# How to describe the arisal reasons of conservation Laws of Classical Mechanics by Newtonian Mechanics?

In the context of conservation laws of Classical Mechanics. that is (1) Conservation of Linear Momentum; (2) Conservation of Angular Momentum; (3) Conservation of Energy, we know that those arise from the following reasons respectively : (1) Homogeneity of Space; (2) Isotropy of space; (3) Homogeneity of Time.

Without going deeper into Classical Mechanics principles, restricting ourselves only to Newtonian Mechanics knowledge, can we describe these three conservation principles and their underlying arisals?

If so then how?

Linear momentum

About conservation of linear momentum when two objects are exerting a force upon each other:

Let it be granted that $$F=ma$$ holds good.
More generally, let it be granted that $$F=ma$$ obtains independent of location of space, and independent of orientiation in space

Let there be two objects, in space, small enough that any gravitational acceleration is negligable. We name these object 1 and object 2.

(Other than that a large poplulation of objects is present, with gravitational acceleration between these objects negligable, thus the population as a whole provides reference of acceleration for the motions of object 1 and object 2.)

Let a cable be stretched between object 1 and object 2. For the benefit of the thought demonstration we treat that cable as massless. One of the objects starts reeling in that cable.

The following sequence is observed:
State A: first the two are both stationary
They start moving towards each other
When the two come in contact the two stick together and in that final state (state B) they don't have a velocity with respect to State A.

The following is of central importance: the resulting motion does not tell us which object was reeling in the cable.

This is because in space the only leverage you have is your own inertial mass. By contrast: on Earth you can attach yourself to the Earth surface, thus securing yourself to something much more massive than yourself; in space the only leverage that you have is your own inertial mass.

In space the amount of leverage that any object has (for the purpose of reeling in another object) is given by F=ma. The amount of opposition to being reeled in is also given by F=ma

Is the process of reeling in mutual? Or is it one-sided? The point is: that is indistinguishable; on the basis of the resulting motion we cannot eliminate either possibility.

Since we cannot eliminate either possibility we turn that around and acknowledge that in our theory of motion no such distinction can be made.

So we have the following:

If it is granted that F=ma obtains independent of location in space, and independent of orientiation in space:

"To every action there is an equal and opposite reaction"

In symbol notation:

$$F_{reaction} = -F_{action}$$

Hence we have for motion with respect to the Common Center Of Mass (CCOM) of object 1 and object 2:

$$m_1 * a_1 = m_2 * -a_2$$

It follows that during the entire time:

$$m_1 * v_1 = m_2 * -v_2$$

It follows that with respect to the CCOM the total linear momentum remains zero.

Angular momentum

As we know, with $$\omega$$ for angular velocity the expression for the conserved angular momentum is $$m r^2 \omega$$

While linear momentum is a property of motion along a line, angular momentum is property of motion in a plane. It can be circular motion or spiralling motion, the point is that the minimum space you need in order to have an angular momentum at all is two spatial dimensions: a plane.

As we know, when we are dealing with two spatial dimensions we are dealing with area. There is a correlation between angular momentum and area.

The first derivation in the Newton's Principia is a derivation of Kepler's law of areas from first principles.

(I reproduce the text from wikipedia, because I am the author of that image, and I wrote that exposition)
During the first interval of time, an object is in motion from point A to point B. Undisturbed, it would continue to point c during the second interval. When the object arrives at B, it receives an impulse directed toward point S. The impulse gives it a small added velocity toward S, such that if this were its only velocity, it would move from B to V during the second interval. By the rules of velocity composition, these two velocities add, and point C is found by construction of parallelogram BcCV. Thus the object's path is deflected by the impulse so that it arrives at point C at the end of the second interval. Because the triangles SBc and SBC have the same base SB and the same height Bc or VC, they have the same area. By symmetry, triangle SBc also has the same area as triangle SAB, therefore the object has swept out equal areas SAB and SBC in equal times.

At point C, the object receives another impulse toward S, again deflecting its path during the third interval from d to D. Thus it continues to E and beyond, the triangles SAB, SBc, SBC, SCd, SCD, SDe, SDE all having the same area. Allowing the time intervals to become ever smaller, the path ABCDE approaches indefinitely close to a continuous curve.

The above geometric reasoning shows there is a conserved quantity. This conserved quantity is proportional to the area of the triangle that are swept out. This explains why the expression for angular momentum, $$m r^2 \omega$$, is a quadratic expression.

Energy

In order to make statements about Energy we need to derive the work-energy theorem.

In the course of the derivation the acceleration is $$a$$ is restated as follows:

$$a = \frac{dv}{dt} = \frac{dv}{ds} \frac{ds}{dt} = v \frac{dv}{ds} \qquad (1)$$

The integral for acceleration from a starting point $$s_0$$ to a final point $$s$$

$$\int_{s_0}^s a \ ds \qquad (2)$$

Use (1) to change the differential from ds to dv. Since the differential is changed the limits change accordingly.

$$\int_{s_0}^s a \ ds = \int_{s_0}^s v \frac{dv}{ds} \ ds = \int_{v_0}^v v \ dv = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2 \qquad (3)$$

So we have:

$$\int_{s_0}^s a \ ds = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2 \qquad (4)$$

Until this point it was all pure mathematics. It's only in the next step that a physics statement is added:

Combine (4) with $$F=ma$$

$$\int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 \qquad (5)$$

This is the work-energy theorem. The physics content of the work-energy theorem is $$F=ma$$.

Given the work-energy theorem it makes sense to define potential energy as the negative of work done.

$$\Delta(E_k) = -\Delta(E_p)$$

So in classical mechanics: in all cases where the integral of force over distance is well defined there is a defined potential energy. Whenever that is the case it follows from the work-energy theorem that in that well defined case Energy is conserved.

[later edit]

About correlation between symmetries and conservation principles:

Such a correlation is applicable in the case of theory of motion.

Two objects can be exerting a force upon one another in several ways (electrostatic force, reeling in a cable, etc.)

If that force has the property that the law describing the amount of force is the same everywhere in space then that force law is a force law for which conservation of linear momentum will hold good.

If that force has the property that the law describing the amount of force is the same in any orientation in space then that force law is a force law for which conservation of angular momentum will hold good.

If that force has the property that the law describing the amount of force is the same at any point in time then that force law is a force law for which conservation of energy will hold good.

Yes, you can, but it's not very deep.

Newton's Laws include: for every force, $$\vec F_0$$, there is an equal and opposite, $$\vec F_R$$, that is:

$$\vec F_R = - \vec F_0$$

Adding all the forces to get the total force:

$$\vec F_T = \vec F_0 + \vec F_R = \vec F_0- \vec F_0= 0$$

Since force is $$\dot{\vec p}$$, the total momentum satisfies:

$$\frac{d\vec p_T}{dt} = \vec F_T=0$$

Momentum is conserved.

Angular momentum is a similar exercise.

With energy, you note that:

$$\vec F =-\nabla V$$

Following some trajectory, you find the work done on a particle:

$$\Delta T= W(s) = \int_{s_0}^{s_1} \vec F \cdot d\vec s = \int_{s_0}^{s_1} -\nabla V \cdot d\vec s = -(V(s_1)-V(s_0))\equiv -\Delta V$$

The change in energy is the sum changes in kinetic and potential energy:

$$\Delta E= \Delta T+\Delta V=-\Delta V+\Delta V=0$$

Energy is conserved.

• I understand your way, but how come we get to describe the (1) Homogeneity of Space; (2) Isotropy of space; (3) Homogeneity of Time, from your work? May 24, 2021 at 21:05