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## Main Question or Discussion Point

As the title says, is energy Galilean invariant?

I'm fairly sure it isn't, since if one considers the simple case of a free particle, such that its energy is ##E=\frac{p^{2}}{2m}##, then under a Galilean boost, it follows that ##E'= \frac{p'^{2}}{2m}=E+\frac{\tilde{p}^{2}}{2m}-\frac{\mathbf{p}\cdot\tilde{\mathbf{p}}}{m}##, where ##\mathbf{p}'=\mathbf{p}-\tilde{\mathbf{p}}##, with ##\tilde{\mathbf{p}}=m\tilde{\mathbf{v}}##, and ##\tilde{\mathbf{v}}## is the relative velocity between the two inertial frames.

I mean, it seems obvious, since momentum is not Galilean invariant (indeed, it transforms as ##\mathbf{p}'=\mathbf{p}-\tilde{\mathbf{p}}##), but I'm having a momentary crisis of confidence in my understanding and so wanted to check!

I'm fairly sure it isn't, since if one considers the simple case of a free particle, such that its energy is ##E=\frac{p^{2}}{2m}##, then under a Galilean boost, it follows that ##E'= \frac{p'^{2}}{2m}=E+\frac{\tilde{p}^{2}}{2m}-\frac{\mathbf{p}\cdot\tilde{\mathbf{p}}}{m}##, where ##\mathbf{p}'=\mathbf{p}-\tilde{\mathbf{p}}##, with ##\tilde{\mathbf{p}}=m\tilde{\mathbf{v}}##, and ##\tilde{\mathbf{v}}## is the relative velocity between the two inertial frames.

I mean, it seems obvious, since momentum is not Galilean invariant (indeed, it transforms as ##\mathbf{p}'=\mathbf{p}-\tilde{\mathbf{p}}##), but I'm having a momentary crisis of confidence in my understanding and so wanted to check!