Newtonian analogue of energy-momentum relation?

[greypilgrim]

Hi.

I've read that there's no Newtonian analogue of the energy-momentum relation
$$E^2-(pc)^2=(mc^2)^2\enspace .$$

Why doesn't
$$E=\frac{p^2}{2m}$$
qualify as such? There's no rest energy in Newtonian physics anyway.

 

[Andrew Mason]

Hi.
There's no rest energy in Newtonian physics anyway.

That is why there is no analogy of the energy-momentum relation to Newtonian physics.

AM

 

[greypilgrim]

That is why there is no analogy of the energy-momentum relation to Newtonian physics.

Well the name "energy-momentum relation" doesn't suggest there to be one. It's just a relation between energy and momentum (and mass) that happens to be true in all inertial frames, as is the Newtonian one. But apparently there's more physical content in the relativistic formula, but I fail to see it.

 

[Andrew Mason]

Well the name "energy-momentum relation" doesn't suggest there to be one. It's just a relation between energy and momentum (and mass) that happens to be true in all inertial frames, as is the Newtonian one. But apparently there's more physical content in the relativistic formula, but I fail to see it.

The energy momentum relation is true in all frames. But energy and momentum are not the same in all inertial frames. Energy and momentum depend on the frame of reference in which they are measured.

AM

 

[greypilgrim]

The energy momentum relation is true in all frames. But energy and momentum are not the same in all inertial frames. Energy and momentum depend on the frame of reference in which they are measured.

Sure, but exactly the same is true with
$$E=\frac{p^2}{2m}$$
or
$$E-\frac{p^2}{2m}=0\enspace ,$$
if you prefer.

 

[Sorcerer]

Hi.

I've read that there's no Newtonian analogue of the energy-momentum relation
$$E^2-(pc)^2=(mc^2)^2\enspace .$$

Why doesn't
$$E=\frac{p^2}{2m}$$
qualify as such? There's no rest energy in Newtonian physics anyway.

My take:

The form
$$E=\frac{p^2}{2m}$$
holds, in the sense that
$$E’=\frac{p’^2}{2m}$$
is also true, but ##E’ ≠ E##. That is just kinetic energy, and kinetic energy is also frame dependent in SR.

Meanwhile, in Newton there is no equivalent relation that is invariant everywhere of the form ##E_0=mc^2## except at most trivially, and I’d argue it is related to the fact that in Newton time is absolute but in SR only the spacetime interval is absolute.

To elaborate, the exact value of ##mc^2##, which is part of the energy-momentum relation, holds everywhere in SR. And that value is related to energy and momentum in a non-trivial way.

 

[PeroK]

Hi.

I've read that there's no Newtonian analogue of the energy-momentum relation
$$E^2-(pc)^2=(mc^2)^2\enspace .$$

Why doesn't
$$E=\frac{p^2}{2m}$$
qualify as such? There's no rest energy in Newtonian physics anyway.

That looks like the analogous equation in my view. It relates mass, energy and momentum. An analogy is not, however, a precisely defined term, hence there is room for quibbling!

 

[Sorcerer]

Hi.

I've read that there's no Newtonian analogue of the energy-momentum relation
$$E^2-(pc)^2=(mc^2)^2\enspace .$$

Why doesn't
$$E=\frac{p^2}{2m}$$
qualify as such? There's no rest energy in Newtonian physics anyway.

My take:

The form
$$E=\frac{p^2}{2m}$$
holds, in the sense that
$$E’=\frac{p’^2}{2m}$$
is also true, but ##E’ ≠ E##. That is just kinetic energy, and kinetic energy is also frame dependent in SR.

Meanwhile, in Newton there is no equivalent relation that is invariant everywhere of the form ##E_0=mc^2## except at most trivially, and I’d argue it is related to the fact that in Newton time is absolute but in SR only the spacetime interval is absolute.

To elaborate, the exact value of ##mc^2##, which is part of the energy-momentum relation, holds everywhere in SR. And that value is related to energy and momentum in a non-trivial way.

