I Planetary orbits - the 2-body problem

[vanhees71]

Thank you.

As regards my 1st question , the Gregory book on CM states "in the 2-body problem the motion of P₁ relative to P₂ is the same as if P₂ were held fixed and P₁had the reduced mass μ instead of its actual mass" Now 1 and 2 can refer to the Sun and Earth or vice versa which is why my question referred to taking the Sun as fixed and then the Earth as fixed ; not rotating around the COM

As for Q3 i was referring to the total mechanical energy . ie. KE+PE

It's a bit strangely formulated, but obviously your book refers to the relative motion, i.e.,
$$\mu \dot{\vec{r}}=-G m_S m_E \frac{\vec{r}}{r^3}, \quad \vec{r}=\vec{r}_E-\vec{r}_S.$$
The details of the kinematics I have given above.

The total mechanical energy is the energy of both Earth and Sun together. It doesn't make sense to attribute "total energy" to the one or the other body, because the potential is an interaction potential, i.e., a two-body quantity.

 

[dyn]

It's a bit strangely formulated, but obviously your book refers to the relative motion, i.e.,
$$\mu \dot{\vec{r}}=-G m_S m_E \frac{\vec{r}}{r^3}, \quad \vec{r}=\vec{r}_E-\vec{r}_S.$$
The details of the kinematics I have given above.

The total mechanical energy is the energy of both Earth and Sun together. It doesn't make sense to attribute "total energy" to the one or the other body, because the potential is an interaction potential, i.e., a two-body quantity.

To calculate the escape velocity of a mass from the Earth , we require that the total energy of the mass , ie KE+PE be equal to zero. Is that not a similar situation where we attribute total energy to one body in an interaction potential ?

 

[vanhees71]

Here you approximate the situation by treating the Earth simply by an external gravitational field, neglecting the motion of the Earth due to its interaction with the object, which is of course justified, because the mass of the Earth is very much larger than that of the object. The same holds for the cestial mechanics Kepler problem in our solar system, because all planets' masses are very much smaller than that of the Sun.

 

[etotheipi]

If you are doing a problem then it's useful to remember that you may convert one-body equations to two-body equations by the transformation $G \mapsto G' = \left(1 + \frac{m_1}{m_2} \right)G$

 

[pbuk]

If you are doing a problem then it's useful to remember that you may convert one-body equations to two-body equations by the transformation $G \mapsto G' = \left(1 + \frac{m_1}{m_2} \right)G$

Stop procrastinating!

 

[dyn]

Thanks for all your replies. Here is my last question !

In #28 it was stated that the total angular momentum of the sun and Earth is a constant because the force between them is a central force.

If i consider the system as being the Earth and sun then there is no external force on that system and hence no external torque on that system ( the gravitational force between them is internal) so the total angular momentum of the 2-body system is constant regardless of whether the force is central or not between the 2 masses.

If i now consider the Earth or sun individually orbiting around the COM. The force on each mass is central so there is no torque on either mass and hence the angular momentum of the Earth is conserved and the angular momentum of the sun is separately conserved. This statement relies on the force being central.

Have i got that totally right ?

 

[jbriggs444]

Thanks for all your replies. Here is my last question !

In #28 it was stated that the total angular momentum of the sun and Earth is a constant because the force between them is a central force.

If i consider the system as being the Earth and sun then there is no external force on that system and hence no external torque on that system ( the gravitational force between them is internal) so the total angular momentum of the 2-body system is constant regardless of whether the force is central or not between the 2 masses.

Correctly reasoned. No external forces, so angular momentum is conserved regardless of the nature of any internal forces.

One might argue that Newton's third law by itself is inadequate for angular momentum to be conserved. One needs an additional principle. For instance, that the direction of a force is always along a line between the interacting objects. Or, in other words, that the force between pairs of interacting objects is always a "central force"

If i now consider the Earth or sun individually orbiting around the COM. The force on each mass is central so there is no torque on either mass and hence the angular momentum of the Earth is conserved and the angular momentum of the sun is separately conserved. This statement relies on the force being central.

