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

In summary, the momentum four vector of a particle is a mathematical quantity that describes its momentum and energy in spacetime. It is composed of four components, including the energy component and the three momentum components. The magnitude of the momentum four vector is directly proportional to the speed of the particle, and as the particle's speed increases, so does the magnitude of its momentum four vector. The speed of a particle as a fraction of the speed of light indicates how close the particle is to reaching the maximum speed. It is calculated by dividing the magnitude of the particle's momentum four vector by its energy component, and it is a relative value that depends on the observer's frame of reference.
  • #1
twinklestar28
21
0

Homework Statement


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.

Homework Equations



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


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?
 
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  • #2


Welcome to PF twinklestar28!

twinklestar28 said:

Homework Statement


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.

Homework Equations



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


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?

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 = (px, py, pz, E/c), where E is the relativistic energy, and (px, py, pz) are the components of the three-momentum, p. In this case, we have px = py = 0, and p = pz = 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.
 
Last edited:
  • #3


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 :)
 
  • #4


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.
 
  • #5


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

Hence, as a fraction of c : v = (7.07/10000) cWould anybody be able to verify this for me?Thanks in advance-Epiclier
EDIT: sorry for the bump
 
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  • #6
epiclier said:
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


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.
 
  • #7


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

I got the final answer as
v=3/5 c
. Sounds good so I'm guessing that's the right answer?

Thanks again

-Epiclier
 
  • #8


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

I got the final answer as
v=3/5 c
. Sounds good so I'm guessing that's the right answer?

Thanks again

-Epiclier

Looks right to me.
 

What is the momentum four vector of a particle?

The momentum four vector of a particle is a mathematical quantity used in special relativity to describe the momentum and energy of a particle in spacetime. It contains four components: the energy component (E) and the three momentum components (px, py, and pz).

How is the momentum four vector related to the speed of a particle?

The magnitude of the momentum four vector (|p|) is directly proportional to the speed of the particle. This means that as the particle's speed increases, so does the magnitude of its momentum four vector.

What is the significance of the speed of a particle as a fraction of the speed of light?

The speed of light is considered to be the maximum speed at which any object can travel in the universe. Therefore, the speed of a particle as a fraction of the speed of light tells us how close the particle is to reaching this maximum speed.

How is the speed of a particle as a fraction of the speed of light calculated?

The speed of a particle as a fraction of the speed of light can be calculated by dividing the magnitude of the particle's momentum four vector (|p|) by its energy component (E). This ratio will always be less than or equal to 1, as the speed of light is the maximum speed.

Is the speed of a particle as a fraction of the speed of light an absolute value?

No, the speed of a particle as a fraction of the speed of light is not an absolute value. It is a relative value that depends on the observer's frame of reference. This means that different observers may measure different fractions of the speed of light for the same particle.

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