Kinetic energy calculation . . .

[shunyadragon]

U mad Shunya?

no

[Kbertsche]

Equation editor is your friend: https://www.codecogs.com/latex/eqneditor.php

True, if you pick a particular reference frame, you can develop the notion of work within something looking a lot like the Newtonian framework, though with a modified inertia.



Where β is the fraction of speed of light, given by



If you take the taylor expansion centered on v = 0 (making it a Laurin series), you get



This isn't my preferred form of this result though, as I like to see the correction to the classical term by a relativistic term of order unity.



Usually the right side is quite small unless we're close to c, at which point this form isn't all that practical. For small velocities we can taylor expand the corrective term into



Note how in this form, its clear that even at 10% of the speed of light, the additional kinetic energy due to relativistic effects is only about 3/4th of a percent!

FYI, you seem to have some typos here.

Your first equation is correct for T = E0 + Ek (total energy = rest energy plus kinetic energy); if you set beta=0, you get the rest energy.

The rest of your energy equations are actually for Ek, the kinetic energy alone, not for the total energy T. If you set beta=0 in them, you get 0, not the rest energy.

[Leonhard]

FYI, you seem to have some typos here.

Your first equation is correct for T = E0 + Ek (total energy = rest energy plus kinetic energy); if you set beta=0, you get the rest energy.

The rest of your energy equations are actually for Ek, the kinetic energy alone, not for the total energy T. If you set beta=0 in them, you get 0, not the rest energy.

Ah yes, correct, thank you. I wrote this quickly after I had gotten home. I can't remember why I used T for the kinetic energy. I think that was a habit from how the notation we used at my campus.

[37818]

Since you merely posted a bunch of equations with no explanation, you left the "topic" open to interpretation.

I interpreted your point to be that Eq. 2 and Eq. 5 differ except in the specific case where v=c, thus (apparently) demonstrating an internal mathematical inconsistency in Einsteinian Relativity Theory. The very fact that you did not explain whatever point it was that you were trying to make is the reason I posted what I did.

The following equations are equal.
Ke = mc²(1 / sqr(1 - v²/c²) - 1) =
mc²(1 / (1 / ((v/c) tan(asin(v/c) / 2))) - 1)) =
mv²/( 1 + sqr(1 - v²/c²) - v²/c²)