Yes. Given this point of view we now have an external force on each object. Conservation of angular momentum (about the chosen center) means that the external force must have a line of action through that center.

 

[vanhees71]

Thanks for all your replies. Here is my last question !

In #28 it was stated that the total angular momentum of the sun and Earth is a constant because the force between them is a central force.

If i consider the system as being the Earth and sun then there is no external force on that system and hence no external torque on that system ( the gravitational force between them is internal) so the total angular momentum of the 2-body system is constant regardless of whether the force is central or not between the 2 masses.

If i now consider the Earth or sun individually orbiting around the COM. The force on each mass is central so there is no torque on either mass and hence the angular momentum of the Earth is conserved and the angular momentum of the sun is separately conserved. This statement relies on the force being central.

Have i got that totally right ?

No. The interaction force must be a central force if total angular momentum is conserved. That's clear without calculuation from Noether's theorem, but can also be easily proven from Newton's postulates:
$$m_E \ddot{\vec{x}}_E=\vec{F}_{12}, \quad m_S \ddot{\vec{x}}_S=\vec{F}_{21}=-\vec{F}_{12}.$$
Now
$$\vec{L}=m_E \vec{x}_E \times \dot{\vec{x}}_E + m_S \vec{x}_S \times \dot{\vec{x}}_S.$$
From this
$$\dot{\vec{L}}=m_E \vec{x}_E \times \ddot{\vec{x}}_E + m_S \vec{x}_S \times \ddot{\vec{x}}_S =( \vec{x}_E-\vec{x}_S) \times \vec{F}_{12}.$$
So if $\dot{\vec{L}}=0$ you must have $\vec{F}_{12}=\lambda (\vec{x}_F-\vec{x}_S)$, i.e., $\vec{F}_{12}$ must be a central force.

 

[jbriggs444]

No. The interaction force must be a central force if total angular momentum is conserved. That's clear without calculuation from Noether's theorem, but can also be easily proven from Newton's postulates:
$$m_E \ddot{\vec{x}}_E=\vec{F}_{12}, \quad m_S \ddot{\vec{x}}_S=\vec{F}_{21}=-\vec{F}_{12}.$$
Now
$$\vec{L}=m_E \vec{x}_E \times \dot{\vec{x}}_E + m_S \vec{x}_S \times \dot{\vec{x}}_S.$$
From this
$$\dot{\vec{L}}=m_E \vec{x}_E \times \ddot{\vec{x}}_E + m_S \vec{x}_S \times \ddot{\vec{x}}_S =( \vec{x}_E-\vec{x}_S) \times \vec{F}_{12}.$$
So if $\dot{\vec{L}}=0$ you must have $\vec{F}_{12}=\lambda (\vec{x}_F-\vec{x}_S)$, i.e., $\vec{F}_{12}$ must be a central force.

There are matters of definition to be cleared up before the above mathematics may be correctly applied. The definition you have adopted here for "central force" appears to be:

1a. A force between two interacting objects is a "central force" if and only if that force is zero or that force is "parallel" to the displacement between the two objects.

2. Two vectors are "parallel" if and only if they are both non-zero and their cross product is zero.

With these definitions in hand, one can successfully demonstrate, as you have done, that conservation of angular momentum for a system consisting of two point-like objects with a non-zero separation requires that the interaction force between the two objects be a "central force".However, the definition that you have adopted for "central force" may not be the one that @dyn is thinking of. There is a competing definition.

1b. A force on an object is said to be a "central force" if the "line of action" of that force passes through the chosen reference axis or reference point.

3. The "line of action" of a force on an object is a line passing through the object parallel to the applied force.

In two-dimensional angular momentum problems it is well known that the reference axis can be chosen freely. Similarly, in three-dimensional angular momentum problems the reference point can be freely chosen. In the absence of external forces, angular momentum of a system is conserved regardless of whether any internal forces have lines of action passing through the chosen reference point or reference axis.Wikipedia offers yet another definition for "central force" which applies in the context of a vector field. That definition need not concern us since it does not involve interacting objects.

