# 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:

1. M. Nakahara, Geometry, Topology & Physics, 2nd edition, 2003, Section 1.1.1.
• Because the correct free value is $(\gamma - 1)mc^2$ (from relativity). In the limit of $v \ll c$ this is well approximated by the Newtonian form. – dmckee --- ex-moderator kitten Sep 21 '13 at 0:18
• but the chapter is talking about classical mechanics there I think. Not to mention that if we're speaking relativistically, then the kinetic energy is then never exactly equal to $1/2mv^2$ except when $v=0$. Are you saying that in classical, it is true by definition that $1/2mv^2$ is the kinetic energy? – Jeff Sep 21 '13 at 0:23
• It's almost a matter of definition. You can define work and energy arbitrarily, so that $work = \kappa \int_\Gamma \vec{F}\cdot\mathrm{d}\vec{r}$, which, through Newton's second law, works out to $1/2 \kappa m v^2$ when that force works on an otherwise free body, where $\kappa$ defines your work units. But, as @dmkee says, relativity and the resultant mass-energy equivalence means your unit system must also think of work on a free body begetting increased inertia, so your units will become very complicated and tangled unless you set $\kappa = 1$. Relativity unifies phenomena such that .... – Selene Routley Sep 21 '13 at 0:32
• You haven't defined what is 'E'. Please state the question properly and provide the exact quote from the text. – guru Sep 21 '13 at 13:30
• Related: physics.stackexchange.com/q/11905/2451 and links therein. – Qmechanic Sep 23 '13 at 11:02

The definition for point particles is:

$T = \frac{p^2}{2m}$, where $T$ denotes my kinetic energy.

1. If you are dealing with classical physics, the momentum for a point-particle equals $\vec{p} = m\vec{v}$, which would give you $T = \frac{mv^2}{2}$.
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.
3. 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".
4. 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 !

• Huh? Your factors of $\gamma$ in #2 are wrong. If you fix the gamma in the equation for p, the gamma in your equation for T is still wrong. – user4552 Sep 23 '13 at 15:12
• @BenCrowell It should be fixed by now ... – Nick Sep 23 '13 at 17:42
• @Nick: But now it's not consistent with the assertion at the beginning that $T=p^2/2m$ in all cases, which I think is false. – user4552 Sep 23 '13 at 19:20
• @BenCrowell that's what I get for doing to many things at the same time, I god rid of the additive constant! – Nick Sep 23 '13 at 21:40
• @Nick: Huh? I don't understand your most recent comment. There is no additive constant. The issue is that $T=p^2/2m$ is false in the relativistic case. – user4552 Sep 23 '13 at 21:45