# Pi, momentum, and kinetic energy

1. Feb 22, 2010

### e2m2a

There is a geometric way one can show the relation between kinetic energy and momentum which is a mathematical curiosity in my opinion. Let the mass of an object be equal to 2 PI. Then:

P = 2 pi v

KE = 1/2 (2 pi ) v sq
or
KE = pi v sq

Hence, graphically, if we set v = r, where r is the radius of a circle, we have a circle whose circumference is momentum, 2 pi v, and whose area is the kinetic energy, pi v sq. Thus, for the case where the mass is 2 pi, we see the kinetic energy is geometrically bounded by the momentum of the object. Interesting.
Wonder what would happen to this circle at relativistic velocities?

Last edited: Feb 22, 2010
2. May 26, 2015

### joe o'rourke

ive been thinking about pi and energy lately.

to simplify things here, ur just describing cosx/2. ie graph cosx...d/dt cosx is -sinx. pi rad of -sinx is (-1, +1 or "total ke" until it goes back down to -1 ie pe). relativistically speaking, the extreme case is analogous to an infinite amp, (staying with cos waves.) ie lim v---> c dt----->0, hence large distance(high amp) and no oscillation.

3. May 26, 2015

### Staff: Mentor

Well, momentum in SR can take the form: where m0 is the invariant mass and gamma is the lorentz factor:

The relativistic relation between kinetic energy and momentum is given by:

I haven't tried turning this relationship into a circle, but you're welcome to try.

4. May 26, 2015

### joe o'rourke

5. May 26, 2015

### joe o'rourke

Also just to add: the classic example of this in relativity is in the expansion of the universe and why we see "light years away" not a distance away at far distances.

6. May 26, 2015

### Staff: Mentor

This has nothing to do with the thread topic.

7. May 27, 2015

### Unified28

I personally don't think there is any significance here. For the cases where v is not equal to r, the properties of a circle disappear. There is no fundamental property here that can be generalized to all cases of v, and is nothing more than a result of simply choosing the variables to get the desired outcome.

8. May 27, 2015

### tommyxu3

I'm a little agree with it, for the special case won't exist if the special amount $\pi$ is replaced with any others.