Wikipedia Article re: Hamiltonians: Could it have been better stated?

  • Context: Graduate 
  • Thread starter Thread starter BadBrain
  • Start date Start date
  • Tags Tags
    article Wikipedia
Click For Summary

Discussion Overview

The discussion revolves around the interpretation of Hamiltonian mechanics as presented in a Wikipedia article, specifically focusing on the relationship between force, potential energy, and kinetic energy in a one-dimensional system. Participants explore the clarity and accuracy of the statements made in the article, debating the implications of the Hamilton equations and their physical interpretations.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant critiques the statement that "the force on the particle equals the rate at which it loses potential energy," suggesting it should instead emphasize the conversion of potential energy into kinetic energy, particularly in a vacuum.
  • Another participant points out that the full context of the statement includes the relationship between force and the gradient of potential energy, arguing that this key point is inadequately emphasized in the article.
  • There is a discussion about the definition of the gradient in vector calculus and its implications for understanding energy conversion, with one participant expressing confusion about the relationship between the gradient and energy conversion rates.
  • One participant highlights a potential misunderstanding regarding the units of measurement for force and power, indicating that they are incommensurate and thus complicating the discussion about energy conversion.
  • Another participant notes the profound implications of Hamilton's equations, suggesting they are foundational to the development of quantum mechanics.

Areas of Agreement / Disagreement

Participants express differing views on the clarity and accuracy of the Wikipedia article's statements regarding Hamiltonian mechanics. There is no consensus on whether the article effectively communicates the relationship between force, potential energy, and kinetic energy.

Contextual Notes

Participants acknowledge limitations in the Wikipedia article's wording, particularly regarding the use of the term "rate" and its potential for confusion. There are unresolved questions about the implications of the gradient in relation to energy conversion.

BadBrain
Messages
196
Reaction score
1
See:

http://en.wikipedia.org/wiki/Hamiltonian_mechanics

I refer specifically to this passage:

"Basic physical interpretation

The simplest interpretation of the Hamilton equations is as follows, applying them to a one-dimensional system consisting of one particle of mass m under time-independent boundary conditions: The Hamiltonian represents the energy of the system (provided that there are NO external forces, or additional energy added to the system), which is the sum of kinetic and potential energy, traditionally denoted T and V, respectively. Here q is the x coordinate and p is the momentum, mv. Then

H = T + V, T = p^2/2m, V = V(q) = V(x)

Note that T is a function of p alone, while V is a function of x (or q) alone.

Now the time-derivative of the momentum p equals the Newtonian force, and so here the first Hamilton equation means that the force on the particle equals the rate at which it loses potential energy with respect to changes in x, its location. (Force equals the negative gradient of potential energy.)

The time-derivative of q here means the velocity: the second Hamilton equation here means that the particle’s velocity equals the derivative of its kinetic energy with respect to its momentum. (Because the derivative with respect to p of p2/2m equals p/m = mv/m = v.)"

***

My problem here is the statement to the effect that "the force on the particle equals the rate at which it loses potential energy with respect to changes in x, its location." I feel that a more correct statement would be that the force on the particle equals the rate at which its potential energy is being converted into kinetic energy (assuming it's falling within a vacuum, eliminating the need to account for frictional atmospheric resistance and consequent draining of gravitational potential energy into heat-generastion through the process of overcoming aerodynamic drag). Furthermore, seeing as we're only speaking of mono-dimemsional motion here, why use the phrase "changes in x, its location", as opposed to the standard term "translational displacement"?

:confused::confused:
 
Physics news on Phys.org
BadBrain said:
My problem here is the statement to the effect that "the force on the particle equals the rate at which it loses potential energy with respect to changes in x, its location." I feel that a more correct statement would be that the force on the particle equals the rate at which its potential energy is being converted into kinetic energy (assuming it's falling within a vacuum, eliminating the need to account for frictional atmospheric resistance and consequent draining of gravitational potential energy into heat-generastion through the process of overcoming aerodynamic drag).
You are ignoring an important part of that statement in the article. The statement in full is "the first Hamilton equation means that the force on the particle equals the rate at which it loses potential energy with respect to changes in x, its location. (Force equals the negative gradient of potential energy.)"

