# How do you calculate the speed of a particle given its Total and Kinetic energy?

1. ### j2dabizo

19
Just curious on this method. I seemed to be getting caught up on the method here.

I'm given a K.E. amd Total Energy of a proton, and I was asked to find the speed. So what is the equations and steps for these.

This is not a homework question, just me trying to wrap my head around the method and equation.

Thanks

2. ### Pengwuino

7,118
Either quantity will tell you the speed. The total energy of a particle is:

$$E = \gamma mc^2$$

where $\gamma = {{1}\over{\sqrt{1-{{v^2}\over{c^2}}}}}$ where 'c' is the speed of light so you can solve for the velocity using this formula.

The kinetic energy for a particle is simply $E=(\gamma - 1)mc^2$ so you can use that as well.

3. ### ZealScience

381
I think you can just solve the mass of the object using E=mc^2, where m is what the mass is at the time but not the rest mass, and you can use E=1/2mv^2. I don't think that rest mass is useful in this case, because in kinetic energy you are not using the rest mass.

4. ### PAllen

5,423
No, this is wrong. If you compute the relativistic mass by E/c^2, then compute speed using 1/2 mv^2, you will get the wrong speed, period. The answer is as Pengwuino describes.

5. ### ZealScience

381
Why doesn't relativistic mass affect kinetic energy? I don't know about it. But when the object is cooled down and have less mass (though a little bit), won't the kinetic decrease? If the kinetic energy change with relativistic mass, then how can mass increase to infinity at speed of light. I'm not professional, just asking.

6. ### PAllen

5,423
The problem is 1/2mv^2 is not valid at relativistic speeds, even if you use relativistic mass for m. This is the danger of the 'relativistic mass' concept. People think it allow you to use Newtonian formulas. It actually allows you to use exactly one basic Newtonian formula: p=mv. The analog of 1/2mv^2 is, if you let m be 'relativistic mass' is (m - m0)c^2, where m is relativistic mass, m0 is rest mass. However, it is much better just to use the formulas as given by Pengwuino.

7. ### ZealScience

381
Yes, those equations are correct definately due to the fact that (γ-1)mc^2 is the total energy minus moving energy which is the kinetic, I can understand that. But it is not convenient to think about something called "rest mass" which the m in the equation refering to, because in my understanding rest mass is not quite sustainable. So what I whnt to do is to neglect the idea of rest mass.

8. ### ZealScience

381
So what is the relativistic kinetic. But does the equation for work still hold? I mean if W=Fx holds, then should W=Fx be added with γ factor?

9. ### PAllen

5,423
The majority of physicists think relativistic mass is relatively useless; and all agree that rest mass or invariant mass are fundamental. Showing that, there is, in fact, no way to avoid rest mass in a formula for kinetic energy in relativity. No such formula exists. You can be really silly and say rest energy instead of rest mass; then you have KE = E - E0. However, E0 is just m0c^2, so what does that get you?

10. ### PAllen

5,423
These formulas work if you use a correct relativistic force. Unfortunately, that force is *not* f=ma, where m is relativistic mass. Please stop. You *cannot* use Newtonian formulas with relativistic mass in place of rest mass. That is precisely why it has fallen out of favor.

11. ### PAllen

5,423
Well, I figured out you could construct a formula that doesn't involve rest mass, if you insist:

KE= mc^2 (1 - 1/gamma) where m is relativistic mass. You won't find this in a book, because no one uses it.

12. ### ZealScience

381
THank you for that. I would be more careful next time dealing with them.

13. ### PAllen

5,423
Going from my last formula, given total and kinetic energy you can get speed, not knowing anything else. Solve for v:

1/gamma = 1 - KE/E