# Newtonian analogue for Lorentz invariant four-momentum norm

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greypilgrim
Hi.

I read that the Lorentz invariance Minkowski norm of the four-momentum
$$E^2-c^2\cdot \mathbf{p}^2=m^2\cdot c^4$$
has no analogue in Newtonian physics. But what about
$$E-\frac{\mathbf{p}^2}{2m}=0\quad ?$$
It might look trivial by the definition of kinetic energy, but it's still a relation between energy and momentum that's invariant under Galilei transforms.

DrStupid
It's not a relation between energy and momentum but between kinetic energy and momentum. It doesn't work with the total energy.

You're right; I think the idea is that the newtonian expression is not the inner product of two four-vectors, since such a product does not really exist in newtonian spacetime; there is no (non-degenerate) metric.

Homework Helper
Gold Member
The natural candidate is ##mu^a## where ##m## is the [rest] mass and ##u^a## is the 4-velocity.
Presumably, there is a mass-shell in energy-momentum space which would look similar to the [timelike but degenerate] Galilean metric.
In https://www.desmos.com/calculator/ti58l2sair, set E=0.
The temporal component would be ##m## (or in standard units of momentum ##mc## where ##c## is a convenient velocity unit with no other significance).
To get the spatial components ##m\vec v##, one would use the spacelike-but-degenerate Galilean metric.
(To do this right, one needs to first write down the postulated structure [e.g., (M,##t_a##, ##h^{ab}##, ...) akin to specifying (M,g) for a spacetime] then formulate the dynamics on it.)
You can do 4-momentum conservation by vector addition.. which amounts to conservation of mass and conservation of spatial-momentum.

Kinetic energy should really be calculated using the Work-energy-theorem.

greypilgrim
It's not a relation between energy and momentum but between kinetic energy and momentum. It doesn't work with the total energy.

So in the relativistic case, the equation is about total energy? Then I'm running into problems with an answer I got in a different thread:
Specifically for the electromagnetic field, the conserved energy is given by:

$E = \gamma mc^2 + q \Phi$

where $\Phi$ is the electric potential.

Say the left side of
$$E^2=c^2\cdot \mathbf{p}^2+m^2\cdot c^4$$
is total energy squared. Consider two identical particles with the same velocity where one is in an electric potential and the other is not. Then their total energies are different, but their momentum is the same. So one of those particles must violate above equation.

DrStupid
So in the relativistic case, the equation is about total energy?

Yes.

Then I'm running into problems with an answer I got in a different thread: [...]

I'm not sure if the term "total energy" makes much sense in this example. It doesn't refer to the total system because the source of the potential is not included.