How to prove that a force is conservative?

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  • #1
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Homework Statement:
A particle of mass m moves in a circular path of radius R in the plane z=0 described by position vector r( t ) = ( Rcos(wt) , Rsin(wt) , 0 ). The particle's motion results from a force ; prove this force is conservative
Relevant Equations:
Newton's 2nd law and for a conservative force , curl F = 0
I have done this question and also seen the model answer but i just want to check my method is ok. I differentiated the position vector twice and multiplied by m to get the force F. I then used the determinant method to find the curl of F which i found to be zero thus proving the force is conservative. What i want to check is this ; i wrote the components of F as functions of time such as Fx = -mRω2cos(ωt) and this method worked . The model solution writes the components of F such as Fx = -mω2x.
Is my method ok ? Is it ok to write the components as functions of time even though the differential operators in curl are derivatives of x , y and z ?
Thanks
 

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  • #2
Orodruin
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You cannot possibly find the curl of the force since you are in essence only given how the force acts on a single path. If this is the exact formulation of the problem then whoever wrote the question needs to go back to the drawing board. What is true is that the force does no work on the particle. This is not the same as a force field being conservative. Nor is the fact that the particle goes in a circle enough to deduce that the force is conservative. (A conservative force would result in work done when moving around any closed path being zero. Having a single example of such a path is not sufficient.)
 
  • #3
Orodruin
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Homework Statement:: A particle of mass m moves in a circular path of radius R in the plane z=0 described by position vector r( t ) = ( Rcos(wt) , Rsin(wt) , 0 ). The particle's motion results from a force ; prove this force is conservative
Relevant Equations:: Newton's 2nd law and for a conservative force , curl F = 0

The model solution writes the components of F such as Fx = -mω2x.
To be more specific: This is only one of many force fields that would lead to the given force along the circle of radius R. This force field is conservative but others need not be. In order to conclude that this is the force field you would need to know that any radius R would give the same expression for the motion
 
  • #4
dyn
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Is it ok to take the curl of the force using the force components as a function of time ?
 
  • #5
Orodruin
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Is it ok to take the curl of the force using the force components as a function of time ?
No. And again, you need to know the field everywhere to take the curl.
 
  • #6
dyn
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The particle is constrained to move in a circle so it can only exist on the circle.

Assuming i can take the curl if Fx = -mω2R cos(ωt) = -mω2x are you saying that curls have to be written in terms of position coordinates not time even though in this case i get the same answer with both methods ?
 
  • #7
Orodruin
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The particle is constrained to move in a circle so it can only exist on the circle.
Which, as I said in the first posts, makes it impossible to know the curl of the field. You need to know the force field everywhere.

Assuming i can take the curl if Fx = -mω2R cos(ωt) = -mω2x are you saying that curls have to be written in terms of position coordinates not time even though in this case i get the same answer with both methods ?
Again, the problem is ill posed. Any answers are irrelevant.
 
  • #8
Orodruin
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But to the more general point, assuming a force field F(x) (consider F and x vector quantities). The equations of motion are then m x’’(t) = F(x(t)), which will make it possible to deduce x(t) given initial conditions. If you only have one solution to these equations then you can only reconstruct the force field along the path. You can say nothing about the field elsewhere.

You can certainly not take the curl of F(x(t)) because it is now a function of t which hides the spatial dependence of the force field.
 
  • #9
dyn
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It seems that my way giving the correct answer was just a coincidence then.
Thanks for your help
 
  • #10
Orodruin
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It seems that my way giving the correct answer was just a coincidence then.
Thanks for your help
There is no correct answer. The question is ill posed. I do not know how many times I can repeat this.
 
  • #11
dyn
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I was just trying to answer the question as it was posed. Thanks for your help
 
  • #12
Orodruin
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I was just trying to answer the question as it was posed. Thanks for your help
Which is futile, since it ill posed.
 
  • #13
dyn
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Which, as I said in the first posts, makes it impossible to know the curl of the field. You need to know the force field everywhere.


Again, the problem is ill posed. Any answers are irrelevant.
On a related topic , what about the spring force ? If the force on a particle is given by F = -kx and the particle is constrained to move in the x-direction is that force conservative ? This force field is not known everywhere but the spring force is considered to be conservative
 
  • #14
valenumr
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I guess you could start with the definition here

And then show that the average work done is zero?
 
