Momentum, position vector dot (scalar) product action

In summary, the conversation discusses the dot (scalar) product of the position vector and momentum vector in classical mechanics, which has the dimension of action. The time derivative of this product also has the dimension of a Lagrangian. However, this "action" or "lagrangian" lacks translational and Galilean (or Lorentz) invariance, causing issues in physical laws of nature. The conversation also touches on the concept of affine space and the proper dimension for a Lagrangian.
  • #1
marfi11
3
0
momentum, position vector dot (scalar) product "action"

Hello,

I was playing with single mass point classical mechanics, when I realized that the dot product of the position vector and momentum vector, p.r , has action dimension. Furthermore, its time derivative, d/dt(p.r) = F.r + p.v, has dimension (and looks like) some kind of "Lagrangian" (p.v suggests double of cinetic energy). I tried to look around the web if I could find some info about this (in classical mechanics), but without success.

What could be the trouble with this "action" / "lagrangian"?
Why nobody mentions this "action" or the quantity p.r, even if only to discard it?

(I also find the quantity F.r interessting, but can't attribute it a general meaning, it suggests to be somekind of double "minus potential energy" ("-2U"), but for potential forces it generally "deforms" the intial potential U by the operator -x∂U/∂x-y∂U/∂y = F.r = "-2Udeform", which keeps invariant the harmonic oscillator potential (in 2-D: U(x,y)= 1/2(x^2+y^2)), but that is a special case, generally we'll get a different "deformed" potential "Udeform" from starting potential U (e.g. central gravitational potential in 2-D Ug(x,y) = -1/(x^2+y^2)^1/2 leads to "Udeform"= -1/2Ug.)

Thank you
I wish a pleasant day
 
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  • #2
The problem with that action is that it lacks translational and Galilean (or Lorentz) invariance. Physical laws of nature (as opposed to particular problems) should not single out a single "point" in space or single out a particular "correct velocity" at which to view things.

p.r would arise in certain problems without these symmetries--and since it has rotational symmetry it would be a problem where we have some isotropy. Perhaps you could find a p.r term in the lagrangian for an orbital mechanics problem. For some reason I think there may be a connection with the virial theorem, with deals with time averages in orbital mechanics: http://en.wikipedia.org/wiki/Virial_theorem . I'm not coming up with a good example of p.r in a Lagrangian at this moment and I have to run, but hopefully that gives you a good starting place.
 
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  • #3
Thank you for your answer Jolb, I'll think about that.

I wish you a nice day

m.
 
  • #4
Also, you should remember that position is not actually a vector. (What is the vector sum of the positions of Paris and New York? Or what is the product of 6 times the position of Seattle?). Position is an affine space, which is sometimes described as a vector space which has lost it's origin.

This is related to Jolbs comment above.
 
  • #5
hello dear DaleSpam
i saw your beautiful answer to the question about r.p , i have search about a person who know about affin space and topology like you. could you guide me in this field to know them carefully ?
 
  • #6
DaleSpam said:
Also, you should remember that position is not actually a vector. (What is the vector sum of the positions of Paris and New York? Or what is the product of 6 times the position of Seattle?). Position is an affine space, which is sometimes described as a vector space which has lost it's origin.
And how do you call the "position vector" P-O?

--
lightarrow
 
  • #7
consider d/dt(mr.r)=(p.r)+(p.r) but (mr.r)=mr2=I the moment of inertia. apparently, (p.r)=(1/2)d/dt I. and as far as I know, the rate of change of the moment of inertia has no significant meaning
 
  • #8
Real lagrangian dimension is not Energy
 
  • #9
The proper dimension for any Lagrangian is that of an energy. After all the action ##\int \mathrm{d} t L## must have the dimension of, well, an action, i.e., energy times time.
 

1. What is momentum?

Momentum is a measure of an object's motion and is defined as the product of an object's mass and velocity. It is a vector quantity, meaning it has both magnitude and direction.

2. How is momentum calculated?

Momentum is calculated by multiplying an object's mass (m) by its velocity (v). This can be expressed as p = m * v. The SI unit for momentum is kilogram-meter per second (kg * m/s).

3. What is the position vector dot product?

The position vector dot product is a mathematical operation that calculates the scalar value of the projection of one vector onto another. In the context of momentum, it is used to determine the amount of force acting on an object in a given direction.

4. How is the position vector dot product used in physics?

In physics, the position vector dot product is used to calculate the work done on an object by a force in a specific direction. It is also used to determine the torque (rotational force) acting on an object.

5. What is the relationship between momentum and the position vector dot product?

There is a direct relationship between momentum and the position vector dot product. The position vector dot product is used to determine the amount of force acting on an object in a given direction, which in turn affects the object's momentum. The larger the force acting on an object, the greater its change in momentum will be.

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