Particle Decay After a Certain Distance

[PatrickStar]

Homework Statement

A certain elementary particle lives only a time [T][/0] = 5 sec (proper time) before disintegrating. What velocity must the particle have if it is to reach the Earth from the Sun before disintegrating? Distance between the Earth and Sun is 1.5x10^11 meters.[/B]

Homework Equations

I don't necessarily know which equations would be most relevant.

The Attempt at a Solution

I have been working on this problem for several hours now and cannot seem to get an answer, I have tried every which way but no luck so far.[/B]

 

[kuruman]

Hi PatrickStar and welcome to PF.

Surely you can do better than this. Maybe you are new, but we do expect to see some initial effort on your part.

I don't necessarily know which equations would be most relevant.

What topics are you covering in class now or have covered recently? Make a list of these equations, eliminate the ones you think are irrelevant and post the rest.

I have been working on this problem for several hours now and cannot seem to get an answer, I have tried every which way but no luck so far.

Show all the ways you have tried. We will steer you away from blind alleys and point out what looks promising.

 

[PatrickStar]

Hi PatrickStar and welcome to PF.

Surely you can do better than this. Maybe you are new, but we do expect to see some initial effort on your part.

What topics are you covering in class now or have covered recently? Make a list of these equations, eliminate the ones you think are irrelevant and post the rest.

Show all the ways you have tried. We will steer you away from blind alleys and point out what looks promising.

Sorry, excuse my lack of effort, I have just been working on this question for nearly 7 hours and have no sight of solving it and am pretty defeated.
I have tried using several equations, such as v=d/t where I subbed out t for the [t][/0]/(sqrt(1-v^2/c^2) to account for the time dilation that would occur when the particle approaches the speed of light but that didn't get me anywhere a long with several other strategies. A push in the right direction would be very much appreciated.

 

[kuruman]

Using d/t is the correct equation. Look at the question from the particle's point of view. The particle lives for 5 s in its own frame, so that's the t in the denominator. What is the distance in the particle's own frame that the particle covers in that amount of time?

 

[PatrickStar]

Using d/t is the correct equation. Look at the question from the particle's point of view. The particle lives for 5 s in its own frame, so that's the t in the denominator. What is the distance in the particle's own frame that the particle covers in that amount of time?

The distance would be the total length, i.e. 1.5x10^11 wouldn't it? Or would it be 5 seconds * its Velocity?

 

[kuruman]

The distance would be the total length, i.e. 1.5x10^11 wouldn't it?

Nope. That's the distance s measured by someone at rest with respect to the Sun-Earth system. Remember, the particle "sees" that distance moving relative to it.

 

[PatrickStar]

Nope. That's the distance s measured by someone at rest with respect to the Sun-Earth system. Remember, the particle "sees" that distance moving relative to it.

So the particle will only see the distance of how fast its traveling V * how long its traveling for t?

 

[kuruman]

So the particle will only see the distance of how fast its traveling V * how long its traveling for t?

That's the left hand side of the equation, v*t. What is the right hand side? In other words, what is an expression for the Earth-Sun distance when viewed from a frame moving at velocity v relative to the Earth-Sun system?

 

[PatrickStar]

That's the left hand side of the equation, v*t. What is the right hand side? In other words, what is an expression for the Earth-Sun distance when viewed from a frame moving at velocity v relative to the Earth-Sun system?

I'm not sure I'm following.

 

[kuruman]

Can you answer the following question?
You have a meter stick that is moving relative to you with velocity 0.8c. You measure the length of the stick while it's moving. What do you get for an answer?

 

[PatrickStar]

Ohh so are you saying that we have to use the length contraction of L`= L/gamma ?

 

[PatrickStar]

Can you answer the following question?
You have a meter stick that is moving relative to you with velocity 0.8c. You measure the length of the stick while it's moving. What do you get for an answer?

Never mind, I just tried using that and didn't really see anything promising so I'm assuming that's not what you were alluding to?

 

[kuruman]

Precisely. The Earth-Sun distance is shortened by gamma in the particle's reference frame.

 

[kuruman]

Never mind, I just tried using that and didn't really see anything promising so I'm assuming that's not what you were alluding to?

Look again. One side of the equation is v*t. The other side is L/γ and γ has v in it. So ...

 

[PatrickStar]

Precisely. The Earth-Sun distance is shortened by gamma in the particle's reference frame.

