Conservative Forces
A conservative force is a force whose work done is independent of the path taken and depends only on the initial and final positions.
Conservative forces are an important aspect of physics. Many forces of nature are conservative like gravitational force, electrostatic force, magnetic force, and elastic force (spring's force).
Before reading this page, make sure you have read Work-Kinetic Energy Theorem and Kinetic Energy. It would be good if you are familiar with potential energy.
Contents
Introduction
As the name suggests, conservative forces hold to the law of conservation of energy. As we saw, one of the most basic understandings of conservation of energy is the conservation of kinetic energy and it gives us an insight as to what actually is meant by the term "conservation." Let's recall what the theorem said:
$\Delta \textrm{KE}=\int\textbf{F}\cdot d\textbf{r}.$
A conservative force may be defined as one for which the work done in moving between two points $A$ and $B$ is independent of the path taken between the two points. The implication of "conservative" in this context is that you could move it from $A$ to $B$ by one path and return to $A$ by another path with no net loss of energy—any closed return path to $A$ takes net zero work.^{[1]} Thus, we define it as follows:
A conservative force is a force whose work done is independent of the path taken and depends only on the initial and final positions.
Path independence of conservative forces:
As the definition says, the conservative forces are path-independent, which is what makes them unique. It is because of this property that the work done by a particle moving in a circle is zero that we will be dealing with the loop property of the conservative forces later. For now, let us see what the definition of path-independency is;
Let $\vec f$ be a vector defined on an open region $D$ in an $n$-dimensional space, and suppose that for two points $A$ and $B$ the line integral $\int_{C}\vec f\cdot dr$ along a path $C$ from $A$ to $B$ in $D$ is the same over all the paths from $A$ to $B.$ Then: $\int_{A}^{B} \vec f\cdot dr$ is path-independent in $D$ and the field $\vec f$ is conservative in $D$. ^{[2]}
Now that we have seen conservative forces are path-independent, let us briefly investigate the cases of gravitational potential energy and the case of friction and get to know the outline behind conservative forces.
Gravitational potential energy
We know that the gravitational force^{[3]} is given by the expression $\vec F=-mg$. Let us now see what happens if we throw a ball in two different paths but with the same start and end points. Let us first throw a ball and find the work done by gravity when the ball reaches the height $H$.
For the next case, let us say we have thrown the same ball with a higher velocity such that the ball passes the height $H$, travels further, reaches a height of $L$, then comes back to $H$, and finally reaches the ground. Thus, we have two different paths in which the same ball has reached the height $H$. Let us compute the work done in each case and see the results:
$\begin{aligned} \text{Case 1: Work Done} = \ \ W_1 &= \int_{0}^{H} -mg \, dx\\ &= -mgH\\\\ \text{Case 2: Work Done} = \ \ W_2 &= \int_{0}^{L} -mg \, dx - \int_{L}^{H} -mg \, dx \\ &= -mgL + mgL - mgH \\ &= -mgH. \end{aligned}$
We observe that $W_1=W_2$. Thus, according to the definition of the conservative forces, we can say that the gravitational potential energy is a 'conservative force' because it is path-independent.
Friction as a non-conservative force
Friction, in contrast to gravitational potential energy, is non-conservative in nature and thus it doesn't just depend upon the start and end points. We will explore now as to the reason behind this. So, let us first consider a body which is sliding from a point $A$ to $B$ under the influence of a frictional force $\vec f$. Let us again try to find two different paths so that we can verify whether the work done is the same in both cases.
Firstly, let us say we slide the body from $0$ to a point $X$ with a constant speed. Next, we move the body at a different, constant speed from $0$ to $2X$ and then back to $X$. Again, as we did in the previous case, let us compute the work done in each case. Then
$\begin{aligned} \text{Case 1: Work Done} = \ \ W_1 &= - \int_{0}^{X} f \, dx\\ &= -fX\\\\ \text{Case 2: Work Done} = \ \ W_2 &= \int_{0}^{2X} -f \, dx - \int_{2X}^{X} -f \, dx \\ &= -2fX + fX - 2fX \\ &= -3fX. \end{aligned}$
Clearly $W_1\neq W_2$. This means friction is not path-independent and thus it is not conservative.