 

[dyn]

I have seen conservation of angular momentum stated as "in any motion of an isolated system , the angular momentum of the system about any fixed point is conserved" or "in the absence of external torques , the total angular momentum of the system is a conserved quantity".

Both these statements imply that if we take the isolated system as the Earth and Sun together ; the total angular momentum of the Earth and Sun is a constant. The type of force (or mathematical structure of it) between these 2 objects does not matter for the total angular momentum to be conserved.

The fact that the force between the 2 masses is central is needed to show that the angular momentum of the Earth and Sun about the COM is individually conserved

 

[etotheipi]

Both these statements imply that if we take the isolated system as the Earth and Sun together ; the total angular momentum of the Earth and Sun is a constant. The type of force (or mathematical structure of it) between these 2 objects does not matter for the total angular momentum to be conserved.

Not quite!

If the two bodies are at positions $\mathbf{r}_a$ and $\mathbf{r}_b$ with respect to some origin $O$ and they exert forces $\mathbf{F}$ and $-\mathbf{F}$ respectively on each other, then $$\mathbf{G}_O = (\mathbf{r}_a - \mathbf{r}_b) \times \mathbf{F} = \frac{d\mathbf{L}_O}{dt}$$where $\mathbf{L}_O = m_a \mathbf{r}_a \times \mathbf{v}_a + m_b \mathbf{r}_b \times \mathbf{v}_b$. The quantity $\mathbf{L}_O$ is only a constant in the non-trivial case if $\mathbf{F} \parallel (\mathbf{r}_a - \mathbf{r}_b)$.

Unlike linear momentum, the angular momentum of an isolated system is not necessarily conserved without this further constraint!

 

[dyn]

Every single statement i have seen from every textbook states that angular momentum is conserved for every isolated system. There is never any mention of what internal forces are involved inside the system

 

[etotheipi]

A lot of forces satisfy the strong form of Newton III i.e. that in addition to $\mathbf{F}_{ab} = - \mathbf{F}_{ba}$ you also have $\mathbf{F}_{ab} = \lambda(\mathbf{r}_a - \mathbf{r}_b)$. But this is an extra condition you need to impose! For example, the Lorentz force is not central.

Example: take two particles who exert forces on each other such that $\mathbf{F}_{ab} \, \bot \, (\mathbf{r}_a - \mathbf{r}_b)$, i.e. perpendicular to the separation vector. If they start from rest, what happens to the angular momentum of that system in the subsequent motion?

 

[hutchphd]

I'm having trouble with this. Do you have a real world example?
My first thought was an electric motor with a battery, attached to a flywheel. But this system conserves angular momentum. A little help here...

 

[etotheipi]

The classic example for this type of asymmetry are the magnetic forces exerted by two particles in relative motion on each other; for instance, consider that at time $t$ a charge $q_1$ is at $(0,0,0)$ with velocity $\parallel \hat{\mathbf{y}}$ and a charge $q_2$ is at $(1,0,0)$ with velocity $\parallel \hat{\mathbf{x}}$.

 

[Ibix]

Note that momentum should be conserved in @etotheipi's scenario if you take into account the momentum of the EM field, the integral of the Poynting vector divided by $c$.

 

[DrStupid]

Every single statement i have seen from every textbook states that angular momentum is conserved for every isolated system. There is never any mention of what internal forces are involved inside the system

And that is correct. @etotheipi is talking about the conclusion from conservation of momentum to conservation of angular momentum. That is something different. That's why his "Not quite!" is not quite correct.

 

[etotheipi]

@etotheipi is talking about the conclusion from conservation of momentum to conservation of angular momentum. That is something different. That's why his "Not quite!" is not quite correct.

