Momentum, position vector dot (scalar) product action

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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
 

Answers and Replies

  • #2
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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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Thank you for your answer Jolb, I'll think about that.

I wish you a nice day

m.
 
  • #4
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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
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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
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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
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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
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Real lagrangian dimension is not Energy
 
  • #9
vanhees71
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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.
 

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