Confusion over the "centrifugal potential energy"

[phantomvommand]

TL;DR

I understand that potential energies are energies relative to reference points. For example, gravitational potential energy is relative to a point at infinity. What is the reference point for centrifugal potential energies?

I understand that potential energies are energies relative to reference points. For example, gravitational/electrical potential energy is relative to a point at infinity. What is the reference point for centrifugal potential energies?

In the case of planetary orbits, must the reference point also be at infinity, so that the reference point of centrifugal potential energy is consistent with that of gravitational potential energy?

Thanks for all the help.

 

[kuruman]

The so-called centrifugal potential is $U_{centr.}=\dfrac{L^2}{2mr^2}$ and results from a transformation of the rotational kinetic energy about the force center. Thus, the reference point is the force center about which the angular momentum $L$ is calculated.

 

[phantomvommand]

The so-called centrifugal potential is $U_{centr.}=\dfrac{L^2}{2mr^2}$ and results from a transformation of the rotational kinetic energy about the force center. Thus, the reference point is the force center about which the angular momentum $L$ is calculated.

Then why does it make sense to sum up the centrifugal potential energy with the GPE to get an "Effective potential", if their reference points are different? Exactly where is the Effective Potential relative to?

 

[vanhees71]

It's not clear to me, what you mean by "centrifugal" energy. So maybe, I misunderstand you by just guessing what you mean.

The usual treatment of planetary orbits (aka "the Kepler problem") is to consider the Sun and a single planet, ignoring the interaction of these with any other objects around. This has the advantage to be not too bad an approximation (because the Sun has much larger mass than all other objects around her) and as well being treatable analytically.

To make the discussion even simpler we can also assume that the Sun's mass is so much larger than the considered planet's mass that we can as well make the Sun fixed at the origin of a Cartesian coordinate system. Then the planet moves simply according to
$$m \ddot{\vec{r}}=-\gamma m M \frac{r}{r^2}.$$
Now first we see that taking the cross product with $\vec{r}$ gives 0 on the right-hand side, which gives
$$m \vec{r} \times \ddot{\vec{r}}=\mathrm{d}_t (m \vec{r} \times \dot{\vec{r}})=0$$
or
$$\vec{L}=m \vec{r} \times \dot{\vec{r}}=\text{const}.$$
That's angular-momentum conservation. Now we make $\vec{L}$ point along the 3-axis. The motion is then in the (12)-plane, and we can introduce polar coordinates
$$\vec{r}=(r \cos \varphi,r \sin \varphi,0).$$
From this
$$\dot{\vec{r}}=\dot{r} (\cos \varphi,\sin \varphi_0)+r \dot{\varphi} (-\sin \varphi,\cos \varphi,0).$$
$$\vec{L}=m \vec{r} \dot{\vec{r}}=m r^2 \dot{\varphi} (0,0,1).$$
Further taking the scalar product of the equation of motion with $\dot{\vec{r}}$, we find the energy-conservation law
$$\frac{m}{2} \dot{\vec{x}}^2-\frac{\gamma m M}{r}=E=\text{const}.$$
Now plugging in our polar coordinates this gives
$$\frac{m}{2} (\dot{r}^2+r^2 \dot{\varphi}^2)-\frac{\gamma mM}{r}=E.$$
Now we use
$$\vec{L}^2=m^2 r^4 \dot{\varphi}^2,$$
to eliminate $\dot{\varphi}$ from the energy-conservation equation:
$$\frac{m}{2} \dot{r}^2 + \frac{L^2}{2 m r^2} - \frac{\gamma m M}{r}=E.$$
Now you can reinterpret this as the equation of motion of a mass point moving along the $r$ axis moving in an effective potential
$$V_{\text{eff}}=\frac{L^2}{2m r^2} - \frac{\gamma m M}{r}.$$
The $r$ coordinate can be taken as the distance of an observer sitting at the planet to the Sun, which is a rotating reference frame, and seen from this frame $L^2/(2m r^2)$ is the potential of the corresponding centrifugal force.

