# Momentum, position vector dot (scalar) product action

## Main Question or Discussion Point

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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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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I wish you a nice day

m.

Dale
Mentor
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.

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 ?

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?

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lightarrow

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

Real lagrangian dimension is not Energy

vanhees71
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.