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Why do some people write the gravitational force as proportional to $mathbfr/|mathbfr|^3$?


Why no basis vector in Newtonian gravitational vector field?Why is Gravitational force proportional to the masses?.Force inversely proportional to the squared distanceHorizontal Gravitational ForceIs gravitational potential energy proportional or inversely proportional to distance?Clarifying some notation, the square of a vector derivativeWhy do objects with mass have gravitational force that is proportional to their mass?Finding Gravitational Force b/w 2 PeopleWhy do people not have gravitational attraction?What is $mathbfl$ (boldfaced ell) in Lorentz force law $mathbf F= Iint (dmathbf l times mathbf B)$?






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








8












$begingroup$


I'm reading Mathematical Aspects of Classical and Celestial Mechanics, Second Edition by Arnold, Kozlov, and Neishtadt. It occurred to me that many people like to use third power when mention the law of universal gravitation.



For example,




$$F_ij=fracgamma m_i m_jr_ij$$ where $r_ij=r_i-r_j, gamma =textconst>0$




Is this just some conversion/tradition/habit, or does it carry any significance? i.e. the absolute value changed sign and derivative?










share|cite|improve this question











$endgroup$







  • 9




    $begingroup$
    One advantage is that you don't need to introduce a unit vector for $vecr_ij$. This is the same as writing $fracgamma m_i m_j fracvecr_ij = fracgamma m_i m_j hatr_ij$.
    $endgroup$
    – Tob Ernack
    Jun 19 at 3:53







  • 1




    $begingroup$
    I think this is a fairly common convention. I've seen it on Biot-Savart as well.
    $endgroup$
    – mathysics
    Jun 19 at 5:01






  • 2




    $begingroup$
    Notice that it is not an absolute value but vector norm.
    $endgroup$
    – Matt
    Jun 19 at 12:02











  • $begingroup$
    It's a convenient form of typing it to avoid the use of unit vectors. Imagine typing an entire book using $hathats$.
    $endgroup$
    – J. Manuel
    Jun 19 at 14:41










  • $begingroup$
    Because all the cool guys are doing it.
    $endgroup$
    – Paracosmiste
    Jun 19 at 14:56

















8












$begingroup$


I'm reading Mathematical Aspects of Classical and Celestial Mechanics, Second Edition by Arnold, Kozlov, and Neishtadt. It occurred to me that many people like to use third power when mention the law of universal gravitation.



For example,




$$F_ij=fracgamma m_i m_jr_ij$$ where $r_ij=r_i-r_j, gamma =textconst>0$




Is this just some conversion/tradition/habit, or does it carry any significance? i.e. the absolute value changed sign and derivative?










share|cite|improve this question











$endgroup$







  • 9




    $begingroup$
    One advantage is that you don't need to introduce a unit vector for $vecr_ij$. This is the same as writing $fracgamma m_i m_j fracvecr_ij = fracgamma m_i m_j hatr_ij$.
    $endgroup$
    – Tob Ernack
    Jun 19 at 3:53







  • 1




    $begingroup$
    I think this is a fairly common convention. I've seen it on Biot-Savart as well.
    $endgroup$
    – mathysics
    Jun 19 at 5:01






  • 2




    $begingroup$
    Notice that it is not an absolute value but vector norm.
    $endgroup$
    – Matt
    Jun 19 at 12:02











  • $begingroup$
    It's a convenient form of typing it to avoid the use of unit vectors. Imagine typing an entire book using $hathats$.
    $endgroup$
    – J. Manuel
    Jun 19 at 14:41










  • $begingroup$
    Because all the cool guys are doing it.
    $endgroup$
    – Paracosmiste
    Jun 19 at 14:56













8












8








8


1



$begingroup$


I'm reading Mathematical Aspects of Classical and Celestial Mechanics, Second Edition by Arnold, Kozlov, and Neishtadt. It occurred to me that many people like to use third power when mention the law of universal gravitation.



For example,




$$F_ij=fracgamma m_i m_jr_ij$$ where $r_ij=r_i-r_j, gamma =textconst>0$




Is this just some conversion/tradition/habit, or does it carry any significance? i.e. the absolute value changed sign and derivative?










share|cite|improve this question











$endgroup$




I'm reading Mathematical Aspects of Classical and Celestial Mechanics, Second Edition by Arnold, Kozlov, and Neishtadt. It occurred to me that many people like to use third power when mention the law of universal gravitation.



