I Planetary orbits - the 2-body problem

[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?

 

[jbriggs444]

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

@vanhees is trying to tell @dyn that he is mistaken.

@vanhees is incorrect. @dyn is perfectly correct.

The difficulty is that the two are using different definitions. They are failing to communicate.

Same with you. You are failing to communicate because you cannot see what the other person is trying to say. It's not that you are wrong. It is that the other person is right as well.

You cannot have a successful discussion until you come to an agreement on the subject matter and definitions.

 

[etotheipi]

No, @dyn was wrong to suggest that angular momentum of an isolated system is conserved regardless of whether the internal forces satisfy the weak or strong versions of lex 3; this is what I pointed out earlier, and also by @vanhees71 in #38.

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.

I don't know why you put so much emphasis on this, it's exactly the same definition as mine up to a co-ordinate transformation. The thing to emphasise is that the central force character of a given interaction is something intrinsic to the force, not to some arbitrary choice of axis in space. That the gravitational field evaluated at $\mathbf{r}$ of a body at point $\mathbf{r}_p$ is central means that $\mathbf{f} \propto \mathbf{r} - \mathbf{r}_p$; of course now define $\mathbf{r}' = \mathbf{r} - \mathbf{r}_p$ and you reduce the problem to standard form.

It's what I said in post #52!

 

[jbriggs444]

No, @dyn was wrong to suggest that angular momentum of an isolated system is conserved regardless of whether the internal forces satisfy the weak or strong versions of lex 3; this is what I pointed out earlier, and also by @vanhees71 in #38.

Absent a definition for "central force", the argument in #38 is utterly irrelevant to the question of whether "central forces" are required for angular momentum conservation.

I pointed this out in #39.

 

[vanhees71]

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

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.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.

For me the standard definition of central force has to distinguish between the more accurate case of (a) an interaction force and (b) the approximation that you consider the motion of a body in the field of a very much more massive body.

For (a), a central force is defined be directed along the relative position, $\vec{r}_1-\vec{r}_2$. For (b) you can choose a point as origin such that the force is along the position vector of the body under consideration.

Usually, you also conservative forces. Then for (a) the interaction potential is $V(|\vec{r}_1-\vec{r}_2|)$ for (b) $V(|\vec{r}|)$.

 

[A.T.]

No, @dyn was wrong to suggest that angular momentum of an isolated system is conserved regardless of whether the internal forces satisfy the weak or strong versions of lex 3;

If the forces between massive bodies don't satisfy lex 3, something else besides the massive bodies must be carrying angular momentum (as pointed out by @Ibix ). The but the angular momentum of an isolated system is conserved regardless.

 

[vanhees71]

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.

My analysis was for an interaction. The relative motion with $\vec{r}=\vec{r}_1-\vec{r}_2$ is described by
$$\mu \ddot{\vec{r}}=\vec{F}_{12}=F(r) \vec{r}/r.$$
Here $\mu=m_1 m_2/(m_1+m_2)$. Then the angular momentum of the relative motion
$$\vec{L}=\mu \vec{r} \times \dot{\vec{r}} = \text{const}.$$

The 3rd law is equivalent to momentum, not angular-momentum conservation.

 

[jbriggs444]

For me the standard definition of central force has to distinguish between the more accurate case of (a) an interaction force and (b) the approximation that you consider the motion of a body in the field of a very much more massive body.

For (a), a central force is defined be directed along the relative position, $\vec{r}_1-\vec{r}_2$. For (b) you can choose a point as origin such that the force is along the position vector of the body under consideration.

Usually, you also conservative forces. Then for (a) the interaction potential is $V(|\vec{r}_1-\vec{r}_2|)$ for (b) $V(|\vec{r}|)$.

Thanks for this. I see this as a disagreement about definitions rather than about substance.

 

[vanhees71]

Then what's your definition?

