# Speed and Speed of Light and Decay

1. Apr 16, 2010

### ihearyourecho

1. The problem statement, all variables and given/known data

When at rest, the Sigma- particle has a lifetime of 0.15 ns before it decays into a neutron and a pion. One particular Sigma- particle is observed to travel 3.2 cm in the lab before decaying.

What was its speed? (Hint: Its speed was not 2/3 c)

2. Relevant equations

V=x/t

3. The attempt at a solution

V=x/t

V=.032m/.15*10^-9s=2.1333*10^8

Since it wants the answer in terms of the speed of light c,

2.1333*10^8 / 3*10^8 = .7111c

EDIT: I just realized that v=x/t^2, but that gives me an implausible answer of 4740740741c, so I still need help

2. Apr 16, 2010

### ihearyourecho

Looking back on the work I did, I'm naive. I know that in this problem I should use either t=t0/(SQRT(1-v2/c2))

OR

L=L0*(SQRT(1-v2/c2))

But I'm not sure which (or both?)

3. Apr 18, 2010

### ihearyourecho

anyone? :)

4. Apr 18, 2010

### Matterwave

You can use either, but the time dilation one should be more straight forward in this case.

EDIT: Oh, and v=x/t^2 is definitely NOT the correct equation. Just check the units!

Last edited: Apr 18, 2010
5. Apr 18, 2010

### ihearyourecho

Okay, wow, yeah, I must have been kind of out of it haha.

So if I use time dilation, t=t0/(SQRT(1-v2/c2))

Is the lifetime of the particle at rest t0? If so, how do I find the other time? If not, how do I find either? Thanks!

6. Apr 18, 2010

### Matterwave

v=x/t=xsqrt(1-v^2/c^2)/t0

Solve for v.

t0 is the at-rest decay time.

7. Apr 18, 2010

### RoyalCat

I've always found time dilation and length contraction to be confusing concepts since invoking them very very often lead you down the wrong path.

The best way to approach these problems, unless you're well-versed in the ways of time dilation and length contraction, is not to cut yourself any slack, and to look at the events and apply the Lorentz Transform.

In this case, you've got two events. The particle begins its motion at $$x'=0, t'=0$$ in the lab frame, and at $$x=0, t=0$$ in its own rest frame.

The second event is the decay of the particle. You are told that the particle's decay time is $$\tau=0.15 ns$$, as measured in its rest frame (!)

Therefore, the decay event in the rest frame happens at: $$x=0, t=\tau$$ (Convince yourself why x is 0!)

Now all you need to do is find the distance and time coordinates in the lab frame, and relate them to the velocity.

For your convenience, the Lorentz Transform, where the primed frame is moving with velocity -v in the +x direction with respect to the unprimed frame:

$$x'=\gamma(x+vt)$$
$$t'=\gamma(t+\frac{vx}{c^2})$$

8. Apr 18, 2010

### Matterwave

Doing it the Lorentz transform way gets you to the exact same answer. Notice that if you set x=0 and take x'/t' you get that x'/t'=v. From here you still have to use the fact that t'=gamma*t to obtain v=x'/gamma*t=3.5cm*sqrt(1-v^2/c^2)/t. It's the exact same equation as before just with minor notation convention differences (x->x', t0->t, and t->t'...a bit confusing haha).

The time-dilation and length contraction equations are just simplified forms which are convenient for specific purposes. t0 is the time measured in the rest frame, or equivalently the frame in which the two events happen at the same position (notice that the time-dilation equation falls out naturally from the Lorentz transforms if you set x=0). It's not incredibly difficult to keep track.

9. Apr 18, 2010

### ihearyourecho

What is gamma in this context? I've been flipping through my book and can't find it.

10. Apr 18, 2010

### Matterwave

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

11. Apr 18, 2010

### RoyalCat

Not saying they're wrong, not at all, it's just that knowing when to apply each can get confusing, especially when you're just getting to know the subject when simultaneity and events occurring in the same point in space are still flaky concepts.

Just pedagogically speaking, I think the transform approach is simpler to keep track of, and promotes better understanding.

As for relativistic notation, these symbols usually hold the following meanings:

$$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$ where $$v$$ is the relative velocity between the two reference frames.

$$\beta = \frac{v}{c}$$ where $$v$$ can refer to any velocity, often you'll see $$\beta '$$ or $$\beta_0$$ or whatever, with the prime or 0 index referring to a particular velocity.

Another thing worth noting is that in the usual relativistic notation, you'll have two reference frame. One which you claim is at rest, the lab frame, or the unprimed frame, S, and another frame which is moving with a velocity v in the +x direction with respect to the lab frame, denoted S', the primed frame.