Based on your other thread I realize this might not be satisfactory, so here's my attempt to clarify, and point out why the way space and time are thought of in special relativity versus Newtonian (aka Galilean) relativity are key.
In Newtonian relativity, the value ##(Δd)^2 = (Δx)^2 + (Δy)^2 + (Δz)^2## is invariant. That means no matter what speed observers are moving at relative to each other, they will all agree on the distance between two points. Observers may disagree on the specific values of the x, y, or z coordinates, but the above value will always be the same. In addition, the value ##Δt## is always the same. These two things are kept separate, and both are invariant.

In special relativity, neither ##(Δd)^2## nor ##Δt## are agreed upon by all observers. However, the value ##(Δs)^2 = (Δt)^2 - [ (Δx)^2 + (Δy)^2 + (Δz)^2]## is agreed upon. Note that the signs chosen here are convention. The following is also agreed upon and is in fact the same thing: ##(Δs)^2 = -(Δt)^2 + (Δx)^2 + (Δy)^2 + (Δz)^2## (Look up sign convention for special relativity for more information on the signs). The point is that the combination of space and time in this way, where a four dimensional "distance" is calculated in such a way that the sign separating time and space is negative, is invariant (this is often called the Minkowski innner product or the Lorentzian inner product, and it is similar to a dot product, which gives you a magnitude; note that space and time are not the only things in special relativity that share this relationship).

So there is an analogue in special relativity to the Euclidean distance, ##(Δd)^2 = (Δx)^2 + (Δy)^2 + (Δz)^2##, in Newtonian relativity: ##(Δs)^2 = -(Δt)^2 + (Δx)^2 + (Δy)^2 + (Δz)^2##.

This is what leads, eventually, to the invariant ##E_0 = mc^2##.
Now what about energy and momentum?

The difference in space and time written about above is what leads to the difference between how Newtonian relativity treats energy and momentum and how special relativity treats energy and momentum. The key reason is that the way measurements are transformed between reference frames (what one observer measures something versus what another observer measures) is different between the two theories.

As a result of the way space and time are described by each theory, you'll get the following differences in energy:

In Newtonian relativity, ##E = U + T##, where U is potential energy and T is kinetic energy. So, if E is total energy, we can write ##T = E - U##. Assuming we're just dealing with point particles (so we can neglect any stored energy inside the object), U is completely dependent upon position, which means with a suitable choice of reference frame, ##E = T##, that is,
$$E = \frac{p^2}{2m}.$$

Now, in special relativity, there is a different relation for kinetic energy (which arises directly from the spacetime interval/space and time transformation discussed at the top of this post). The value for T is:
$$T = \frac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}} - mc^2$$

If you use the binomial theorem, you will find that this new equation for kinetic energy is approximately equal to ##\frac{1}{2}mv^2## if v << c. This link shows this to be true.

Anyway, it is also true that ##T = E - U## in special relativity.

Looking at the two parts of ##T = E - U## as written for special relativity, it is then clear that
$$E = \frac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}}$$
and
$$U = mc^2$$

But you cannot choose a reference frame in which ##U = mc^2## will be zero, because m is not dependent upon space or time, nor speed; nor is c. So no matter what reference frame you choose, that pesky ##mc^2## will always be there.And as I alluded to, that comes from the difference in how space and time is treated in special relativity. If you want to see, I'm sure somewhere on this forum you can find a derivation of the energy and momentum equations in special relativity starting from the Loretnz transformation equations (the way values are transformed in special relativity), and there are probably two or three threads per month about how the Lorentz transformation equations are derived.

So ultimately, I'd say the answer to your question about why there is not an equivalent relation ##E_0 = mc^2## in Newtonian relativity is because special relativity treats space and time differently. ##mc^2## is a magnitude in the same way that ##(Δs)^2## is, and it will be the same regardless of which reference frame you choose. As Newton does not have the same type of spacetime interval magnitudes, it will not have the same energy-momentum magnitude (again, speaking of the Lorentzian inner product). In particular, ##mc^2## is the Lorentzian inner product of 4-momentum with itself, as described here. (and 4-momentum is the special relativity relation between momentum and energy)

 

[Nugatory]

Hi.

