion  , and thus:
^2=-9) .
But in the Lab frame,
^2-c^2\triangle t'^2) .
Now, using the invariant  ,
^2-c^2\triangle t'^2=-9)  
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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Comments |
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, .
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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| | 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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|  | 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...
=^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}))
in the first term in the denominator, leaving only
in the denominator,
but 
so we simplify
<^2 ((3)(10^8)(\frac{m}{s}))^2}{(30m)^2})
after canceling we see that
v<(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.
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:
/(\sqrt{1-v^2/c^2}))
)
 = v^2(10^{-16}))
==================> *10^{-16}=10^{-14})
)
) = v)
 = v)
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|  | 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. | |
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