Though checking whether a force is path-independent or not may seem a good idea to check whether a force is conservative, things may get messier for complicated ones. Thus we need to develop some other simpler methods to identify these types of forces, which is what will be achieved in the next section.
Identifying Conservative Forces
Sravanth and Julian
Until now, we have a brief understanding of conservative force; now we will try to find some methods to determine whether a force is conservative. In this section, we will try to understand this, with the help of four different methods. To understand this, you must be familiar with Stokes' theorem, partial derivatives, and Green's functions in physics. Let's see how we can Identify a conservative force:
Method 1: Test different paths keeping endpoints the same
This method is nothing but checking whether the field is path-independent or not. According to this method, if we prove that work done does not change with the path taken, the force is conservative. This method doesn't require a proof as the method itself is what the core of the conservative forces is. So we just need to ensure that
$\displaystyle \int_{C_1} {\overrightarrow {f} \cdot \overrightarrow {dr}} = \int_{C_2} {\overrightarrow {f} \cdot \overrightarrow {dr}}.$
The above image specifies 2 different paths $C_1$ and $C_2$. Our vector field is conservative if and only if the above expression holds good. It simply shows that the work done doesn't change no matter what path you take as long as you start and end at the same point.
Method 2: Loop property of conservative forces
Let us say we have a closed loop/path, with the start and end points being $A$ and $B$, so according to this method, if the work done in moving from $A$ to $B$ and back to $A$ from $B$ is zero, then the force is conservative. Mathematically it says
$\displaystyle \oint_{\color{#D61F06}{\text{Any Closed Path}}} {\overrightarrow {f} \cdot \overrightarrow {dr}} = 0 \quad \forall \overrightarrow{f} \in \color{#3D99F6}{\text{ Conservative Fields}}.$
Advantage of 2 over 1: We can't be sure in 1 that $\overrightarrow{f}$ is conservative even if it follows 1 as the path or points we have taken may be a special case when it obeys the condition. Now, let's see the proof of the method 2.
Let us say we have two points $A$ and $B$ which are simply connected (see figure) on the loop $C$. Let us say the path from $A$ to $B$ is $C_1$ and the path from $B$ to $A$ is $C_2$. But if we reverse the direction of the path $C_2$ from $A$ to $B$, its sign changes to become $-C_2$, so we have
$\begin{aligned} \oint_{C} \vec{f} \cdot dr &= \int_{C_1} \vec{f} \cdot dr + \int_{C_2} \vec{f} \cdot dr\\ &= \int_{A}^{B} \vec{f} \cdot dr - \int_{A}^{B} \vec{f} \cdot dr\\ &= 0.\ _\square \end{aligned}$
Method 3: Every conservative force can be expressed as $\vec f = \nabla F.$
According to this method, if a vector field $\vec f$ can be expressed as $\vec f = \nabla F$ for a differentiable function $F$, then it is conservative. In other words, if $\vec f =\nabla F,$ then the value of the line integral $\int_C \vec f \vec{dr}$ is path-independent. More formally, it can be defined as follows:
If $\mathbf f=M\mathbf i + N\mathbf j + P\mathbf k$ is a vector field whose three components are continuous in an open region $D$ in space, then if $\vec{f}$ can be expressed as
$\overrightarrow{f} = \nabla F \text{ where F is a } {\color{#D61F06}{\text{scalar}}},$
then $\vec{f}$ is a conservative force.
Let us see how this can be proved:
Let $\mathbf f=M\mathbf i + N\mathbf j + P\mathbf k$. Now, assume that we have $2$ points in space $A$ and $B$, and if $\overrightarrow f$ is a gradient field then
$\int_{C} \overrightarrow f\cdot dr = F(B) - F(A).$
Let us say the path is a smooth curve $C$, and let us also say we have a point $B_0$ with coordinates $(x_0,y,z)$ near $B$ with coordinates $(x,y,z)$, which are connected by a line segment $L$. Also, let us say the path from $A$ to $B_0$ is another curve $C_0$.
