- #1

- 991

- 1

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

Thanks

Thanks

- Thread starter bernhard.rothenstein
- Start date

- #1

- 991

- 1

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

Thanks

Thanks

- #2

- 228

- 1

- #3

- 991

- 1

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

- #4

jtbell

Mentor

- 15,735

- 3,892

- #5

- 228

- 1

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

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

- #6

- 5,779

- 1,078

- #7

- 991

- 1

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?

- Replies
- 15

- Views
- 3K