# Relativistic Kinetic Energy

1. Mar 25, 2007

### Leonardo Sidis

I am reading "Relativity" by Einstein right now, and I came across a formula he gave for kinetic energy in accordance with his theory of relativity. He says that "the kinetic energy of a material point of mass m is no longer given by the well-known expression

.5m(v^2)

but by the expression

[m(c^2)]/[sqrt(1-(v^2)/(c^2))]

I'm assuming this expression was found by the Lorentz transformation, but he doesn't show that. He goes on to explain how this formula suggests the impossibility of speeds faster than c, since the fraction approaches infinity as v approaches c.

The reason why I am confused is that the two expressions are extremely different. If the mass is 2 kg and the velocity is 3 m/s, then by the first expression, the KE is 9 J. By the second expression, you get about
180,000,000,000,000,000 J. What am I missing?

(also, what is a material point?)

Thanks for any help on this

Last edited: Mar 25, 2007
2. Mar 25, 2007

### bernhard.rothenstein

The expression you propose is not correct.
It is if I remember well it is
K=mcc[(1-vv/cc)^-1/2 -1]

3. Mar 25, 2007

### Leonardo Sidis

If the expression I gave above is incorrect then there is a typo in the book because that is exactly what the book has in it - I double-checked.

4. Mar 25, 2007

### masudr

The form given in the book includes the rest energy term $mc^2$.

If you know how to obtain the Taylor series for a function, try this: expand the form given for the energy in powers of v/c, and then subtract off the rest energy term. Take a look at what you have left, and remember that for non relativistic speeds, v/c is usually small, and so higher powers of v/c are even smaller.

5. Mar 25, 2007

### anantchowdhary

Actually the kinetic energy of a body is given as

$$K.E=(\gamma - 1)m_0c^2$$

6. Mar 25, 2007

### bernhard.rothenstein

Ke

Compring my answers with yours some people will blame me for using m for the rest mass and you for using m(0). That is the situation when we do not use all the same notation for the same phyhsical quantity. I like more m(0).:rofl:

7. Mar 25, 2007

### anantchowdhary

isn't $$m_0$$ standard notation????

8. Mar 25, 2007

### pmb_phy

There really is no standard notation for mass. Workers in different fields use different symbols according to the objective of their work. There's even a paper by Einstein which was published in 1907 in which he used neither for proper mass (aka rest mass). He used $\mu$. This is what I prefer myself. The invariants in relativity, i.e. the proper quantities, often have the greek symbol on place of the latin symbol. So instead of s for distance the symbol $\sigma$ is used. Instead of t for time the symbol $\tau$ is used. Thus the extention of this idea to "m". Instead of m the symbol $\mu$ is used.

Pete

9. Mar 25, 2007

### robphy

Unfortunately, this idea does not extend to $$\beta$$ and $$\gamma$$ [unless one interprets these quantities as invariants that depend on two 4-velocities].

10. Mar 25, 2007

### MeJennifer

I prefer rest mass over relativistic mass because relativistic mass is not Lorentz invariant.

But I realize that, in the macro world, relativistic mass is used more often than rest mass.

For example measuring the rest mass of a spaceship even when it is at rest is not as simple as it appears to be. For instance if we were to use a balance we would have to take into account the composition of the material on each side of the balance.

11. Mar 25, 2007

### bernhard.rothenstein

mass on the balance

It is not enough that the masses on the pans of the balance equate each other? Please explain.

12. Mar 25, 2007

### MeJennifer

The restmass and relativistic mass differential depends on the atomic numbers of the elements of the composition.

Take for instance the mass of a gold bar, the restmass and relativistic mass are not identical even when that gold bar does not appear to be moving. The electrons do move and in the case of gold some of them reach relativistic speeds.

Obviously, the difference is very small but nevertheless it is not zero!

Last edited: Mar 25, 2007
13. Mar 25, 2007

### Leonardo Sidis

I think we're kind of getting off topic here. So, the reason why the book's expression for relativistic KE gives us a value 20 quadrillion times greater in my example than the non-relativistic KE is not that the expression in the book is incorrect, but that it includes the rest energy? Why should this account for such a huge difference though?

14. Mar 25, 2007

### masudr

Yes.

Because $m_0 c^2 >> m_0 v^2$ at non relativistic speeds!

15. Mar 25, 2007

### pmb_phy

I don't follow you. Would you prefer proper time over time because the former is a Lorentz invariant? Would you prefer proper length over length? Would you prefer proper energy over energy? What does it mean for you to prefer one over the other??

Pete

16. Mar 25, 2007

### pmb_phy

Neither of those are intrinsic properties of a body so it really doesn't apply here. I think you know where I was going with what I said anyway.

Pete

17. Mar 25, 2007

### masudr

I'm not sure what MeJennifer meant (of course), but I'd just like to add that if I made a similar statement, I would mean that in situations where I can use either notions of mass correctly, I would prefer to use the rest mass.

My reasoning would be as follows: wherever I can I would prefer to use tensorial quantities because to me it feels like physical systems/properties can be represented best by tensors on spacetime. I know that in certain situations using non-tensors can simplify matters, and even in some cases where we are forced to consider non-tensors (e.g. Christoffel symbols), but if I can, I'd use the tensors.

18. Mar 25, 2007

### pmb_phy

In a situation in which you could use both then if you were in a position to use Energy or proper energy which would you choose?

Do you see where I'm going here? These questions/selections are far to vauge to generalities and one is never replaceable with the other, ... , at least I can't think of one. How you you MJ and masudr?

Pete

Last edited: Mar 25, 2007
19. Mar 25, 2007

### masudr

I'm not entirely sure of the difference... as far as I know energy is the zeroth component of the energy-momentum 4-vector

$$p^\mu = m_0 \frac{dx^\mu}{d\tau}$$

So what do you mean by energy and proper energy? As in, what's the difference between them?

20. Mar 25, 2007

### pmb_phy

You are quite correct in what Einstein's book says on this. Consider this: If the rest energy of the electron is constant then it can be dropped from the equation for the conservation of energy leaving only $\gamma$m0c2 which is what Einstein has in his book. The meaningful quantities now are the changes in kinetic energy.

Hope that helps

Pete