8
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ Don't forget about the uncertainty relationships of position and momentum or energy and time :) $\endgroup$ – Aaron Stevens Sep 22 '18 at 0:54
7
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Thats an interesting notion, but I'm not entirely convinced. If this is the case, then how would one integrate $\mathbf F_\text{net}$ over both $dt$ and $dy$, for example, to express simultaneous symmetries in $dx$ and $dz$, producing some sort of a conserved $x$-$z$-only momentum? $\endgroup$ – Trevor Kafka Sep 22 '18 at 14:25
  • 1
    $\begingroup$ Sure, you could do that if you wanted. What you describe would indeed be conserved. $\endgroup$ – Dale Sep 23 '18 at 1:42
  • $\begingroup$ But how? I can't think of how to construct an integral to integrate over a 3-vector over only two dimensions in a manner analogous to a dot product. $\endgroup$ – Trevor Kafka Sep 23 '18 at 2:09
  • 1
    $\begingroup$ Then I don’t think that line of reasoning is worth pursuing further. If it is not immediately obvious to you then pursuing it further is not likely to help, particularly since it is something that has no physical motivation. $\endgroup$ – Dale Sep 23 '18 at 10:03
  • $\begingroup$ Why wouldn't it have physical motivation? It would be a conserved quantity, as you said... $\endgroup$ – Trevor Kafka Sep 26 '18 at 5:33
1
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ Sure, the units work out. They must work out in any equation. Not sure this addresses the question though, especially since it doesn't apply to the Noether's theorem bit. $\endgroup$ – Trevor Kafka Sep 21 '18 at 22:33
1
$\begingroup$
  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).

$\endgroup$
  • $\begingroup$ I'm not sure I understand what you mean by "upside-down" velocity. Could you elaborate? $\endgroup$ – balu Sep 24 '18 at 18:13
  • $\begingroup$ I added an eq. (C). $\endgroup$ – Qmechanic Sep 24 '18 at 19:40
  • $\begingroup$ Thank you! Now I see what you mean. I still think you might want to add some more comments and/or maybe structure your post in a different way. Right now, your answer mostly just reads like a list of facts and observations and I'm missing a clear and unequivocal statement about whether OP is right or wrong. $\endgroup$ – balu Sep 24 '18 at 23:17
0
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ I wouldn't say the correspondences are nonsensical. They make sense, even if you don't think they are at the same level as Noether's theorem. $\endgroup$ – Aaron Stevens Sep 22 '18 at 0:24
  • 1
    $\begingroup$ @AaronStevens For you, i have added some reasons to my answer. or if you have any objection, please write your own answer in the answer section. thank you! $\endgroup$ – Aman pawar Sep 22 '18 at 0:34
  • 2
    $\begingroup$ Thank you for the reply. I still don't think his claims are nonsensical. Perhaps we are using different definitions of the word. To me saying something is nonsensical means that it does not make sense. This is independent of its correctness. The OP has layed out their reasoning and there is nothing that does not make sense. You might disagree, but the line of reasoning is easy to follow and makes sense. $\endgroup$ – Aaron Stevens Sep 22 '18 at 0:42
  • 1
    $\begingroup$ @AaronStevens Just go with each steps more physically, then you will realize that making such a correspondences directly from a pair of different equations which are looking similar is very nonsensical approach in physics. $\endgroup$ – Aman pawar Sep 22 '18 at 1:08
  • 1
    $\begingroup$ 1) could you please point out where you see an error? 2) Conservation laws have everything to do with changes. Whenever ∆Q=0, we say the quantity Q is conserved. $\endgroup$ – Trevor Kafka Sep 22 '18 at 1:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.