The usual form of the Lorentz Transform relates the position and velocity of a particle in the primed frame to those in the unprimed frame. If you're ever in doubt on a special relativity question, just go back to the basics, and a bit of algebra later, the answer should pop out:

For the position of an event in the primed frame, based on the position and time coordinates of that event in the unprimed frame:

$$x'=\gamma(x-vt)$$

And for the time of an event in the primed frame, based on the position and time coordinates of that event in the unprimed frame:

$$t'=\gamma(t-\frac{vx}{c^2})$$

In my previous post I gave the form of the transformation where the primed frame is moving with velocity v in the -x direction, which as you can see, simply meant that I used -v instead of v in the above equations.

Last edited: Apr 18, 2010
12. Apr 18, 2010

### ihearyourecho

If I were to do it this way, how do I isolate v?

v=xSQRT(1-v2/c2)/t0
I've tried to get v down to one side, but I'm stuck. Or was I supposed to set x/t=xSQRT(1-v2/c2)/t0

This stuff is just confusing to me, and I'm not really clear on what the other Royal was getting at, so I'm coming back to this..

13. Apr 18, 2010

### Matterwave

First square both sides of the equation, and then you should be able to manipulate it easier.

$$v^2=\frac{x^2(1-\frac{v^2}{c^2})}{t_0^2}=\frac{x^2}{t_0^2}(1-\frac{v^2}{c^2})$$

Can you go from there?

@Royalcat: I know what you were saying =), I wanted to reassure ihearyourecho that the two methods are in fact equivalent.

14. Apr 18, 2010

### RoyalCat

All you've got left from here on out is a bit of algebra.

$$v=\frac{x}{t_0}\sqrt{1-\frac{v^2}{c^2}}$$

$$\frac{t_0^2}{x^2}v^2=1-\frac{v^2}{c^2}$$

$$v^2(\frac{t_0^2}{x^2}+\frac{1}{c^2})=1$$

And you can finish this up and see if the answer makes sense.
I found $$v=0.58c$$, though I didn't double-check my calculator input.

Good night and good luck. :)

15. Apr 18, 2010

### ihearyourecho

Okay, can we come at this from another angle?

On another site, someone has posted this as the work for the problem (and the answer is right). Can you tell me how they figured out a step? I feel as if it seems easier than everything you all have been doing (no offense, and hell, it might be the same thing).

Here is the work:
t = 0.15 ns, L = 3.2 cm
(I know these are given)
t' = t/√[1 - (v/c)2]
(I know this is the equation for time dilation)
speed v = L/t' = L√[1 - (v/c)2]/t
(I understand that v=x/t)
∴v = Lc/√[L2 + (ct)2] = 1.74*108 m/s
(How did this step happen?)

Also, @Matter, I know how to get to that step you're talking about, but then v is still in the right hand side of the equation (unless it's a different v?) I'm just confused with the multiple variables, and the equation they used seemed to follow the time dilation equation (which I'm used to) the most, and had the least number of variables that confused me haha.

Anywho, I've got the right answer, but I still want to find out how it happened. If not, it's not a huge deal; I've got the basic gist of time dilation and length contraction down, which is all I will need to know for the test.

Thanks again

16. Apr 18, 2010

### ihearyourecho

Oh that algebra works. Darn that distributing the v! That's the part that was confusing me (not including the rest of the problem haha) Thanks for clearing up the algebra though. And yes, your answer is right.

17. Apr 18, 2010

### Matterwave

The other solution is exactly the same (notice the same equation arises), except they solved for v all in 1 step (and probably did different manipulations so that the form looks different but is in fact equivalent - you can tell because Royalcat's answer is exactly the same as the other answer).

Royalcat gave a pretty good derivation for v.

If you need it more detailed, starting from:
$$v^2=\frac{x^2}{t_0^2}(1-\frac{v^2}{c^2})$$

You need to distribute the x^2/t0^2 term:
$$v^2=\frac{x^2}{t_0^2}-\frac{x^2 v^2}{c^2 t_0^2}$$

Bring the term with v to the other side:
$$v^2+\frac{x^2 v^2}{c^2 t_0^2}=\frac{x^2}{t_0^2}$$

Factor out the v^2:
$$v^2(1+\frac{x^2}{c^2 t_0^2})=\frac{x^2}{t_0^2}$$

Now just divide through by the coefficient on v^2 and you will have isolated v^2. Then just squareroot and you have your answer.