What is \(\gamma\) for a proton having a mass energy of 938.3 MeV accelerated through an effective potential of \(1.0 \mathrm{TV}\) (teravolt)?

The relativistic gamma factor, denoted as \(\gamma\), is a crucial concept in the realm of special relativity. It represents the factor by which time, length, and relativistic mass change for an object moving relative to an observer. The formula for the gamma factor is \(\gamma = \frac{1}{\sqrt{1 - (v/c)^2}}\), where \(v\) is the velocity of the object and \(c\) is the speed of light in a vacuum.

As an object's speed increases towards the speed of light, the gamma factor increases significantly, indicating substantial time dilation and length contraction. In our textbook exercise, we calculated the gamma factor for a proton accelerated to a significant fraction of the speed of light. Understanding this concept helps us comprehend how particles behave at high speeds, fundamental for applications such as particle accelerators and astrophysics.

Proton acceleration is a process used in physics to increase the energy and, consequently, the speed of protons. This is often achieved using electric fields, such as the teravolt potential mentioned in our exercise. When protons are accelerated through these high voltages, they gain kinetic energy. The energy gained can be calculated using the formula \(E_{gained} = qV\), where \(q\) is the charge of the proton and \(V\) is the potential difference.

Accelerating protons to high speeds enables us to observe relativistic effects and is a technique widely used in particle physics research, including in the Large Hadron Collider (LHC). By understanding how protons accelerate and the resulting energies involved, we delve into fundamental questions about matter and the universe.

In physics, it's often necessary to convert between different units of energy. This is especially true when dealing with relativistic calculations, where energy can be described in terms of electron volts (eV) or joules (J). The conversion is straightforward with the use of conversion factors. 1 eV, which is the energy gained by an electron when it is accelerated through an electric potential of 1 volt, is equivalent to \(1.6 \times 10^{-19} J\).

In our exercise, we converted the energy gained by a proton, initially given in teravolts, to mega electron volts (MeV), which is a unit more commonly used in particle physics. This conversion allows for easier comparison of energy values when analyzing particle collisions or decays. Remembering these conversion factors, \(1\,\text{eV} = 1.6 \times 10^{-19} J\) and \(1\,\text{MeV} = 10^6\,\text{eV}\), is essential when performing energy calculations in the realm of high-energy and nuclear physics.