Understanding Kinetic Energy and Lorentz Factor

This question asks us to find the Lorentz factor, often represented by the Greek letter gamma ($\gamma$), for a proton given its kinetic energy. The kinetic energy is provided as 3752 MeV.

In physics, especially when dealing with particles moving at speeds close to the speed of light, we need to use relativistic mechanics. The Lorentz factor ($\gamma$) is a key component in these calculations. It relates measurements in different inertial frames of reference.

The relationship between a particle's total energy ($E$), its rest energy ($E_0$), its kinetic energy ($KE$), and the Lorentz factor ($\gamma$) is given by the following equations:

• Total Energy: $E = \gamma E_0$
• Total Energy also equals Rest Energy plus Kinetic Energy: $E = E_0 + KE$

By setting these two expressions for total energy equal, we get:

$\gamma E_0 = E_0 + KE$

We can rearrange this formula to solve for the Lorentz factor ($\gamma$):

$\gamma = \frac{E_0 + KE}{E_0}$

This can also be written as:

$\gamma = 1 + \frac{KE}{E_0}$

Calculating the Proton's Lorentz Factor

To calculate $\gamma$, we need two values:

1. The given kinetic energy ($KE$) of the proton, which is 3752 MeV.
2. The rest energy ($E_0$) of a proton.

The rest mass of a proton ($m_p$) is approximately $938$ MeV/$c^2$. Therefore, the rest energy ($E_0$) of a proton is approximately:

$E_0 = m_p c^2 \approx 938$ MeV

Step-by-Step Calculation:

1. Identify the known values:
   • Kinetic Energy ($KE$) = 3752 MeV
   • Rest Energy of proton ($E_0$) $\approx$ 938 MeV
2. Use the formula for the Lorentz factor:

   $\gamma = 1 + \frac{KE}{E_0}$

3. Substitute the values into the formula:

   $\gamma = 1 + \frac{3752 \text{ MeV}}{938 \text{ MeV}}$

4. Perform the division:

   $\frac{3752}{938} \approx 3.999$

5. Complete the addition:

   $\gamma \approx 1 + 3.999$

   $\gamma \approx 4.999$

6. Approximate the result:

   The calculated value of $\gamma$ is very close to 5.

Conclusion

Therefore, the Lorentz factor ($\gamma$) for a proton with a kinetic energy of 3752 MeV is approximately 5.