I was trying to do some studying regarding the definitions of energy, and I've hit a road block. I know that since velocity is frame-dependent, an object's kinetic energy as observed in two different frames will yield two different results. My initial expectation was that changes in kinetic energy of an object should be the same regardless of the reference frame (as a change is kinetic energy is a scalar and therefore invariant under coordinate transformations). I tried to work this out however in the following manner: Say you have an object that is considered "moving" in two frames: S and S'. The object is initially moving at v1 in S and v'1 in S'. It experiences a force which changes the velocity to v2 in S and v'2 in S'. v is the velocity of the object moving in frame S v' is the velocity of the object moving in a frame S' which is moving with respect to frame S. This movement is characterized by the constant velocity vector vc. I assume v'=v+vc. Now the "generic" differential I'm working with here is that d(mu2) (I have omitted the 1/2 factor out of the diffferential while I work for simplicity's sake and u here is just a "generic" velocity that is not related to the velocity of the scenario). Now, assuming that the mass of the object does not change, I get that the differential in frame S is md(v2) and the differential in frame S' is md(v'2). What I tried to do next was to expand the magnitude of v' and got that v'2=v2+vc2+2(dot_prod(v,v')). Applying the differential over this I get: d(v'2)=d(v2)+2(d(dot_prod(v,v')) (Note: The vc2 contribution is zero since this value is constant) Does the dot product contribution zero-out? Is my math correct? Did I do something wrong? When I do the integration for these terms from the first set of velocities to the second set, the dot product term will not cancel out and thus indicates that the change in kinetic energy as seen in one reference frame differs from that a reference frame moving at uniform velocity with respect to the former (which to me physically makes no sense).