Formula of speed

[DimaMazin]

v=c_*(m'²+2m'm)^1/2/ (m'+m)

v - speed

c - light speed

m' - mass of kinetic energy(relative mass)

m - gravitation mass

for object with gravitation mass: m'=[1/(1-v²/c²)^1/2 -1]m

[Bignose]

At least the units are right (all too often this isn't the case when someone posts a formula here), but unless it comes from a derivation, it doesn't really have any meaning.

 

that is, how is it any different from: [math]v = c \frac{\sqrt{m'^4+2m'^2m^2}}{m'^2+m^2}[/math]? units cancel correctly again, but I just completely made it up and it really doesn't have much meaning at all.

 

So, where does your idea come from, and what does it really mean?

Edited December 6, 2010 by Bignose

[Janus]

Let's see:

 

[math]v = c \frac{\sqrt{m'^2+2m'm}}{m'+m}[/math]

 

and

 

[math]m'= \left ( \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} -1 \right ) m [/math]

 

At first, to keep things from being unwieldy, we'll say that

 

[math] \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/math]

 

So we can say that

 

[math]m' = m \gamma - m [/math]

 

Putting this into the first equation we get

 

[math]v = c \frac{\sqrt{\left( m \gamma - m \right)^2+2\left (m \gamma - m \right ) m}}{m \gamma - m+m}[/math]

 

[math]v = c \frac{\sqrt{\left( m \gamma - m \right)^2+2\left (m \gamma - m \right ) m}}{m \gamma }[/math]

 

[math]v = c \frac{\sqrt{m^2\gamma^2-2m^2 \gamma + m^2 + 2m^2 \gamma -2m^2}}{m \gamma}[/math]

 

[math]v = c \frac{\sqrt{m^2\gamma^2 - m^2 }}{m \gamma}[/math]

 

[math]v^2 = c^2 \frac{m^2\gamma^2 - m^2 }{m^2 \gamma^2}[/math]

 

[math]v^2 = c^2 \frac{\gamma^2 - 1 }{ \gamma^2}[/math]

 

[math]\frac{v^2}{c^2} = \frac{\gamma^2 - 1 }{ \gamma^2}[/math]

 

[math]\frac{v^2}{c^2} = 1-\frac{ 1 }{ \gamma^2}[/math]

 

[math] \frac{v^2}{c^2} = 1- \left ( 1- \frac {v^2}{c^2} \right )[/math]

 

[math] \frac{v^2}{c^2} = \frac {v^2}{c^2}[/math]

 

[math] v^2 = v^2 [/math]

 

[math]v = v[/math]

 

Hardly an earth-shattering revelation. And not very surprising, considering that your initial equation solving for v relied on a term (m'), that itself relied on the value of v.

[DimaMazin]

At least the units are right (all too often this isn't the case when someone posts a formula here), but unless it comes from a derivation, it doesn't really have any meaning.

 

that is, how is it any different from: [math]v = c \frac{\sqrt{m'^4+2m'^2m^2}}{m'^2+m^2}[/math]? units cancel correctly again, but I just completely made it up and it really doesn't have much meaning at all.

 

So, where does your idea come from, and what does it really mean?

It comes from a derivation:

total energy²=(mc²)²+(cp)²

Einstein's kinetic energy=m'c²

This formulae I have used.If you have a defined gravitation mass and a defined turned mass into energy then you have a defined speed.Thank you for your discussion.

 

 

 

Let's see:

 

[math]v = c \frac{\sqrt{m'^2+2m'm}}{m'+m}[/math]

 

and

 

[math]m'= \left ( \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} -1 \right ) m [/math]

 

At first, to keep things from being unwieldy, we'll say that

 

[math] \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/math]

 

So we can say that

 

[math]m' = m \gamma - m [/math]

 

Putting this into the first equation we get

 

[math]v = c \frac{\sqrt{\left( m \gamma - m \right)^2+2\left (m \gamma - m \right ) m}}{m \gamma - m+m}[/math]

 

[math]v = c \frac{\sqrt{\left( m \gamma - m \right)^2+2\left (m \gamma - m \right ) m}}{m \gamma }[/math]

 

[math]v = c \frac{\sqrt{m^2\gamma^2-2m^2 \gamma + m^2 + 2m^2 \gamma -2m^2}}{m \gamma}[/math]

 

[math]v = c \frac{\sqrt{m^2\gamma^2 - m^2 }}{m \gamma}[/math]

 

[math]v^2 = c^2 \frac{m^2\gamma^2 - m^2 }{m^2 \gamma^2}[/math]

 

[math]v^2 = c^2 \frac{\gamma^2 - 1 }{ \gamma^2}[/math]

 

[math]\frac{v^2}{c^2} = \frac{\gamma^2 - 1 }{ \gamma^2}[/math]

 

[math]\frac{v^2}{c^2} = 1-\frac{ 1 }{ \gamma^2}[/math]

 

[math] \frac{v^2}{c^2} = 1- \left ( 1- \frac {v^2}{c^2} \right )[/math]

 

[math] \frac{v^2}{c^2} = \frac {v^2}{c^2}[/math]

 

[math] v^2 = v^2 [/math]

 

[math]v = v[/math]

 

Hardly an earth-shattering revelation. And not very surprising, considering that your initial equation solving for v relied on a term (m'), that itself relied on the value of v.

You have connected the universal formula with the local formula. Good job in the appropriate place.

[DimaMazin]

Universal formula of momentum

 

p = c (m'²+2m'm)^1/2