Momentum four vector of a particle and its speed as a fraction of the speed of light

twinklestar28

1. The problem statement, all variables and given/known data
Write the momentum four vector of a particle that has:

Rest mass = 80 Gev/c^2
Energy = 100 Gev

and moves along the z axis.What is its speed as a fraction of the speed of light.

2. Relevant equations

E^2 = p^2c^2 + m^2C^4


3. The attempt at a solution

Momentum four vector - (0, 0, (3*10^10)v, 1.1*10^15) in units of Gev/c

The speed i got was v/c = 1.22*10^-34 but when I double check it, it doesn't seem to match. I'm not sure where I'm going wrong, is my four momentum vector wrong?

cepheid

Welcome to PF twinklestar28!

Your four-vector confuses me a bit. I'm used to seeing the timelike component listed first, but I am assuming that you are using a convention where the timelike component of the vector is the fourth and last one. In any case, using that convention the momentum four-vector is just given by P = (p_x, p_y, p_z, E/c), where E is the relativistic energy, and (p_x, p_y, p_z) are the components of the three-momentum, p. In this case, we have p_x = p_y = 0, and p = p_z = γmv, where v is the velocity (which is entirely in the z-direction). So your four-vector becomes

P = (0, 0, γv(80 GeV/c²), (100 GeV/c) ).

There is no sense converting anything into SI units in any part of this problem. I suspect that's the problem you're having with calculating v/c: the unit conversions. I recommend leaving everything in terms of GeV and just carrying the various factors of c along for the ride in the calculation. In fact, most physicists would use a unit system where c = 1, and everything (mass, momentum, energy) is measured in eV. The pesky factors of c would then disappear from the calculation entirely. In fact, we could write E² = p² + m² in that unit system. But for now, I'll keep all the pesky factors of c in there, to avoid confusing you (I hope). To calculate v/c, start with the equation that you wrote for the relativistic energy:$$E^2 = p^2c^2 + m^2c^4$$Solve for p:$$pc = \sqrt{E^2 - m^2c^4} = \sqrt{(100~\textrm{GeV})^2 - (80~\textrm{GeV}/c^2)^2 \cdot c^4}$$Notice that the c's cancel entirely from the second term under the square root, just leaving us with:$$pc = \sqrt{(100~\textrm{GeV})^2 - (80~\textrm{GeV})^2} = 60~\textrm{GeV}$$

So, p = (60 GeV/c). Using this, and the fact that p = γmv, you should be able to solve for v/c, which in turn tells you γ. Again, DON"T plug in values for c in SI units. Just carry the factors of c along for the ride in your algebra.

twinklestar28

Thanks guys! that was a great help. I was getting confused because i thought the four momentum vector had to be in the same units so I was trying to rearrange everything. Thanks a bunch :)

Chestermiller

What is the equation for the energy of a particle in its rest frame of reference, in terms of its rest mass and the speed of light? (Hint: Einstein)

What is the equation for the energy of a particle as reckoned from a frame of reference that is not its rest frame, in terms of its rest mass, the speed of light, and the relativity factor γ? (Hint: Planck relationship to include kinetic energy)

One of these energies is your 80 Gev, and the other is your 100 Gev.

epiclier

I calculated that the velocity was equal to 2.12E5 ms-1.

Hence, as a fraction of c : v = (7.07/10000) c


Would anybody be able to verify this for me?


Thanks in advance


-Epiclier



EDIT: sorry for the bump

cepheid

Did you get assigned the same problem as the original poster, or are you just trying it for practice?

Anyway, your answer incorrect, sorry. As I mentioned above, there should never ever be any need to convert to SI units (eg m/s) in this problem, and you should either get γ or v/c directly out as a result. Which one depends on whether you use Chestermiller's method or mine. His is by far simpler.

epiclier

Ah, I was just doing it for practise. Completely messed it up though.

I got the final answer as . Sounds good so I'm guessing thats the right answer?

Thanks again

-Epiclier

cepheid

Looks right to me.