This "first Hamilton equation" to which the article is referring is
[tex]\dot p = -\frac {\partial \mathcal H}{\partial q}[/tex]
When the Hamiltonian can be separated into a form H(p,q,t)=T(p,t)+U(q,t) the above reduces to
[tex]\dot p = -\frac {\partial U}{\partial q}[/tex]
In words, the time derivative of the generalized momentum is the negative gradient of the potential energy with respect to generalized position. Specializing to ordinary momentum and position leads to force being the negative gradient of potential energy.

Could the wikipedia article have been better written? Of course. But your improvement misses the key point that relates force with gradient of potential. The wikipedia article should have hammered this home rather than making it a short parenthetic remark.
 
D H said:
You are ignoring an important part of that statement in the article. The statement in full is "the first Hamilton equation means that the force on the particle equals the rate at which it loses potential energy with respect to changes in x, its location. (Force equals the negative gradient of potential energy.)"

This "first Hamilton equation" to which the article is referring is
[tex]\dot p = -\frac {\partial \mathcal H}{\partial q}[/tex]
When the Hamiltonian can be separated into a form H(p,q,t)=T(p,t)+U(q,t) the above reduces to
[tex]\dot p = -\frac {\partial U}{\partial q}[/tex]
In words, the time derivative of the generalized momentum is the negative gradient of the potential energy with respect to generalized position. Specializing to ordinary momentum and position leads to force being the negative gradient of potential energy.

Could the wikipedia article have been better written? Of course. But your improvement misses the key point that relates force with gradient of potential. The wikipedia article should have hammered this home rather than making it a short parenthetic remark.

OK. So, here, (http://en.wikipedia.org/wiki/Gradient) Wikipedia says:

"In vector calculus, the gradient of a scalar field is a vector field that points in the direction of the greatest rate of increase of the scalar field, and whose magnitude is the greatest rate of change."

***

I guess what I'm not understanding here is: why is the vector field which points in the direction of the greatest rate of increase of the scalar field, and whose magnitude is the greatest rate of change in the same, not proportional to the rate of conversion of gravitational potential energy into gravitational kinetic energy, assuming the particle is moving within a vaccuum? (I here recognize my earlier confusion of the proportionality of the rate of energy conversion along this vector with identity of such rate of conversion.)

EDIT:

I guess what I'm really saying is, is there redistribution (obviously not loss, but redistribution) of gravitational potential energy of the particle along vectors other than the negative gradient sufficient to prevent an increase of mass in the particle as it moves directly into the negative gradient?
 
Last edited:
BadBrain said:
OK. So, here, (http://en.wikipedia.org/wiki/Gradient) Wikipedia says:

"In vector calculus, the gradient of a scalar field is a vector field that points in the direction of the greatest rate of increase of the scalar field, and whose magnitude is the greatest rate of change."
That's wikipedia for ya. A typical problem with wiki articles is not that they are wrong; they are just poorly written. That is an abuse of the term "rate" and it is going to lead to confusion. "Rate" typically means with respect to time.

I guess what I'm not understanding here is: why is the vector field which points in the direction of the greatest rate of increase of the scalar field, and whose magnitude is the greatest rate of change in the same, not proportional to the rate of conversion of gravitational potential energy into gravitational kinetic energy, assuming the particle is moving within a vaccuum? (I here recognize my earlier confusion of the proportionality of the rate of energy conversion along this vector with identity of such rate of conversion.)
And right here is where that confusion bites you. That vector field you speak of has units of energy over length, or force. On the other hand, your "rate of conversion of gravitational potential energy into gravitational kinetic energy" has units of energy over time, or power. You have incommensurate units!

You are missing something very profound in those seemingly simple Hamilton's equations.
[tex]\begin{aligned}<br /> \phantom{-}\frac{dq}{dt} = \frac{\partial \mathcal H}{\partial p} \\ \\<br /> -\frac{dp}{dt} = \frac{\partial \mathcal H}{\partial q}<br /> \end{aligned}[/tex]
These equations relate the total derivatives with respect to time of the generalized position / generalized momentum with the partial derivatives of energy with respect to the conjugate variable. This is incredibly profound. It is the basis of quantum mechanics; it would have been very hard to develop quantum mechanics without Hamilton's reformulation of Newtonian mechanics.
I guess what I'm really saying is, is there redistribution (obviously not loss, but redistribution) of gravitational potential energy of the particle along vectors other than the negative gradient sufficient to prevent an increase of mass in the particle as it moves directly into the negative gradient?
Increase of mass?? :confused:
 
D H said:
That's wikipedia for ya. A typical problem with wiki articles is not that they are wrong; they are just poorly written. That is an abuse of the term "rate" and it is going to lead to confusion. "Rate" typically means with respect to time.