Huh? What do you mean by "conclusion from conservation of momentum to conservation of angular momentum"? @dyn's post was wrong and yours is too. It should not be hard to see that angular momentum can be generated inside an isolated system without the central force constraint!

As @Ibix did allude to, for example to recover angular momentum conservation for problems involving the electromagnetic field it is necessary to write $M^{ik} = \int_{\phi(\Omega)} x^i dP^k - x^k dP^i$ with $P^i$ the four-momentum of matter + field contained in the hypersurface $\Omega$

 

[vanhees71]

There are matters of definition to be cleared up before the above mathematics may be correctly applied. The definition you have adopted here for "central force" appears to be:

1a. A force between two interacting objects is a "central force" if and only if that force is zero or that force is "parallel" to the displacement between the two objects.

We talk about closed systems and interaction forces. Then this is the standard definition in any textbook I know.

2. Two vectors are "parallel" if and only if they are both non-zero and their cross product is zero

With these definitions in hand, one can successfully demonstrate, as you have done, that conservation of angular momentum for a system consisting of two point-like objects with a non-zero separation requires that the interaction force between the two objects be a "central force".However, the definition that you have adopted for "central force" may not be the one that @dyn is thinking of. There is a competing definition.

1b. A force on an object is said to be a "central force" if the "line of action" of that force passes through the chosen reference axis or reference point.

This is for the approximation where you describe the Sun as a fixed (infinitely massive) object and describe its action on the planet by an external gravitational field, which is a central force in the limit that you can consider the Sun as a spherical symmetric mass distribution.

3. The "line of action" of a force on an object is a line passing through the object parallel to the applied force.

In two-dimensional angular momentum problems it is well known that the reference axis can be chosen freely. Similarly, in three-dimensional angular momentum problems the reference point can be freely chosen. In the absence of external forces, angular momentum of a system is conserved regardless of whether any internal forces have lines of action passing through the chosen reference point or reference axis.

I'm not sure what you mean by this and which "reference axis" you are referring to. Angular momentum conservation refers to SO(3) symmetry around some point. If you have rotation symmetry only around some preferred axis then only the component of the angular momentum around this axis is conserved.

Wikipedia offers yet another definition for "central force" which applies in the context of a vector field. That definition need not concern us since it does not involve interacting objects.

 

[jbriggs444]

We talk about closed systems and interaction forces. Then this is the standard definition in any textbook I know.

Definition of what? I do not think that we are communicating successfully.

This is for the approximation where you describe the Sun as a fixed (infinitely massive) object and describe its action on the planet by an external gravitational field, which is a central force in the limit that you can consider the Sun as a spherical symmetric mass distribution.

Again, we seem not to be communicating. By the definition I have given, the interaction force between Sun and Earth may or may not be central. Which it is depends on what reference axis or reference point one chooses to use.

I'm not sure what you mean by this and which "reference axis" you are referring to. Angular momentum conservation refers to SO(3) symmetry around some point. If you have rotation symmetry only around some preferred axis then only the component of the angular momentum around this axis is conserved.

The first notion of angular momentum to which students are exposed is in two dimensions. The third dimension is ignored. In this version, one can think of the reference point as, instead, a reference axis perpendicular to the plane.

If linear momentum is conserved and if angular momentum is conserved around any given axis then angular momentum is conserved about all parallel axes. This is an easy consequence of the parallel axis theorem. This result carries over into three dimensions, of course.

Accordingly, angular momentum is conserved regardless of whether a particular force pair happens to have a line of action that passes through a chosen reference axis.

If one defines "central force" based on whether the line of action for a force pair passes through the reference axis, it follows that whether an internal force pair is "central" or not is irrelevant to angular momentum conservation.