Then if we have to travel from $A$ to $B,$ we have to traverse the path from $A$ to $B_0$ and then from $B_0$ to $B$, or in short we have to travel along $C_0$ and then $L$. Thus,
$F(x,y,z) = \int_{C_0} \overrightarrow f\cdot dr + \int_{L} \overrightarrow f\cdot dr.$
Differentiating this, we get
$\frac{\partial f}{\partial x} F(x,y,z) = \frac{\partial f}{\partial x} \int_{C_0} \overrightarrow f\cdot dr + \frac{\partial f}{\partial x}\int_{L} \overrightarrow f\cdot dr.$
As it is the second term which depends on $x$, we arrive at
$\frac{\partial}{\partial x} F(x,y,z) = \frac{\partial f}{\partial x}\int_{L} \overrightarrow f\cdot dr.$
We can parametrize the path $L$ as $\mathbf r(t) = t \mathbf i + y\mathbf j + z\mathbf k,$ where the value of $t$ is $x_0\leq t \leq x$. Thus we have $\frac{d\mathbf r}{dt} = \mathbf i, \frac{\mathbf f\cdot d\mathbf r}{dt} = M,$ and $\int_L \mathbf f\cdot d\mathbf r = \int_{x_0}^x M(t,y,z)\, dt.$ Substituting these in the above integral gives
$\frac{\partial}{\partial x} F(x,y,z) = \frac{\partial f}{\partial x}\int_{x_0}^x M(t,y,z)\, dt = M(x,y,z).$
We can do the same for the other two partial derivatives $\frac{\partial f}{\partial z} =Pz$ and $\frac{\partial f}{\partial y} = N,$ concluding that
$\mathbf F = \nabla f.\ _\square$
Here's another way to prove it: click here.
Method 4: Curl of a conservative field is 0.
The curl of a vector field say $\vec f =M\mathbf i + N\mathbf j + P\mathbf k$ is defined as
$\nabla\times \vec f = \left(\dfrac{\partial P}{\partial y} - \dfrac{\partial N}{\partial z}\right)\mathbf i + \left(\dfrac{\partial M}{\partial z} - \dfrac{\partial P}{\partial x}\right)\mathbf j + \left(\dfrac{\partial N}{\partial x} - \dfrac{\partial M}{\partial y}\right)\mathbf k.$
And according to this method, if the curl of a field is zero, then it is conservative. A formal statement is given below:
If $\nabla\times \vec f = 0$ at each and every point in a simply connected region of $D$ in space, then
$\oint_C \vec f\cdot d\vec{r}=0\implies \vec f \text{ is conservative}.$
And thus it implies that it is a conservative force over the region of space.
Here's one way to prove it using Stokes' theorem:
We have a theorem from a branch of mathematics called "topology" which states, "every smooth simple closed curve $C$ in a simply connected open region $D$ is the boundary of a smooth two-sided surface $S$ that also lies in $D$."^{[4]}
Thus, using Stokes' theorem, we have
$\oint_C \vec f d\vec r =\iint_S\nabla \times\vec f \cdot\mathbf n \ d\sigma = 0.$
Gravitational Force
This section doesn't speak anything about the topic, please edit it as soon as possible.
Now, speaking about gravity, it is another conservative force because the work done by it is independent of the path travelled. The work done by it is is calculuated by the formula $W = F × S$ (where $F$ is force and $S$ is displacement).