For example,




$$F_ij=fracgamma m_i m_jr_ij$$ where $r_ij=r_i-r_j, gamma =textconst>0$




Is this just some conversion/tradition/habit, or does it carry any significance? i.e. the absolute value changed sign and derivative?







newtonian-gravity vectors notation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jun 19 at 15:27









knzhou

51.5k13 gold badges144 silver badges250 bronze badges




51.5k13 gold badges144 silver badges250 bronze badges










asked Jun 19 at 3:49









user9976437user9976437

44713 bronze badges




44713 bronze badges







  • 9




    $begingroup$
    One advantage is that you don't need to introduce a unit vector for $vecr_ij$. This is the same as writing $fracgamma m_i m_j fracvecr_ij = fracgamma m_i m_j hatr_ij$.
    $endgroup$
    – Tob Ernack
    Jun 19 at 3:53







  • 1




    $begingroup$
    I think this is a fairly common convention. I've seen it on Biot-Savart as well.
    $endgroup$
    – mathysics
    Jun 19 at 5:01






  • 2




    $begingroup$
    Notice that it is not an absolute value but vector norm.
    $endgroup$
    – Matt
    Jun 19 at 12:02











  • $begingroup$
    It's a convenient form of typing it to avoid the use of unit vectors. Imagine typing an entire book using $hathats$.
    $endgroup$
    – J. Manuel
    Jun 19 at 14:41










  • $begingroup$
    Because all the cool guys are doing it.
    $endgroup$
    – Paracosmiste
    Jun 19 at 14:56












  • 9




    $begingroup$
    One advantage is that you don't need to introduce a unit vector for $vecr_ij$. This is the same as writing $fracgamma m_i m_j fracvecr_ij = fracgamma m_i m_j hatr_ij$.
    $endgroup$
    – Tob Ernack
    Jun 19 at 3:53







  • 1




    $begingroup$
    I think this is a fairly common convention. I've seen it on Biot-Savart as well.
    $endgroup$
    – mathysics
    Jun 19 at 5:01






  • 2




    $begingroup$
    Notice that it is not an absolute value but vector norm.
    $endgroup$
    – Matt
    Jun 19 at 12:02











  • $begingroup$
    It's a convenient form of typing it to avoid the use of unit vectors. Imagine typing an entire book using $hathats$.
    $endgroup$
    – J. Manuel
    Jun 19 at 14:41










  • $begingroup$
    Because all the cool guys are doing it.
    $endgroup$
    – Paracosmiste
    Jun 19 at 14:56







9




9




$begingroup$
One advantage is that you don't need to introduce a unit vector for $vecr_ij$. This is the same as writing $fracgamma m_i m_j fracvecr_ij = fracgamma m_i m_j hatr_ij$.
$endgroup$
– Tob Ernack
Jun 19 at 3:53





$begingroup$
One advantage is that you don't need to introduce a unit vector for $vecr_ij$. This is the same as writing $fracgamma m_i m_j fracvecr_ij = fracgamma m_i m_j hatr_ij$.
$endgroup$
– Tob Ernack
Jun 19 at 3:53





1




1




$begingroup$
I think this is a fairly common convention. I've seen it on Biot-Savart as well.
$endgroup$
– mathysics
Jun 19 at 5:01




$begingroup$
I think this is a fairly common convention. I've seen it on Biot-Savart as well.
$endgroup$
– mathysics
Jun 19 at 5:01




2




2




$begingroup$
Notice that it is not an absolute value but vector norm.
$endgroup$
– Matt
Jun 19 at 12:02





$begingroup$
Notice that it is not an absolute value but vector norm.
$endgroup$
– Matt
Jun 19 at 12:02













$begingroup$
It's a convenient form of typing it to avoid the use of unit vectors. Imagine typing an entire book using $hathats$.
$endgroup$
– J. Manuel
Jun 19 at 14:41




$begingroup$
It's a convenient form of typing it to avoid the use of unit vectors. Imagine typing an entire book using $hathats$.
$endgroup$
– J. Manuel
Jun 19 at 14:41












$begingroup$
Because all the cool guys are doing it.
$endgroup$
– Paracosmiste
Jun 19 at 14:56




$begingroup$
Because all the cool guys are doing it.
$endgroup$
– Paracosmiste
Jun 19 at 14:56










3 Answers
3






active

oldest

votes


















15












$begingroup$

This is not about gravity but about making maths easier; the same idea comes up whenever you have a force in the radial direction. The reason that $bf r/r^3$ is a little easier to work with than $hatbf r/r^2$ is the following.



First, one can write it out in rectangular coordinates easily:
$$
fracbf rr^3 = frac1r^3 left( beginarrayc x\y\z endarray right)
$$



Secondly, because $hatbf r$ has a square root hidden in it which makes some manipulations trickier:
$$
hatbf r = (x^2 + y^2 + z^2)^-1/2(x bf i + y bf j + z bf k)
$$

Indeed, when doing things like differentiation, the first step is often to write $hatbf r$ as $bf r/r$.



