Finding the potential function of a vector field

In summary, the conversation discusses finding the potential function of a vector field with components 2xy-yz, x^2-xz, and 2z-xy. The speaker first checks for a rotation using the Jacobi matrix and finds that it is not zero, which is necessary for a potential function to exist. They then integrate the first row of the vector field to find the potential function, which is a scalar function. The conversation then goes on to discuss taking partial derivatives with respect to y and z to determine the constants C(y,z) and C(z). The final potential function is determined to be V=x^2y-xyz+z^2, with a generic constant C added at the end.
  • #1
sylent33
39
5
Homework Statement
Find the pontential function of a vector field if possible
Relevant Equations
Jacobi Matrix
Hello! So I need to find the potential function of this Vector field

$$
\begin{matrix}
2xy -yz\\
x^2-xz\\
2z-xy
\end{matrix}
$$

Now first I tried to check if rotation is not ,since that is mandatory for the potentialfunction to exist.For that I used the jacobi matrix,and it was not symmetric hence rot cannot be 0.

Now to get the corresponding potentialfunction I first integrated the first row.It should look like this.

$$ x^2y-yzx+C(y,z) $$

Now this should be my corresponding x component (if we look at it like a vector) Now to find y I derrived this in respect to y and I got this

$$ x^2-zx + Cy(y,z) = x^2-zx $$

We can cancel a few things out and are left with

$$ Cy(y,z) = 1 $$

Now how do I get my y (or z) constant from this.I have no y or z standing anywhere and I think I made a mistake somwhere.Am I doing something wrong or should I simply procced like I would usually (which is to integrate both sides)Thank you!
 
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  • #2
First of all the Curl (or rotation as you call it) of this vector field is zero (I used wolfram curl calculator to check it https://www.wolframalpha.com/input/?i=curl+calculator&assumption={"F",+"CurlCalculator",+"vectorfunctioncomponent2"}+->"x^2-xz"&assumption={"F",+"CurlCalculator",+"vectorfunctioncomponent3"}+->"2z-xy"&assumption={"F",+"CurlCalculator",+"vectorfunctioncomponent1"}+->"2xy-yz"), and it should be zero if we want to find a valid potential function.
Second you are almost right in what you doing in the first half, you find the potential function to be ##V=x^2y-yzx+C(y,z)## ( the potential function is a scalar function, it doesn't have x,y,z components)

Now in what it follows, I think what we can conclude, by taking the partial derivative of V with respect to y, is that $$\frac{\partial C(y,z)}{\partial y}=0$$ and hence we can conclude that ##C(y,z)## is independent of y, so it is actually ##C(z)##.

Now what you should take the partial derivative of V with respect to z and set it equal to ##2z-xy##. In a similar way as before you should be able to determine ##C(y,z)## or ##C(z)##.
 
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  • #3
Delta2 said:
First of all the Curl (or rotation as you call it) of this vector field is zero (I used wolfram curl calculator to check it https://www.wolframalpha.com/input/?i=curl+calculator&assumption={"F",+"CurlCalculator",+"vectorfunctioncomponent2"}+->"x^2-xz"&assumption={"F",+"CurlCalculator",+"vectorfunctioncomponent3"}+->"2z-xy"&assumption={"F",+"CurlCalculator",+"vectorfunctioncomponent1"}+->"2xy-yz"), and it should be zero if we want to find a valid potential function.
Second you are almost right in what you doing in the first half, you find the potential function to be ##V=x^2y-yzx+C(y,z)## ( the potential function is a scalar function, it doesn't have x,y,z components)

Now in what it follows, I think what we can conclude, by taking the partial derivative of V with respect to y, is that $$\frac{\partial C(y,z)}{\partial y}=0$$ and hence we can conclude that ##C(y,z)## is independent of y, so it is actually ##C(z)##.

Now what you should take the partial derivative of V with respect to z and set it equal to ##2z-xy##. In a similar way as before you should be able to determine ##C(y,z)## or ##C(z)##.
Okay so first your totally right it should be zero,I kind of rushed it there my calculations say its 0 as well.Now about the rest.I've calculated like this.

So we know that V looks like this $$ V = x^2 y- xyz +C(y,z) $$ Now if we take the partial derivation of V in respect to y I get this

$$ x^2 -xz +Cy(y,z) = x^2 -xz $$ and after canceling I get this

$$Cy(y,z) = 1 $$ Now I am not sure how you get that it is 0? Would you be able to post your work in detail?