And right here is where that confusion bites you. That vector field you speak of has units of energy over length, or force. On the other hand, your "rate of conversion of gravitational potential energy into gravitational kinetic energy" has units of energy over time, or power. You have incommensurate units!

You are missing something very profound in those seemingly simple Hamilton's equations.
[tex]\begin{aligned}<br /> \phantom{-}\frac{dq}{dt} = \frac{\partial \mathcal H}{\partial p} \\ \\<br /> -\frac{dp}{dt} = \frac{\partial \mathcal H}{\partial q}<br /> \end{aligned}[/tex]
These equations relate the total derivatives with respect to time of the generalized position / generalized momentum with the partial derivatives of energy with respect to the conjugate variable. This is incredibly profound. It is the basis of quantum mechanics; it would have been very hard to develop quantum mechanics without Hamilton's reformulation of Newtonian mechanics.

I see it now. Thanks!

I've never seen the equations in quite those forms before.



D H said:
Increase of mass?? :confused:

I was looking for lost energy, and, so far as I could see, lost energy meant increased mass. I wasn't expecting to actually find an increase in mass, that was just my form of asking where did the lost energy go?

I now see that no energy is lost.

Thanks!
 
BadBrain said:
I see it now. Thanks!

I've never seen the equations in quite those forms before.
That's the same form as the wiki article, and many texts. I just changed the Newtonian fluxion into Liebniz d/dt form. As a recap, the wiki article says "The Hamilton equations are generally written as follows:"

[tex]\begin{aligned}<br /> \dot p = -\frac{\partial \mathcal H}{\partial q} \\ \\<br /> \dot q = \phantom{-}\frac{\partial \mathcal H}{\partial p}<br /> \end{aligned}[/tex]

And here's how I wrote Hamilton's equations:

[tex]\begin{aligned}<br /> \phantom{-}\frac{dq}{dt} = \frac{\partial \mathcal H}{\partial p} \\ \\<br /> -\frac{dp}{dt} = \frac{\partial \mathcal H}{\partial q}<br /> \end{aligned}[/tex]

Obviously the same two equations.

I like the Leibniz d/dt form as opposed to the Newtonian dot form because expressing it the way I did makes it painstakingly clear that you have a time derivative on one side and a gradient on the other.


Thanks!
Glad I could help.
 
D H said:
That's the same form as the wiki article, and many texts. I just changed the Newtonian fluxion into Liebniz d/dt form. As a recap, the wiki article says "The Hamilton equations are generally written as follows:"

[tex]\begin{aligned}<br /> \dot p = -\frac{\partial \mathcal H}{\partial q} \\ \\<br /> \dot q = \phantom{-}\frac{\partial \mathcal H}{\partial p}<br /> \end{aligned}[/tex]

And here's how I wrote Hamilton's equations:

[tex]\begin{aligned}<br /> \phantom{-}\frac{dq}{dt} = \frac{\partial \mathcal H}{\partial p} \\ \\<br /> -\frac{dp}{dt} = \frac{\partial \mathcal H}{\partial q}<br /> \end{aligned}[/tex]

Obviously the same two equations.

I like the Leibniz d/dt form as opposed to the Newtonian dot form because expressing it the way I did makes it painstakingly clear that you have a time derivative on one side and a gradient on the other.



Glad I could help.


The Leibnizian form is how I was taught calculus: to be honest with you, I've never heard of "Newtonian dot form" in my life. Honestly never knew it existed. Obviously the source of my confusion.

Thanks again!
 
Anyways, I posted this little piece on another website about how well this discussion has gone for me:

"For whatever it's worth (probably very little), I got into an argument on the Physics Forum tonight, and here's about how it went for me:




:blushing:
 
Last edited by a moderator:

Similar threads

  • · Replies 31 ·
2
Replies
31
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 43 ·
2
Replies
43
Views
5K
Replies
5
Views
1K
  • · Replies 32 ·
2
Replies
32
Views
8K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K