Overall, the move to replace $hatbf r$ by $bf r/r$ reduces the number of different symbols in the mathematical expression and that is the main reason why it is regularly done. Having completed some algebra, one may then choose to present the result in terms of $hatbf r$ so as to draw attention to the overall scaling of the result, for example in order to make it clear that the gravitational law is an inverse square not an inverse cube law.






share|cite|improve this answer











$endgroup$




















    13












    $begingroup$

    enter image description here



    beginequation
    mathbfF_ij boldsymbol=mathrm kdfracm_i m_jVert mathbfr_iboldsymbol- mathbfr_jVert^2 mathbfn_ij
    tag1label1
    endequation

    where $:mathbfn_ij:$ the unit vector along the vector $:left(mathbfr_iboldsymbol- mathbfr_jright)$. But
    beginequation
    mathbfn_ij boldsymbol=dfracleft(mathbfr_iboldsymbol- mathbfr_jright)Vert mathbfr_iboldsymbol- mathbfr_jVert
    tag2label2
    endequation

    so the third power
    beginequation
    mathbfF_ij boldsymbol=mathrm kdfracm_i m_jVert mathbfr_iboldsymbol- mathbfr_jVert^3 left(mathbfr_iboldsymbol- mathbfr_jright)
    tag3label3
    endequation






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      So the authors presumably stated early on that, in their text, bold face quantities are all vectors. The only scalar in the OP's equation are the masses?
      $endgroup$
      – DJohnM
      Jun 19 at 6:31










    • $begingroup$
      @DJohnM ...the masses $:m_i,m_j:$ and the constant $:mathrm k$...
      $endgroup$
      – Frobenius
      Jun 19 at 6:48






    • 1




      $begingroup$
      @DJohnM I don't think I've seen boldface used for anything other than vectors.
      $endgroup$
      – David Richerby
      Jun 19 at 13:19










    • $begingroup$
      But why (3) and not (1). Is it a traditionhabit?
      $endgroup$
      – J. Manuel
      Jun 19 at 15:59










    • $begingroup$
      @J.Manuel You see both in the literature, and I would wager (1) is actually more common as it makes manifest the inverse square relationship. (3) can often be easier to work with however.
      $endgroup$
      – gabe
      Jun 19 at 17:58


















    3












    $begingroup$

    It is a matter of convenience:



    Let $vecr_ij$ be a distance vector with magnitude $r_ij$ along the line connecting the masses $m_i$ and $m_j$. Then:



    (I): $vecr_ij$ squared is a scalar whose value equals its magnitude squared. Proof: $vecr_ij^2=vecr_ij cdot vecr_ij=r_ij r_ij cos0=r_ij^2$.



    (II): $vecr_ij$ can be written as $vecr_ij=r_ij hatr_ij$, with $hatr_ij$ being a unit vector having the same direction as $vecr_ij$. Proof: This goes as an axiom ;-)



    If ones writes Newton’s law of gravity as
    $$F_ij=fracγm_im_jvecr_ij^2=fracγm_im_jr_ij^2 tag1$$



    Then (1) is an incomplete description of the gravitational force. Equation (1) only represents the magnitude of the gravitational force, as can be noticed by the fact that the right side of it is just a scalar.



    It is a well-established observation that gravity is a force whose direction is along the line connecting the masses $m_i$ and $m_j$. As a force, gravity must be written as a vector, and therefore the right form would be



    $$vecF_ij=fracγm_im_jr_ij^2 hatr_ij tag2$$



    From (II), equation (2) can be written as



    $$vecF_ij=fracγm_im_jr_ij^2 fracvecr_ijr_ij=fracγm_im_jr_ij^3 vecr_ij tag3$$



    Typing vectors and unit vectors with arrows and “hats” atop (as in current case) is a little cumbersome. To solve for typing vectors with arrows, one can select other options as boldfacing them for example. To avoid unit vectors one uses the convention $fracvecrr$, i.e., the choice of (3) is a convenient form of writing these sort of equations, especially for works containing a huge amount of text.



    It is not only easier (if one uses a WYSIWYG text editor of typing machine) to type equation (3) using these conventions, but visually cleaner as well. Just check:



    beginequation
    mathbfF_ij boldsymbol=mathrm γdfracm_i m_jmathbfr_ij^3 mathbfr_ij
    tag4
    endequation



    As was pointed out by @AndrewSteane, there are even mathematical advantages by using this notation, hence, it is just more convenient to use the $fracbf rbf r^3$ notation.



    The problem with this, however, is that people (almost) always is introduced to Newton’s law of gravity in its non-vector form (equation 1) and is expecting to find the inverse square law in a first glance, therefore, it may look odd for beginners until one gets used to it some point in future ;-)






    share|cite|improve this answer











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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      15












      $begingroup$

      This is not about gravity but about making maths easier; the same idea comes up whenever you have a force in the radial direction. The reason that $bf r/r^3$ is a little easier to work with than $hatbf r/r^2$ is the following.



