# Homework Help: Help with relativity particle physics Problem

1. Dec 22, 2012

### DunWorry

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

The D meson has a mass of 1865 $MeV / c^{2}$ and a lifetime in its rest frame of 4.1 x 10^ -13 seconds. A D meson is produced and decays into a Kaon with mass 0.494 GeV/c^2 and a Pion with mass 0.14 GeV/c^2

A) what is the typical speed of D meson that travels 1mm before decaying?

B) What is the momentum of D meson that has an energy of 5000 MeV?

2. Relevant equations

Relevant equations t = t$_{0}$$\gamma$ and L = $\frac{L_{0}}{\gamma}$
Where Gamma is the lorentz factor

and E$^{2}$ = c$^{2}$p$^{2}$ + m$^{2}$c$^{4}$

3. The attempt at a solution

For part A I thought it was just a simple time dilation and length contraction, but you have the time in the particles rest frame, and the length in the Lab frame.

For part B I tried using E$^{2}$ = c$^{2}$p$^{2}$ + m$^{2}$c$^{4}$

So p = $\sqrt{\frac{E^{2} - m^{2}c^{4}}{c^{2}}}$

p = $\sqrt{\frac{(5000x10^{6})^{2} - (1865x10^{6})^{2}(c^{4}) }{c^2}}$

But I got math error, am I propagating the 1865 MeV incorrectly? I tried converting everything into joules and kg but still didn't get correct answer of 4640 MeV

Thanks

2. Dec 23, 2012

### SteamKing

Staff Emeritus
The mass of the D meson is 1865 MeV/c^2, but I don't see where you have accounted for the c^2 factor in your formula for p.

3. Dec 23, 2012

### soothsayer

Yes, remember that a mass of 1865 MeV/c2 literally means the mass is (1865/c2) MeV/(m2/s2), so you can't just plug 1875 MeV into your equations without factoring that added factor of c-2, which will nicely cancel out with your relativistic energy equations (which is one of the reasons that particle masses are expressed in that way)

You could also do the problem in natural units (c = 1).

4. Dec 23, 2012

### DunWorry

Ok so if I do it p = $\sqrt{\frac{(5000x10^{6})^{2} - \frac{(1865x10^{6})}{(3x10^{8})^{2}}(3x10^{8})^{4}}{(3x10^{8})^{2}}}$

I get an answer of 15.46, is this in units of kgm/s? the answer is 4640 MeV/C?

5. Dec 23, 2012

### vela

Staff Emeritus
It may help you to rewrite your expression slightly as
$$p = \sqrt{\frac{E^2 - (mc^2)^2}{c^2}}.$$ You should be able to see that mc2 has units of MeV and that p will have units of MeV/c. You don't need to plug in a value for c anywhere.

For part a), pick a frame and work in it. Show us what you get.

6. Dec 23, 2012

### DunWorry

Sorry I'm not understanding this very well, I think I am confused with how to deal with the fact that the mass is given in terms of MeV/C^2 and I am unsure how to put this into the m$^{2}$C$^{4}$ part of the formula and this is why I am getting math error or 15.46. How am I supposed to deal with the mass in terms of MeV/C^2? I thought I did it correctly in the 2nd attempt but I got 15...

as for part A) well I thought its either t=t$_{0}$$\gamma$ or L = $\frac{L_{0}}{\gamma}$ But since we do not have t, or L$_{0}$, we also do not have its velocity which is what we are trying to find out so we do not know the value of $\gamma$

I just tried rewriting $\frac{t}{t_{0}}$ = $\frac{L_{0}}{L}$ = $\gamma$ and try to solve for gamma so I could try and find the velocity.

I did $\frac{L}{t_{0}}$ = $\gamma$ where L = 1mm and t0 = 4.1x10$^{-13}$ but I got the wrong answer

7. Dec 23, 2012

### vela

Staff Emeritus
The mass is 1865 MeV/c2, so mc2 = (1865 MeV/c2)c2 = 1865 MeV. The factors of c cancel.

You need to use the fact that distance equals speed times time, so if you work in the lab frame, you have 1 mm = vt, where t is the time in the lab frame, which you can relate to t0 via γ. Remember that γ only depends on v, so really you have only one unknown, namely v.

8. Dec 23, 2012

### DunWorry

Ah ok I solved part A now thanks for that.

for part B, I see, so I put in the numbers as P = $\sqrt{\frac{(5000x10^{6})^{2} - (1865x10^{6})^{2}}{(3x10^{8})^{2}}}$ this should give me the answer in eV/C? I get 15 as before.... but it isnt the correct answer it should be 4640 MeV/C, what am I doing wrong?

9. Dec 23, 2012

### vela

Staff Emeritus
Don't plug in a value for c in the denominator. That c becomes part of the units MeV/c in the final answer.

10. Dec 23, 2012

### DunWorry

Ah right thankyou! so essentially, the top row gives you units of MeV, whereas the denominator gives you units of c. Since the momentum is given in units of MeV/c you dont have to numerically divide by c. Are there any good places for me too look to develop more understanding for these units of MeV/C^2 etc?

11. Dec 24, 2012

### vela

Staff Emeritus
You pretty much summed it up in what you said: you don't divide by c numerically because c is in the units. There's nothing more to it than that.