GREPhysics.NET: GR9677 #48

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GR9677 #48

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Special Relativity $\Rightarrow$ }Half Life

The halflife of the mesons is given. Since only half of the mesons reach point B 15 meters away, one presume that it takes 1 halflife of proper time to get there.

The proper time is $t_0=2.5E-8$ . The length $L=15=vt$ is in the lab frame, and since the time dilation equation gives $t=t_0\gamma$ , one has $L = 15 = vt_0\gamma=vt_0/\sqrt{1-(v/c)^2}\Rightarrow L/t_0 = \frac{v}{\sqrt{1-(v/c)^2}}$ .

Now, the gory arithmetics. No calculators allowed; 12 years of American public school mathematics wasted! $15/2.5E-8=15/25E-9=3/5E9=6E8=\kappa=\frac{v}{1-(v/c)^2}=\frac{vc}{\sqrt{c^2-v^2}}$ . Multiply things out to get $\kappa^2(c^2-v^2)=v^2c^2 \Rightarrow \kappa^2 c^2 = v^2 (c^2+\kappa^2)$ . Plug in numbers to get $v=\frac{6E8c}{\sqrt{9E16+36E16}}=\frac{6E8c}{\sqrt{45E16}}=\frac{6c}{3\sqrt{5}}$ , which is choice (C). Whew!

See below for user comments and alternate solutions!

Alternate Solutions

tdl17 2019-01-03 21:16:15	Since the particles only took one half-life, which is $2.5\\\\\\\\times10^{-8}$ seconds, to travel 15m, it must be traveling close to the speed of light. \\\\r\\\\n\\\\r\\\\nAssume the speed of the particles is some fraction of the speed of light and let this fraction be $a$ , we now have $v = ac$ . \\\\r\\\\n\\\\r\\\\nUsing $v=\\\\\\\\frac{L\\\\\\\\sqrt{1-\\\\\\\\frac{v^2}{c^2}}}{t_0}$ , we can replace v with ac and from there, you can solve for a and that should get you to $a = \\\\\\\\frac{2}{\\\\\\\\sqrt{5}}c$ , which is choice C. Reply to this comment
Mexicana 2007-10-03 18:06:00	The second easiest way to solve this, after 'banana's solution' (which is reeeally nice btw!) is just to simply solve for $v$ , getting $v=\frac{L/\tau c}{\sqrt{1+(L/\tau c)^2)}}$ and then just do the ratio $L/\tau c$ , which is reeally easy to calculate since the lifetime is given in units of $10^{-8}$ and $c=3\times 10^{8}$ .... Hence, the nice ratio of $L/\tau c = 2$ exact. Reply to this comment

Comments

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NEC

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2019-06-10 11:34:08

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tdl17
2019-01-03 21:16:15

Since the particles only took one half-life, which is $2.5\\\\\\\\times10^{-8}$ seconds, to travel 15m, it must be traveling close to the speed of light. \\\\r\\\\n\\\\r\\\\nAssume the speed of the particles is some fraction of the speed of light and let this fraction be $a$ , we now have $v = ac$ . \\\\r\\\\n\\\\r\\\\nUsing $v=\\\\\\\\frac{L\\\\\\\\sqrt{1-\\\\\\\\frac{v^2}{c^2}}}{t_0}$ , we can replace v with ac and from there, you can solve for a and that should get you to $a = \\\\\\\\frac{2}{\\\\\\\\sqrt{5}}c$ , which is choice C.

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Alternate Solution - Unverified

psitae
2016-10-27 14:27:54

probably very close to the speed of light --> $\\frac{2}{\\sqrt 5} c$

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bcomnes
2011-11-02 15:06:43

I found the solution confusing as written. This is my attempt to make it more clear to myself.rnrnGiven: rnDistance traveled 15m.rnHalf life of Pion is $2.5 \times 10^{-8}$ seconds.rnrnWhat we do:rnThe particles move relativistically, hinted at given the subject matter.rnrnSince the pion is moving, its clock runs slower than the lab clock. After traveling 15m, it has decayed less than another pion sample that is not moving, so it appears to have moved faster than the speed of light in order to traverse 15m and only decay 1 half life. The pion only decays for $2.5 \times 10^{-8}$ s since only one half of the pions show up. This is our proper time $t_0$ .rnrnDefine the speed of the pion in the lab frame. $\kappa = \frac{L}{t_0}= \frac{15m}{2.5 \times 10^{-8}s$ Notice that 15/2.5 = 6 and we get $\kappa = 6 \times 10^8$ or $\kappa = 2c$ rnrnWe want the speed of the pion that the pion actually feels (or in the pions frame), $v = \frac{L}{t}$ .rnrnRemembering our basic time dilation formula $t = \frac{t_0}{\sqrt{1-v^2/c^2}} = \gamma t_0$ , all we need to do now is solve for v.rnrnPlug in: $v = \frac{L}{t} = \frac{L}{ \gamma t_0} = \frac{\kappa}{\gamma} = \kappa \sqrt{1-v^2/c^2}$ Skipping some algebra $v = \frac{\kappa}{\sqrt{1+\kappa^2/c^2}} = \frac{2c}{\sqrt{1+4}}$ rnrnYou need to remember $\gamma = \frac{1}{\sqrt{1-v^2/c^2}$ of course, but also you could remember where, $\kappa$ is the speed of the particle that the rest frame sees and v is the actual speed of the moving particle as the particle experiences, that $\kappa = \gamma v$ , or conversely define $\gamma' = \frac{1}{\sqrt{1-\kappa^2/c^2}$ and remember that $v = \gamma' \kappa.$

