Determining Particle Type from Energy: Relativistic Approach

[DukeLuke]

Homework Statement

Determine how accurately you will need to measure the speed of a particle that has energy 2 GeV to determine whether it is a pion (mass = 0.14 Gev/c^2) or a Kaon (mass = 0.49 Gev/C^2). *This question is in a relativity section, so I'm assuming relativistic equations are neccesary*

Homework Equations

<br /> E^2 = (p^2)(c^2) + (m^2)(c^4)<br />
<br /> p = (\gamma)mv<br />

The Attempt at a Solution

I started by substituting p= (\gamma)mv into the other equation so velocity can be related to mass, but I'm lost as to what the basic approach to this problem would be.

 

[Dick]

That looks like a good start. I'd now try to solve for v in terms of E and m and get the value of v for a kaon and a pion and see what the difference is. It looks to me like it becomes a quadratic equation in v^2.

 

[SimonZ]

use gamma r = E/(mc^2) = 1/sqrt(1 - v^2/c^2)
so v/c = sqrt[1 - (mc^2/E)^2]
for pion, mc^2/E = 0.14/2 = 0.07
v/c = 0.9975

 

[DukeLuke]

That looks like a good start. I'd now try to solve for v in terms of E and m and get the value of v for a kaon and a pion and see what the difference is. It looks to me like it becomes a quadratic equation in v^2.

Thanks, the only thing is when I solved it I didn't end up with a quadratic.

 

[Dick]

Thanks, the only thing is when I solved it I didn't end up with a quadratic.

What DID you get?

 

[DukeLuke]

I started by solving for p and substituing p= (\gamma mv) for it.

<br /> \frac{mv}{\sqrt{1 - v^2/c^2}} = \sqrt{\frac{E^2 - m^2c^4}{c^2}}<br />



then I squared both sides and moved gamma and c^2 to opposite sides

<br /> m^2v^2c^2 = (E^2 - m^2c^4) (1 - v^2/c^2)<br />



multiply out right hand side

<br /> m^2v^2c^2 = E^2 + m^2c^2v^2 -m^2c^4 - \frac{E^2v^2}{c^2}<br />



group v^2 on same side and factor it out

<br /> v^2 (c^2m^2 + E^2/c^2 - m^2c^2) = E^2 - m^2c^4<br />



solve for v

<br /> v = \sqrt{\frac{E^2 -m^2c^4}{c^2m^2 + E^2/c^2 - m^2c^2}}<br />

 

[Dick]

That looks ok to me. You can cancel the two m^2*c^2 terms. It is even easier than I thought.