KE at speeds close to c

  • #1
julianwitkowski
133
0
I'm curious as to the reasons why KE = ½ m ⋅ v2 only works at speeds much less than c?

Also, how does the equation change?

Thank you!
 

Answers and Replies

  • #3
julianwitkowski
133
0
This thread may help.
Is it because you have to compensate for the increasing mass of the object which is proportional to the Lorentz factor, and then the integrals give you the the accurate change in mass over the entire distance with acceleration/deceleration and other factors compensated for?
 
  • #4
2,809
604
Is it because you have to compensate for the increasing mass of the object which is proportional to the Lorentz factor, and then the integrals give you the the accurate change in mass over the entire distance with acceleration/deceleration and other factors compensated for?

I prefer not to use relativistic mass at all and in fact its not needed. Physicists don't use it too. Its just that in both Newtonian and Relativistic mechanics, linear momentum and kinetic energy depend on the reference frame, but with different forms. In relativity we have [itex] KE=(\gamma-1)mc^2 [/itex] and [itex] \vec p=\gamma m \vec v [/itex]. By m, I mean rest mass and this is the only concept of mass I use.
These forms are dictated by consistency with special relativity and actually they can be derived. See e.g. this paper!
 
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