bcomnes
2011-11-02 15:08:31

I found the solution confusing as written. This is my attempt to make it more clear to myself.

Given:
Distance traveled 15m.
Half life of Pion is $2.5 \times 10^{-8}$ seconds.

What we do:
The particles move relativistically, hinted at given the subject matter.

Since the pion is moving, its clock runs slower than the lab clock. After traveling 15m, it has decayed less than another pion sample that is not moving, so it appears to have moved faster than the speed of light in order to traverse 15m and only decay 1 half life. The pion only decays for $2.5 \times 10^{-8}$ s since only one half of the pions show up. This is our proper time $t_0$ .

Define the speed of the pion in the lab frame. $\kappa = \frac{L}{t_0}= \frac{15m}{2.5 \times 10^{-8}s$ Notice that 15/2.5 = 6 and we get $\kappa = 6 \times 10^8$ or $\kappa = 2c$

We want the speed of the pion that the pion actually feels (or in the pions frame), $v = \frac{L}{t}$ .

Remembering our basic time dilation formula $t = \frac{t_0}{\sqrt{1-v^2/c^2}} = \gamma t_0$ , all we need to do now is solve for v.

Plug in: $v = \frac{L}{t} = \frac{L}{ \gamma t_0} = \frac{\kappa}{\gamma} = \kappa \sqrt{1-v^2/c^2}$ Skipping some algebra $v = \frac{\kappa}{\sqrt{1+\kappa^2/c^2}} = \frac{2c}{\sqrt{1+4}}$

You need to remember $\gamma = \frac{1}{\sqrt{1-v^2/c^2}$ of course, but also you could remember where, $\kappa$ is the speed of the particle that the rest frame sees and v is the actual speed of the moving particle as the particle experiences, that $\kappa = \gamma v$ , or conversely define $\gamma' = \frac{1}{\sqrt{1-\kappa^2/c^2}$ and remember that $v = \gamma' \kappa.$

bcomnes
2011-11-02 15:09:10

Sorry to double post but the first post was totally garbled.

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NEC

CarlBrannen
2010-10-07 14:50:48

The speeds are in multiples of c so use c=1. We have a messy half life time and a clean distance in the problem so work in meters. To convert the half-life from seconds to meters use $c = 3.0\times 10^{-8} m/s$ to get 7.5 meters.

The equation for movement is distance = velocity x time, so we have $15m = v t$ . In the rest frame of the meson the time is 7.5 meters. But time is stretched by the gamma factor so we have:
$15 = v 7.5 /\sqrt{1-v^2}.$ then
$2\sqrt{1-v^2} = v$ then
$4(1-v^2) = v^2$ so $v^2 = 4/5$

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NEC

shafatmubin
2009-11-03 13:55:26

For the physics GRE, consider memorising the gamma-v relativistic table:

v = 0.999c -> gamma ~23
v = 0.995c -> gamma ~10
v = 0.99c -> gamma~7
v = 0.98c -> gamma~6
v = 0.97c -> gamma~5
v = 0.96c -> gamma~4
v = 0.95c -> gamma~3
v = 0.9c -> gamma~2.2
v = 0.85 -> gamma~2
v = 0.8c -> gamma = 5/3=1.67

I spent a while on Mathematica to figure these out. But it was worth the effort - I could guess the answer by looking at the gamma required (gamma=7/3 -> v~0.9 i.e v=Sqrt(4/5) )

jack238
2010-04-04 10:08:06

How could you quickly tell that gamma is 7/3?

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NEC

ams379
2009-10-07 10:46:45

I found an easier way to do this problem.
Just use L=L_o/ $\gamma$
where L=vt=v( $2.5x10^-8$ )= $L_0$ $\sqrt{1-v^2/c^2}$
then just plug in the values:
(2/ $\sqrt{5}$ ( $3x10^8$ )( $2.5x10^-8$ )= 15 $\sqrt{1-(2/$ \sqrt{5} $)/c^2}$
15/ $\sqrt{5}$ =15/ $\sqrt{5}$

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ams379
2009-10-07 10:29:28

I found an easier way to do this problem. Just use the length contraction equation. L=Lo Gamma
Where L=vt =(2.5x10^-8)v=15 $\sqrt{1-v^2/c^2}$
and just plug in the numbers.
(2.5x10^-8)(2/ $\sqrt{5}$ (3x10^8)=15 $\sqrt{1-4/5}$
15/ $\sqrt{5}$ =15/ $\sqrt{5}$

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shartacus
2008-08-06 12:52:58

I think we can save ourselves some hairy arithmetic by using a little logic: Choices (D) and (E) are obviously not correct and can be eliminated.

