Exploring the Advantages of Using Momentum Over Velocity in Particle Physics

[jackiepollock]

TL;DR Summary

We know that to calculate the kinetic energy of an object we usually use the formula k=1/2mv^2. On my textbook for A-Level Physics, it says, for a subatomic particle travelling at non-relativistic speeds, the formula p^2/2m is particularly useful for calculating its kinetic energy. Why is this formula used instead of the other one in this circumstance?

 

[PeroK]

It's fairly obvious that if you know the mass of a particle and anyone of its velocity/speed, momentum/magnitude of momentum or kinetic energy, then you can immediately calculate the others. There's clearly no immediate calculational advantage of $p^2/2m$ against $\frac 1 2 mv^2$ in non-relativistic, non-quantum-mechanical physics.

In SR (Special Relativity) energy and momentum are unified in the energy-momentum four-vector (which is a beautiful and profound idea, in my opinion). And, the magnitude of the energy-momentum four-vector is the particle's mass (even more profound). Although $E = mc^2$ is the most famous equation in physics, even better is:$$E^2 = p^2c^2 + m^2c^4$$which relates the energy, momentum and mass of a particle.

When you come to study relativistic collisions, it turns out that using energy and momentum is usually much simpler - mainly because of the above equation. And, if you use equations for $v$ instead, the maths tends to get very messy.

Moreover, in quantum mechanics you have the HUP (Heisenberg Uncertainty Principle) involving position ($x$) and momentum ($p$): $$\sigma_x \sigma_p \ge \frac \hbar 2$$ and generally it's the momentum of the particle that is more calculationally useful than velocity.

 

[Gaussian97]

Well, I think there could be two reasons why the last expression can be particularly useful, but neither one is really applicable in this case. In the first instance, both expressions are exactly the same, you have just posted a proof, if you know the velocity of the particle but not its momentum the first expression is better, why should you first compute the momentum to compute the energy? The same is true the other way, if you know the momentum but not the velocity, use the other one...
Again, they are exactly the same thing!

Now, it is true that for more advanced physics one will prefer using the momentum instead of the velocity:
The first reason is that the concept of energy in most situations is related to a more powerful concept, the Hamiltonian. But the math behind this requires the Hamiltonian to be expressed as a function of the position and the momentum, not the velocity.

The second reason, more practical, is that for relativistic particles the velocities cannot be as large as you want, in appropriate units they cannot be larger than 1. Therefore if you need to specify the velocity of a particle you need to give numbers like 0.9999 or 0.99999, which are not very comfortable to work with. On the other hand, the momentum can be arbitrarily large and therefore saying 70MeV or 200MeV is much easier.

 

[vanhees71]

The deeper reason to use $E_{\text{kin}}=\vec{p}^2/(2m)$ instead of $E_{\text{kin}}=m \vec{v}^2/2$, which are of course the same in classical mechanics, is that in atomic physics you have to switch to quantum theory, and there you usually do not work with velocities but with momenta (and, e.g., if magnetic fields are involved with the socalled "canonical momentum" rather than with the usual "kinematic momentum").

 

[physicsworks]

Perhaps even a deeper reason is that in atomic physics conservation laws which follow from symmetry principles for a closed system become extremely important, because they allow one to say a lot about basic features of atomic processes even without knowing details of underlying interactions. And, of course, going from classical to quantum mechanical description, conserved quantities such as energy, momentum and angular momentum are a basis for classification of quantum states.

 

[jackiepollock]

It's fairly obvious that if you know the mass of a particle and anyone of its velocity/speed, momentum/magnitude of momentum or kinetic energy, then you can immediately calculate the others. There's clearly no immediate calculational advantage of $p^2/2m$ against $\frac 1 2 mv^2$ in non-relativistic, non-quantum-mechanical physics.

In SR (Special Relativity) energy and momentum are unified in the energy-momentum four-vector (which is a beautiful and profound idea, in my opinion). And, the magnitude of the energy-momentum four-vector is the particle's mass (even more profound). Although $E = mc^2$ is the most famous equation in physics, even better is:$$E^2 = p^2c^2 + m^2c^4$$which relates the energy, momentum and mass of a particle.

When you come to study relativistic collisions, it turns out that using energy and momentum is usually much simpler - mainly because of the above equation. And, if you use equations for $v$ instead, the maths tends to get very messy.

Moreover, in quantum mechanics you have the HUP (Heisenberg Uncertainty Principle) involving position ($x$) and momentum ($p$): $$\sigma_x \sigma_p \ge \frac \hbar 2$$ and generally it's the momentum of the particle that is more calculationally useful than velocity

Thank you!

 

[jackiepollock]

Well, I think there could be two reasons why the last expression can be particularly useful, but neither one is really applicable in this case. In the first instance, both expressions are exactly the same, you have just posted a proof, if you know the velocity of the particle but not its momentum the first expression is better, why should you first compute the momentum to compute the energy? The same is true the other way, if you know the momentum but not the velocity, use the other one...
Again, they are exactly the same thing!

Now, it is true that for more advanced physics one will prefer using the momentum instead of the velocity:
The first reason is that the concept of energy in most situations is related to a more powerful concept, the Hamiltonian. But the math behind this requires the Hamiltonian to be expressed as a function of the position and the momentum, not the velocity.

The second reason, more practical, is that for relativistic particles the velocities cannot be as large as you want, in appropriate units they cannot be larger than 1. Therefore if you need to specify the velocity of a particle you need to give numbers like 0.9999 or 0.99999, which are not very comfortable to work with. On the other hand, the momentum can be arbitrarily large and therefore saying 70MeV or 200MeV is much easier.

Thank you so much!