  • #15
Office_Shredder
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On a related topic , what about the spring force ? If the force on a particle is given by F = -kx and the particle is constrained to move in the x-direction is that force conservative ? This force field is not known everywhere but the spring force is considered to be conservative

The spring force describes a force that only depends on the position of the particle. The force here might depend on both the position and the velocity (in fact, to keep it moving in a circle it probably depends on the velocity), along with a bunch of other things that we don't know about.
 
  • #16
Orodruin
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On a related topic , what about the spring force ? If the force on a particle is given by F = -kx and the particle is constrained to move in the x-direction is that force conservative ? This force field is not known everywhere but the spring force is considered to be conservative
It is known everywhere within its domain. The motion is one-dimensional.
 
  • #17
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But for motion constrained to a circle the force field is known everywhere within that domain which can be considered one-dimensional. I don't understand why the spring force is conservative but the force constraining the particle to move in a circle is not
 
  • #18
Orodruin
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But for motion constrained to a circle the force field is known everywhere within that domain which can be considered one-dimensional. I don't understand why the spring force is conservative but the force constraining the particle to move in a circle is not
Your case is not constrained to move on a circle. The force field keeps the particle moving in a circle.
 
  • #19
valenumr
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Can I raise my hand and ask a question?

I'm rusty in some ways on basic physics, but is a conservative force something that basically ensures the work-energy relation?
 
  • #20
Office_Shredder
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Can I raise my hand and ask a question?

I'm rusty in some ways on basic physics, but is a conservative force something that basically ensures the work-energy relation?

Yes.

But for motion constrained to a circle the force field is known everywhere within that domain which can be considered one-dimensional. I don't understand why the spring force is conservative but the force constraining the particle to move in a circle is not

No, a better analogy would be if you were told that a particle happened to be oscillating back and forth in a harmonic fashion, and then asked if the force acting on it was conservative. It could be a spring force acting on the particle, it could be a pendulum, or it could be something totally different that you haven't thought of.
 
  • #21
jbriggs444
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I'm rusty in some ways on basic physics, but is a conservative force something that basically ensures the work-energy relation?
The work energy theorem always works. You do not need a conservative force to make it so. A proper definition of a "conservative" force needs some background.

[Picking a Google'd reference at random: https://www.iitrpr.ac.in/MA101/MA101-Lecturenotes(2019-20)-Module 13.pdf]

A "field" is an assignment of a value to every point in a space. If the assigned value for each point is a scalar then you have a scalar field. If the assigned value for each point is a vector then you have a vector field.

For instance, if we consider the temperature for every point on the surface of the Earth, that would be a "scalar field". Every point has a temperature and that temperature is a scalar. Similarly for the altitude of each point in a park. That would be a "scalar field."

Or we could consider the acceleration of gravity in the space surrounding the Earth. Gravity has a magnitude and a direction at each point in the space, so that would be a "vector field".

Vector fields can be "conservative". The notion does not apply to scalar fields. There are [at least] two ways from here to define what it means for a vector field to be conservative.


The first definition for "conservative" uses a concept known as the "gradient". Suppose you have a scalar field. Like the altitude of every point in a park. From this scalar field, you construct a vector field. At every point in the park you come up with a vector. The direction of the vector is the direction that points downhill. The magnitude of the vector is how steep the hill slopes in that direction. [I can never remember whether the gradient points uphill or down -- it does not really matter until you need to get the sign conventions right].

This notion of "gradient" is very much like the notion of a regular derivative. We're doing multi-variable calculus!

Definition: A vector field is said to be "conservative" if it is the gradient of some scalar field. That scalar field is the "potential" associated with the vector field.


The other [equivalent] definition uses the concept known as a "path integral". Say you have a vector field ##\vec{F(x)}## defined for every point in your space. You add up ##F(x) \cdot ds## for every incremental interval ##ds## along the path. Basically you are computing the work done along the path.

Definition: If the work done around every closed loop is zero then the vector field is said to be conservative.
 
  • #22
vela
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But for motion constrained to a circle the force field is known everywhere within that domain which can be considered one-dimensional. I don't understand why the spring force is conservative but the force constraining the particle to move in a circle is not
To elaborate on what @Orodruin said, the gravitational force admits circular orbits as a solution, but in general, the trajectory of a body in the field isn't circular. The force field doesn't constrain the motion of a body to a circle.

Suppose you have a non-conservative force field where on the circle of radius R the force just happens to coincide with what Newton's law of gravitation would give but elsewhere the force fields differ. From just the information about what the force is on the circle, there's no way to tell the difference between the non-conservative field and the gravitational field, so you can't draw any conclusions.
 