 

[jbriggs444]

My analysis was for an interaction. The relative motion with $\vec{r}=\vec{r}_1-\vec{r}_2$ is described by
$$\mu \ddot{\vec{r}}=\vec{F}_{12}=F(r) \vec{r}/r.$$
Here $\mu=m_1 m_2/(m_1+m_2)$. Then the angular momentum of the relative motion
$$\vec{L}=\mu \vec{r} \times \dot{\vec{r}} = \text{const}.$$

The 3rd law is equivalent to momentum, not angular-momentum conservation.

The question at hand is about the meaning of the phrase "central force". You cannot resolve that disagreement with equations. Especially not with equations that do not contain the phrase "central force".

 

[vanhees71]

You can ONLY resolve this with formulae. Perhaps we have to go again to the very basics, formulating a general closed two-body system
$$m_1 \ddot{\vec{x}}_1=\vec{F}_{12}(\vec{r}_1,\vec{r}_2), \quad m_2 \ddot{\vec{x}}_2=-\vec{F}_{12}(\vec{r}_1,\vec{r}_2).$$
Then it's easy to see that the total momentum
$$\vec{P}=m_1 \dot{\vec{x}}_1 + m_2 \dot{\vec{x}}_2=\text{const}.$$
Then by definition the force is a central force, if
$$\vec{F}_{12}=F(|\vec{x}_1-\vec{x}_2|) \frac{\vec{x}_1-\vec{x}_2}{|\vec{x}_1-\vec{x}_2|}.$$
Then it's very easy to see that the total angular momentum,
$$\vec{J}=m_1 \vec{x}_1 \times \dot{\vec{x}}_1 + m_2 \vec{x}_2 \times \dot{\vec{x}}_2=\text{const}.$$

 

[etotheipi]

If the forces between massive bodies don't satisfy lex 3, something else besides the massive bodies must be carrying angular momentum (as pointed out by @Ibix ). The but the angular momentum of an isolated system is conserved regardless.

Not necessarily! Yes, for electromagnetic interactions you can write the four momentum $P^i$ of the fields + the matter and conserve the associated angular momentum by setting the divergence of the integrand equal to zero.

But the Newtonian formalism does not prevent you from considering any other types of non-central forces for which the $\mathbf{L}_O = \sum_a \mathbf{r}_a \times \mathbf{p}_a$ is not an integral of the motion

 

[jbriggs444]

You can ONLY resolve this with formulae.

You can NEVER resolve matters of definition without using the definition(s).

 

[A.T.]

But the Newtonian formalism does not prevent you from considering any other types of non-central forces...

Maybe @dyn is not talking about anything that the Newtonian formalism doesn't explicitly prevent. Just saying that angular momentum is conserved in the real world.

 

[etotheipi]

Maybe @dyn is not talking about anything that the Newtonian formalism doesn't explicitly prevent. Just saying that angular momentum is conserved in the real world.

Perhaps, but the point is definitely worth stating! If you are trying to prove that this $\mathbf{L}$ thing is conserved for some system you're studying, at some point you're going to need to use that $(\mathbf{r}_a - \mathbf{r}_b) \times \mathbf{F}_{ab} = \mathbf{0}$ or similar. So it is important, as for any result, to understand what are the necessary assumptions. A priori there is no reason why this condition on the internal forces must hold!

 

[DrStupid]

What do you mean by "conclusion from conservation of momentum to conservation of angular momentum"?

Deriving conservation of angular momentum from conservation of momentum (or Newton's laws of motion).

@dyn's post was wrong and yours is too.

Are you really telling me that angular momentum is not conserved? Then please provide a proper reference.

 

[etotheipi]

Deriving conservation of angular momentum from conservation of momentum (or Newton's laws of motion).

In fact I never mentioned or wrote anything along those lines, so I don't know what you are referring to. But if you want then it's indeed possible to show that $\mathbf{L}$ is an integral using Newton's laws, yeah.

You can also just do it from symmetry principles, e.g. see post #51.

Are you really telling me that angular momentum is not conserved? Then please provide a proper reference.