I tried solving for v in that case and I got stuck. My process is as follows:
vt = L/gamma
vt = L * sqrt(1-v^2/c^2)
vt/L = sqrt(1-v^2/c^2)
(v^2 * t^2)/L^2 = 1-v^2/c^2
then to v^2/c^2 = 1 - (v^2 * t^2)/L^2
but after that I just feel like I am going no where

 

[kuruman]

but after that I just feel like I am going no where

Not so. Multiply things out and you'll get a fourth order polynomial in v that looks like Av⁴+Bv²+C = 0. Replace v² with y and you have a quadratic equation in y that you can solve. Then v is the appropriate square root of y.

 

[kuruman]

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

L/gamma is not L * sqrt(1-v^2/c^2). gamma = sqrt(1-v^2/c^2)

On edit: It actually is. Sorry for confusing you. gamma = 1/sqrt(1-v^2/c^2). I must have had a brain lapse.

 

[PatrickStar]

L/gamma is not L * sqrt(1-v^2/c^2). gamma = sqrt(1-v^2/c^2)

So it would be vt = L/(1/sqrt(1-v^2/c^2)) or is it just L/(sqrt(1-v^2/c^2)) ?

 

[PatrickStar]

L/gamma is not L * sqrt(1-v^2/c^2). gamma = sqrt(1-v^2/c^2)

As of now, I have worked it out to y^2/c^2 - y + L^2/t^2 = 0 after subbing y = v^2. Does that look right so far?
I went with vt = L / (1/sqrt(1-v^2/c^2))

 

[kuruman]

An easy way to remember which way it goes is that γ = sqrt(1-v^2/c^2) is greater than 1. So if you want a shorter length you need to put it in the denominator.

 

[kuruman]

I went with vt = L / (1/sqrt(1-v^2/c^2))

You shouldn't have. See post #20.

On edit: I take it back, you should have. I missed the extra 1.

 

[PatrickStar]

You shouldn't have. See post #20.

Yeah, I wrote it like that but have actually been solving it as if L / (1/sqrt(1-v^2/c^2))
Which has brought me to a stand still at y^2/c^2 - y + L^2/t^2 = 0 after subbing y = v^2. And not knowing how to progress from here

 

[PatrickStar]

You shouldn't have. See post #20.

On edit: I take it back, you should have. I missed the extra 1.

I think I have the right answer of 7.7*10^4.

 

[kuruman]

If that is in m/s it is too slow. Can you show the algebraic expression you used? Also, I apologize for confusing you a bit, but I got confused and turned around, see post #17.

 

[PatrickStar]

If that is in m/s it is too slow. Can you show the algebraic expression you used? Also, I apologize for confusing you a bit, but I got confused and turned around, see post #17.

So I went from vt = L/(1/sqrt(1-V^2/C^2)) to
sqrt(1-V^2/C^2) * vt = L
then divided by t
the squared both sides giving me
v^2 * (1 - V^2/C^2) = L^2/t^2
then multiplied out the v^2
set y = v^2
y - y^2/c^2 = L^2/t^2
is that right so far?

 

[kuruman]

$$\frac{L}{\frac{1}{\sqrt{1-v^2/c^2}}} =L~\sqrt{1-v^2/c^2}$$
You put the radical on the wrong side before you squared.

 

[PatrickStar]

$$\frac{L}{\frac{1}{\sqrt{1-v^2/c^2}}} =L~\sqrt{1-v^2/c^2}$$
You put the radical on the wrong side before you squared.

But now I got 79372 as the answer.
I started with vt = L*sqrt(1-v^2/c^2)
then squared both sides
then divided by t^2 and (1-v^2/c^2)
giving me
v^2 / (1-v^2/c^2) = L^2 / t^2
then separated the v^2 into
v^2 - v^2 / v^2 / c^2 = L^2 / t^2
which then I cancled the V^2's
v^2 - c^2 = L^2 / t^2
then added the c to both sides
plugged in L = 1.5 * 10^11 and t = 5 and c = 3 * 10^8
and solved for v giving me 79372

 

[kuruman]

I started with vt = L*sqrt(1-v^2/c^2)
then squared both sides

At this point you get $$v^2t^2=L^2(1-v^2/c^2)$$
Remove the parentheses by distributing L², then move the v² over to other side and then factor it out because you have two v² terms on the left. Then solve for v.

 

[PatrickStar]

At this point you get $$v^2t^2=L^2(1-v^2/c^2)$$
Remove the parentheses by distributing L², then move the v² over to other side and then factor it out because you have two v² terms on the left. Then solve for v.

So I did that and got 2.9999 * 10^8. Which is coincidentally the speed of light? or very close to it?