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Given:

$F(x,y,z)=(x-y,-x-y+z,y+z)$

Find a potential energy that corresponds to this force field. Check your answer by taking its gradient.

I've already shown that this force field is conservative by

$$\nabla \times F =0$$

Now, I used $\nabla U=-F$ to find the potential function. I did so by

$$U=-\int F \cdot \vec{dr}$$ $$=-\left [\int (x-y)dx + \int(-x-y+z)dy + \int (y+z)dz \right ]$$ $$=\frac{1}{2}(-x^2+y^2-z^2)+2xy-2yz$$

Now I need to check it by taking its gradient, but its not resulting in the original force field.

$$\nabla U=-F$$ $$F=-\nabla U$$ $$=- \left [\frac{\partial U}{\partial x}+\frac{\partial U}{\partial x}+\frac{\partial U}{\partial x} \right ]$$ $$=- \left [ (-x+2y)+(y+2x-2z)+(-z-2y)\right ]$$ $$=(x-2y)+(-2x-y+2z)+(2y+z)$$

What did I miss?

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  • $\begingroup$ What is $\overrightarrow{r}$? How do you perform the line integral $-\int F\cdot \overrightarrow{dr}$? $\endgroup$ – Sam Weatherhog Oct 16 '16 at 22:36
  • $\begingroup$ $\vec{r}=dx \hat x + dy \hat y +dz \hat z$ $\endgroup$ – whatwhatwhat Oct 16 '16 at 22:37
  • $\begingroup$ Where did you find this definition $U=-\int F \cdot \overrightarrow{dr}$? $\endgroup$ – Sam Weatherhog Oct 16 '16 at 22:48
  • $\begingroup$ there a problem with factor $2$ $\endgroup$ – hamam_Abdallah Oct 16 '16 at 22:49
  • $\begingroup$ I just integrated both sides of $\nabla U = -F$, but it's also in my classical mechanics book as the work done from point A to point B. $\endgroup$ – whatwhatwhat Oct 16 '16 at 23:03
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You started out OK by setting up $U=-\int\mathbf F\cdot d\mathbf r$, but then you appear to have confused this line integral with an indefinite integral. Perhaps you made this mistake because of leaving out an important detail: the path along which the line integral is to be taken (the integral’s bounds, as it were). I.e., it’s properly something like $U=-\int_\Gamma\mathbf F\cdot d\mathbf r$, which is like a definite integral from elementary calculus—it produces a number, not a function. This error is akin to saying that if $f={dF\over dx}$, then $F(x)=\int_a^b f(t)\,dt$.

You can in fact use a line integral to find a function $U:\mathbb R^n\to\mathbb R$ such that $\nabla U=-\mathbf F$, but you have to proceed a little differently. Recall from elementary calculus that $F(x)=\int_0^x f(t)\,dt$ is an antiderivative of $f$, that is, that ${dF\over dx}=f$. Note the difference between this integral and the one in the previous paragraph: the upper bound is variable. We can do a similar thing for a conservative vector field: Let $U(0)=0$ and set $U(\mathbf r)=-\int_{\Gamma_{\mathbf r}}\mathbf F\cdot d\mathbf r$, where $\Gamma_{\mathbf r}$ is a differentiable path from the origin to $\mathbf r$. Since $\mathbf F$ is conservative, the value of this integral depends only on its endpoints, not on the specific path taken, so this function is well-defined (as long as the integral exists, of course). A convenient choice for $\Gamma$ is the line segment joining the point $\mathbf r$ to the origin, which we can parameterize in the obvious way as $\gamma: t\mapsto t\,\mathbf r$ for $t\in[0,1]$. We then get $$U(\mathbf r)=-\int_0^1F(\gamma(t))\cdot\gamma'(t)\,dt=-\int_0^1\mathbf F(t\,\mathbf r)\cdot\mathbf r\,dt.\tag{*}$$ Applying this to your example, $$\begin{align} U(x,y,z) &= -\int_0^1\mathbf F(tx,ty,tz)\cdot(x,y,z)\,dt \\ &= -\int_0^1x(tx-ty)+y(-tx-tz+tz)+z(ty+tz)\,dt \\ &=-\int_0^1(x^2-2xy-y^2+2yz+z^2)\,t\,dt \\ &= -\frac12(x^2-2xy-y^2+2yz+z^2).\end{align}$$ You can verify for yourself that $\nabla U=-\mathbf F$. If we take $U(0)$ to be some value other than $0$, this amounts to changing the constant of integration, i.e., (*) becomes $$U(\mathbf r)=U(0)-\int_0^1\mathbf F(t\,\mathbf r)\cdot\mathbf r\,dt.$$

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Since you know that $F$ is conservative, you know that $F=\nabla U$ for some $U$. Using the definition of $\nabla U$ this tells you that:

$$ \begin{align} \frac{\partial U}{\partial x}&= x-y \\ \frac{\partial U}{\partial y}&= -x-y+z \\ \frac{\partial U}{\partial z}&=y+z \end{align} $$ You can begin integrating with any of these. If we start with the first we get: $$ \begin{align} \frac{\partial U}{\partial x}&= x-y \\ \implies U&=\frac{1}{2}x^2-xy+h(y,z) \end{align} $$ for some function $h(y,z)$. Differentiating this with respect to $y$ and comparing to $\frac{\partial U}{\partial y}= -x-y+z$ yields: $$ \frac{\partial U}{\partial y}=-x+\frac{\partial h}{\partial y} $$ So $\frac{\partial h}{\partial y}=-y+z$ and hence $h=-\frac{1}{2}y^2+zy+g(z)$ for some function $g(z)$. Finally differentiating $U$ with respect to $z$ and comparing to $\frac{\partial U}{\partial z}=y+z$ yields: $$ \frac{\partial U}{\partial z}=y+\frac{dg}{dz} $$ so $\frac{dg}{dz}=z$. This gives us $U=\frac{1}{2}x^2-\frac{1}{2}y^2+\frac{1}{2}z^2-xy+yz+C$ for some constant $C$. You want $-U$ for your solution.

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  • $\begingroup$ Wait, it appears that $\nabla U$ already gives the correct answer, without the need to take the negative of it. Why is this? $\endgroup$ – whatwhatwhat Oct 16 '16 at 22:56
  • $\begingroup$ In your question you wanted $\nabla U=-F$. If you want $-F$ then you need to take $-U$ above. $\endgroup$ – Sam Weatherhog Oct 18 '16 at 23:05
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we have

$\frac{\partial U}{\partial x}=-x+y$

so $U=\frac{-x^2}{2}+xy+C_1(y,z)$

$\frac{\partial U}{\partial y}=x+y-z$

so $U=xy+\frac{y^2}{2}-zy+C_3(x,z)$

$\frac{\partial U}{\partial z}=-y-z$

so $U=-yz-\frac{z^2}{2}+C_3(x,y)$

and finally

$U=-\frac{x^2}{2}+xy-yz+\frac{y^2}{2}-\frac{z^2}{2}+C$

$C$ is a constant.

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