When I studied physics in junior high and high school, we always took it for granted that potential energy was equal to kinetic energy. In Lagrangian terms, $T = V$at least on average. But I realized that assumption might not always be true. I have been studying the Euler-Lagrange equation and been very confused by it. But I read in Gravity by James Hartley that Newton's law can be expressed as the Euler-Lagrange equation.
My Question
If a Lagrangian satisfies the Euler-Lagrange equation, does that mean potential energy equals kinetic energy?
EDIT:
In other words, suppose I have a mass held above the surface of the earth with $PE = mgh$. That's the amount of energy that upon release gradually gets converted to kinetic energy. So therefore $PE = KE$ in that case. That's what I mean. I am saying I have always assumed the system I am considering is conservative. So when I looked at the Euler Lagrange equation, it looked trivial because had (without knowing) assumed every system satisfied it. So I am trying to see if I have the story straight this time.
Best Answer
Potential energy is always measured relative to something, just as kinetic energy is measured in a particular frame of reference. That being the case, I could always make $T=V$ by choosing the reference in a certain way - but it will not generally be true.
For example, I can set $T=0$ at the top of a tower, or at ground level. Both are valid - it's really up to me. Or I can say $T=0$ at infinite distance from the earth.
And when I observe a falling object, that object will appear to be decelerating relative to a frame of reference that is moving towards the earth faster than the object.
Instead, you can say that
$$T + V = \mathrm{constant}$$
Because as you lose potential energy, you gain kinetic energy (assuming those are the only two forms that energy can take in your system). If you add friction, it's of course possible that the sum $T+V$ will decrease as the system warms up (unless you define thermal energy as a form of either potential or kinetic energy, which you could...)
But no, $T=V$ is not in general true.