A bird sitting on the top of a tree flew down to the ground in a semicircular path, such that the final position is exactly below the initial position (or the angle of inclination of the top of the tree to its final position is $0^{\circ}$. If the mass of the bird is $10kg$, its acceleration on downward journey is $5m/s$. It is found that the semicircle formed by the perpendicular line from the initial and final position of the bird and the semicircular path travelled by it has an area of $77m^2$ calculate the work done by it. [Take $\pi$ as $\dfrac {22}{7}$
$\begin{aligned} \text {Given}:-\\ & \text{Mass of the bird} = 10kg\\ & \text{Acceleration} = 5m/s\\ & \text{Area of semircircle formed} = 77m^2 \end{aligned}$
$\begin{aligned} & \text{Area of semicircle} = \dfrac {1}{2} \pi r^2\\ & 77 = \dfrac{1}{2} × \dfrac {22}{7} × r^2\\ & r^2 = \dfrac {77×7×2}{22}\\ & r^2 = {\dfrac {7\cancel{77} × 7 × \cancel {2}}{\cancel{2}\cancel {22}}}\\ & r^2 = 49\\ & r = \sqrt{49}\\ & r = 7m\\ \text {Total distance travelled by bird} & = \text {Diameter of the semi-circle}\\ & = (2 × r)\\ & = (2 × 7)\\ & = 14m \end{aligned}$
$\begin{aligned} \text {Work done by the bird} & = m × a × S\\ & = 10 × 5 × 14\\ & = \boxed {700J} \end{aligned}$
So, from the following example, we see that work done by gravity is independent of the path taken, so gravity is conservative.
Electrostatic Force
To be Copied from electric potential
Magnetic Force
Deepraj
Elastic Force
Rajdeep
Old Content
A Conservative force is a force whose work done is independent of the path taken and depends only on the initial and final position.
In other words, we can say that the Work Done by this force along a closed path is 0.
A particle is moving under the action of force. What would the work done be (zero or non-zero), if the particle returns to it's starting position after some time, if the force acted would be :
- Conservative
- Non - Conservative
This is the answer to the question, with a detailed solution. Please explain your steps so that people can follow along.
Method 1 : The above given image specifies 2 different paths $c_1$ and $c_2$. Our vector field is conservative if and only if $\displaystyle \int_{c_1} {\overrightarrow {f} . \overrightarrow {dr}} = \int_{c_2} {\overrightarrow {f} . \overrightarrow {dr}}$ It simply shows that the work done is doesn't change no matter what path you take as long as you start and end at the same points.
Method 2 : Now, Using the above theory if the start and end points of the path are same the work done should be 0 . If it is the force is conservative if not it is non - conservative. $\displaystyle \oint_{\color{#D61F06}{\text{Any Closed Path}}} {\overrightarrow {f} . \overrightarrow {dr}} = 0 \quad \forall \overrightarrow{f} \in \color{#3D99F6}{\text{ Conservative Fields}}$
Advantage of 2 over 1 : We can't be sure in 1 that $\overrightarrow{f}$ is conservative even if it follows 1 as the path or points we have taken may be a special case when it obeys the condition.
Method 3 :
If $\overrightarrow{f}$ can be expressed as $\overrightarrow{f} = \nabla F \text{ where F is a } \color{#D61F06}{\text{Scalar}}$ then $\overrightarrow{f}$ is a conservative force.
Let $\mathbf f=M\mathbf i + N\mathbf j + P\mathbf k$. Now, assume that we have we have $2$ points in space $A$ and $B$, and if $\overrightarrow f$ is a gradient field then;
$\int_{C} \overrightarrow f\cdot dr = F(B) - F(A)$
Let us say the path is a smooth curve $C$, let us say we have a point $B_0$ with co-ordinates $(x_0,y,z)$ near $B$ with co-ordinates $(x,y,z)$, which are connected by a line segment $L$. Also, let us say the path from $A$ to $B_0$ is another curve $C_0$.
