Kinetic Energy approximation both Einstein's and Newton's

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Discussion Overview

The discussion centers around the transition between Newtonian and relativistic definitions of kinetic energy, specifically the conditions under which one should use each formula. Participants explore the accuracy of both approaches at varying velocities and seek models that can effectively represent kinetic energy across the spectrum from low to high velocities.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that the Einstein equation for kinetic energy is accurate for all speeds, while the Newtonian equation becomes increasingly inaccurate at higher velocities.
  • Others argue that there is no specific point where one formula transitions to the other; rather, the Newtonian expression simply becomes less accurate as speed increases.
  • A participant presents a series expansion of the total energy formula, indicating that the Newtonian kinetic energy term appears as a part of the relativistic energy expression.
  • Some contributions mention specific velocity thresholds, such as the claim that Newton's equation remains accurate within 1% for speeds less than 0.1c.
  • One participant challenges the idea that Newton's equation can be made equivalent to Einstein's by incorporating relativistic mass, arguing that this approach leads to incorrect results for kinetic energy.

Areas of Agreement / Disagreement

Participants express differing views on the accuracy and applicability of Newtonian versus Einsteinian kinetic energy formulas, with no consensus reached on a specific transition point or the validity of using relativistic mass in Newton's framework.

Contextual Notes

Some claims depend on specific definitions and assumptions about velocity ranges and the context of use, which remain unresolved in the discussion.

TWest
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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 Ek = 1/2MV2 Now at high energies/high volecities it turns into the relativity version of it which is Ek = MC2G-MC2. 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?
 
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The Einstein version is good for all speeds. It just depends on how accurate you need to be.
 
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.
 
If you take the formula for the total energy of a moving mass:

E_t = \frac{mc^2}{sqrt{1-\frac{v^2}{c^2}}}

and expand it into a series, you get :

E_t = mc^2+ \frac{mv^2}{2}+ m \frac{3v^4}{8c^2}...

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.
 
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TWest said:
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 Ek = 1/2MV2 Now at high energies/high volecities it turns into the relativity version of it which is Ek = MC2G-MC2. 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?

Weidner And Sells Elementary Modern Physics gives the guide that that calculations using Ek=mv2/2 will be accurate within 1% when v/s < 0.1.
 
TWest said:
Basically, at low energies we use the Newtonian definition of kinetic energy which is Ek = 1/2MV2 Now at high energies/high volecities it turns into the relativity version of it which is Ek = MC2G-MC2. 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?

Einstein's equation:

E_k = \frac{m c^2}{1-v^2/c^2} - mc^2

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.
 
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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.
 
eljaala said:
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.

This is simply wrong and is a classic example of the pitfalls of the obsolete notion of relativistic mass. The relativistic mass is: mγ. Putting into Newton's kinetic energy you get:

(1/2) m γ v^2

compare to the correct:

(γ - 1) mc^2

Noting that γ = 1/ sqrt(1-v^2/c^2), algebra will show you these are not the same. You can also plug numbers into 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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