Is the Cauchy momentum equation the general form of Newton II?

[vco]

Is it appropriate to say that within classical physics the general form of Newton II is the Cauchy momentum equation?

This equation applies to an arbitrary continuum body. Therefore it is more general than the common form of Newton II which applies basically to point masses and centers of mass of rigid bodies only. However, I haven't heard anyone referring to the Cauchy momentum equation as the general form of Newton II.

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

 

[Orodruin]

No. NII applies to any system, not just point masses. The Cauchy momentum equations result from additional modelling of the forces acting on a continuum.

 

[vco]

No. NII applies to any system, not just point masses. The Cauchy momentum equations result from additional modelling of the forces acting on a continuum.

Does it really apply to a continuum as such? Because NII is usually expressed as F = ma, which kind of implies we are dealing with a point mass rather than an infinitesimal volume.

And shouldn't a continuum be considered the "default" system in mechanics, just like in electromagnetism (e.g. Maxwell's equations)?

 

[Orodruin]

Does it really apply to a continuum as such?

Yes.

Because NII is usually expressed as F = ma, which kind of implies we are dealing with a point mass rather than an infinitesimal volume.

No. First of all there is nothing in NII that implies a point mass. Second, the more appropriate formulation is F = dp/dt. What makes you think total momentum is not a property of an extended system? It is an extensive quantity.

And shouldn't a continuum be considered the "default" system in mechanics, just like in electromagnetism (e.g. Maxwell's equations)?

There is no ”default” system. It is all up to modelling after Newton’s laws.

 

[vco]

No. First of all there is nothing in NII that implies a point mass. Second, the more appropriate formulation is F = dp/dt. What makes you think total momentum is not a property of an extended system? It is an extensive quantity.

I understand that momentum is a property of all systems regardless of their type.

But why is momentum often defined as the "product of the mass and velocity of an object", $p=mv$. This implies that we are restricting ourselves either to a rigid body (using the velocity of the center of mass) or a point mass.

The aforementioned definition for momentum does not apply to a deformable continuum as such. The total momentum for a continuum would be something like $p = \int \rho v \mathrm{d}V$, from which we get $p=mv$ for special cases such as point masses. Right?

 

[Orodruin]

But why is momentum often defined as the "product of the mass and velocity of an object", p=mvp=mvp=mv. This implies that we are restricting ourselves either to a rigid body (using the velocity of the center of mass) or a point mass.

No it does not. For the particular case when you restrict yourself to a rigid body, it becomes p=mv.

The Cauchy momentum equations are derived from NII after also applying additional modelling of forces acting on a continuum. NII is more fundamental.

Nothing in NII a priori requires you to define momentum as p=mv, this is a special case based on modelling of a rigid body. In some sense, NII defines what a force is: it is something that causes a change in the momentum of a system over time. It is then up to you to model that system.

 

[Orodruin]

Also note that p=mv is also true for a continuum when v is taken to be the center of mass motion.

 

[vco]

Nothing in NII a priori requires you to define momentum as p=mv, this is a special case based on modelling of a rigid body. In some sense, NII defines what a force is: it is something that causes a change in the momentum of a system over time. It is then up to you to model that system.

Thanks.

However, I am struggling to find an online reference where Newtonian momentum is NOT defined as p = mv, i.e. the particular case for rigid bodies. Is this merely because of the appealing simplicity and wide applicability of the rigid-body assumption?

For example Wikipedia states that p = mv, but does not specify that the equation has a limited applicability. EDIT: Actually, the Wikipedia article has a section for continua.

 

[Orodruin]

.e. the particular case for rigid bodies.

Again, it is also true for continua where v is the center of mass motion.

 

[vanhees71]

The Cauchy equation is indeed the local momentum-balance equation, and this is indeed nothing else a direct application of Newton's Lex Secunda, which in full glory indeed reads (for a point particle $\vec{F}=\dot{\vec{p}}$.

The Cauchy equation refers to a material (macroscopically small microscopically large) fluid volume-element. In Euler notation the equation reads
$$\partial_t g_i+\partial_{j} S_{ji}=\vec{f},$$
where $\vec{g}$ is the momentum density, $S_{ji}=S_{ij}$ the momentum-flux densit, and $\vec{f}$ the force density.

The momentum density $S_{ji}$ gives the flux of the momentum component in direction $i$ (momentum per unit time and unit area) through a surface with normal vector in direction $j$. Part of this flux is due to the material particles moving through the surface and another part is due to the stresses $-\sigma_{ij}$ at the surface of the fluid volume element:
$$S_{ji}=\rho v_i v_j - \sigma_{ij}.$$
For a isotropic ideal fluid, e.g., $\sigma_{ij}=-p \delta_{ij}$.

From this you can derive usual fluid equation of motion (Euler for ideal, Navier-Stokes for viscous fluids):
$$g_i=\rho v_i \; \Rightarrow \; \partial_t g_i =v_i \partial_t \rho + \rho \partial_t v_i,$$
and
$$\partial_j S_{ji} =v_i \partial_j (\rho v_j) + \rho v_j \partial_j v_i - \partial_{j} \sigma_{ij}$$
Together with the continuity equation for mass (local mass-conservation law)
$$\partial_t \rho + \partial_j (\rho v_j)=0,$$
you get
$$\rho \mathrm{D}_t v_i = \rho [\partial_t v_i +(\vec{v} \cdot \vec{\nabla}) v_i]=\partial_j \sigma_{ji} + f_i.$$
For the ideal fluid you have
$$\partial_j \sigma_{ji}=-\delta_{ji} \partial_j p=-\partial_i p.$$
Then
$$\rho \mathrm{D}_t \vec{v}=-\vec{\nabla} p + \vec{f}.$$
Note that the signs of the stress tensor and pressure are conventions.