But now what I've done is I have integrated both sides with the constant being dy, so we get this

$$ Cy(y,z) = y + C(z) $$ Meaning that V should look like this now

$$ V = x^2y -xyz +y +C(z) $$

Now we derive it partially with respect to z and set it equal to 2z - xy Should look ike this

$$-yx+Cz(z) = 2z - xy $$ The yx should cancel out giving us

$$ Cz(z) = 2z $$ Integrate that with dz being the constant we get

$$ Cz(z) = z^2 $$ Now V and this is final should look like this

$$ V = x^2y -xyz+y+z^2$$

Pretty sure I've got the algorithm down but how you partial derivation of V in respect to y is = 0 I do not undestand.
 
  • #4
All good(you are correct about C(z)) except I don't understand why you keep saying that ##C_y(y,z)=1##. From ##x^2-xz+C_y(y,z)=x^2-xz## we get that ##C_y(y,z)=0## after cancellation. Isn't it simple algebra when you have ##A+C=A ,(A=x^2-xz,C=C_y(y,z))##that C=0??
The final potential function is ##V=x^2y-xyz+z^2##. You can check that $$\frac{\partial V}{\partial x}=2xy-yz$$
$$\frac{\partial V}{\partial y}=x^2-zx$$ and $$\frac{\partial V}{\partial z}=2z-xy$$

P.S We both forgot to add a generic constant C (independent of x,y,z) when integrating ##C_z(z)=2z## so it will actually be ##C(z)=z^2+C## and the generic potential ##V=x^2-xyz+z^2+C##.

As a general rule always add a constant C independent of x,y,z at the end of a potential function, that's what gauge theory tell us!
 
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  • #5
Delta2 said:
All good(you are correct about C(z)) except I don't understand why you keep saying that ##C_y(y,z)=1##. From ##x^2-xz+C_y(y,z)=x^2-xz## we get that ##C_y(y,z)=0## after cancellation. Isn't it simple algebra when you have ##A+C=A ,(A=x^2-xz,C=C_y(y,z))##that C=0??
The final potential function is ##V=x^2y-xyz+z^2##. You can check that $$\frac{\partial V}{\partial x}=2xy-yz$$
$$\frac{\partial V}{\partial y}=x^2-zx$$ and $$\frac{\partial V}{\partial z}=2z-xy$$

P.S We both forgot to add a generic constant C (independent of x,y,z) when integrating ##C_z(z)=2z## so it will actually be ##C(z)=z^2+C## and the generic potential ##V=x^2-xyz+z^2+C##.

As a general rule always add a constant C independent of x,y,z at the end of a potential function, that's what gauge theory tell us!
Oh wait its + not *, of course its 0.Very silly mistake.Also what you showed me here is exactly what I need to know,how to check if the potentialfunction I calculated right or not.So if I derive the final potentialfunction (partially) in respect to x y and z I need to get the Vectorfield I started with.Also just as because I find it interesting why do we have to add the +C at the very end of V? Since we have determined x y z the constant C cannot be in respect of x y and z.If not any of those 3 than in respect to what? What other constant could potentially show up here?
 
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  • #6
It is just the gauge invariance constant... If you haven't been introduced to gauge theory yet, then you can see it as a constant to ensure compatibility with initial condition. For example if you had the initial condition V(0,0,0)=10 this constant would be equal to 10.

Just like when you solve an ODE you find in the general solution one or more constants that you can determine by the initial conditions, the same thing is here.
 
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  • #7
OKay of "gauge theory" I hafve never heared,but that explanation is what I was looking for thanks!
 
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Related to Finding the potential function of a vector field

1. What is a potential function?

A potential function, also known as a scalar field, is a mathematical function that describes the potential energy of a physical system. In the context of vector fields, a potential function describes the relationship between the direction and magnitude of the vectors in the field.

2. Why is finding the potential function of a vector field important?

Finding the potential function of a vector field allows us to simplify complex vector equations and make predictions about the behavior of the field. It also helps us understand the underlying physical principles that govern the behavior of the vector field.

3. How do you find the potential function of a vector field?

To find the potential function of a vector field, we use a mathematical technique called vector calculus. This involves taking the gradient of the vector field, which is a mathematical operation that describes the rate of change of the field in different directions.

4. What are the applications of finding the potential function of a vector field?

The potential function of a vector field has many practical applications, including in physics, engineering, and economics. It can be used to model and predict the behavior of electric and magnetic fields, fluid flow, and economic supply and demand.

5. Are there any limitations to finding the potential function of a vector field?

While finding the potential function of a vector field can be useful, it is not always possible. In some cases, the vector field may not have a potential function, or the process of finding it may be too complex. Additionally, the potential function may not fully describe the behavior of the vector field in all situations.

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