      First, one can write it out in rectangular coordinates easily:
      $$
      fracbf rr^3 = frac1r^3 left( beginarrayc x\y\z endarray right)
      $$



      Secondly, because $hatbf r$ has a square root hidden in it which makes some manipulations trickier:
      $$
      hatbf r = (x^2 + y^2 + z^2)^-1/2(x bf i + y bf j + z bf k)
      $$

      Indeed, when doing things like differentiation, the first step is often to write $hatbf r$ as $bf r/r$.



      Overall, the move to replace $hatbf r$ by $bf r/r$ reduces the number of different symbols in the mathematical expression and that is the main reason why it is regularly done. Having completed some algebra, one may then choose to present the result in terms of $hatbf r$ so as to draw attention to the overall scaling of the result, for example in order to make it clear that the gravitational law is an inverse square not an inverse cube law.






      share|cite|improve this answer











      $endgroup$

















        15












        $begingroup$

        This is not about gravity but about making maths easier; the same idea comes up whenever you have a force in the radial direction. The reason that $bf r/r^3$ is a little easier to work with than $hatbf r/r^2$ is the following.



        First, one can write it out in rectangular coordinates easily:
        $$
        fracbf rr^3 = frac1r^3 left( beginarrayc x\y\z endarray right)
        $$



        Secondly, because $hatbf r$ has a square root hidden in it which makes some manipulations trickier:
        $$
        hatbf r = (x^2 + y^2 + z^2)^-1/2(x bf i + y bf j + z bf k)
        $$

        Indeed, when doing things like differentiation, the first step is often to write $hatbf r$ as $bf r/r$.



        Overall, the move to replace $hatbf r$ by $bf r/r$ reduces the number of different symbols in the mathematical expression and that is the main reason why it is regularly done. Having completed some algebra, one may then choose to present the result in terms of $hatbf r$ so as to draw attention to the overall scaling of the result, for example in order to make it clear that the gravitational law is an inverse square not an inverse cube law.






        share|cite|improve this answer











        $endgroup$















          15












          15








          15





          $begingroup$

          This is not about gravity but about making maths easier; the same idea comes up whenever you have a force in the radial direction. The reason that $bf r/r^3$ is a little easier to work with than $hatbf r/r^2$ is the following.



          First, one can write it out in rectangular coordinates easily:
          $$
          fracbf rr^3 = frac1r^3 left( beginarrayc x\y\z endarray right)
          $$



          Secondly, because $hatbf r$ has a square root hidden in it which makes some manipulations trickier:
          $$
          hatbf r = (x^2 + y^2 + z^2)^-1/2(x bf i + y bf j + z bf k)
          $$

          Indeed, when doing things like differentiation, the first step is often to write $hatbf r$ as $bf r/r$.



          Overall, the move to replace $hatbf r$ by $bf r/r$ reduces the number of different symbols in the mathematical expression and that is the main reason why it is regularly done. Having completed some algebra, one may then choose to present the result in terms of $hatbf r$ so as to draw attention to the overall scaling of the result, for example in order to make it clear that the gravitational law is an inverse square not an inverse cube law.






          share|cite|improve this answer











          $endgroup$



          This is not about gravity but about making maths easier; the same idea comes up whenever you have a force in the radial direction. The reason that $bf r/r^3$ is a little easier to work with than $hatbf r/r^2$ is the following.



          First, one can write it out in rectangular coordinates easily:
          $$
          fracbf rr^3 = frac1r^3 left( beginarrayc x\y\z endarray right)
          $$



          Secondly, because $hatbf r$ has a square root hidden in it which makes some manipulations trickier:
          $$
          hatbf r = (x^2 + y^2 + z^2)^-1/2(x bf i + y bf j + z bf k)
          $$

          Indeed, when doing things like differentiation, the first step is often to write $hatbf r$ as $bf r/r$.



          Overall, the move to replace $hatbf r$ by $bf r/r$ reduces the number of different symbols in the mathematical expression and that is the main reason why it is regularly done. Having completed some algebra, one may then choose to present the result in terms of $hatbf r$ so as to draw attention to the overall scaling of the result, for example in order to make it clear that the gravitational law is an inverse square not an inverse cube law.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jun 19 at 9:13

























          answered Jun 19 at 7:25









          Andrew SteaneAndrew Steane

          7,4801 gold badge8 silver badges42 bronze badges




          7,4801 gold badge8 silver badges42 bronze badges























              13












              $begingroup$

              enter image description here



              beginequation
              mathbfF_ij boldsymbol=mathrm kdfracm_i m_jVert mathbfr_iboldsymbol- mathbfr_jVert^2 mathbfn_ij
              tag1label1
              endequation

              where $:mathbfn_ij:$ the unit vector along the vector $:left(mathbfr_iboldsymbol- mathbfr_jright)$. But
              beginequation
              mathbfn_ij boldsymbol=dfracleft(mathbfr_iboldsymbol- mathbfr_jright)Vert mathbfr_iboldsymbol- mathbfr_jVert
              tag2label2
              endequation

              so the third power
              beginequation
              mathbfF_ij boldsymbol=mathrm kdfracm_i m_jVert mathbfr_iboldsymbol- mathbfr_jVert^3 left(mathbfr_iboldsymbol- mathbfr_jright)
              tag3label3
              endequation






              share|cite|improve this answer











              $endgroup$












              • $begingroup$
                So the authors presumably stated early on that, in their text, bold face quantities are all vectors. The only scalar in the OP's equation are the masses?
                $endgroup$
                – DJohnM
                Jun 19 at 6:31










