Center of mass velocity greater than 1c?

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?

mathman

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.

jtbell
Mentor

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##.

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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##.

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.

jtbell
Mentor

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

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

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).

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, 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).

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.

mathman

How did you calculate that btw?

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.

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.

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

jtbell
Mentor

Well I tried it both ways now and got V_cm ~ .76c so that must be right.

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.

The center of mass velocity is simply the average of the two velocities.

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.

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mathman

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.

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.

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.

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.

jtbell
Mentor

I don't understand your objection to calculating the center of mass velocity as the average.

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.

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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.

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.

mathman

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.

jtbell
Mentor

My comments were always in the context of a non-relativistic situation

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