Kinetic energy transformation

bernhard.rothenstein · Mar 31, 2007

Please let me know a reference where the transformation of the kinetic energy is performed.
Thanks

country boy · Apr 2, 2007

Since the kinetic energy is the total energy minus the rest energy, and since the rest energy is invariant, won't any reference suffice that gives the transformation of total energy? That is, K = E - mc^2, m is invariant, and E transforms in the usual way.

bernhard.rothenstein · Apr 2, 2007

kinetic energy

country boy said: ↑

Since the kinetic energy is the total energy minus the rest energy, and since the rest energy is invariant, won't any reference suffice that gives the transformation of total energy? That is, K = E - mc^2, m is invariant, and E transforms in the usual way.

Thanks. There is a poblem. K is a physical quantity for which we do not define a proper value whereas for E we do.

jtbell · Apr 2, 2007

If by "proper" you mean the value in an object's own rest frame, then the "proper kinetic energy" of any object is zero!

country boy · Apr 2, 2007

bernhard.rothenstein said: ↑

Thanks. There is a poblem. K is a physical quantity for which we do not define a proper value whereas for E we do.

I suggest that you will need to calculate K in each frame after transforming E. Kinetic energy is not a component of a 4-tensor, since it is just the additional energy imparted by motion. However, it can be written

K = mc^2 [(1/sqrt(1-(v/c)^2))-1]

robphy · Apr 2, 2007

Geometrically, the relativistic kinetic energy is difference between the projections of two 4-momenta-related-by-a-boost onto one of those 4-momenta.

bernhard.rothenstein · Apr 2, 2007

robphy said: ↑

Geometrically, the relativistic kinetic energy is difference between the projections of two 4-momenta-related-by-a-boost onto one of those 4-momenta.

Thanks. For a less sophisticated audience I would present the problem as:
Start with the expression of the kinetic energy in I
K=mcc[(1/g(u))-1] (1)
where g(u) stands for gamma as a function of the speed of the bullet in I,
Express the right side of (1) as a function of u' the speed of the bullet in I' via the composition law of parallel speeds in order to obtain
K=mcc[(1+u'V/cc)/g(V)g(u'))-1] (2)
The transformation equation (2) leads to the following consequences
a. For u'=0
K=mcc[(1/g(V))-1)
b.For V=0 (u'=u)
K=mcc[(1/g(u))-1]
whereas for u'=0 and V=0
K=0
all in good accordance with phyhsical reality!
Is there some thing wrong in my derivation?
Use please soft words and hard arguments:rofl:

Log in or Sign up

Kinetic energy transformation

bernhard.rothenstein

country boy

bernhard.rothenstein

jtbell

Staff: Mentor

country boy

robphy

bernhard.rothenstein

Kinetic energy (Replies: 6)

Mass Energy Equivalence and Kinetic Energy (Replies: 9)

Kinetic energy conservation in matter/energy transform (Replies: 15)

Relativistic kinetic energy (Replies: 2)

Kinetic energy at C/2 (Replies: 20)