              • $begingroup$
                @DJohnM ...the masses $:m_i,m_j:$ and the constant $:mathrm k$...
                $endgroup$
                – Frobenius
                Jun 19 at 6:48






              • 1




                $begingroup$
                @DJohnM I don't think I've seen boldface used for anything other than vectors.
                $endgroup$
                – David Richerby
                Jun 19 at 13:19










              • $begingroup$
                But why (3) and not (1). Is it a traditionhabit?
                $endgroup$
                – J. Manuel
                Jun 19 at 15:59










              • $begingroup$
                @J.Manuel You see both in the literature, and I would wager (1) is actually more common as it makes manifest the inverse square relationship. (3) can often be easier to work with however.
                $endgroup$
                – gabe
                Jun 19 at 17:58















              13












              $begingroup$

              enter image description here



              beginequation
              mathbfF_ij boldsymbol=mathrm kdfracm_i m_jVert mathbfr_iboldsymbol- mathbfr_jVert^2 mathbfn_ij
              tag1label1
              endequation

              where $:mathbfn_ij:$ the unit vector along the vector $:left(mathbfr_iboldsymbol- mathbfr_jright)$. But
              beginequation
              mathbfn_ij boldsymbol=dfracleft(mathbfr_iboldsymbol- mathbfr_jright)Vert mathbfr_iboldsymbol- mathbfr_jVert
              tag2label2
              endequation

              so the third power
              beginequation
              mathbfF_ij boldsymbol=mathrm kdfracm_i m_jVert mathbfr_iboldsymbol- mathbfr_jVert^3 left(mathbfr_iboldsymbol- mathbfr_jright)
              tag3label3
              endequation






              share|cite|improve this answer











              $endgroup$












              • $begingroup$
                So the authors presumably stated early on that, in their text, bold face quantities are all vectors. The only scalar in the OP's equation are the masses?
                $endgroup$
                – DJohnM
                Jun 19 at 6:31










              • $begingroup$
                @DJohnM ...the masses $:m_i,m_j:$ and the constant $:mathrm k$...
                $endgroup$
                – Frobenius
                Jun 19 at 6:48






              • 1




                $begingroup$
                @DJohnM I don't think I've seen boldface used for anything other than vectors.
                $endgroup$
                – David Richerby
                Jun 19 at 13:19










              • $begingroup$
                But why (3) and not (1). Is it a traditionhabit?
                $endgroup$
                – J. Manuel
                Jun 19 at 15:59










              • $begingroup$
                @J.Manuel You see both in the literature, and I would wager (1) is actually more common as it makes manifest the inverse square relationship. (3) can often be easier to work with however.
                $endgroup$
                – gabe
                Jun 19 at 17:58













              13












              13








              13





              $begingroup$

              enter image description here



              beginequation
              mathbfF_ij boldsymbol=mathrm kdfracm_i m_jVert mathbfr_iboldsymbol- mathbfr_jVert^2 mathbfn_ij
              tag1label1
              endequation

              where $:mathbfn_ij:$ the unit vector along the vector $:left(mathbfr_iboldsymbol- mathbfr_jright)$. But
              beginequation
              mathbfn_ij boldsymbol=dfracleft(mathbfr_iboldsymbol- mathbfr_jright)Vert mathbfr_iboldsymbol- mathbfr_jVert
              tag2label2
              endequation

              so the third power
              beginequation
              mathbfF_ij boldsymbol=mathrm kdfracm_i m_jVert mathbfr_iboldsymbol- mathbfr_jVert^3 left(mathbfr_iboldsymbol- mathbfr_jright)
              tag3label3
              endequation






              share|cite|improve this answer











              $endgroup$



              enter image description here



              beginequation
              mathbfF_ij boldsymbol=mathrm kdfracm_i m_jVert mathbfr_iboldsymbol- mathbfr_jVert^2 mathbfn_ij
              tag1label1
              endequation

              where $:mathbfn_ij:$ the unit vector along the vector $:left(mathbfr_iboldsymbol- mathbfr_jright)$. But
              beginequation
              mathbfn_ij boldsymbol=dfracleft(mathbfr_iboldsymbol- mathbfr_jright)Vert mathbfr_iboldsymbol- mathbfr_jVert
              tag2label2
              endequation

              so the third power
              beginequation
              mathbfF_ij boldsymbol=mathrm kdfracm_i m_jVert mathbfr_iboldsymbol- mathbfr_jVert^3 left(mathbfr_iboldsymbol- mathbfr_jright)
              tag3label3
              endequation







