Why is kinetic energy only "often $(1/2)mv^2$"? I am reading the first few pages of Nakahara and refreshing my memory on physics I learned a while ago as a physics math undergrad. Nakahara defines a field $F$ to be conservative if it's the gradient of some potential $V(x)$. (standard)  Then he defines the total energy as $(1/2)mv^2+V(x)$. He then states that $E$ is "often" the kinetic+potential energy, hence deserving the name "total energy." I was taught that kinetic is defined as $(1/2)mv^2$. So what does he mean here?
Addum: The exact quote of Ref. 1 is: 

The function $E$, which is often the sum of the kinetic energy and the potential energy, is called the energy.

References:


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*M. Nakahara, Geometry, Topology & Physics, 2nd edition, 2003, Section 1.1.1.

 A: The definition for point particles is:
$T = \frac{p^2}{2m}$, where $T$ denotes my kinetic energy.

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*If you are dealing with classical physics, the momentum for a point-particle equals $\vec{p} = m\vec{v}$, which would give you $T = \frac{mv^2}{2}$.

*In the theory of special relativity we notice that objects become heavier when they move, the momentum becomes: $\vec{p} = {m\vec{v}\gamma}$, where $\gamma = \frac{1}{\sqrt{1-(v/c)^2}}$. Hence the kinetic energy becomes:
$T={mc^2\gamma}$, this is derived below.

*If you try to make a non-relativistic electromagnetic Lagrangian, it's not possible to incorporate the magnetic field in terms of a potential energy. So the solution is to define the momentum equal to $\vec{p} = m\vec{v}-\frac{q\vec{A}}{c}$, where A is the vector potential so that $\vec{B} = \nabla\times\vec{A}$. This would give an kinetic energy (or kinetic term in your Hamiltonian) equal to: $T = \frac{(m\vec{v}-\frac{q\vec{A}}{c})^2}{2m}$.
One often calls $m\vec{v}$ the kinematic momentum and $\vec{p}$ the canonical momentum (since this is the momentum that follows from a Lagrangian). It are the canonical momenta that are called "thé momenta".

*And last but not least, in quantum mechanics you work on a different kind of mathematical structure (in Hilbert-space in stead of the classical phase-space). By equivalence with wave-theory the momentum is derived as $\hat{p} = \frac{\hbar}{i}\frac{\partial}{\partial x}$, where we have an operator $\hat{p}$ which works on that Hilbertspace.


Note: from dj_mummy
Note that the given relations for kinetic energy all assume the particle was originally at rest at a given time $t=t_0$. The kinetic energy is derived by using the work-energy theorem (difference in kinetic energy = work done on the point particle). Using the laws of classical physics (Newton's second law) we get:
$T = \int\limits_{x_0}^{x_1}\vec{F}\cdot d\vec{x} = \int\limits_{t_0}^{t_1}\vec{F}\cdot \vec{v}\,dt = \int\limits_{t_0}^{t_1}\left(m\frac{d\vec{v}}{dt}\right)\cdot \vec{v}\,dt= m\int\limits_{t_0}^{t_1} \vec{v}\cdot d\vec{v} = \frac{m}{2}v^2$. Where $\vec{v}$ is the velocity at time $t_1$.
This gives us the kinetic energy in the case of classical physics! For the relativistic particle we define a 4-momentum $p^\alpha = m\frac{du^\alpha}{d\tau}$. Where $u^\alpha$ is the four-velocity and $\tau$ is the proper time. The derivation is actually done here and we get the formula above !

