DB said:
I was reading about Kinetic Energy and why it is 1/2 mv^2. I couldn't really understand what it was saying, it was mentioning a constant of proportionality being:
\frac{kg * m^2/s^2}{m^2/s^2} and then of course you're left with mass. It was very confusing, so basically what I'm asking is for a basic mathematical explanation of why Ke=1/2mv^2
Thanks in advance.
DB I read all the answers here, and nowhere did I see what I would have put. I would have answered you by explaining to you an experiment that Galileo did several hundred years, and that anyone can do, and then tell you the answer to your question lies somewhere in this experiment.
Of course all the approaches above are nice too.
But, if I had to develop a mind based on all the knowledge that can be gleaned from just one experiment, this would be it:
http://www.glenbrook.k12.il.us/gbssci/phys/Class/newtlaws/u2l1b.html
http://id.mind.net/~zona/mstm/physics/mechanics/forces/galileo/galileoInertia.html
And this site is my favorite:
http://www.beyondbooks.com/psc91/4a.asp?pf=on
This experiment gives you motivation to define total system energy E as the sum of the kinetic energy of the ball, with the potential energy of the ball, and figure out that if there was no 'friction' that the total system energy would be time invariant, and hence conserved, and that you would have built a perpetual motion machine.
This would mean that something is invariant. But what?
The answer is the system's total energy E.
Define: E=T+U
dE/dt=0 (total system energy constant in time; invariant)
T=kinetic energy
U = potential energy = mgh
Define the speed of the center of inertia of the ball in the rest frame of the experimental device v. The following quantity is conserved:
mgh + (1/2mv^2)
where the inertial mass m of the falling ball is also treated as invariant. In other words dm/dt=0
Your goal is to formulate true statements that are frame independent.
One of the many things that Galileo inferred from this experiment, is that if he reduced what he called 'friction' (I don't know the italian word he used), the ball almost rose to the same height it was released from, all else constant. He knew that air friction slowed it down, but he also noticed that
if he smoothed the surface down (imagine the surface is made out of ice instead), it went higher. The key thing to note, is that if what are called frictional forces are removed from the set up, so that they don't act on the ball after it is relased, the only external force that acts upon the ball is the force called gravity. So that brings us to gravity.
We cannot remove gravity now can we.
Gravity is not a frictional force now is it?
The Earth is under the experiment, and in fact is what gets the ball to fall in the very first place.
The weight of the ball is given by:
W= Mg
Where g is the local acceleration due to gravity (which can be measured very easily), and M is the 'mass' of the ball.
If we mix up Galileo's thoughts with Newton's we get the following:
F = \frac{GMm}{r^2} = mg
Where M is the mass of the earth, m is the mass of the ball which we watch roll back and forth, g is acceleration due to gravity, r is the distance between the center of inertia of the earth, and the center of inertia of the ball, and G is the Newtonian Gravitational constant.
F is the 'force' of gravity.
But, if Newton's law is true, the g isn't actually a constant now is it. It's value really would depend on just what h is. The higher you dropped the ball, the lower would be its rate of free fall, so that if you released it high enough, it wouldn't fall at all, it would float.
So really, that brings us to the fact that the force of gravity above is a conservative force, meaning that it can be written as the gradient of a scalar function U, and mathematically, the curl of the gradient of U must be zero for a conservative force.
Let F denote the force of gravity.
We have:
\vec F = -\nabla U
But where on Earth did this come from?
It is a purely mathematical fact that:
Theorem:
\frac{\hat R}{R^2} = \nabla(\frac{-1}{R})
You prove this.
In order to formulate gravitational field theory, you start off defining gravity as a central force.
This means that if two bodies interact gravitationally, that the direction of the force of object one on object two is along the straight line path from the center of inertia of object one to the center of inertia of object two.
That is the symbol \hat R that you see above. That is a unit vector which points from the CM of body one, to the CM of body two.
Define the gravitational field of an object of mass M as follows:
\vec \Gamma = GM\frac{\hat R}{R^2} = GM \nabla (\frac{-1}{R} )
Then an object of mass m will experience a gravitational force given by:
\vec F = m \Gamma
So that the gravitational force o an object with mass M, on an object with mass m is given by:
\vec F = m \vec \Gamma = GMm\frac{\hat R}{R^2} = GMm \nabla (\frac{-1}{R})
The only problem now is that the inertial mass of the Earth is a strictly positive quantity, and the inertial mass of the falling body is a strictly positive quantity, and the way we have the force formula above set up, the force between the two objects is repulsive instead of attractive, which is wrong since gravity is assumed to always pull things together not push them apart.
So all we have to do now, is just throw in a minus sign right?
Instead define the gravitational field as follows:
\vec \Gamma = -GM\frac{\hat R}{R^2} = -GM \nabla (\frac{-1}{R} )
Now we have the gravitational force on an object with a mass of m, given by
\vec F = m \vec \Gamma
And the force is attractive.
Let's step back a moment, to the gravitational field.
\vec \Gamma = -GM\frac{\hat R}{R^2} = -GM \nabla (\frac{-1}{R} )
Suppose that the density of an object is represented by \rho
Density is mass divided by volume, so we can express the field using the integral calculus as follows:
\vec \Gamma = -G \int \rho d\tau \frac{\hat R}{R^2}
Where d tau is a differential volume element, so that this is a volume integral.
Now use the mathematical theorem above, to write the gravitational field created by a body with inertial mass M as follows:
\vec \Gamma = -G \int \rho d\tau \nabla (\frac{-1}{R})
For mathematical reasons alone, you can pull the gradient symbol all the way out front and write this as:
\vec \Gamma = \nabla -G \int \rho d\tau (\frac{-1}{R})
And now let's pull out that stray minus sign which we inserted ad hoc, to make the force of gravity purely attractive, and write:
\vec \Gamma = - \nabla G \int \rho d\tau (\frac{-1}{R})
Thus, we have:
- \vec \Gamma = \nabla G \int \rho d\tau (\frac{-1}{R})
So...
\vec F = m\Gamma = - m (-\vec \Gamma ) = - (-m\vec \Gamma )
And -m times Gamma is:
- m\vec \Gamma = m\nabla G \int \rho d\tau (\frac{-1}{R})
Now, if we assume that m isn't affected by the gradient operator, we can allow the gradient operator to act upon it, so that we have this:
- m\vec \Gamma = \nabla G m\int \rho d\tau (\frac{-1}{R})
Now we can finally see what U is.
U = G m\int \rho d\tau (\frac{-1}{R})
Notice that U is a scalar function, not a vector function.
