Does the sign matter for proportionality?












3












$begingroup$


This question arose from Physics, where the force on an object attached on a spring is proportional to the displacement to the equilibrium (that is, the rest position). Also, if the displacement to the equilibrium is positive, the force will be negative, as it tries to pull the object back (i.e. if you pull a string, the force is opposite to your direction of pull).



Therefore, it can be said that:



$$F propto -x$$
Where $F$ is the force and $x$ is the displacement from equilibrium



Is this the same as:
$$F propto x$$
According to the relation of proportionality, it should be, but my friend says that putting the second one is not correct.



Both are equivalent right?






ratio







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  • $begingroup$
    It depends on definitions/context. Using the definition on wikipedia, yes they are equivalent.
    $endgroup$
    – David Peterson
    3 hours ago
















3












$begingroup$


This question arose from Physics, where the force on an object attached on a spring is proportional to the displacement to the equilibrium (that is, the rest position). Also, if the displacement to the equilibrium is positive, the force will be negative, as it tries to pull the object back (i.e. if you pull a string, the force is opposite to your direction of pull).



Therefore, it can be said that:



$$F propto -x$$
Where $F$ is the force and $x$ is the displacement from equilibrium



Is this the same as:
$$F propto x$$
According to the relation of proportionality, it should be, but my friend says that putting the second one is not correct.



Both are equivalent right?






ratio







share|cite|improve this question









New contributor




Sebi S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    It depends on definitions/context. Using the definition on wikipedia, yes they are equivalent.
    $endgroup$
    – David Peterson
    3 hours ago














3












3








3





$begingroup$


This question arose from Physics, where the force on an object attached on a spring is proportional to the displacement to the equilibrium (that is, the rest position). Also, if the displacement to the equilibrium is positive, the force will be negative, as it tries to pull the object back (i.e. if you pull a string, the force is opposite to your direction of pull).



Therefore, it can be said that:



$$F propto -x$$
Where $F$ is the force and $x$ is the displacement from equilibrium



Is this the same as:
$$F propto x$$
According to the relation of proportionality, it should be, but my friend says that putting the second one is not correct.



Both are equivalent right?






ratio







share|cite|improve this question









New contributor




Sebi S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




This question arose from Physics, where the force on an object attached on a spring is proportional to the displacement to the equilibrium (that is, the rest position). Also, if the displacement to the equilibrium is positive, the force will be negative, as it tries to pull the object back (i.e. if you pull a string, the force is opposite to your direction of pull).



Therefore, it can be said that:



$$F propto -x$$
Where $F$ is the force and $x$ is the displacement from equilibrium



Is this the same as:
$$F propto x$$
According to the relation of proportionality, it should be, but my friend says that putting the second one is not correct.



Both are equivalent right?






ratio




ratio






share|cite|improve this question









New contributor




Sebi S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









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Check out our Code of Conduct.









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edited 15 mins ago









Brian

1,529516




1,529516






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asked 3 hours ago









Sebi SSebi S

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Sebi S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






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  • $begingroup$
    It depends on definitions/context. Using the definition on wikipedia, yes they are equivalent.
    $endgroup$
    – David Peterson
    3 hours ago


















  • $begingroup$
    It depends on definitions/context. Using the definition on wikipedia, yes they are equivalent.
    $endgroup$
    – David Peterson
    3 hours ago
















$begingroup$
It depends on definitions/context. Using the definition on wikipedia, yes they are equivalent.
$endgroup$
– David Peterson
3 hours ago




$begingroup$
It depends on definitions/context. Using the definition on wikipedia, yes they are equivalent.
$endgroup$
– David Peterson
3 hours ago










2 Answers
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$begingroup$

Yes, proportionality does not tell anything about the sign of the proportionality constant. This is probably done in physics so that the proportionality constant $k$ can be considered to be always positive. This has a physical interpretation , so it is convenient for it to be positive.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Mathematically they are equivalent, but physycist may want to differentiate between a force in the same direction as the displacement, and a force opposite to the displacement, as they lead to physically different behavior of the system. Thus the minus sign matters.






    share|cite|improve this answer









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      2 Answers
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      2 Answers
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      active

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      2












      $begingroup$

      Yes, proportionality does not tell anything about the sign of the proportionality constant. This is probably done in physics so that the proportionality constant $k$ can be considered to be always positive. This has a physical interpretation , so it is convenient for it to be positive.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Yes, proportionality does not tell anything about the sign of the proportionality constant. This is probably done in physics so that the proportionality constant $k$ can be considered to be always positive. This has a physical interpretation , so it is convenient for it to be positive.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Yes, proportionality does not tell anything about the sign of the proportionality constant. This is probably done in physics so that the proportionality constant $k$ can be considered to be always positive. This has a physical interpretation , so it is convenient for it to be positive.






          share|cite|improve this answer









          $endgroup$



          Yes, proportionality does not tell anything about the sign of the proportionality constant. This is probably done in physics so that the proportionality constant $k$ can be considered to be always positive. This has a physical interpretation , so it is convenient for it to be positive.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 3 hours ago









          John DoeJohn Doe

          12.3k11341




          12.3k11341























              2












              $begingroup$

              Mathematically they are equivalent, but physycist may want to differentiate between a force in the same direction as the displacement, and a force opposite to the displacement, as they lead to physically different behavior of the system. Thus the minus sign matters.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Mathematically they are equivalent, but physycist may want to differentiate between a force in the same direction as the displacement, and a force opposite to the displacement, as they lead to physically different behavior of the system. Thus the minus sign matters.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Mathematically they are equivalent, but physycist may want to differentiate between a force in the same direction as the displacement, and a force opposite to the displacement, as they lead to physically different behavior of the system. Thus the minus sign matters.






                  share|cite|improve this answer









                  $endgroup$



                  Mathematically they are equivalent, but physycist may want to differentiate between a force in the same direction as the displacement, and a force opposite to the displacement, as they lead to physically different behavior of the system. Thus the minus sign matters.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 3 hours ago









                  Adam LatosińskiAdam Latosiński

                  9709




                  9709






















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