I've read that there's no Newtonian analogue of the energy-momentum relation
$$E^2-(pc)^2=(mc^2)^2\enspace .$$

This entire thread would go more smoothly if you could provide a better reference than "I've read that..."; that way we'd know what qualifies as an "analogue" and what does not.

 

[haushofer]

Hi.

I've read that there's no Newtonian analogue of the energy-momentum relation
$$E^2-(pc)^2=(mc^2)^2\enspace .$$

Why doesn't
$$E=\frac{p^2}{2m}$$
qualify as such? There's no rest energy in Newtonian physics anyway.

My take: on the level of the underlying Lie algebra, conserved quantities are given by Casimir-operators (which are quadratic in the generators and commute with all the generators). The Poincare algebra contains two such Casimirs, which you can easily find in any textbook; one of them is ##P_{\mu}P^{\mu}##. The non-relativistic analogue of the Poincare algebra is the Bargmann algebra,

https://en.wikipedia.org/wiki/Representation_theory_of_the_Galilean_group

which contains three "Casimirs". One of them corresponds to the non-relativistic dispersion relation you mention. The other relevant one is not really a Casimir, but a (actually: the) central extension ##Z##. This corresponds to the mass.

So you see, heuristically in going from the Poincare to the Bargmann algebra the Poincare Casimir ##P_{\mu}P^{\mu}## "breaks up" into the Casimir operator ##ZH - \frac{1}{2}P_i P^i## and the central extension ##Z##.

See also

https://www.nikhef.nl/pub/services/biblio/theses_pdf/thesis_R_Andringa.pdf

section 3.2 onwards.

 

[Dale]

Why doesn't

E=p22mE=p22m​

E=\frac{p^2}{2m}
qualify as such?

Because it is not frame invariant in Newtonian physics (except for the trivial form)

 

[DrStupid]

I've read that there's no Newtonian analogue of the energy-momentum relation
$$E^2-(pc)^2=(mc^2)^2\enspace .$$
Why doesn't
$$E=\frac{p^2}{2m}$$
qualify as such?

These relations have different meaning. In the upper formula E is the total energy and in the lower it is the kinetic energy of translation.

[Moderator's note: The rest of this post has been spun off into a separate thread; see link below.]

https://www.physicsforums.com/threads/rest-energy-in-classical-mechanics.960432/

 

[PeterDonis]

Thread closed for moderation.

 

[PeterDonis]

A subthread off the main topic has been moved to a separate thread (see the link in post #12). Thread reopened.

 

[sweet springs]

I've read that there's no Newtonian analogue of the energy-momentum relation

E2−(pc)2=(mc2)2.​

E^2-(pc)^2=(mc^2)^2\enspace .

Why doesn't

E=p22m​

E=\frac{p^2}{2m}
qualify as such? There's no rest energy in Newtonian physics anyway.

1
[tex]E=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}=mc^2+\frac{1}{2}mv^2+m\ o(\frac{v^2}{c^2})[/tex]
[tex]\frac{p^2}{2m}=\frac{mv^2}{2}(1+o(\frac{v^2}{c^2}))[/tex]
Say v/c << 1, kinetic energy K is
[tex]K:=E-mc^2=\frac{p^2}{2m}+o(\frac{v^2}{c^2})[/tex]

2
[tex](E-pc)(E+pc)=m^2c^4[/tex]
introducing kinetic energy K,
[tex](mc^2+K-pc)(mc^2+K+pc)=m^2c^4[/tex]
[tex]K^2 +2 mc^2 K -p^2c^2=0[/tex] as K>0
[tex]K=-mc^2+\sqrt{m^2c^4+p^2c^2}[/tex]
[tex]\frac{K}{mc^2}=\sqrt{1+\frac{p^2}{m^2c^2}}-1=\frac{1}{2}\frac{p^2}{m^2c^2}+o((\frac{p^2}{m^2c^2})^2)[/tex]
First term of RHS is usual non relativistic kinetic energy expression by momentum.