Prove that for any relative velocity \(v\) between two observers, a beam of light sent from one to the other will approach at speed \(c\) (provided that \(v\) is less than \(c\), of course).

We are given the relative velocity between two observers, represented by \(v\). The speed of light \(c\) is a constant value. We will use the Lorentz transformation formulas, which relate the coordinates and time of an event in one inertial frame (observer's frame) to the coordinates and time in another inertial frame (light's frame) moving with a velocity \(v\) relative to the first frame. The Lorentz transformation formulas are given by: \[x' = \gamma(x - vt)\] \[t' = \gamma(t - \frac{v}{c^2}x)\] where, - \(x'\) and \(t'\) are the position and time in the light's frame - \(x\) and \(t\) are the position and time in the observer's frame - \(\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\) is the Lorentz factor

In the light's frame, the speed of light is \(c\). To find the speed of light in the observer's frame, we need to find the coordinate differences in both frames. To do this, we must take the derivative of both the coordinates with respect to time in their respective frames. Using the Lorentz transformation formulas, we differentiate both equations with respect to time: \[\frac{dx'}{dt'} = \frac{d}{dt'}(\gamma(x - vt))\] \[\frac{dt'}{dt'} = \frac{d}{dt'}(\gamma(t - \frac{v}{c^2}x))\] Now, we use the chain rule to differentiate with respect to time: - \(\frac{dx'}{dt'} = \gamma(\frac{dx}{dt} - v)\) - \(\frac{dt'}{dt'} = 1\)

The speed of light in the observer's frame is given by the derivative of position with respect to time in that frame (\(\frac{dx}{dt}\)). Using the expression for \(\frac{dx'}{dt'}\), we can solve for \(\frac{dx}{dt}\). \[\frac{dx'}{dt'} = \gamma(\frac{dx}{dt} - v)\] Since we know that the speed of light in the light's frame (\(x'\)) is \(c\): \[\frac{dc}{dt'} = \gamma(\frac{dx}{dt} - v)\] Solving for \(\frac{dx}{dt}\): \[\frac{dx}{dt} = \frac{dc}{\gamma dt'} + v\] Since the speed of light is constant, \(\frac{dc}{dt'} = 0\). Thus, in the observer's frame, the speed of light is: \[\frac{dx}{dt} = v\]

Using the Lorentz transformation, we have shown that the speed of light remains constant, and is equal to the relative velocity between the two observers (\(v\)). Therefore, a beam of light sent from one observer to another will approach at the speed of light \(c\), provided that the relative velocity is less than \(c\).