Frames of Reference: A Skateboarder's View - Comments

[kuruman]

kuruman submitted a new PF Insights post

Frames of Reference: A Skateboarder's View

Continue reading the Original PF Insights Post.

 

[Greg Bernhardt]

Looking forward to the followup!

 

[rcgldr]

Technically, the mass of Earth and slide should be included in conservation of mechanical energy in either frame (closed system free from external forces). From a frame where the Earth and box are not initially moving, v₀ = 0, then vf_earth will be slightly below 0 (slightly negative) and vf_box will be slightly below $\sqrt{2 \ g \ h}$. Since the Earth is much more massive than the box, the box gets most of the increase in kinetic energy.

From the skateboarders frame, v₀ will be slightly above (less magnitude) $-\sqrt{2 \ g \ h}$ and vf_earth will be slightly below (more magnitude) $-\sqrt{2 \ g \ h}$ . In this frame, the Earth gains kinetic energy, while the box loses kinetic energy.

The problem is similar to a box sliding down a frictionless wedge on a horizontal frictionless surface, except in this case the mass of the wedge (earth + slide) is huge.

 

[Laurie K]

Don't forget about the properties of a Tautochrone curve, which is based on a cycloid arc.

https://en.wikipedia.org/wiki/Tautochrone_curve#.22Virtual_gravity.22_solution

 

[ObjectivelyRational]

I kind of agree with rcgldr:

Kuruman states:
"It should be clear to the reader that in either reference frame momentum and energy conservation gives the exact solution while energy conservation only results in an approximation, albeit a good one."

What happens if you include the kinetic energy of the Earth in the energy conservation calculations?

 

[vanhees71]

As usual, the apparent paradox is much easier to understand when analyzed using the action principle. In the original frame we have
$$L=\frac{m}{2} \dot{\vec{x}}^2-mgy,$$
where $y$ points vertically up. Further we have the holonomous constraint
$$y=f(x)$$
and thus the action reads
$$L=\frac{m}{2} [1+f^{\prime 2}(x)] \dot{x}^2-mgf(x).$$
Now it's clear, why seen from the skater's reference frame the energy is not conserved. We have
$$x=x'+u t, \quad y=y'$$
and thus
$$L=\frac{m}{2} [1+f^{\prime 2}(x'+u t)](\dot{x}'+u)^2-mgf(x'+ut).$$
Since now the Lagrangian is explicitly time dependent, energy is no longer conserved from the skater's point of view.

 

[Greg Bernhardt]

Great Insight Kuruman!

 

[ObjectivelyRational]

Can you clarify what you mean by

energy is no longer conserved from the skater's point of view

This seems to imply actual conservation of energy is violated in a particular frame of reference... so I must be missing a subtlety of interpretation.

BTW is the energy of the Earth taken into account? There is energy transferred between the two bodies.

GENERALLY: Is there any reason we can assume that the Earth can be ignored when dealing with an interaction/force involving the Earth?

 

[ObjectivelyRational]

Kuruman:

I like the approach with momentum of both the Earth and the block, however, since the locus of interaction is also at the surface of the Earth (a point of contact via a shaped ramp) it seems that angular momentum comes into the picture.

Some of the momentum of the block would indeed be transferred as final linear momentum of the Earth, the Earth would also be imparted angular momentum, since the block end up traveling tangentially to the Earth's surface. Assuming the Earth is rigid, the ratio of the angular to linear momentum transferred involves the moment of inertia of the Earth which of course depends on the density distribution of the Earth.

I wonder what is the moment of inertia of the Earth and how much of a role does that play in the answer for v₀... it may be more complex than what is written in the section "a second opinion"

EDIT: The concept of "transferring" linear momentum into angular momentum is not correct. The idea that some of the energy is stored in both rotational energy of the Earth and linear kinetic energy of the Earth is likely valid. Sorry for the sloppiness here.

 

[rcgldr]

The reason that Earth's energy can be ignored from a Earth based frame, is that that v₀ = 0, and Δv is very small. The increase in angular energy of the Earth in this frame is

1/5 m_earth Δv_earth²

From a different frame, v₀ $\ne$ 0, and the increase of energy of the Earth is

1/5 m_earth ((v₀ + Δv_earth)² - v₀²) = 1/5 m_earth (2 v₀ Δv_earth + Δv_earth²)

For an "exact" solution, the Earth needs to be considered as part of a closed system that includes earth, slide, and box (the skateboarder is effectively an outside observer).

Using mass of Earth = 5.972e24 kg, radius of Earth = 6.371e6 m, mass of box 1kg:

change in velocity of box
$\Delta vb = \sqrt{\frac{2(2.3888e24)^2}{2.3888e24+2.3888e24^2}} \ \ \sqrt{g \ h}$
$\Delta vb = 0.99999999999999999999999979069 \ \sqrt{2 \ g \ h}$

change in velocity of earth
$\Delta ve = - \sqrt{\frac{2}{2.3888e24 + 2.3888e24^2}} \ \sqrt{g \ h}$
$\Delta ve = -0.00000000000000000000000041862 \ \sqrt{2 \ g \ h}$

For v₀ = 0, I get

Δenergyofearth =  0.00000000000000000000000041862 kg g h
Δenergyofbox   =  0.99999999999999999999999958138 kg g h
Δenergy        =  1.00000000000000000000000000000 kg g h

for v₀ = - 0.99999999999999999999999979069 sqrt(2 g h), I get

Δenergyofearth =  1.99999999999999999999999958138 kg g h
Δenergyofbox   = -0.99999999999999999999999958138 kg g h
Δenergy        =  1.00000000000000000000000000000 kg g h

 

[A.T.]

