Special Relativity length contraction and velocity

Eruestan
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Homework Statement


Λ particle has a proper life-time τ = 2×10−10 s. After being born in the cloud chamber (a
device to track energetic particles) of physics laboratory it left there a a 300cm long trail. Find
the speed of this particle in the laboratory frame.

Homework Equations



Gamma(Lorentz Factor) = (1 - v^2/c^2)^-1/2
v=l/t
τ=tgamma

The Attempt at a Solution


Every time I try a solution, I seem to find I need more information, say the length traveled in the frame of the particle. Or get a speed above c! Urgently need help
 
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Looks like you have all the right equations. Show what you did with them. You should end up with a single equation where the only unknown is the velocity.
 
so I tried gamma = (1- v^2/c^2)^-1/2) and rearranged to get
v = c(1- 1/gamma^2)^1/2
and tried subbing in either
gamma = τ/t

so v = c(1 - t^2/τ^2)^1/2
but I didn't really know where to go from there
 
Start here:
Distance = v*time

You're given the distance in the lab frame. What's the time in the lab frame in terms of the life-time τ? (Write an expression, don't try to give a numerical answer.)
 
so v=l/t
and t=gamma/τ
so v = lτ/gamma

and gamma=(1-v^2/c^2)^-1/2

so v=lτ(1-v^2/c^2)^1/2

v^2=(lτ)^2 (1-v^2/c^2)

I rearranged it to get

v^2(1 +(lτ)^2/c^2) = (lτ)^2

So

v^2= (lτ)^2 / (1 + (lτ)^2/c^2))

Is that correct?
 
Eruestan said:
so v=l/t
OK.
and t=gamma/τ
so v = lτ/gamma
Redo this.
 
oops that was a very silly mistake!
So instead I should have got
v = lgamma/τ

so v=l/τ(1 - v^2/c^2)^1/2

so v^2 = l^2/τ^2(1 - v^2/c^2)

but now I'm a little stuck. How do I factorise out v from this? Or have I used a wrong substitution?
 
Eruestan said:
So instead I should have got
v = lgamma/τ
Better. (At least time is in the denominator!) But your gamma is in the wrong position. Looking more carefully at your first post, I see that you have the time dilation formula backwards. (I should have caught that earlier.) Remember that the given half-life is the proper time.

You're almost there. One more time.
 
Yes that was a very silly mistake! Ah so I have! I always get them the wrong way round, something I need to get right for definite.

So now I have

v^2 = l^2/τ^2(1-(l/τc)^2)

Is that more like it? When I tried to do it a few other ways I see that my problem was always factorising the v, although I found I needed more info (as I tried to do it with length contraction), it would have helped if I didn't have the equations the wrong way round!
 
  • #10
Eruestan said:
So now I have

v^2 = l^2/τ^2(1-(l/τc)^2)
Almost. Do the algebra one more time. What's your starting equation before you isolated the v? (And be sure to use brackets so that it's clear what's on top or on bottom of any fraction.)
 
  • #11
okay so
v = l/τgamma
= (l/τ)(1 - v^2/c^2)^1/2
so V^2 = (l/τ)^2 - (lv/τc)^2
so v^2 + (lv/τc)^2 = (l/τ)^2
so v^2(1 + (l/τc)^2) = (l/τ)^2
so v^2 = l^2/(τ^2(1 + (l/τc)^2)

Is that correct?
 
  • #12
also I've just seen some parts cancel so I'm left with

v^2 = l^2/(τ^2 + l^2/c^2)
 
  • #13
Eruestan said:
okay so
v = l/τgamma
= (l/τ)(1 - v^2/c^2)^1/2
so V^2 = (l/τ)^2 - (lv/τc)^2
so v^2 + (lv/τc)^2 = (l/τ)^2
so v^2(1 + (l/τc)^2) = (l/τ)^2
so v^2 = l^2/(τ^2(1 + (l/τc)^2)

Is that correct?
Looks good.

I'd add a bracket, just for added clarity:
v^2 = l^2/[τ^2(1 + (l/τc)^2)]
 
  • #14
Eruestan said:
also I've just seen some parts cancel so I'm left with

v^2 = l^2/(τ^2 + l^2/c^2)
Good!
 
  • #15
that looks a lot more like the other formulas we've seen so assuming it is correct, thankyou very much!
 
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