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Jun 19 at 13:41

























              answered Jun 19 at 4:36









              FrobeniusFrobenius

              6,1331 gold badge16 silver badges30 bronze badges




              6,1331 gold badge16 silver badges30 bronze badges











              • $begingroup$
                So the authors presumably stated early on that, in their text, bold face quantities are all vectors. The only scalar in the OP's equation are the masses?
                $endgroup$
                – DJohnM
                Jun 19 at 6:31










              • $begingroup$
                @DJohnM ...the masses $:m_i,m_j:$ and the constant $:mathrm k$...
                $endgroup$
                – Frobenius
                Jun 19 at 6:48






              • 1




                $begingroup$
                @DJohnM I don't think I've seen boldface used for anything other than vectors.
                $endgroup$
                – David Richerby
                Jun 19 at 13:19










              • $begingroup$
                But why (3) and not (1). Is it a traditionhabit?
                $endgroup$
                – J. Manuel
                Jun 19 at 15:59










              • $begingroup$
                @J.Manuel You see both in the literature, and I would wager (1) is actually more common as it makes manifest the inverse square relationship. (3) can often be easier to work with however.
                $endgroup$
                – gabe
                Jun 19 at 17:58
















              • $begingroup$
                So the authors presumably stated early on that, in their text, bold face quantities are all vectors. The only scalar in the OP's equation are the masses?
                $endgroup$
                – DJohnM
                Jun 19 at 6:31










              • $begingroup$
                @DJohnM ...the masses $:m_i,m_j:$ and the constant $:mathrm k$...
                $endgroup$
                – Frobenius
                Jun 19 at 6:48






              • 1




                $begingroup$
                @DJohnM I don't think I've seen boldface used for anything other than vectors.
                $endgroup$
                – David Richerby
                Jun 19 at 13:19










              • $begingroup$
                But why (3) and not (1). Is it a traditionhabit?
                $endgroup$
                – J. Manuel
                Jun 19 at 15:59










              • $begingroup$
                @J.Manuel You see both in the literature, and I would wager (1) is actually more common as it makes manifest the inverse square relationship. (3) can often be easier to work with however.
                $endgroup$
                – gabe
                Jun 19 at 17:58















              $begingroup$
              So the authors presumably stated early on that, in their text, bold face quantities are all vectors. The only scalar in the OP's equation are the masses?
              $endgroup$
              – DJohnM
              Jun 19 at 6:31




              $begingroup$
              So the authors presumably stated early on that, in their text, bold face quantities are all vectors. The only scalar in the OP's equation are the masses?
              $endgroup$
              – DJohnM
              Jun 19 at 6:31












              $begingroup$
              @DJohnM ...the masses $:m_i,m_j:$ and the constant $:mathrm k$...
              $endgroup$
              – Frobenius
              Jun 19 at 6:48




              $begingroup$
              @DJohnM ...the masses $:m_i,m_j:$ and the constant $:mathrm k$...
              $endgroup$
              – Frobenius
              Jun 19 at 6:48




              1




              1




              $begingroup$
              @DJohnM I don't think I've seen boldface used for anything other than vectors.
              $endgroup$
              – David Richerby
              Jun 19 at 13:19




              $begingroup$
              @DJohnM I don't think I've seen boldface used for anything other than vectors.
              $endgroup$
              – David Richerby
              Jun 19 at 13:19












              $begingroup$
              But why (3) and not (1). Is it a traditionhabit?
              $endgroup$
              – J. Manuel
              Jun 19 at 15:59




              $begingroup$
              But why (3) and not (1). Is it a traditionhabit?
              $endgroup$
              – J. Manuel
              Jun 19 at 15:59












              $begingroup$
              @J.Manuel You see both in the literature, and I would wager (1) is actually more common as it makes manifest the inverse square relationship. (3) can often be easier to work with however.
              $endgroup$
              – gabe
              Jun 19 at 17:58




              $begingroup$
              @J.Manuel You see both in the literature, and I would wager (1) is actually more common as it makes manifest the inverse square relationship. (3) can often be easier to work with however.
              $endgroup$
              – gabe
              Jun 19 at 17:58











              3












              $begingroup$

              It is a matter of convenience:



              Let $vecr_ij$ be a distance vector with magnitude $r_ij$ along the line connecting the masses $m_i$ and $m_j$. Then:



              (I): $vecr_ij$ squared is a scalar whose value equals its magnitude squared. Proof: $vecr_ij^2=vecr_ij cdot vecr_ij=r_ij r_ij cos0=r_ij^2$.



