Special Relativity length contraction and velocity

In summary, we are given a Λ particle with a proper life-time τ = 2×10−10 s and a 300cm long trail in the cloud chamber of a physics laboratory. Using the equations Gamma(Lorentz Factor) = (1 - v^2/c^2)^-1/2, v=l/t, and τ=tgamma, we can find the speed of the particle in the laboratory frame by first finding an equation where the only unknown is the velocity. By using distance = v*time and the time in the laboratory frame in terms of the life-time τ, we can ultimately get the equation v^2 = l^2/(τ^2 + l^2/c^2), where v represents the
  • #1
Eruestan
8
0

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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  • #2
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.
 
  • #3
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
 
  • #4
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.)
 
  • #5
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?
 
  • #6
Eruestan said:
so v=l/t
OK.
and t=gamma/τ
so v = lτ/gamma
Redo this.
 
  • #7
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?
 
  • #8
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.
 
  • #9
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!
 

FAQ: Special Relativity length contraction and velocity

What is the concept of length contraction in special relativity?

Length contraction is the phenomenon in which an object's length appears shorter when it is moving at a high velocity relative to an observer. This effect is a consequence of Einstein's theory of special relativity, which states that the laws of physics are the same for all inertial observers.

How does length contraction affect the measurement of an object's length?

Length contraction affects the measurement of an object's length by causing it to appear shorter in the direction of motion. This means that the measured length of an object will be different for observers in different inertial frames of reference, depending on their relative velocities.

Does length contraction only occur for objects moving at very high velocities?

Yes, length contraction only becomes noticeable for objects moving at speeds close to the speed of light. At everyday speeds, the effect is too small to be measured. However, as an object's velocity approaches the speed of light, the amount of length contraction increases significantly.

Can length contraction be observed in everyday life?

No, length contraction is not directly observable in everyday life because the velocities of objects in our daily experiences are far below the speed of light. However, we can indirectly observe length contraction in particle accelerators, where particles are accelerated to very high speeds and their lengths are measured.

How does length contraction relate to time dilation in special relativity?

Length contraction and time dilation are two sides of the same coin in special relativity. As an object's velocity increases, it experiences both length contraction and time dilation. This means that time appears to pass more slowly for a moving object compared to an observer at rest, while the object's length appears shorter to the observer.

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