# Calculating v/c of a proton and electron

1. Oct 21, 2009

### chenying

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

Calculate the speed parameter v/c of a particle with a momentum of 12.5 MeV/c if the paricle is

proton

2. Relevant equations

E$$_{}2$$ = m$$_{}2$$c$$_{}4$$ + p$$_{}2$$c$$_{}2$$

3. The attempt at a solution

Btw, the numbers above are powers.

Ok, so i followed what my teacher did:

E$$_{}2$$ = m$$_{}2$$c$$_{}4$$ + p$$_{}2$$c$$_{}2$$

m is so small for a proton that it is 0, so that is neglected

E$$_{}2$$ = p$$_{}2$$c$$_{}2$$

E = p*c = mv$$\gamma$$ * c

Multiply the right side by c/c to get:

E = p*c = mv/c$$\gamma$$ * c^2

then my teacher moved mc^2 to the other side to get

E/mc^2 = v/c*$$\gamma$$

Finding the value of E/mc^2, and solving for v/c, I got the value of 1, which makes sense.

That is, however, not the right answer.

Can anyone tell me what I did wrong?

2. Oct 21, 2009

### turin

No. Did you mean momentum = 12.5 GeV/c? What is the mass of the proton (in MeV/c2)?

3. Oct 21, 2009

### chenying

No, I meant 12. MeV.

and I do not know what you mean by the mass of a proton in MeV/c^2

4. Oct 21, 2009

### turin

In this case, the approximation that mp=0 is quite bad. Look up "proton rest energy". The first website that google found for me gave the the mass of the proton in MeV/c2. Also, the Particle Data Group is an authoritative reference for this kind of thing, and they also have a website.

5. Oct 21, 2009

### chenying

Well, the value I found for MeV/c2 is 938.27.

So I multiplied by c to get MeV/c, which is 2.814811e11

Why is the assumption that mass of proton is 0 bad? I did the same process for the electron and I got 1, which was correct.

Last edited: Oct 21, 2009
6. Oct 21, 2009

### w3390

First of all, it should be a glaring red flag if you get the ratio of the speed of a proton to the speed of light to be 1. This is saying that the proton is traveling at the speed of light which is impossible. When I went through and solved for the velocity of the proton, I found it to be much smaller than the speed of light.

On to the problem: you have a problem that is asking for the ratio of the speed of the proton to the speed of light. There are two main unknowns, the relativistic energy of the proton and the speed of the proton. Therefore, I would suggest using two equations. I personally would use these two equations:

1. (u/c)=(pc)/E, where u is the velocity of the proton
2. E=(mc^2)/(sqrt(1-(u^2/c^2)))

Plugging equation 2 into equation 1 eliminates the energy and you can do some algebra to solve for u. By the way, when doing this algebra, do not plug any numbers in until you have simplified completely. If you do things correctly, you will never have to plug in the value of c.

Responding to your most recent post, do not multiply the 938.27 by c. You will see that the c^2 will cancel and you won't have to worry about. Try out what I have just said and see if it makes sense.

7. Oct 21, 2009

### chenying

Thanks, that helped a lot.

But where did you get equation 1? Is it a ratio or was it derived?

8. Oct 21, 2009

### chenying

does the p in equation 1 represent?

9. Oct 21, 2009

### w3390

Equation 1 can be found by multiplying the equation for relativistic momentum by c^2 and comparing it to the equation for relativistic energy. The p in the equation represents momentum.

10. Oct 21, 2009

### chenying

So then what is the p in the equation?

And I did the algebra, and tried to solve for v, but i keep running into trouble.

I keep getting a value of 0 whenever I use the algebra.........

11. Oct 21, 2009

### w3390

Once again, the p in the equation is momentum. I cannot see what you have done to get zero. Maybe post the steps leading up to the problem and I may be able to tell you where you went wrong.

12. Oct 21, 2009

### chenying

$$\frac{v}{c}$$ = $$\frac{pc}{E}$$

$$\frac{v}{c}$$ = $$\frac{mv\gamma}{mc^2\gamma}$$

in which case becomes

$$\frac{1}{}$$ = $$\frac{1}{c}$$

i realized originally i made a mistake but now i got something different......

13. Oct 21, 2009

### w3390

Okay, first of all just leave the p as it is throughout all of the algebra. Since you know what p is you can just leave it as p in your equations. So instead of mv(gamma), it should be pc. Then you can solve for v by multiplying by c. This gives you v=(pc^2)/E. Take it from here without substituting anything in for p. Also, write gamma as
1/(sqrt(1-(v^2/c^2))). You should then manipulate the equation until you are able to easily square out the square root. Then solve for v.

14. Oct 21, 2009

### turin

The MeV and c are unit factors, not physical quantities. (Well, they are also physical quantities, but we are using them as units in this problem). You cannot compare MeV/c2 to MeV/c any more than you can compare, say, velocity and acceleration. The numerical value of 938.27 does not change, regardless of multiplication by c, because c is already included in the unit. So, the thing to do is to compare the momentum to mc.

mc = (938.27 MeV/c2)c = 938.27 MeV/c

p = 12.5 MeV/c

So, p << mc, which means that the mass is nonnegligible. You cannot assume m=0. In fact, doing so is tantamount to assuming v=c, period. Whoever told you to use this approximation in order to solve the problem was just plain wrong. Use:

pc = γβmc2

where γ=1/sqrt(1-β2), and β=v/c.

15. Oct 21, 2009

### chenying

um..... I solved for v and got 3e8.............. I have no idea what I'm doing wrong......

sorry if im frustrating you btw.

16. Oct 21, 2009

### chenying

Turin:

I did it your way and I also got that v = 3e8......

17. Oct 22, 2009

### turin

We cannot help you effectively if you don't show us your work. Let's try an intermediate calculation: determine β2. Also, to give you a clue about what's going on, you should realize that the numerical value of β is actually the numerical value of speed in units of c.

BTW, don't worry about frustrating me. If frustration were a problem, I would have stopped coming here years ago. :)

edit:
Oh, I just realized that β is precisely what the problem asks you to calculate. Anyway, though, I hope that you will begin to realize that it's all just a matter of your unit system, and using c as a unit is particularly convenient in relativity problems.

Last edited: Oct 22, 2009