# Center of mass velocity greater than 1c?

1. Sep 28, 2012

### randombill

center of mass velocity greater than 1c???

Basically I have two electrons moving in opposite directions towards each other.
One is moving 2.5x10^8 m/s and the other is 2.0x10^8 m/s and I'm trying to use the
following equation for finding the center of mass velocity:

V_cm = ((p1 + p2)c^2)/(E1 + E2)

I calculated the momentum's and energies to be,

p1 = 456280 eV/c
p2 = 769846 eV/c
E1 = 174216 eV
E2 = 413132 eV

and the V_cm = 2.08756 c?

Is that supposed to happen, shouldn't it be less than 1?

2. Sep 28, 2012

### mathman

Re: center of mass velocity greater than 1c???

I am not at all familiar with the formula you are using. However, the c.m. (since they are going in opposite directions) is moving at 0.25 x 10^8 m/s in the same direction as the faster electron.

3. Sep 28, 2012

### randombill

Re: center of mass velocity greater than 1c???

How did you calculate that btw?

Basically I'm going off these two pages...

http://www.rafimoor.com/english/PECE.htm
http://teachers.web.cern.ch/teacher...ch/mbitu/applications_of_special_relativi.htm

I'm trying to calculate a 1-d elastic collision.

4. Sep 28, 2012

### Staff: Mentor

Re: center of mass velocity greater than 1c???

Your p1 and p2 are correct, but E1 and E2 are not. How did you calculate E1 and E2?

Reality check: the energy in eV must be larger than the magnitude of the momentum in eV/c, because $E^2 = (pc)^2 + (mc^2)^2$.

Last edited: Sep 28, 2012
5. Sep 28, 2012

### randombill

Re: center of mass velocity greater than 1c???

I used:

E_kin eV = (mass kg c^2(gamma -1))/(1.609E-19eV) (1)

Am I suppose to use the total energy including the kinetic energy so it would be:

E_total eV = (mass kg c^2(gamma +1)/(1.609E-19eV)) (2)

Or I'll just use $E^2 = (pc)^2 + (mc^2)^2$ instead. (3)

I got V_cm = .46c with equation (2), haven't tried (3) yet.

6. Sep 28, 2012

### Staff: Mentor

Re: center of mass velocity greater than 1c???

Yes, you need to use the total energy in v = pc2/E.

No, it's $E = \gamma m c^2$ not $E = (\gamma + 1) m c^2$. And then your numeric factor to convert from joules to eV.

It's useful to remember that for an electron, mc2 = 5.11 x 105 eV = 511 keV.

When I know either the energy or the momentum, I always use $E^2 = (pc)^2 + (mc^2)^2$ to get the other one (provided I know the energy equivalent of the mass, of course, which is what Real Physicists [TM] always use anyway).

Last edited: Sep 29, 2012
7. Sep 29, 2012

### randombill

Re: center of mass velocity greater than 1c???

Well I tried it both ways now and got V_cm ~ .76c so that must be right. Using the momentum relation maybe the physicist's way of doing it but in the computer world using square roots are frowned upon. I'm writing this with C++ btw. The other way is much faster (without square root) and I would also guess more accurate although that is something else entirely different from this discussion. Thanks anyways; let me know if you think the velocity is wrong.

8. Sep 29, 2012

### mathman

Re: center of mass velocity greater than 1c???

The calculation ignores relativity. The electrons have the same mass, so the momentum is determined by the velocities. The center of mass velocity is simply the average of the two velocities. This assumes the problem is one dimensional.

9. Sep 29, 2012

### randombill

Re: center of mass velocity greater than 1c???

Its interesting to do it that way but is it correct?

10. Sep 29, 2012

### Staff: Mentor

Re: center of mass velocity greater than 1c???

No, that's not it. I suspect that you didn't take into account that the two electrons are moving in opposite directions in your frame, so the momentum of one of them must be negative, whereas the energy is always positive.

No, it's not. I checked my answer by using the relativistic "velocity addition" formula to calculate the velocities of the two electrons in the c.o.m. frame. They come out equal in magnitude and opposite in direction, as expected.

Last edited: Sep 29, 2012
11. Sep 30, 2012

### mathman

Re: center of mass velocity greater than 1c???

In the center of mass system (two equal masses) the velocity should be equal and in opposite direction - I don't understand your objection to calculating the center of mass velocity as the average.

12. Sep 30, 2012

### randombill

Re: center of mass velocity greater than 1c???

Ok I fixed it now I have the velocity of the first electron at .83c and -.66c for the second one and the final velocities are -.66c and .83c respectively. That makes more sense.

13. Sep 30, 2012

### Staff: Mentor

Re: center of mass velocity greater than 1c???

The arithmetic average of the velocities in the "lab frame" simply does not give v_cm in a relativistic situation. Calculate v_cm using the equation in post #1. The result does not equal the average of the velocities, (0.833c + (-0.667c))/2 = 0.083c.

Alternatively, use 0.083c as the velocity of a new reference frame, and use the relativistic "velocity addition" formula to calculate the velocities of the two electrons in that frame. They turn out not to be equal in magnitude (in opposite directions, of course), in that frame.

Last edited: Sep 30, 2012
14. Sep 30, 2012

### randombill

Re: center of mass velocity greater than 1c???

I've tried using "fraud mechanics" to get answers that are quasi relativistic but the problem lies within the energy-momentum relation and the fact that that equation works. Getting around that and getting a decent answer is impossible. Same goes for Newtonian mechanics with the work-energy theorem and the conservation of momentum, there is no way to simply fudge the formulas and get a decent result.

15. Oct 1, 2012

### mathman

Re: center of mass velocity greater than 1c???

My comments were always in the context of a non-relativistic situation and I tried to make it clear.

In a relativistic situation, obviously it is somewhat more complicated.

16. Oct 1, 2012

### Staff: Mentor

Re: center of mass velocity greater than 1c???

No wonder I was confused. This is the relativity forum, after all.