# Relativistic Kinetic Energy

1. Jul 28, 2008

### rhz_prog

I saw that the Relativistic Kinetic Energy calculation for these two sources, seems to be different :

http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken]
see : Conservation of Energy
EK = gamma*m*c^2

While here :
http://en.wikipedia.org/wiki/Kinetic_energy
see : Relativistic kinetic energy of rigid bodies
EK = gamma*m*c^2 - m*c^2

Which one is right ? Or did I misunderstand something ?

Last edited by a moderator: May 3, 2017
2. Jul 28, 2008

### TeTeC

I haven't read the first link, but from what I know I'd say that the second definition is the right one. $E = \gamma mc^{2}$ is the total energy, not the kinetic energy. You can Taylor expand the first expression $E = \gamma mc^{2} - mc^{2}$ with respect to the variable v/c. The first term in the expression, $\frac{1}{2}mv^{2}$ represents the Newtonian kinetic energy, which will be the dominant term if v/c is small.

3. Jul 28, 2008

### yenchin

Where does the first link say that? All I can see under "conservation of energy" is:

E_final = γmc^2 + E_L

?

4. Jul 28, 2008

### Staff: Mentor

That page does not refer to $\gamma m c^2$ as kinetic energy, but rather, simply as "energy". In fact, the word "kinetic" does not appear on that page at all!

Last edited by a moderator: May 3, 2017
5. Jul 28, 2008

### yuiop

Your expressions are better written as:

http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken]
see : Conservation of Energy
TE = gamma*m*c^2

While here :
http://en.wikipedia.org/wiki/Kinetic_energy
see : Relativistic kinetic energy of rigid bodies
KE = gamma*m*c^2 - m*c^2

where TE is Total Energy and KE is Kinetic Energy.
The second expression can also be written as:

KE = TE - RE

where RE is Rest Energy or rest mass energy.

Total Energy can can also be found from this relationship:

$TE = \sqrt{(gamma*m*v*c)^2+(m*c^2)^2} = \sqrt{(pc)^2+(mc^2)^2}$

which can be written as:

$TE = \sqrt{ME^2+RE^2}$

where ME is Momentum Energy.

By rearranging this becomes :

$RE = \sqrt{TE^2-ME^2}$

Since rest energy is usually an invariant, the quantity $\sqrt{TE^2-ME^2}$ is the same when switching from one reference frame to another. In fact, in a perfectly elastic collision, the quantity $\sqrt{TE^2-ME^2}$ is the same for a given particle before and after the collision.

Hope that helps.

Last edited by a moderator: May 3, 2017