Are all constant fields conservative? [closed]

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Are all constant fields conservative? Can there be some constant vector fields which are not conservative?




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closed as off-topic by Jendrik Stelzner, Theoretical Economist, Adrian Keister, Gibbs, Strants Aug 29 at 14:52


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    Are all constant fields conservative? Can there be some constant vector fields which are not conservative?




    multivariable-calculus vector-analysis vector-fields constants




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    closed as off-topic by Jendrik Stelzner, Theoretical Economist, Adrian Keister, Gibbs, Strants Aug 29 at 14:52


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jendrik Stelzner, Theoretical Economist, Adrian Keister, Gibbs, Strants
    If this question can be reworded to fit the rules in the help center, please edit the question.














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      Are all constant fields conservative? Can there be some constant vector fields which are not conservative?




      multivariable-calculus vector-analysis vector-fields constants




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      Are all constant fields conservative? Can there be some constant vector fields which are not conservative?





      multivariable-calculus vector-analysis vector-fields constants





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      asked Aug 29 at 8:47









      Joe

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      closed as off-topic by Jendrik Stelzner, Theoretical Economist, Adrian Keister, Gibbs, Strants Aug 29 at 14:52


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jendrik Stelzner, Theoretical Economist, Adrian Keister, Gibbs, Strants
      If this question can be reworded to fit the rules in the help center, please edit the question.




      closed as off-topic by Jendrik Stelzner, Theoretical Economist, Adrian Keister, Gibbs, Strants Aug 29 at 14:52


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jendrik Stelzner, Theoretical Economist, Adrian Keister, Gibbs, Strants
      If this question can be reworded to fit the rules in the help center, please edit the question.




















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          The answer is affirmative. A conservative field is a vector field which is the gradient of some function. So, if $mathbf v$ is a constant vector field, that is $mathbfv(x_1,ldots,x_n)=(a_1,ldots,a_n)$, you can take$$F(x_1,ldots,x_n)=a_1x_1+cdots+a_nx_n.$$Then $mathbfv=nabla F$.






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          • I do not understand what $x_1$ and $a_1$ means? Can you please give a proof of it in three dimensions, so that a high school student (like me) can understand easily?
            – Joe
            Aug 29 at 9:06










          • @Joe If you want a $3$-dimensional proof, just take $n=3$. You wrote that $mathbf v$ is constant and so I took $mathbfv(x,y,z)=(a,b,c)$. Then $mathbfv=nabla f$, where $f(x,y,z)=ax+by+cz$.
            – José Carlos Santos
            Aug 29 at 9:08










          • Thanks.... I understand.
            – Joe
            Aug 29 at 9:11










          • @Joe I'm glad I could help.
            – José Carlos Santos
            Aug 29 at 9:11

















          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          3
          down vote



          accepted










          The answer is affirmative. A conservative field is a vector field which is the gradient of some function. So, if $mathbf v$ is a constant vector field, that is $mathbfv(x_1,ldots,x_n)=(a_1,ldots,a_n)$, you can take$$F(x_1,ldots,x_n)=a_1x_1+cdots+a_nx_n.$$Then $mathbfv=nabla F$.






          share|cite|improve this answer




















          • I do not understand what $x_1$ and $a_1$ means? Can you please give a proof of it in three dimensions, so that a high school student (like me) can understand easily?
            – Joe
            Aug 29 at 9:06










          • @Joe If you want a $3$-dimensional proof, just take $n=3$. You wrote that $mathbf v$ is constant and so I took $mathbfv(x,y,z)=(a,b,c)$. Then $mathbfv=nabla f$, where $f(x,y,z)=ax+by+cz$.
            – José Carlos Santos
            Aug 29 at 9:08










          • Thanks.... I understand.
            – Joe
            Aug 29 at 9:11










          • @Joe I'm glad I could help.
            – José Carlos Santos
            Aug 29 at 9:11














          up vote
          3
          down vote



          accepted










          The answer is affirmative. A conservative field is a vector field which is the gradient of some function. So, if $mathbf v$ is a constant vector field, that is $mathbfv(x_1,ldots,x_n)=(a_1,ldots,a_n)$, you can take$$F(x_1,ldots,x_n)=a_1x_1+cdots+a_nx_n.$$Then $mathbfv=nabla F$.






