Conservative vector field or not?

In summary, to show that a vector field F=(P,Q,R) is conservative, it is necessary to show that the curl of the vector field is zero and that the domain is simply connected. This can be proven using Stoke's theorem. In R2, it is sufficient to show that DP/DY = DQ/DX, but in R3, the mixed partials must also be equal for all three variables. Alternatively, one can find a scalar potential function F(x,y,z) such that grad(F) = Fvec. This involves integrating P, Q, and R with respect to their respective variables and adding unknown "constant" functions. Lastly, it is important to note that the field must be checked for all three mixed
  • #1
kasse
384
1
To show that a vector field F=(P,Q,R) is conservative, is it enough to show that DP/DY = DQ/DX?
 
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  • #2
Hi,

I believe you have to show that the curl of your vector field is the zero vector (this is really equating the mixed partials as you'll see when doing it). Moreover, the domain of the vector field must be simply connected. A simple proof of this is by using Stoke's theorem I believe.
 
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  • #3
I mean only showing that a certain vector field in R3 is conservative.
 
  • #4
What is the definition of conservative (and just so we're clear: I know what it is, I'm not asking for my benefit)?
 
  • #5
matt grime said:
What is the definition of conservative (and just so we're clear: I know what it is, I'm not asking for my benefit)?

A vector field Fvec is conservative if it has a potential function f, so that grad f = Fvec

My book says that in R2, you only have to check that DP/Dy=DQ/Dx. However, it doesn't say anything aboud conservative vector fields in R3.

Is it enough that DP/Dy=DQ/Dx here as well?
 
  • #6
Clearly not. R^3 has x,y, and z. You can't just ignore the z. Where did the R go to? I suggest you operate with a nicer definition of conservative, for R^3, such as its curl is 0.
 
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  • #7
matt grime said:
Clearly not. R^3 has x,y, and z. You can't just ignore the z. Where did the R go to? I suggest you operate with a nicer definition of conservative, for R^3, such as its curl is 0.

We haven't learned about conservative fields in R^3 yet. Only in R^2, and then it's sufficent that DP/dy=DQ/Dy.

But the curl-definition is better.
 
  • #8
I mean only showing that a certain vector field in R3 is conservative.

You can try figuring out the original scalar potential based on what your P,Q and R are:If grad(Phi(x,y,z)) = F = (P, Q, R) then:

P = dPhi/dx
Q = dPhi/dy
R = dPhi/dz

Integrate P with respect to x, Q with respect to y and R with respect to z (remember to add "constant" functions that are unknown but functions of the other two variables... so when integrating P w.r.t. x, you have to add G(y,z) to your answer).

It's likely you can eyeball the answer and figure out what the potential is without actually solving methodically the final part

Obviously this isn't as clean as taking the curl, but it's possibly the method you're intended to use
 
  • #9
for R3
you need to check all
x: d/dy , d/dz
y: d/dx, d/dz
z: d/dx, d/dy
 
  • #10
To say that <P(x,y), Q(x,y), R(x,y)> is a conservative vector field (that's really physics terminology- I prefer saying that "P(x,y)dx+ Q(x,y)dy+ R(x,y)dz is an exact differential) means that there exist some function F(x,y,z) such that
[tex]dF= \frac{\partial F}{\partial x}dx+ \frac{\partial F}{\partial y}dy+ \frac{\partial F}{\partial z}dz= P(x,y)dx+ Q(x,y)dy+ R(x,y)dz[/tex]

IF that is true, then look at the mixed second derivatives:
[tex]\frac{\partial^2F}{\partial x\partial y}= \frac{\partial Q}{\partial x}[/tex]
must be equal to
[tex]\frac{\partial^2 F}{\partial y\partial x}= \frac{\partial P}{\partial y}[/tex]

[tex]\frac{\partial^2 F}{\partial z\partial x}= \frac{\partial P}{\partial z}[/tex]
must be equal to
[tex]\frac{\partial^2 F}{\partial x\partial z}= \frac{\partial R}{\partial x}[/tex]

and
[tex]\frac{\partial^2 F}{\partial y\partial z}= \frac{\partial R}{\partial y}[/tex]
must be equal to
[tex]\frac{\partial^2 F}{\partial z\partial y}= \frac{\partial Q}{\partial z}[/tex].

To be sure the field is "conservative" you must check all three of those equations.
 

1. What is a conservative vector field?

A conservative vector field is a type of vector field in which the line integral of the vector field over any closed path is equal to zero. This means that the work done by the vector field on any particle moving along a closed path is equal to zero, regardless of the path taken.

2. How can I determine if a vector field is conservative or not?

To determine if a vector field is conservative, you can use the fundamental theorem of calculus. If the vector field can be written as the gradient of a scalar function, then it is conservative. Alternatively, you can also calculate the line integral of the vector field over a closed path and see if it equals zero.

3. What are the applications of conservative vector fields?

Conservative vector fields have many applications in physics, engineering, and mathematics. They are used in the study of fluid dynamics, electromagnetism, and gravity. In engineering, they are used in the design of machines and structures, as well as in optimization problems.

4. What is the significance of conservative vector fields?

The significance of conservative vector fields lies in the fact that they represent a type of force field that conserves energy. This means that the work done by the force is path-independent and the total energy of the system is conserved. This makes them useful in many real-world applications and also simplifies mathematical calculations.

5. Can a vector field be partially conservative?

Yes, a vector field can be partially conservative. This means that the vector field is conservative in some regions, but not in others. In such cases, the vector field can be decomposed into a conservative part and a non-conservative part. The conservative part can be calculated using the methods mentioned in the answer to question 2.

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