              (II): $vecr_ij$ can be written as $vecr_ij=r_ij hatr_ij$, with $hatr_ij$ being a unit vector having the same direction as $vecr_ij$. Proof: This goes as an axiom ;-)



              If ones writes Newton’s law of gravity as
              $$F_ij=fracγm_im_jvecr_ij^2=fracγm_im_jr_ij^2 tag1$$



              Then (1) is an incomplete description of the gravitational force. Equation (1) only represents the magnitude of the gravitational force, as can be noticed by the fact that the right side of it is just a scalar.



              It is a well-established observation that gravity is a force whose direction is along the line connecting the masses $m_i$ and $m_j$. As a force, gravity must be written as a vector, and therefore the right form would be



              $$vecF_ij=fracγm_im_jr_ij^2 hatr_ij tag2$$



              From (II), equation (2) can be written as



              $$vecF_ij=fracγm_im_jr_ij^2 fracvecr_ijr_ij=fracγm_im_jr_ij^3 vecr_ij tag3$$



              Typing vectors and unit vectors with arrows and “hats” atop (as in current case) is a little cumbersome. To solve for typing vectors with arrows, one can select other options as boldfacing them for example. To avoid unit vectors one uses the convention $fracvecrr$, i.e., the choice of (3) is a convenient form of writing these sort of equations, especially for works containing a huge amount of text.



              It is not only easier (if one uses a WYSIWYG text editor of typing machine) to type equation (3) using these conventions, but visually cleaner as well. Just check:



              beginequation
              mathbfF_ij boldsymbol=mathrm γdfracm_i m_jmathbfr_ij^3 mathbfr_ij
              tag4
              endequation



              As was pointed out by @AndrewSteane, there are even mathematical advantages by using this notation, hence, it is just more convenient to use the $fracbf rbf r^3$ notation.



              The problem with this, however, is that people (almost) always is introduced to Newton’s law of gravity in its non-vector form (equation 1) and is expecting to find the inverse square law in a first glance, therefore, it may look odd for beginners until one gets used to it some point in future ;-)






              share|cite|improve this answer











              $endgroup$

















                3












                $begingroup$

                It is a matter of convenience:



                Let $vecr_ij$ be a distance vector with magnitude $r_ij$ along the line connecting the masses $m_i$ and $m_j$. Then:



                (I): $vecr_ij$ squared is a scalar whose value equals its magnitude squared. Proof: $vecr_ij^2=vecr_ij cdot vecr_ij=r_ij r_ij cos0=r_ij^2$.



                (II): $vecr_ij$ can be written as $vecr_ij=r_ij hatr_ij$, with $hatr_ij$ being a unit vector having the same direction as $vecr_ij$. Proof: This goes as an axiom ;-)



                If ones writes Newton’s law of gravity as
                $$F_ij=fracγm_im_jvecr_ij^2=fracγm_im_jr_ij^2 tag1$$



                Then (1) is an incomplete description of the gravitational force. Equation (1) only represents the magnitude of the gravitational force, as can be noticed by the fact that the right side of it is just a scalar.



                It is a well-established observation that gravity is a force whose direction is along the line connecting the masses $m_i$ and $m_j$. As a force, gravity must be written as a vector, and therefore the right form would be



                $$vecF_ij=fracγm_im_jr_ij^2 hatr_ij tag2$$



                From (II), equation (2) can be written as



                $$vecF_ij=fracγm_im_jr_ij^2 fracvecr_ijr_ij=fracγm_im_jr_ij^3 vecr_ij tag3$$



                Typing vectors and unit vectors with arrows and “hats” atop (as in current case) is a little cumbersome. To solve for typing vectors with arrows, one can select other options as boldfacing them for example. To avoid unit vectors one uses the convention $fracvecrr$, i.e., the choice of (3) is a convenient form of writing these sort of equations, especially for works containing a huge amount of text.



                It is not only easier (if one uses a WYSIWYG text editor of typing machine) to type equation (3) using these conventions, but visually cleaner as well. Just check:



                beginequation
                mathbfF_ij boldsymbol=mathrm γdfracm_i m_jmathbfr_ij^3 mathbfr_ij
                tag4
                endequation



                As was pointed out by @AndrewSteane, there are even mathematical advantages by using this notation, hence, it is just more convenient to use the $fracbf rbf r^3$ notation.



                The problem with this, however, is that people (almost) always is introduced to Newton’s law of gravity in its non-vector form (equation 1) and is expecting to find the inverse square law in a first glance, therefore, it may look odd for beginners until one gets used to it some point in future ;-)






                share|cite|improve this answer











                $endgroup$















                  3












                  3








                  3





                  $begingroup$

                  It is a matter of convenience:



                  Let $vecr_ij$ be a distance vector with magnitude $r_ij$ along the line connecting the masses $m_i$ and $m_j$. Then:



                  (I): $vecr_ij$ squared is a scalar whose value equals its magnitude squared. Proof: $vecr_ij^2=vecr_ij cdot vecr_ij=r_ij r_ij cos0=r_ij^2$.