Assume for the moment that the mesons are traveling at about c. Then, ignoring relativity for the moment, the time for the beam to reach the detector is $\frac{x}{c}=\frac{15m}{3e8m/s}=5e-8s$

Notice that this is $\2t_0$ . This means that $\gamma$ , the relativistic factor, must be at least 2, since time in the frame of the mesons must be slowed by at least a factor of two for half of them to reach the detector. Gamma stays close to 1 until v is about .8c or .9c, where it increases asymptotically. Thus, choices (A) and (B) are too low. Only (C) remains.

In a nutshell, relativistic effects are large, so pick the choice that is closest to c without being greater than or equal to c.

Hope you guys followed my logic. This is the messed up way I think of things.

resinoth
2015-09-18 19:24:11

Exactly. The dilation factor is 2, which corresponds to about 0.85c. Just learn a simple table of these values.

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NEC

Mexicana
2007-10-03 18:06:00

The second easiest way to solve this, after 'banana's solution' (which is reeeally nice btw!) is just to simply solve for $v$ , getting $v=\frac{L/\tau c}{\sqrt{1+(L/\tau c)^2)}}$ and then just do the ratio $L/\tau c$ , which is reeally easy to calculate since the lifetime is given in units of $10^{-8}$ and $c=3\times 10^{8}$ .... Hence, the nice ratio of $L/\tau c = 2$ exact.

HaveSpaceSuit
2008-10-16 21:13:37

Mexicana, I think that there is a unit discrepancy in your solution. It looks like it is unit-less. I think that factor of c in the numerator is not supposed to be there. Correct me if I'm wrong.

grace
2010-11-05 20:55:20

I think you miss time c.

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Alternate Solution - Unverified

mm
2007-09-22 23:16:31

Plug each choice into

$L/t_0 = \frac{v}{\sqrt{1-(v/c)^2}}$

, and you can reach the answer.

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NEC

banana
2005-12-07 18:17:12

Much easier: The interval is the same in all frames. Thus: $-t_0^2=-t^2+x^2\quad\Rightarrow\quad t^2=t_0^2+x^2=7.5^2+15^2=281.25\quad\Rightarrow\quad v=\frac{x}{t}=\frac{15}{\sqrt{281.25}}=\frac{2}{\sqrt{5}}$

Gaffer
2007-10-27 09:47:59

Great soln! The math is easier if you think of it like:
The RHside
$\ 7.5^2 +15^2 = (15/2)^2 +15^2 = 15^2(1/4+1) = t^2$
$\ t = 15*sqrt{5}/2$
Then $\ v = x/t = (15/15 ) (2/sqrt{5})$

Basically, don't multiply big numbers unless you have too. This saves time.

jw111
2008-11-05 12:53:37

Similar way(c=1)

the time for travel is

t = 15/v

and t is also the half life seen on the ground

t = 7.5r

so 15/v=7.5r -> 2/v = r

$\frac{4}{v^2}=\frac{1}{1-v^2}$

$4=5v^2$

bac
2012-03-30 12:02:39

After you get to $v=\frac{15}{\sqrt{281.25}$ , here is how I was able to very quickly eyeball the answer. (A), (D), and (E) are obviously out.rnrnSquaring the answers (B) and (C), you get $\frac{2}{5}$ and $\frac{4}{5}$ , respectively. Which one does $v^2$ look more like? $v^2=\frac{225}{281.25}$ is closest to $\frac{4}{5}$ , so (C) is the correct answer.

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NEC

mg
2005-11-27 16:05:44

The arithmetic simplifies considerably if one realizes that $\frac{x}{t_0}$ =2c. Then one has v=2c $\sqrt{1-(v/c)^2}$

fredluis721
2019-06-10 11:38:18

Thanks for the information on this. I really enjoy the write up. tile installer

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NEC

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I think we can save ourselves some hairy arithmetic by using a little logic: Choices (D) and (E) are obviously not correct and can be eliminated. Assume for the moment that the mesons are traveling at about c. Then, ignoring relativity for the moment, the time for the beam to reach the detector is $\frac{x}{c}=\frac{15m}{3e8m/s}=5e-8s$ Notice that this is $\2t_0$ . This means that $\gamma$ , the relativistic factor, must be at least 2, since time in the frame of the mesons must be slowed by at least a factor of two for half of them to reach the detector. Gamma stays close to 1 until v is about .8c or .9c, where it increases asymptotically. Thus, choices (A) and (B) are too low. Only (C) remains. In a nutshell, relativistic effects are large, so pick the choice that is closest to c without being greater than or equal to c. Hope you guys followed my logic. This is the messed up way I think of things.

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