Technically, the mass of Earth and slide should be included in conservation of mechanical energy in either frame (closed system free from external forces).

The simple approach in the ramp's frame is not based on the conservation of total energy, just on the work theorem applied to the block: Since the ramp does no work on the block, the block's energy should be constant.

However, this assumes that the ramp's frame is inertial, which it actually isn't as the lack of momentum conservation shows. So even if the ramp does no work on the block in the ramp's frame, the inertial force present in that non-inertial frame does a tiny bit of work on the block.

 

[vanhees71]

Well, I think the "naive-mechanics explanation" has been very well explained in the article. For me it's very difficult to think in terms of forces, but the action-principle approach makes it immediately clear. It's due to the constaint of the block to the ramp which makes the energy non-conserved in the skater's rest frame, because in this frame the Lagrangian becomes explcitly time dependent. The momentum is neither conserved in the restframe of the ramp nor in the skater's restframe, because in neither frame the independent variable #x# is cyclic (see the posting above, where I corrected some typos). Of course nobody would solve this problem in the skater' frame, since it's immediately clear that in the rf. of the ramp the Lagrangian doesn't depend explicitly on time, and thus that the Hamiltonian in these cordinates is conserved.

 

[ObjectivelyRational]

Well, I think the "naive-mechanics explanation" has been very well explained in the article. For me it's very difficult to think in terms of forces, but the action-principle approach makes it immediately clear. It's due to the constaint of the block to the ramp which makes the energy non-conserved in the skater's rest frame, because in this frame the Lagrangian becomes explcitly time dependent. The momentum is neither conserved in the restframe of the ramp nor in the skater's restframe, because in neither frame the independent variable #x# is cyclic (see the posting above, where I corrected some typos). Of course nobody would solve this problem in the skater' frame, since it's immediately clear that in the rf. of the ramp the Lagrangian doesn't depend explicitly on time, and thus that the Hamiltonian in these cordinates is conserved.

What happens if you include the Earth in the Lagrangian? Would it matter that the calculation was made in the skater's rest frame?

 

[A.T.]

The momentum is neither conserved in the restframe of the ramp nor in the skater's restframe

The skater is an inertial observer. Why wouldn't momentum be conserved in his frame?

in the rf. of the ramp the Lagrangian doesn't depend explicitly on time, and thus that the Hamiltonian in these cordinates is conserved.

Even if the mass of ramp+Earth was comparable to the block's mass?

 

[vanhees71]

Why should momentum be conserved? It's not conserved in either frame, because there are forces acting on the block (gravitation from Earth and the constraining forces from the ramp. The latter forces become obviously time dependent in the skater's rest frame, and thus the forces are not conservative in this frame. Of course the skater's frame is an inertial frame, but that doesn't imply momentum or energy conservation if there are nonconservative forces acting on the block.

 

[A.T.]

Of course the skater's frame is an inertial frame, but that doesn't imply momentum or energy conservation if there are nonconservative forces acting on the block.

By "momentum conservation" I meant total momentum of the isolated system, not the momentum of the block only, which obviously not conserved.

 

[vanhees71]

Sure, if you have a closed system, the Lagrangian must be Galileo invariant, and all 10 conservation laws must hold. However, in this case it's a very complicated Lagrangian you can't write down to begin with. That's why we use the effective description with constraints (which are idealized too of course) to simplify the problem. I don't see, what's the problem to understand the calculation concerning the Lagrangian I gave above. I thought it might help to understand the problem, expressed in the article in terms of the Newtonian formulation (which I find much more complicated than the Lagrange formalism).

 

[rcgldr]

Sure, if you have a closed system, the Lagrangian must be Galileo invariant, and all 10 conservation laws must hold. However, in this case it's a very complicated ...

Post #10 includes the Newtonian math assuming a closed (isolated) system.

 

[ObjectivelyRational]

The conclusions I get from this discussion is that although particular reference frames may be used to simplify calculations,

1. there are no incorrect reference frames as such.

2. When adopting a nonstandard reference frame one must be careful not to ignore the reality of the whole physical system (e.g. the Earth) and one must be careful not to blindly apply the computational shortcuts and assumptions conveniently used in the calculations in the standard frame, which no longer apply when attempting to solve the problem in the nonstandard reference frame.

 

[tade]

kuruman submitted a new PF Insights post

Frames of Reference: A Skateboarder's View

Continue reading the Original PF Insights Post.

kuruman, do you skate? To complement theory with experiment haha

 

[kuruman]

kuruman, do you skate? To complement theory with experiment haha

No, but I admire those who do as long as they don't collide with me (it happened once).

 

[PeroK]

This is a good insight. An important point is that we tend to see the universal gravitational field as allowing us to apply the Galilean transformation. If the problem concerned a pendulum the answer would be simpler to see. A pendulum, for example, analysed in a frame where the fixed point is moving can't obey the conservation of mechanical energy.

We could also consider an observer moving vertically when an object is dropped. Again, it's clear that mechanical energy cannot be conserved in the moving frame. E.g. The observer could be moving down at the final speed of the object, hence the object ends up with zero KE in the moving frame. In this case, the observer could calculate the vertical displacement in his reference frame, not in the Earth's frame, and that should sort the energy equation.

We have a special case, therefore, that when the force is vertical and the moving reference frame is horizontal, then the vertical displacement in the two frames is the same. The mistake is to generalise this special case.

Another way to look at this problem is to say that first in the Earth frame the gravitational force is partially opposed by a normal force, so the nett (gravitational) force is tangential to the slope. We could imagine that this is the only force we have.

Then, we can resolve this nett tangential force into horizontal and vertical components. The vertical component is no problem for a reference frame moving horizontally, but the horizontal component is a problem if you try to model it as a distance-based potential.