# What is the relativistic equation for finding kinetic energy?

1. Feb 9, 2012

### Ralphonsicus

Let's say, I wanted to find the kinetic energy of a ball travelling at 99% the speed of light, what is the equation used for that calculation?

And also, do photons have kinetic energy?

Thanks.

2. Feb 9, 2012

### Pengwuino

The formula for a particle of mass m has a kinetic energy is given by $(\gamma - 1)mc^2$ where $\gamma = {{1}\over{\sqrt{1-{{v^2}\over{c^2}}}}}$ where c is the speed of light.

The energy of a photon with frequency $f$ is $E_{photon} = hf$ where h is Planck's constant.

3. Feb 9, 2012

### PAllen

mc^2(γ - 1)

where γ = 1/(√(1- v^2/c^)

4. Feb 9, 2012

### Staff: Mentor

Here you go http://bit.ly/xZN1YS
I don't think so because they are massless.

5. Feb 9, 2012

### PAllen

I missed the question about photons. What Pengwino says is correct, but (and we simul-posted, else I wouldn't have bothered) adding a little more, and disagreeing with Ryan_m_b:

Since a photon is massless it has no rest energy. Therefore all of its energy is kinetic. For a massive particle, you can say the frame dependent energy has a minimum - the rest energy; the frame dependent additional energy is kinetic. For a photon, there is no minimum - you can redshift to arbitrarily close to zero energy by choice of frame, consistent with its having no rest energy and all kinetic energy.

6. Feb 9, 2012

### Staff: Mentor

I tried to make it clear I wasn't sure good to learn though, cheers.

7. Feb 9, 2012

### tom.stoer

$$E^2 = (mc^2)^2 + p^2c^2$$

From this equation the kinetic energy can be determined directly

$$E_\text{kin} = E - mc^2 = \sqrt{(mc^2)^2 + p^2c^2} - mc^2$$

For photons we have m=0 and therefore

$$E_\text{kin} = E = pc$$

For m>0 one gets the equations with v<c mentioned above, of course