o, if we have to travel from $A$ to be, we have to traverse the path from $A$ to $B_0$ and then from $B_0$ to $B$, or in short, we have to travel along $C_0$ and then $L$. Thus,
$F(x,y,z) = \int_{C_0} \overrightarrow f\cdot dr + \int_{L} \overrightarrow f\cdot dr$
Differentiating this we get:
$\frac{\partial f}{\partial x} F(x,y,z) = \frac{\partial f}{\partial x} \int_{C_0} \overrightarrow f\cdot dr + \frac{\partial f}{\partial x}\int_{L} \overrightarrow f\cdot dr$
As, it is the second term which depends on $x$, we arrive at:
$\frac{\partial}{\partial x} F(x,y,z) = \frac{\partial f}{\partial x}\int_{L} \overrightarrow f\cdot dr$
We can Parametrize the path $L$ as $\mathbf r(t) = t \mathbf i + y\mathbf j + z\mathbf k$ where the value of $t$ is: $x_0\leq t \leq x$. Thus we have, $d\mathbf r/dt = \mathbf i$ ; $\mathbf f\cdot d\mathbf r/dt = M$ and $\int_L \mathbf f\cdot d\mathbf r = \int_{x_0}^x M(t,y,z) dt$, substituting them in the above integral gives:
$\frac{\partial}{\partial x} F(x,y,z) = \frac{\partial f}{\partial x}\int_{x_0}^x M(t,y,z) dt = M(x,y,z)$
We can do the same for the other two partial derivatives $\partial f/\partial z =Pz$ and $\partial f/\partial y = N$ concluding that: $\mathbf F = \nabla f$
Here's another way to prove it: click here.
Let us consider the kinetic energy $F=\frac{1}{2}m\dot{r}\cdot\dot{r}$ , the position $r$ is governed by the second law of Netwon. Precisely, $F(r\dot{r})=\dot{p}$. Friction forces depend on $\dot{r}$and $p=m\dot{r}$ is the momentum of the particle.
Let the mass be constant, hence the change of kinetic energy $\frac{dT}{dt}=\dot{p}\cdot\dot {r}=F\cdot\dot{r}$.If the particle travels from position $r_1$ at time $t_1$ to position $r_2$ at time $t_2$ then the changes of kinetic energy will be $T(t_2)-T(t_1)=\int_{t_1}{t_2}\frac {dT}{dt}\cdot dt=\int_{r_1}{r_2}F\cdot dr$
Conservation force depends only on position $r$, not on the velocity $\dot {r}$. The work done is independent of the path. For a closed path it vanishes $\oint F\cdot r\implies\nabla\times F=0$ Total Kinetic energy of many particles can be written as $T=\frac{1}{2}M\cdot {R}+\frac {1}{2}\sum_i m_i\dot{r^2}$, which splits the kinetic energy into kinetic energy of the center, with an internal energy and the system is moving around the center of mass. For a single particle, the change in total kinetic energy $T(t_2)-T (t_1)=\sum_i\int F_i^{ext}\cdot dr_i+\sum_{i\neq j}F_{ij}\cdot dr_i$
External conservative forces: $F_i^{ext}=-\nabla_i V_i (r_1,\cdots,r_N)$ Internal conservative forces: $F_{ij}=-\nabla V_{ij}(r_1,\cdots,r_N)$ Here $\nabla_{ij}\equiv\frac {\partial}{\partial r_i}$, according to the third law of Newton $F_{ij}=-F_{ji}$ together, this parallel to $r_i-r_j$. This law holds for both Gravitational and Coulomb forces. Center of mass $R=\frac {1}{M}\sum_{i=1}^N m_i x_i$
Gravitational force of a moving particle that a moving particle of mass $m$ experiences the field $F=-Gm\sum_i\frac {M_i}{|r-r_i|}(r-r_i)$. If we fix the particles with the mass$M_i$ at position $r_i$.
References
- Hyperphysics.phy-astr.gsu.edu, U. Potential energy. Retrieved from http://hyperphysics.phy-astr.gsu.edu/hbase/pegrav.html#cfo
- George B. Thomas Jr., U. Thomas' Calculus. Pearson: Indian Subcontinent adaptation 2014.
- Image from Wikipedia under the Creative Commons attribution for reuse and modification, U. Wikipedia.org. Retrieved from https://commons.m.wikimedia.org/wiki/File:Conservative_Force_Gravity_Example.svg#mw-jump-to-license
- George B. Thomas Jr., U. Thomas' Calculus. Pearson: Indian Subcontinent adaptation 2014.