For a closed system we have energy conservation and this is deeply tied into the notion of the physics being unaltered under shifts in time. A salient point here is that we have fixed a frame in which to measure the energy. It is not true that the energy is the same measured in any inertial frame, what is true is that it is conserved.<\/p>\n
A simple example<\/strong><\/p>\n Let us consider a free particle of mass \\(m\\) in one dimension. Let us pick some coordinates \\((x,t)\\). With respect to this frame we have that the kinetic energy<\/p>\n \\(E = \\frac{1}{2 m}p^{2}\\),<\/p>\n where the momentum \\(p \\) is given by \\(m v = m \\dot{x} \\). The dot is the derivative with respect to time. (I am using the usuall abuses of notaion here, this should not confuse)<\/p>\n Now let us consider a Galilean transformation<\/p>\n \\(x’ = x {-} ut\\) Then we calculate how the momentum transforms<\/p>\n \\(p = m \\frac{d}{dt}\\left( x’ – ut \\right) = m \\left( v’ +u \\right)= p’ + mu\\).<\/p>\n Or \\(p’ = p – mu\\).<\/p>\n This is exactly what you would expect. So on to the energy…<\/p>\n \\(E = \\frac{1}{2m}p^{2} = \\frac{1}{2} \\left(p’ + mu \\right)^{2}\\) so<\/p>\n \\(E’ = E + \\frac{1}{2}mu^{2} -up\\).<\/p>\n Clearly \\(E’ \\neq E\\)<\/p>\n .<\/p>\n Recap<\/strong><\/p>\n Energy conservation should not be confused with energy being observer independent, it clearly is not.<\/p>\n","protected":false},"excerpt":{"rendered":" For a closed system we have energy conservation and this is deeply tied into the notion of the physics being unaltered under shifts in time. A salient point here is that we have fixed a frame in which to measure the energy. It is not true that the energy is the same measured in any … Continue reading Energy is not frame independent<\/span>
\n\\(t’ =t\\).<\/p>\n
\n\\(= E’ + \\frac{1}{2}mu^{2} + up’ = E’ – \\frac{1}{2}mu^{2}+ up\\)<\/p>\n