 

[etotheipi]

Yes the symmetry transformation $\delta \mathbf{r} = \delta \boldsymbol{\theta} \times \mathbf{r}$ and $\delta \mathbf{v} = \delta \boldsymbol{\theta} \times \mathbf{v}$ with $\delta \boldsymbol{\theta} = \delta \theta \mathbf{\hat{n}}$ induces a change in Lagrangian of \begin{align*}
\delta \mathcal{L} = \dfrac{\partial \mathcal{L}}{\partial \mathbf{r}} \cdot \delta \mathbf{r} + \dfrac{\partial \mathcal{L}}{\partial \mathbf{v}} \cdot \delta \mathbf{v} &= \dot{\mathbf{p}} \cdot \delta \boldsymbol{\theta} \times \mathbf{r} + \mathbf{p} \cdot \delta \boldsymbol{\theta} \times \mathbf{v} \\

&= \delta \boldsymbol{\theta} \cdot \left( \mathbf{r} \times \dot{\mathbf{p}} + \mathbf{v} \times \mathbf{p} \right) \\

&= \delta \boldsymbol{\theta} \cdot \dfrac{d}{dt} \left( \mathbf{r} \times \mathbf{p} \right)

\end{align*}so letting $\hat{\mathbf{n}}$ be arbitrary implies $\dfrac{d}{dt} \left( \mathbf{r} \times \mathbf{p} \right) = 0$. Of course, if the angular momentum about $O$ is $\mathbf{L}_O = \mathbf{r} \times \mathbf{p}$ then about some axis $(O,\hat{\mathbf{n}})$ you may write\begin{align*}
\mathbf{L}_O \cdot \hat{\mathbf{n}} = \mathbf{r} \times \mathbf{p} \cdot \hat{\mathbf{n}} &= \hat{\mathbf{n}} \times \mathbf{r} \cdot \mathbf{p} \\

&= |\mathbf{r}| \sin{\varphi} \hat{\boldsymbol{\phi}} \cdot \mathbf{p} \\
&= \rho \hat{\boldsymbol{\phi}} \cdot \mathbf{p}

\end{align*}where $\rho = |\mathbf{r}| \sin{\varphi}$ is the distance from the axis $(O,\hat{\mathbf{n}})$ to the particle and $\hat{\boldsymbol{\phi}} \cdot \mathbf{p}$ is the azimuthal component of the momentum; $\mathbf{L}_O \cdot \hat{\mathbf{n}}$ is also an integral!

 

[etotheipi]

If one defines "central force" based on whether the line of action for a force pair passes through the reference axis, it follows that whether an internal force pair is "central" or not is irrelevant to angular momentum conservation.

I never seen this definition, usually a central force defined as one for which $\mathbf{f} = f(r) \hat{\mathbf{r}}$ where $\hat{\mathbf{r}}$ is a unit vector from the source of the field to the field point. Nothing a priori to do with an arbitrary axis in space, but sometimes convenient to consider an axis $(O, \hat{\mathbf{n}})$ passing through the source point to simplify the computations

 

[jbriggs444]

I never seen this definition, usually a central force defined as one for which $\mathbf{f} = f(r) \hat{\mathbf{r}}$ where $\hat{\mathbf{r}}$ is a unit vector from the source of the field to the field point. Nothing a priori to do with an arbitrary axis in space, but sometimes convenient to consider an axis $(O, \hat{\mathbf{n}})$ passing through the source point to simplify the computations

If this definition is used then the analysis by @vanhees71 vanishes in puff of irrelevance. We are dealing with an interaction force. There is no external field.

 

[etotheipi]

If this definition is used then the analysis by @vanhees71 vanishes in puff of irrelevance. We are dealing with an interaction force. There is no external field.

Why so? The analysis by @vanhees71 in #38 looks good! Both the Earth and the Sun generate central force fields, e.g. the force field of the Sun $\mathbf{f} = \tilde{f}(r) (\mathbf{r} - \mathbf{r}_{\odot}) = f(r) \hat{\mathbf{r}}$ where $\hat{\mathbf{r}} = \dfrac{\mathbf{r} - \mathbf{r}_{\odot}}{|\mathbf{r} - \mathbf{r}_{\odot}|}$. It is this central force property of the gravitational interaction which ensures angular momentum conservation, i.e. if $\mathbf{f}$ was not parallel to $\hat{\mathbf{r}}$ then $\mathbf{L}_O$ about any $O \in \mathbf{R}^3$ would no longer be an integral of the motion for two bodies moving under mutual gravitation in each other's fields.

