- #1
bernhard.rothenstein
- 991
- 1
Globalization makes that we can become aware of what other physicists, located at different remote places, have achieved.
Kard1,2 is the author of the derivations of the fundamental equations of relativistic dynamics without using conservation laws. He starts by defining the momentum of a particle in I as
p=mu (1)
and as
p’=m’u’ (2)
in I’, the particle moving in the positive direction of the OX(O’X’) axes.
Combining (1) and (2) and taking into account the addition law of relativistic velocities he obtains
p/m=(p’/m’)(1+V/u’)/(1+Vu’/c2) (3)
Equation (3) suggests considering
p=g(V)p’(1+V/u’)=g(V)(p’+m’V)
and
m=g(V)m’(1+Vu’/c2)=g(V)(m’+Vp’/c2). (4)
Isotropy of space requires that g(V)=g(-V) the inverse of (4) being
m’=g(V)(m-Vp/c2). (5)
Combining (4) and (5) we obtain
g(V)=(1-V2/c2)-1/2 (6)
and so
p=(p’+Vm’)/(1-V2/c2)1/2 (7)
m=(m’+Vp’/c2)/(1-V2/c2)1/2 (8)
and the way is paved to all the transformation equations we encounter in relativistic dynamics.
Considering a collision from I and I’ and using the transformation equations derived above (and others we could derive from them) we see that they lead to results in accordance with conservation of momentum and mass (energy).
Is the derivation circular?
IMHO no!
Is it time saving?
IMHO yes!
1Leo Karlov, “Paul Kard and Lorentz-free special relativity,” Phys.Educ. 24 165 (1989)
2Paul Kard, “Foundation of the concepts of relativistic mass and energy,” EESTI NSV Teaduste Akademia Toimetised 25 Koide Fuusika Matematika 1976 No.1 75-77 (in Russian)
Kard1,2 is the author of the derivations of the fundamental equations of relativistic dynamics without using conservation laws. He starts by defining the momentum of a particle in I as
p=mu (1)
and as
p’=m’u’ (2)
in I’, the particle moving in the positive direction of the OX(O’X’) axes.
Combining (1) and (2) and taking into account the addition law of relativistic velocities he obtains
p/m=(p’/m’)(1+V/u’)/(1+Vu’/c2) (3)
Equation (3) suggests considering
p=g(V)p’(1+V/u’)=g(V)(p’+m’V)
and
m=g(V)m’(1+Vu’/c2)=g(V)(m’+Vp’/c2). (4)
Isotropy of space requires that g(V)=g(-V) the inverse of (4) being
m’=g(V)(m-Vp/c2). (5)
Combining (4) and (5) we obtain
g(V)=(1-V2/c2)-1/2 (6)
and so
p=(p’+Vm’)/(1-V2/c2)1/2 (7)
m=(m’+Vp’/c2)/(1-V2/c2)1/2 (8)
and the way is paved to all the transformation equations we encounter in relativistic dynamics.
Considering a collision from I and I’ and using the transformation equations derived above (and others we could derive from them) we see that they lead to results in accordance with conservation of momentum and mass (energy).
Is the derivation circular?
IMHO no!
Is it time saving?
IMHO yes!
1Leo Karlov, “Paul Kard and Lorentz-free special relativity,” Phys.Educ. 24 165 (1989)
2Paul Kard, “Foundation of the concepts of relativistic mass and energy,” EESTI NSV Teaduste Akademia Toimetised 25 Koide Fuusika Matematika 1976 No.1 75-77 (in Russian)