                  (II): $vecr_ij$ can be written as $vecr_ij=r_ij hatr_ij$, with $hatr_ij$ being a unit vector having the same direction as $vecr_ij$. Proof: This goes as an axiom ;-)



                  If ones writes Newton’s law of gravity as
                  $$F_ij=fracγm_im_jvecr_ij^2=fracγm_im_jr_ij^2 tag1$$



                  Then (1) is an incomplete description of the gravitational force. Equation (1) only represents the magnitude of the gravitational force, as can be noticed by the fact that the right side of it is just a scalar.



                  It is a well-established observation that gravity is a force whose direction is along the line connecting the masses $m_i$ and $m_j$. As a force, gravity must be written as a vector, and therefore the right form would be



                  $$vecF_ij=fracγm_im_jr_ij^2 hatr_ij tag2$$



                  From (II), equation (2) can be written as



                  $$vecF_ij=fracγm_im_jr_ij^2 fracvecr_ijr_ij=fracγm_im_jr_ij^3 vecr_ij tag3$$



                  Typing vectors and unit vectors with arrows and “hats” atop (as in current case) is a little cumbersome. To solve for typing vectors with arrows, one can select other options as boldfacing them for example. To avoid unit vectors one uses the convention $fracvecrr$, i.e., the choice of (3) is a convenient form of writing these sort of equations, especially for works containing a huge amount of text.



                  It is not only easier (if one uses a WYSIWYG text editor of typing machine) to type equation (3) using these conventions, but visually cleaner as well. Just check:



                  beginequation
                  mathbfF_ij boldsymbol=mathrm γdfracm_i m_jmathbfr_ij^3 mathbfr_ij
                  tag4
                  endequation



                  As was pointed out by @AndrewSteane, there are even mathematical advantages by using this notation, hence, it is just more convenient to use the $fracbf rbf r^3$ notation.



                  The problem with this, however, is that people (almost) always is introduced to Newton’s law of gravity in its non-vector form (equation 1) and is expecting to find the inverse square law in a first glance, therefore, it may look odd for beginners until one gets used to it some point in future ;-)






                  share|cite|improve this answer











                  $endgroup$



                  It is a matter of convenience:



                  Let $vecr_ij$ be a distance vector with magnitude $r_ij$ along the line connecting the masses $m_i$ and $m_j$. Then:



                  (I): $vecr_ij$ squared is a scalar whose value equals its magnitude squared. Proof: $vecr_ij^2=vecr_ij cdot vecr_ij=r_ij r_ij cos0=r_ij^2$.



                  (II): $vecr_ij$ can be written as $vecr_ij=r_ij hatr_ij$, with $hatr_ij$ being a unit vector having the same direction as $vecr_ij$. Proof: This goes as an axiom ;-)



                  If ones writes Newton’s law of gravity as
                  $$F_ij=fracγm_im_jvecr_ij^2=fracγm_im_jr_ij^2 tag1$$



                  Then (1) is an incomplete description of the gravitational force. Equation (1) only represents the magnitude of the gravitational force, as can be noticed by the fact that the right side of it is just a scalar.



                  It is a well-established observation that gravity is a force whose direction is along the line connecting the masses $m_i$ and $m_j$. As a force, gravity must be written as a vector, and therefore the right form would be



                  $$vecF_ij=fracγm_im_jr_ij^2 hatr_ij tag2$$



                  From (II), equation (2) can be written as



                  $$vecF_ij=fracγm_im_jr_ij^2 fracvecr_ijr_ij=fracγm_im_jr_ij^3 vecr_ij tag3$$



                  Typing vectors and unit vectors with arrows and “hats” atop (as in current case) is a little cumbersome. To solve for typing vectors with arrows, one can select other options as boldfacing them for example. To avoid unit vectors one uses the convention $fracvecrr$, i.e., the choice of (3) is a convenient form of writing these sort of equations, especially for works containing a huge amount of text.



                  It is not only easier (if one uses a WYSIWYG text editor of typing machine) to type equation (3) using these conventions, but visually cleaner as well. Just check:



                  beginequation
                  mathbfF_ij boldsymbol=mathrm γdfracm_i m_jmathbfr_ij^3 mathbfr_ij
                  tag4
                  endequation



                  As was pointed out by @AndrewSteane, there are even mathematical advantages by using this notation, hence, it is just more convenient to use the $fracbf rbf r^3$ notation.



                  The problem with this, however, is that people (almost) always is introduced to Newton’s law of gravity in its non-vector form (equation 1) and is expecting to find the inverse square law in a first glance, therefore, it may look odd for beginners until one gets used to it some point in future ;-)







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited yesterday

























                  answered Jun 19 at 14:30









                  J. ManuelJ. Manuel

                  1,1103 silver badges22 bronze badges




                  1,1103 silver badges22 bronze badges



























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