Connection between Noether's Theorem and classical definitions of energy / momentum

Question

In classical mechanics, change in momentum $\Delta \mathbf p$ and change in kinetic energy $\Delta T$ of a particle are defined as follows in terms of the net force acting on the particle $\mathbf F_\text{net}$, where in each case the integrations are done over the path taken by the particle through spacetime. $$\begin{align} \Delta T &= \int \mathbf F_\text{net} \cdot d\mathbf x \\ \Delta \mathbf p &= \int \mathbf F_\text{net} dt \end{align}\tag{1}$$

This suggests some sort of correspondence.

$$\begin{align} \mathbf x &\longleftrightarrow T \\ t & \longleftrightarrow \mathbf p \end{align}\tag{2}$$

Noether's theorem provides an association between physical symmetries and conserved quantities.

$$\begin{align} \text{symmetry in time} &\longleftrightarrow \text{conservation of energy} \\ \text{symmetry in position} &\longleftrightarrow \text{conservation of momentum} \end{align}\tag{3}$$

Additionally, when studying special relativity, there is a similar suggested correspondence between the components of the position four-vector $\mathbf X$ and the energy-momentum four-vector $\mathbf P$. Here, $E$ represents total energy $E = mc^2 + T + \mathcal O \left( v^3/c^3 \right)$

$$\begin{array} \ \mathbf X = \begin{bmatrix} ct \\ x \\ y \\ z \end{bmatrix} & \mathbf P = \begin{bmatrix} E/c \\ p_x \\ p_y \\ p_z \end{bmatrix} \end{array}\tag{4}$$ Thus, comparing components, and discarding factors of $c$ the following correspondence is suggested. $$ \begin{align} t &\longleftrightarrow E \\ \mathbf x &\longleftrightarrow \mathbf p \end{align}\tag{5} $$

Is there an underlying relationship between these correspondences I have pointed out? And, if so, why are the ones for the classical definitions of kinetic energy and momentum swapped compared to the ones arising from Noether's theorem and special relativity?

Answer 1

I think that you have the correspondences in classical mechanics a little inside out.

When you are integrating over some variable you are effectively getting rid of it. For example, if you are calculating an average velocity you will integrate over time to get rid of the fluctuation in time. Similarly, in multivariate probability distributions, if you want to get rid of one variable then you “marginalize” it which is just an integral over the variable you want to get rid of.

So when you integrate over space you get rid of space leaving only time, and when you integrate over time you get rid of time leaving only space. Thus the correspondences are the same both classically and relativistically and through Noether’s theorem. In each case energy corresponds with time and momentum corresponds with space

Answer 2

If you look a bit closely, what you are working out is simply the dimensional correspondences in the units. The unit of energy is force $\times$ distance, the unit of momentum is force $\times$ time (your first pair of relations). Velocity has units of distance/time, therefore by multiplying or dividing by the speed of light, you can convert one pair of dimensional relationships to another (giving the second pair of relations). If there is any deeper 'meaning' to this, it is that special relativity shows how and why $c$ must enter into equations of position and momentum.

Answer 3

1. OP's Newtonian observation (1) even generalizes to special relativity. The 4-force $${\bf F}~=~\frac{d{\bf P}}{d\tau}~=~\gamma \begin{bmatrix} {\bf f}\cdot \color{red}{{\bf u}/c} \cr {\bf f}\end{bmatrix}\tag{A}$$ is related to the 3-force $${\bf f}~=~\frac{d{\bf p}}{dt}, \qquad {\bf p} ~=~m_0 \gamma {\bf u}, \tag{B} $$ via what superficially appears in eq. (A) to be an "upside-down" version $$ \begin{bmatrix} \color{red}{\bf u} \cr \color{red}{c}\end{bmatrix} \tag{C}$$ of the 4-velocity $${\bf U}~=~\frac{d{\bf X}}{d\tau}~=~\gamma \begin{bmatrix} \color{red}{c} \cr \color{red}{\bf u}\end{bmatrix}, \qquad {\bf X}~=~\begin{bmatrix} ct \cr {\bf x}\end{bmatrix}, \tag{D}$$ [if we ignore the obvious mismatch (2) between a 3-vector & a 1-vector]. But (i) the 3-force ${\bf f}$ is not a Lorentz-covariant object, and (ii) ${\bf f}$ and ${\bf u}$ are not completely independent, so OP's observation (1) does not imply that an "upside-down" 4-velocity/4-position (2) has any (covariant) significance.

2. On the other hand, in point mechanics, indeed the Noether charge $$Q~=~ {\bf P} \cdot \delta{\bf X}\tag{E}$$ is a Minkowski 4-product between the 4-momentum and the spacetime symmetry 4-vector $\delta{\bf X}$ of the action $S$, cf. e.g eq. (7) in my Phys.SE answer here. This confirms the usual duality (5).

Answer 4

Physical Thinking might help over mathematical approaches:

The Noether theorem is the one and only which states the correspondences that the energy is nothing but a conserved charge of time translation symmetry and the momentum is the conserve charge of space translational symmetry. These correspondences are the direct consequences of the Noether theorem.

For your classical correspondences:

1) The first correspondences have no physical meaning. 2) The equations above gives change in kinetic energy or momentum and not the absolute values. And just because you got two identical looking equations does't mean that you can play with it to make any physical correspondences. 3) The concept of charge is only defined for a symmetry.

Therefore, the first conclusions are completely wrong and have no physical meaning.