How to check if a given force is conservative or not ? Is there any mathematical way of doing the same?

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    $\begingroup$ This is a standard result from vector calculus written out in a long-hand format using only symbols understood by students who have calculus but not the multi-dimensional infrastructure. $\endgroup$ – dmckee --- ex-moderator kitten Dec 5 '19 at 16:03
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/31672/2451 and links therein. $\endgroup$ – Qmechanic Dec 8 '19 at 13:08
  • $\begingroup$ Well because i am in class 11 and just know basic maths so i cannot understand the 2nd answer and the first answers is very less mathematical and more theoretical $\endgroup$ – Tushar Dec 8 '19 at 13:31

Definition of Conservative Force :- The work done by the force doesn’t depend on the path taken but only on endpoints.

Consider the image at bottom :-

If the work done (electrical work, gravitational work etc.) in moving something from the point A to point B through the path shown (purple path) is represented as $$ W_{A~to~B} = \int_{A}^{B} \mathbf{F} \cdot d\mathbf{l}$$

And work done in moving the same thing from B to A through some other path (shown by red in figure) is represented as $$ W_{B~to~A} = \int_{B}^{A} \mathbf{F} \cdot d\mathbf{l} $$.

If we consider $\mathbf{F}$ as a conservative force then : $$ W_{A~to~B} = - W_{B~to~A} $$

(As you have said that your in 11th standard therefore I tried to keep things as simple as possible but here I have to make a little point: F is being considered a field going from A to B)

$$ W_{A~to~B} + W_{B~to~A} =0$$ $$ \int_{A}^{B} \mathbf{F} \cdot d\mathbf{l} + \int_{B}^{A} \mathbf{F} \cdot d\mathbf{l} = 0$$

$$ \oint \mathbf{F} \cdot d\mathbf{l} = 0$$

And this is precisely the mathematical definition of Conervative Force , i.e. a force field (I cannot wrote this without using the word field) is conservative if it’s line integral ( the integral the dot product of force field with the element of path) around any closed loop is zero.

However, it is more useful or customary to write it as $$ \nabla \times \mathbf{F} = 0$$ as a necessary condition for F to be a conservative force field. This form can be derived using Stokes Law which you will learn in your Vector Calculus courses.

P.S. : Field is just a thing which is defined at every point in space. Whenever I said force field I meant that force is caused due to a field.

Hope it helps.

enter image description here


This is the curl of the force.
Via the Helmholtz-decomposition theorem (which you should learn about in Electrodynamics), if the curl of a vector field is zero, then this field possesses a scalar potential. Scalar potential fields are conservative force fields.

  • $\begingroup$ Can you simplify it to me, because i am in xi th standard and electrodynamics is not in our course ? $\endgroup$ – Tushar Dec 5 '19 at 16:17
  • $\begingroup$ @Tushar refer it up in Feynman lectures volume 2, there is clear explanation $\endgroup$ – SK Dash Dec 5 '19 at 17:07
  • $\begingroup$ @Tushar Not sure how to make it simpler than this... Have a look at Knights answer, this one goes more in-depth $\endgroup$ – AtmosphericPrisonEscape Dec 8 '19 at 14:45

Let the given force be $F = F_x \hat{i} + F_y \hat{j} + F_z\hat{k}$ , where $F_x$ ,$F_y$ and $F_z$ are functions of any of these three variables $x$, $y$, $z$ .

For this force to be conservative, it must satisfy these conditions:

1) $ \frac{\partial F_x}{\partial y} = \frac{\partial F_y}{\partial x}$

2) $ \frac{\partial F_x}{\partial z} = \frac{\partial F_z}{\partial x}$

3) $ \frac{\partial F_y}{\partial z} = \frac{\partial F_z}{\partial y}$

If these are satisfied then the given force is conservative.

PROOF: Do let me know if I'm wrong somewhere. Let $\overrightarrow{F}$ = $(F_x , F_y, F_z)$ and $\overrightarrow{dr}$ = $(dx, dy, dz)$ . Using $W = \int\overrightarrow{F} . \overrightarrow{dr}$, we get $W = \int (F_xdx + F_ydy + F_zdz)$ .

Let's say we integrate this from some point Q to some point P. This means that if the force is conservative, then what matters to us only the final and the initial position. So, the value for the above integral will equate to some function so,

$W = \int (F_xdx + F_ydy + F_zdz) = C(x_P, y_P, z_P) - C(x_Q, y_Q, z_Q)$ where $C(x, y, z)$ is that some function that satisfies the integral.

Generally, $x=x(t), y=y(t)$ and $z=z(t)$, that is, they're a function of time.

We can write $dx = \frac{dx(t)}{dt}dt$, $dy = \frac{dy(t)}{dt}dt$, $dz = \frac{dz(t)}{dt}dt$.

Now, we can write $$W = \int (F_x\frac{dx(t)}{dt}dt + F_y\frac{dy(t)}{dt}dt + F_z\frac{dz(t)}{dt}dt) = C(x_P, y_P, z_P) - C(x_Q, y_Q, z_Q)$$

Now if, $F_x = \frac{\partial{C}}{\partial{x}}$, $F_y = \frac{\partial{C}}{\partial{y}}$ and $F_z = \frac{\partial{C}}{\partial{z}}$ then,

$$W = \int (\frac{\partial{C}}{\partial{x}}\frac{dx(t)}{dt}dt + \frac{\partial{C}}{\partial{y}}\frac{dy(t)}{dt}dt + \frac{\partial{C}}{\partial{z}}\frac{dz(t)}{dt}dt) = C(x_P, y_P, z_P) - C(x_Q, y_Q, z_Q)$$

Using the chain rule for partial derivatives, we get:

$$\int\frac{dC}{dt}dt = \int{dC} = C(x_P, y_P, z_P) - C(x_Q, y_Q, z_Q)$$

Do not forget that I'm integrating the function from the point Q to P.

As our "if" assumption works, this means that if a force is conservative then it must satisfy $F_x = \frac{\partial{C}}{\partial{x}}$, $F_y = \frac{\partial{C}}{\partial{y}}$ and $F_z = \frac{\partial{C}}{\partial{z}}$ .

Using that, we can write ${F_x}{\partial{x}} = \partial{C} = {F_y}{\partial{y}}$ . From here, we get $$ \frac{\partial F_x}{\partial y} = \frac{\partial F_y}{\partial x}$$.

Similarly, we can also show $ \frac{\partial F_x}{\partial z} = \frac{\partial F_z}{\partial x}$ and $ \frac{\partial F_y}{\partial z} = \frac{\partial F_z}{\partial y}$ .

  • $\begingroup$ Well my teacher ( on youtube ) had also told the same method but do you have any proof for it because i cant find it on google. $\endgroup$ – Tushar Dec 8 '19 at 13:55
  • $\begingroup$ I guess you're referring to Physics Wallah, isn't it? I first learnt about this result from him (I'm in 12th grade). Anyway, I've edited the answer to incorporate the proof as well. I hope you understand! $\endgroup$ – Pratham Hullamballi Dec 8 '19 at 16:02
  • $\begingroup$ You are absolutely correct $\endgroup$ – Tushar Dec 9 '19 at 1:05
  • $\begingroup$ Well i cannot understant your proof, i think i might have to wait till class 12. Well thanks for your effort. $\endgroup$ – Tushar Dec 9 '19 at 1:08

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