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About F=ma and vectors

  1. Aug 11, 2011 #1
    Let F be a force vector and a an acceleration vector.
    A vector v(x,y,z) in 3 dimensional space xyz is the set of
    objects ((x1,y1,z1),(x2,y2,z2)) with conditions : x2-x1=x , y2-y1=y , z2-z1=z

    Writing vectorial F=ma , the equality means an equality between the sets.
    When a ball falls in a gravitional field it follows the vectorial law F=ma.

    So now i have an equality between 2 sets with infinite elements,
    but only one ball ? Did i miss something ?
  2. jcsd
  3. Aug 11, 2011 #2
    I tried duplicating my beach ball once by putting it in a gravitational field but it did not work. However F=MA means that the force acting on the ball is equal to the product of the mass and acceleration of the same ball. In free fall I think the proper view is that the force and acceleration is zero.

    That should answer the physics side of it. I am terrible at math though and it sounds like a math question.
  4. Aug 11, 2011 #3


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    Why do you have ((x1,y1,z1),(x2,y2,z2))? What sets are you talking about?

    What are these infinite elements?

    The force of gravity is a vector pointing to the center of the earth.
  5. Aug 11, 2011 #4
    Think of a vector as an arrow.
    A vector is the set of couples of points (A,B) so that t(A)=B with t a translation.
    (A being the beginpoint and B the endpoint)
    A vector (C,D) is the set of couples (A,B) equipollent with (C,D)
    (C being the beginpoint and D the endpoint)
    (A being the beginpoint and B the endpoint)
    As you can chose any beginpoint for A you have a set with infinite elements. ( In 3D space infinite^3 )
    Last edited: Aug 11, 2011
  6. Aug 11, 2011 #5

    D H

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    When you use nonsense definitions you may well get nonsense results.
  7. Aug 12, 2011 #6
    Dear DH,
    Please have a look at :
    Vectors are isomorph to Translation.
    A vector v is a set of tuples of points (A,B) where A and B follow the condition B-A=v.
    (This definition is independent of choise of a basis).
    When you choose a basis ex,ey,ez a vector v(Dx,Dy,Dz) is a set of tuples of points (A,B) where A has coordinates (x,y,z) and B has coordinates (x+Dx,y+Dy,z+Dz).
    I see no mathematical problem with this definition.
  8. Aug 12, 2011 #7
    Where is the question or query? All you've done there by setting those conditions is found the vector 1->2. You can find the distance easily from there, but...
  9. Aug 12, 2011 #8
    Maybe my question was not clear enough so i refrase.
    What is a vector,vector equation?
    If you don't like my definition of vector above. I will give you another one.
    "A vector is an objects which transforms under a coordinate transformation in exactly the same way as the position vector".
    Please also note a vector is not a point in space. (although you can assign a direction and magnetude to a point in space)
    And a vector is not a representative of a vector.
  10. Aug 13, 2011 #9
    This leads me to another question :
    What is the relationship between :
    1) The object : Vectors
    2)a) The group : Translation in time , translation in space, rotation in space
    2)b) Noether's theorem
  11. Aug 15, 2011 #10
    By lack of response on previous question additional thoughts,you have bounded vectors and you have unbounded/free vectors. You have position vectors. You have displacement vectors where displacement vector space can be different from position vector space (check general relativity)... Don't now about acceleration vector space.... You have vectors as a 1-dimensional array/matrix. Hmm, which vectors are the vectors in the physical vector equation F=ma :-)
  12. Aug 15, 2011 #11


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    Most of us would define "vector" as "a member of a vector space". Since you are talking about non-relativistic classical mechanics, we can take the vector space to be specifically the set [itex]\mathbb R^3[/itex] (i.e. the set of 3-tuples (x,y,z)), with addition and scalar multiplication defined in the usual way. The motion of a particle is described by a function [itex]x:\mathbb R\rightarrow\mathbb R^3[/itex]. I would define the force (in the single-particle theory) as a function [itex]\mathbb R^3\times\mathbb R^3\times\mathbb R\rightarrow\mathbb R[/itex] and write Newton's second law as [tex]mx''(t)=F(x(t),x'(t),t).[/tex] This can be viewed as a single equality between functions (mx'' is equal to the function that takes t to [itex]F(x(t),x'(t),t)[/itex]) or as infinitely many equalities between real numbers (for each t, mx''(t) is equal to F(x(t),x'(t),t)).

    The simplest way to represent space mathematically is to just use the vector space [itex]\mathbb R^3[/itex]. This choice ensures that points in space are vectors. You could argue that this choice is "ugly" because the point 0 has algebraic properties not shared by any of the other points. Things can be made "prettier" by using an affine space or a manifold instead of a vector space, but the price is pretty high. I mean, the mathematics gets much more difficult to learn, and things only get a little bit prettier.
    Last edited: Aug 15, 2011
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