how to prove prove m = m0 / sqrt(1 - v^2/c^2)?
If you are asking, you don't understand it (Einstein quote).
http://en.wikipedia.org/wiki/Lorentz_factor
See the section on derivation.
To prove the equation \(m = \frac{{m_0}}{{\sqrt{1 - \frac{{v^2}}{{c^2}}}}}\), we will work through a derivation using the principles of special relativity.
Special relativity is a theory that describes the relationship between space and time in the presence of relative motion between observers. One of its key concepts is the Lorentz transformation, which relates coordinates and time between different inertial reference frames.
Let's begin the derivation:
Step 1: Start with the definition of relativistic momentum.
In special relativity, the momentum is defined as \(p = \frac{{m_0v}}{{\sqrt{1 - \frac{{v^2}}{{c^2}}}}}\), where \(m_0\) is the rest mass of an object, \(v\) is its velocity, and \(c\) is the speed of light in a vacuum.
Step 2: The equation for momentum can also be expressed as \(p = mv\), where \(m\) represents the relativistic mass.
Step 3: Equate the expressions for momentum.
Setting the expressions in steps 1 and 2 equal to each other, we have \(\frac{{m_0v}}{{\sqrt{1 - \frac{{v^2}}{{c^2}}}}}\) = \(mv\).
Step 4: Simplify the equation.
Next, we can simplify the equation by multiplying both sides by \(\sqrt{1 - \frac{{v^2}}{{c^2}}}\):
\(m_0v\) = \(mv\sqrt{1 - \frac{{v^2}}{{c^2}}}\).
Step 5: Cancel out common factors.
Note that \(v\) is present on both sides of the equation. Dividing both sides by \(v\) allows us to cancel it out:
\(m_0\) = \(m\sqrt{1 - \frac{{v^2}}{{c^2}}}\).
Step 6: Solve for \(m\).
Finally, divide both sides of the equation by \(\sqrt{1 - \frac{{v^2}}{{c^2}}}\):
\(m\) = \(\frac{{m_0}}{{\sqrt{1 - \frac{{v^2}}{{c^2}}}}}\).
Therefore, we have derived the equation \(m = \frac{{m_0}}{{\sqrt{1 - \frac{{v^2}}{{c^2}}}}}\). This equation relates the relativistic mass (\(m\)) of an object with its rest mass (\(m_0\)) and its velocity (\(v\)).
Note: It is essential to understand that the concept of relativistic mass is no longer in common use. The modern understanding of relativity, as described by Einstein's general theory of relativity, focuses on the concept of invariant mass (rest mass). The equation derived here is a historical artifact and may not be needed for understanding modern physics.