Two frames \(R\) and \(\tilde{R}\) are moving relative to each other without rotation. A particle has kinetic energy \(T\) measured in \(R\) and \(\tilde{T}\) measured in \(\widetilde{R}\). Show that $$ T=\tilde{T}+\boldsymbol{u} \cdot \boldsymbol{p}+c $$ where \(\boldsymbol{p}=m \boldsymbol{v}\) is the momentum of the particle measure in \(R, \boldsymbol{u}\) is the velocity of \(\tilde{R}\) relative to \(R\), and \(c\) is a constant.

Short Answer: To prove the relationship between the kinetic energy of a particle in two different reference frames R and R̃, we first define the kinetic energy in both frames and then relate its velocities using the Galilean transformation. In Frame R, kinetic energy is represented by T, and in Frame R̃, it is represented by \(\widetilde{T}\). The Galilean transformation states that the particle's velocity in one frame (R̃) is equal to the sum of its velocity in the other frame (R) and the relative velocity between both frames (u). We then express the kinetic energy in Frame R̃ in terms of velocities in Frame R, expand the equation, and introduce the momentum. Finally, we introduce a constant (c) to account for the difference between the two frames and obtain the final result: \(T = \widetilde{T} + \boldsymbol{u} \cdot \boldsymbol{p} + c\), representing the relationship between the kinetic energy measured in the two reference frames.