Taking the time-derivative of the energy-conserving equation indeed gives (after deviding by $\dot{r}$)
$$m \ddot{r}=\frac{L^2}{m r^3}-\frac{\gamma m M}{r^2}.$$
The first term can be rewritten as
$$F_Z=\frac{L^2}{m r^3}=m r \dot{\varphi}^2,$$
and this is indeed the well-known expression of the centrifugal force. That's why the first term in $V_{\text{eff}}$ is called the "potential of the centrifugal force".

The 2nd term is of course the gravitational attraction of the Sun. So from the point of view of the "corotating" observer in addition to the "true" force (gravitation) you have the centrifugal force as an "inertial force".

 

[phantomvommand]

It's not clear to me, what you mean by "centrifugal" energy. So maybe, I misunderstand you by just guessing what you mean.

The usual treatment of planetary orbits (aka "the Kepler problem") is to consider the Sun and a single planet, ignoring the interaction of these with any other objects around. This has the advantage to be not too bad an approximation (because the Sun has much larger mass than all other objects around her) and as well being treatable analytically.

To make the discussion even simpler we can also assume that the Sun's mass is so much larger than the considered planet's mass that we can as well make the Sun fixed at the origin of a Cartesian coordinate system. Then the planet moves simply according to
$$m \ddot{\vec{r}}=-\gamma m M \frac{r}{r^2}.$$
Now first we see that taking the cross product with $\vec{r}$ gives 0 on the right-hand side, which gives
$$m \vec{r} \times \ddot{\vec{r}}=\mathrm{d}_t (m \vec{r} \times \dot{\vec{r}})=0$$
or
$$\vec{L}=m \vec{r} \times \dot{\vec{r}}=\text{const}.$$
That's angular-momentum conservation. Now we make $\vec{L}$ point along the 3-axis. The motion is then in the (12)-plane, and we can introduce polar coordinates
$$\vec{r}=(r \cos \varphi,r \sin \varphi,0).$$
From this
$$\dot{\vec{r}}=\dot{r} (\cos \varphi,\sin \varphi_0)+r \dot{\varphi} (-\sin \varphi,\cos \varphi,0).$$
$$\vec{L}=m \vec{r} \dot{\vec{r}}=m r^2 \dot{\varphi} (0,0,1).$$
Further taking the scalar product of the equation of motion with $\dot{\vec{r}}$, we find the energy-conservation law
$$\frac{m}{2} \dot{\vec{x}}^2-\frac{\gamma m M}{r}=E=\text{const}.$$
Now plugging in our polar coordinates this gives
$$\frac{m}{2} (\dot{r}^2+r^2 \dot{\varphi}^2)-\frac{\gamma mM}{r}=E.$$
Now we use
$$\vec{L}^2=m^2 r^4 \dot{\varphi}^2,$$
to eliminate $\dot{\varphi}$ from the energy-conservation equation:
$$\frac{m}{2} \dot{r}^2 + \frac{L^2}{2 m r^2} - \frac{\gamma m M}{r}=E.$$
Now you can reinterpret this as the equation of motion of a mass point moving along the $r$ axis moving in an effective potential
$$V_{\text{eff}}=\frac{L^2}{2m r^2} - \frac{\gamma m M}{r}.$$
The $r$ coordinate can be taken as the distance of an observer sitting at the planet to the Sun, which is a rotating reference frame, and seen from this frame $L^2/(2m r^2)$ is the potential of the corresponding centrifugal force.

Taking the time-derivative of the energy-conserving equation indeed gives (after deviding by $\dot{r}$)
$$m \ddot{r}=\frac{L^2}{m r^3}-\frac{\gamma m M}{r^2}.$$
The first term can be rewritten as
$$F_Z=\frac{L^2}{m r^3}=m r \dot{\varphi}^2,$$
and this is indeed the well-known expression of the centrifugal force. That's why the first term in $V_{\text{eff}}$ is called the "potential of the centrifugal force".

The 2nd term is of course the gravitational attraction of the Sun. So from the point of view of the "corotating" observer in addition to the "true" force (gravitation) you have the centrifugal force as an "inertial force".