Yes in the general case this $\mathbf{L} = \sum_a \mathbf{r}_a \times \mathbf{p}_a$ is not an integral of the motion, unless you also assume the internal forces are central.

How many times do I need to repeat the same thing until you understand?

 

[DrStupid]

In fact I never mentioned or wrote anything along those lines

That mean this is not from you:

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 $\mathbf{F} \parallel (\mathbf{r}_a - \mathbf{r}_b)$.

There must be something wrong with the forum software.

Yes

Your reference is missing as well.

 

[etotheipi]

Your reference is missing as well.

It's really not hard! This is ridiculous, it's literally a high school level argument\begin{align*}
\frac{d\mathbf{L}_O}{dt} &= \sum_a \frac{d}{dt} \left( \mathbf{r}_a \times \mathbf{p}_a \right) \\

&= \sum_a \left( \underbrace{\mathbf{v}_a \times \mathbf{p}_a}_{= \mathbf{0}} + \mathbf{r}_a \times \dot{\mathbf{p}}_a \right) \\ \\

&= \sum_a \mathbf{r}_a \times \mathbf{F}_a \\

&= \sum_b \sum_{a \neq b} \mathbf{r}_a \times \mathbf{F}_{ab} \\

&= \sum_b \sum_{a<b} (\mathbf{r}_a - \mathbf{r}_b) \times \mathbf{F}_{ab}
\end{align*}Now you may only generally conclude that $\dfrac{d\mathbf{L}_O}{dt} = \mathbf{0}$ if $\mathbf{F}_{ab} \parallel (\mathbf{r}_a - \mathbf{r}_b)$, i.e. if the internal forces are central.

 

[DrStupid]

It's really not hard!

It seems it is. Please refer to the Physics Forums Global Guidelines for acceptable references.

 

[berkeman]

It seems it is. Please refer to the Physics Forums Global Guidelines for acceptable references.

You need a reference for high school level math? Why? Do you have a reference that shows the reply is mistaken?

 

[dyn]

I just want to say thank you to everyone who replied in this thread. I appreciate your time and respect all your opinions and arguments

 

[fresh_42]

Extraordinary claims require extraordinary evidence!

... which is what the guidelines meant to communicate. However, there is an unspoken opposite of it:

Simple truths require pencil and paper to check.

E.g. I have no reference for, however, claim that ...

$$
\mathfrak{A(g)}=\{\alpha :\mathfrak{g}\longrightarrow \mathfrak{g}\,|\,\forall \,X,Y\in \mathfrak{g}\, : \,[\alpha (X),Y]+[X,\alpha (Y)]=0\} \text{ is a Lie algebra}
$$

... or at least none I would easily find. Nevertheless, all it needs to prove this statement are some applications of the Jacobi identity. Literally, everybody should be able to do this. And if someone wants proof, then I could either write down the few (boring) steps, as has been done in the above case, or simply say:
"A closer look at why $\mathfrak{Der(g)}$ is a Lie algebra shows, that the terms can be paired in such a way, that the required condition for $\mathfrak{A(g)}$ is a separate part of it."

You cannot demand references for some specific, nevertheless easy calculations.

 

[DrStupid]

You need a reference for high school level math?

No, I asked @etotheipi to provide a reference for his claim that angular momentum is not conserved. The high school level math he posted instead just shows that conservation of angular momentum does not follow from Newton's laws of motion without additional conditions. I already mentioned in #47 that this is something different and does not mean that angular momentum is not conserved. Thus, I'm still waiting for the reference.

 

[vanhees71]

This is really ridiculous. It is very clear, what a central interaction force is in the literature and for which forces angular momentum is conserved, namely precisely for central forces. I don't know, why the forum more and more has the tendency to discuss pseudo-problems and why we get more and more into a mode where you have to argue like a lawyer rather than a scientist. I have given the really simple proof, and @etotheipi has done so too. He is right in saying that this is high school level. If you need a reference take an arbitrary standard textbook on mechanics.

 

[fresh_42]

There is no need to discuss the distributive law or vector addition endlessly.

Thread closed.