[And simple co-ordinate transformation $\mathbf{r}' = \mathbf{r} - \mathbf{r}_{\odot}$ reduces to standard form]

 

[jbriggs444]

Why so? The analysis by @vanhees71 in #38 looks good! Both the Earth and the Sun generate central force fields, e.g. the force field of the Sun $\mathbf{f} = \tilde{f}(r) (\mathbf{r} - \mathbf{r}_S) = f(r) \hat{\mathbf{r}}$ where $\hat{\mathbf{r}} = \dfrac{\mathbf{r} - \mathbf{r}_S}{|\mathbf{r} - \mathbf{r}_S|}$. It is this central force property of the gravitational interaction which ensures angular momentum conservation, i.e. if $\mathbf{f}$ was not parallel to $\hat{\mathbf{r}}$ then $\mathbf{L}_O$ about any $O \in \mathbf{R}^3$ would no longer be an integral of the motion

[And simple co-ordinate transformation $\mathbf{r}' = \mathbf{r} - \mathbf{r}_S$ reduces to standard form]

Since both are moving, neither generates a static force field at all.

 

[etotheipi]

Since both are moving, neither generates a static force field at all.

It is not required to be time-independent, the gravitational fields of both bodies in the unconstrained motion are time-dependent! This makes no difference to the central character of these force fields and $\mathbf{f}(r,t) \propto \hat{\mathbf{r}}(t)$ for all time

 

[jbriggs444]

It is not required to be time-independent, the gravitational fields of both bodies in the unconstrained motion are time-dependent! This makes no difference to the central character of these force fields

In the definition of a vector field one learns when the concept is introduced, a time-varying field is not a field.

Please, for the love of God, can we keep this discussion at a level where @dyn has a prayer of learning and participating?

Again, I reiterate that the definition for "central force" that is in play is the one that @dyn is using:

A force is "central" if its line of action passes through the chosen reference axis.

You cannot successfully argue that this is not the definition with any fancy mathematical derivations. Definitions are what they are. They do not follow from the math. They precede it.

 

[etotheipi]

In the definition of a vector field one learns when the concept is introduced, a time-varying field is not a field.

Why not? How then do you propose to describe time-varying electric fields, magnetic fields, velocity fields of water, etc.?

At every point $p \in \mathbf{R}^3$ at any time $t$ there is associated a vector $\mathbf{g}(\mathbf{r}, t) = \mathbf{g}_{\odot}(\mathbf{r},t) + \mathbf{g}_{\oplus}(\mathbf{r},t)$, i.e. you may consider $\mathbf{g} : \mathbf{R}^3 \times \mathbf{R} \longrightarrow \mathbf{R}^3$?

I think, you are slightly missing the bigger picture

 

[jbriggs444]

Why not? How then do you propose to describe time-varying electric fields, magnetic fields, velocity fields of water, etc.?

You ask why the topic is taught as it is taught?

Very well, I shall answer: Because it is simple. We do not want to drown students in too much generality.

At every point $p \in \mathbf{R}^3$ at any time $t$ there is associated a vector $\mathbf{g}(\mathbf{r}, t) = \mathbf{g}_{\odot}(\mathbf{r},t) + \mathbf{g}_{\oplus}(\mathbf{r},t)$, i.e. you may consider $\mathbf{g} : \mathbf{R}^3 \times \mathbf{R} \longrightarrow \mathbf{R}^3$?

I think, you are slightly missing the bigger picture

I think you are missing the actual problem being discussed and the participant we are trying to reach.

 

[etotheipi]

I think you are missing the actual problem being discussed

Tell me the problem! I don't see any problem?