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Buzz Bloom

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- Since L = T - V, and T equals the kinetic energy (KE) of a particle whose trajectory is to be calculated, how is KE defined since some of its motion will be due to the expanding universe?

**Summary:**Since L = T - V, and T equals the kinetic energy (KE) of a particle whose trajectory is to be calculated, how is KE defined since some of its motion will be due to the expanding universe?

My understanding may well be wrong, but it is the following.

if a particle is stationary at point Y with respect to (WRT) comoving coordinates (CC) , then its motion WRT a reference point X which is also stationary WRT CC is entirely due to Hubble's law, and this motion does not contribute to the particle's KE relative to X. If the distance between X and Y is D, then the the velocity of the particle as seen by an observer at X is H

_{0}D. (H

_{0}is Hubble's constant.) However, this quantity does not contribute to the KE of the particle as seen by the observer at X.

Is the above correct? If it is not, please post an explanation of how this KE is related to the calculation of a trajectory using Lagrangian mechanics. If it is correct, then is the following also correct?

If a particle of mass M is moving at velocity v relative to the point it occupies which is stationary WRT CC then the particle has

KE = (1/2)Mv

^{2}

as seen by an observer stationary WRT CC at any point.

Is this correct? If not please post any help you might have WRT this topic which you have conveniently available.

If it is correct, I would appreciate some help in understanding how to use L = T-V to set up an equation for the trajectory of a particle in an expanding universe satisfying Hubble's law. To keep the problem simple, assume no gravitating bodies and that any changes in H

_{0}are insignificant.