          share|cite|improve this answer




















          • I do not understand what $x_1$ and $a_1$ means? Can you please give a proof of it in three dimensions, so that a high school student (like me) can understand easily?
            – Joe
            Aug 29 at 9:06










          • @Joe If you want a $3$-dimensional proof, just take $n=3$. You wrote that $mathbf v$ is constant and so I took $mathbfv(x,y,z)=(a,b,c)$. Then $mathbfv=nabla f$, where $f(x,y,z)=ax+by+cz$.
            – José Carlos Santos
            Aug 29 at 9:08










          • Thanks.... I understand.
            – Joe
            Aug 29 at 9:11










          • @Joe I'm glad I could help.
            – José Carlos Santos
            Aug 29 at 9:11












          up vote
          3
          down vote



          accepted







          up vote
          3
          down vote



          accepted






          The answer is affirmative. A conservative field is a vector field which is the gradient of some function. So, if $mathbf v$ is a constant vector field, that is $mathbfv(x_1,ldots,x_n)=(a_1,ldots,a_n)$, you can take$$F(x_1,ldots,x_n)=a_1x_1+cdots+a_nx_n.$$Then $mathbfv=nabla F$.






          share|cite|improve this answer












          The answer is affirmative. A conservative field is a vector field which is the gradient of some function. So, if $mathbf v$ is a constant vector field, that is $mathbfv(x_1,ldots,x_n)=(a_1,ldots,a_n)$, you can take$$F(x_1,ldots,x_n)=a_1x_1+cdots+a_nx_n.$$Then $mathbfv=nabla F$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 29 at 8:51









          José Carlos Santos

          120k16101182




          120k16101182











          • I do not understand what $x_1$ and $a_1$ means? Can you please give a proof of it in three dimensions, so that a high school student (like me) can understand easily?
            – Joe
            Aug 29 at 9:06










          • @Joe If you want a $3$-dimensional proof, just take $n=3$. You wrote that $mathbf v$ is constant and so I took $mathbfv(x,y,z)=(a,b,c)$. Then $mathbfv=nabla f$, where $f(x,y,z)=ax+by+cz$.
            – José Carlos Santos
            Aug 29 at 9:08










          • Thanks.... I understand.
            – Joe
            Aug 29 at 9:11










          • @Joe I'm glad I could help.
            – José Carlos Santos
            Aug 29 at 9:11
















          • I do not understand what $x_1$ and $a_1$ means? Can you please give a proof of it in three dimensions, so that a high school student (like me) can understand easily?
            – Joe
            Aug 29 at 9:06










          • @Joe If you want a $3$-dimensional proof, just take $n=3$. You wrote that $mathbf v$ is constant and so I took $mathbfv(x,y,z)=(a,b,c)$. Then $mathbfv=nabla f$, where $f(x,y,z)=ax+by+cz$.
            – José Carlos Santos
            Aug 29 at 9:08










          • Thanks.... I understand.
            – Joe
            Aug 29 at 9:11










          • @Joe I'm glad I could help.
            – José Carlos Santos
            Aug 29 at 9:11















          I do not understand what $x_1$ and $a_1$ means? Can you please give a proof of it in three dimensions, so that a high school student (like me) can understand easily?
          – Joe
          Aug 29 at 9:06




          I do not understand what $x_1$ and $a_1$ means? Can you please give a proof of it in three dimensions, so that a high school student (like me) can understand easily?
          – Joe
          Aug 29 at 9:06












          @Joe If you want a $3$-dimensional proof, just take $n=3$. You wrote that $mathbf v$ is constant and so I took $mathbfv(x,y,z)=(a,b,c)$. Then $mathbfv=nabla f$, where $f(x,y,z)=ax+by+cz$.
          – José Carlos Santos
          Aug 29 at 9:08




          @Joe If you want a $3$-dimensional proof, just take $n=3$. You wrote that $mathbf v$ is constant and so I took $mathbfv(x,y,z)=(a,b,c)$. Then $mathbfv=nabla f$, where $f(x,y,z)=ax+by+cz$.
          – José Carlos Santos
          Aug 29 at 9:08












          Thanks.... I understand.
          – Joe
          Aug 29 at 9:11




          Thanks.... I understand.
          – Joe
          Aug 29 at 9:11












          @Joe I'm glad I could help.
          – José Carlos Santos
          Aug 29 at 9:11




          @Joe I'm glad I could help.
          – José Carlos Santos
          Aug 29 at 9:11


          -

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