Nonrelativistic Lab-frame Velocity as a function of kinetic energy per unit mass

[fliptomato]

Hi everyone--I'm a bit stuck trying to follow a calculation in an article (Nucl. Phys. B360 (1991) p. 145-179) regarding the lab-frame relative velocity of two colliding particles as a function of the kinetic energy per unit mass. (I'll include references to equations in the article, but the article is not required for this particular question.)

Suppose we have two particles of mass $$m$$ colliding with one another with energies $$E_1, E_2$$ and 3-momenta $$\mathbf{p_1}, \mathbf{p_2}$$. Define $$p_1 = |\mathbf{p_1}|, p_2 = |\mathbf{p_2}|$$

Define (eq. 3.3)
$$s = 2m^2 + 2E_1E_2 - 2p_1p_2\cos\theta$$

Which, I believe is the same as the center of mass energy $$(p_1^\mu+p_2^\mu)^2$$.

Now define the kinetic energy per unit mass in the lab frame, (3.20)
$$\epsilon = \frac{(E_{1,\mathrm{lab}}-m)+(E_{2,\mathrm{lab}}-m)}{2m}$$

First question: Why can we write
$$\epsilon=\frac{s-4m^2}{4m^2}$$

Second question: Why is is true that the lab velocity is given by
$$v_\mathrm{lab} = \frac{2\epsilon^{1/2}(1+\epsilon)^{1/2}}{1+2\epsilon}$$


Thanks very much for any assistance!

Flip

 

[Meir Achuz]

Since s includes cos\theta in the general case, and \epsilon does not, the equation
\epsilon=(s-4m^2)/4m^2 cannot be true in general. It only holds for the special case of one particle being initialy at rest. That is the meaning of the subscript _lab.
The equation is easy to derive in this case.
The equation for v_lab is also for one particle initially at rest.
If you start with v=p/E, the equation is easily derived.
The calculation is fully relativistic.

 

[charng]

I understand that it can be challenging to follow calculations in scientific articles, especially when there are equations involved. However, I will do my best to explain the concepts and answer your questions.

First, let's define some terms for clarity. The lab-frame velocity refers to the velocity of the particles as measured in the laboratory frame, which is the frame of reference of the observer conducting the experiment. Kinetic energy per unit mass is a measure of the energy of a particle relative to its mass, and it is often used in particle physics to compare the energies of different particles.

Now, to answer your first question, we can write the kinetic energy per unit mass as \epsilon = \frac{(E_{1,\mathrm{lab}}-m)+(E_{2,\mathrm{lab}}-m)}{2m} because in the lab frame, the total energy of a particle is given by its rest mass (m) plus its kinetic energy (E_{\mathrm{lab}}). This equation simply takes into account the kinetic energy contributions from both particles in the collision.

For your second question, the lab velocity is given by v_{\mathrm{lab}} = \frac{2\epsilon^{1/2}(1+\epsilon)^{1/2}}{1+2\epsilon} because it takes into account the conservation of energy and momentum in the collision. The term \epsilon^{1/2} represents the kinetic energy per unit mass of the particles, and the term (1+\epsilon)^{1/2} is related to the total energy of the particles in the lab frame. This equation is derived using the equations for conservation of energy and momentum, and it allows us to calculate the lab-frame velocity based on the kinetic energy per unit mass of the particles.

I hope this helps to clarify the concepts and equations involved. If you have any further questions or need clarification, please do not hesitate to ask. As scientists, it is important to continuously seek understanding and ask questions when something is not clear. Good luck with your calculations!