Kinetic Energy approximation both Einstein's and Newton'sby TWest Tags: approximation, einstein, energy, kinetic, newton 

#1
Jul311, 04:35 PM

P: 48

Something has been bothering me for quite some time and when you cannot figure it out yourself ask a higher power, which happens to be this forum.
Basically, at low energies we use the newtonian definition of kinetic energy which is E_{k} = 1/2MV^{2} Now at high energies/high volecities it turns into the relativity version of it which is E_{k} = MC^{2}GMC^{2}. Now my question is at what point does this change from Newton's to Einstein's? Secondly, can you point me in the direction of a model that is accurate at low velocity and high? 



#2
Jul311, 04:44 PM

Mentor
P: 40,907

The Einstein version is good for all speeds. It just depends on how accurate you need to be.




#3
Jul311, 04:45 PM

P: 2,083

There's no point where one formula turns into another. The Einstein equation is accurate. The Newtonian one is inaccurate. It's just that at higher and higher energies, the Newtonian expression gets more and more inaccurate. For example at a low energy, the Newtonian expression may be off by only .05%, while at a higher energy, it may be off by 50%, or 100% or more. As you get closer and closer to the speed of light, the inaccuracy becomes more and more extreme.
There's no point where one equation turns into the other though. 



#4
Jul311, 04:51 PM

Emeritus
Sci Advisor
PF Gold
P: 2,352

Kinetic Energy approximation both Einstein's and Newton's
If you take the formula for the total energy of a moving mass:
[tex]E_t = \frac{mc^2}{sqrt{1\frac{v^2}{c^2}}}[/tex] and expand it into a series, you get : [tex]E_t = mc^2+ \frac{mv^2}{2}+ m \frac{3v^4}{8c^2}...[/tex] The first term is the Energy equivalence of the rest mass. The second term is the Newtonian definition of KE. And the additional terms make up the KE difference due to relativity. As to when to use Newton vs. Einstein: Einstein's formula alwaysgive the more accurate answer for any value of v. It really just depends on how accurate you need your answer to be for the particular problem at hand as to whether or not you can get away with using Newton's formula. 



#5
Jul311, 04:54 PM

P: 329





#6
Jul311, 05:00 PM

Mentor
P: 40,907

If you want to play around with some numbers to compare the two formulas, here's a site that provides a calculator: Relativistic Kinetic Energy




#7
Jul311, 08:36 PM

P: 3,967

[tex]E_k = \frac{m c^2}{1v^2/c^2}  mc^2 [/tex] is accurate at low velocity and high. Newtons equation is an approximation which is fairly accurate at low velocities. At 0.2c there is error of 2% in Newtons equation and the error gets much larger as velocities get nearer c. Technically, Newton's equation is inaccurate for any velocity greater than zero. 



#8
Dec1611, 09:54 AM

P: 1

Newton's equation of kinetic energy is as accurate as Einstein's one at any velocity, if you add the increasing mass of the moving objet into newton's equation.




#9
Dec1611, 10:25 AM

Sci Advisor
PF Gold
P: 4,863

(1/2) m γ v^2 compare to the correct: (γ  1) mc^2 Noting that γ = 1/ sqrt(1v^2/c^2), algebra will show you these are not the same. You can also plug numbers in to see that they are not the same. The only common formula for which relativistic mass 'works' is momentum. It fails for force, it fails for kinetic energy. [edit: if you take the series expansion of the wrong and right formulas above, the first order difference 1/8 v^4/c^2. That is, they differ in even the first correction to Newtonian formula.] 


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