Yes, the centrifugal potential energy is exactly the one you have mentioned here. Why are we allowed to add up 2 different potential energies (GPE and CPE) to give the effective potential energy, when the 2 potential energies are based off different reference points? The GPE is with respect to a point at infinity, while the CPE is with respect to the mass about which the planet orbits.

 

[vanhees71]

What do you mean by "reference point"? The reference point here is the origin, i.e., the Sun, and that's just chosen, because of the rotational symmetry of the gravitational force/potential around the origin. There's not more to the effective potential than the derivation I've given above. It's just an effective description in terms of a one-dimensional one-particle motion, which is achieved by exploiting all the conservation laws (hidden behind this is to make use of the standard symmetries of Galileo-Newtonian spacetime).

Also the potential of a force is always only determined up to an additive constant, because the force is the gradient of the potential, i.e., this additive constants cancels anyway in the physically observable quantities (i.e., the trajectory of the planet around the Sun given initial conditions).

 

[phantomvommand]

What do you mean by "reference point"? The reference point here is the origin, i.e., the Sun, and that's just chosen, because of the rotational symmetry of the gravitational force/potential around the origin. There's not more to the effective potential than the derivation I've given above. It's just an effective description in terms of a one-dimensional one-particle motion, which is achieved by exploiting all the conservation laws (hidden behind this is to make use of the standard symmetries of Galileo-Newtonian spacetime).

Also the potential of a force is always only determined up to an additive constant, because the force is the gradient of the potential, i.e., this additive constants cancels anyway in the physically observable quantities (i.e., the trajectory of the planet around the Sun given initial conditions).

The reference point refers to the point of 0 Potential energy, since potential energy is all about relative positions. The GPE's reference point is at infinity, yet the centrifugal PE's reference point is the sun? How can you add them up to form the effective PE if they are based off different reference points?

And as you have written nicely, the CPE is really a form of KE, but it has somehow been re-interpreted as a form of PE, and I am still fine with this. What I find weird is when you add the CPE to the GPE, to form the effective potential, because this means you are adding 2 different potential energies with 2 different '0' points?

If you see the CPE as a KE, I guess the above problem doesn't exist. But when I look at it as a potential energy, I find the incosistency between the 0 PE points confusing.

 

[vanhees71]

If you mean this by "reference point", for the effective potential it's also $r \rightarrow \infty$, because obviously $V_{\text{eff}}(r) \rightarrow 0$ for $r \rightarrow \infty$.

As I said, there's no problem adding potentials. You can add arbitrary constants to any potential without changing the physics. The solution of the equations of motion (given the forces) are unique for given initial values $\vec{r}(0)=\vec{r}_0$ and $\dot{\vec{r}}(0)=\vec{v}_0$. If you change the "reference point" of the potential all that changes is the value of $E$, which however cancels in the solutions of the equations of motion when expressing everything in terms of the initial values.

 

[sophiecentaur]

How can you add them up to form the effective PE if they are based off different reference points?

I can understand why this bothers you but ask yourself why you can't use two different reference points for two different energies, if you choose to.
I ask myself, though, why not just consider all the Energies involved in the Orbital Energy in the conventional way? Isn't the Centrifugal Energy just a way of stating one Kinetic Energy part of the Orbital energy. It refers to a term in that equation that, in Polar Co ordinates, gives the normal and tangential kinetic velocities. The argument was presented that way because it suited the author but I think it's little more than just an alternative way of thinking that, in the author's view, would make for an interesting conversation. It certainly sparked off this thread.

I was thinking: If the rocket engine gives a burst and adds KE to the rocket, does it know that the nearby planet is is affecting the rocket? How does it know how its input energy has been shared out?

 

[sophiecentaur]

Exactly where is the Effective Potential relative to?

I just pictured, in my head, two different springs with different k constants, pulling a mass in different directions. The Potential energy of each spring is calculated in terms of its own anchor point as reference. The potential field can be plotted over a range of xy co ordinates and the path of the mass can be calculated using those potential values.

 

[jbriggs444]

There is no need to even have a "reference point" as such -- a point where the potential energy is taken to be zero. The potential energy can be strictly positive everywhere. Or strictly negative everywhere. The math does not care. Only differences are physically meaningful.