  • #23
dyn
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To elaborate on what @Orodruin said, the gravitational force admits circular orbits as a solution, but in general, the trajectory of a body in the field isn't circular. The force field doesn't constrain the motion of a body to a circle.

Suppose you have a non-conservative force field where on the circle of radius R the force just happens to coincide with what Newton's law of gravitation would give but elsewhere the force fields differ. From just the information about what the force is on the circle, there's no way to tell the difference between the non-conservative field and the gravitational field, so you can't draw any conclusions.
Yes , that makes sense, But what about the case of spring ? We know the force on the x-axis ; F = -kx but we don't know the force anywhere else but the spring force is considered conservative
 
  • #24
vela
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The spring stretches or compresses only along its length, so the potential energy function only depends on one variable.
 
  • #25
valenumr
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The work energy theorem always works. You do not need a conservative force to make it so. A proper definition of a "conservative" force needs some background.

[Picking a Google'd reference at random: https://www.iitrpr.ac.in/MA101/MA101-Lecturenotes(2019-20)-Module 13.pdf]

A "field" is an assignment of a value to every point in a space. If the assigned value for each point is a scalar then you have a scalar field. If the assigned value for each point is a vector then you have a vector field.

For instance, if we consider the temperature for every point on the surface of the Earth, that would be a "scalar field". Every point has a temperature and that temperature is a scalar. Similarly for the altitude of each point in a park. That would be a "scalar field."

Or we could consider the acceleration of gravity in the space surrounding the Earth. Gravity has a magnitude and a direction at each point in the space, so that would be a "vector field".

Vector fields can be "conservative". The notion does not apply to scalar fields. There are [at least] two ways from here to define what it means for a vector field to be conservative.


The first definition for "conservative" uses a concept known as the "gradient". Suppose you have a scalar field. Like the altitude of every point in a park. From this scalar field, you construct a vector field. At every point in the park you come up with a vector. The direction of the vector is the direction that points downhill. The magnitude of the vector is how steep the hill slopes in that direction. [I can never remember whether the gradient points uphill or down -- it does not really matter until you need to get the sign conventions right].

This notion of "gradient" is very much like the notion of a regular derivative. We're doing multi-variable calculus!

Definition: A vector field is said to be "conservative" if it is the gradient of some scalar field. That scalar field is the "potential" associated with the vector field.


The other [equivalent] definition uses the concept known as a "path integral". Say you have a vector field ##\vec{F(x)}## defined for every point in your space. You add up ##F(x) \cdot ds## for every incremental interval ##ds## along the path. Basically you are computing the work done along the path.

Definition: If the work done around every closed loop is zero then the vector field is said to be conservative.
Thanks for the thorough response. I suppose I was thinking that what I considered "non-conservative forces" would be things that invoke thermodynamic type "errors" from the ideal mechanics of things.

So I do then wonder... Is there such a thing, truly, as a non-conservative force in reality?
 
  • #26
dyn
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No. And again, you need to know the field everywhere to take the curl.

Yes , that makes sense, But what about the case of spring ? We know the force on the x-axis ; F = -kx but we don't know the force anywhere else but the spring force is considered conservative
The top quote states that we need to know the field everywhere to take the curl to find if a force is conservative. For a spring we only know the force on the x-axis so we don't know the field everywhere thus we can't take the curl
 
  • #27
Office_Shredder
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Dyn, I think you're missing a more key point here. Even restricted to just stuff on the circle, can you tell me given a particle at a position, what the force on the particle is?
 
  • #28
Orodruin
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For a spring we only know the force on the x-axis so we don't know the field everywhere thus we can't take the curl
Again, the motion is one-dimensional. Curl does not exist in one dimension. The constraint to one dimension is necessarily by other forces that are not being considered.
 
  • #29
Orodruin
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The work energy theorem always works. You do not need a conservative force to make it so. A proper definition of a "conservative" force needs some background.

[Picking a Google'd reference at random: https://www.iitrpr.ac.in/MA101/MA101-Lecturenotes(2019-20)-Module 13.pdf]

A "field" is an assignment of a value to every point in a space. If the assigned value for each point is a scalar then you have a scalar field. If the assigned value for each point is a vector then you have a vector field.

For instance, if we consider the temperature for every point on the surface of the Earth, that would be a "scalar field". Every point has a temperature and that temperature is a scalar. Similarly for the altitude of each point in a park. That would be a "scalar field."

Or we could consider the acceleration of gravity in the space surrounding the Earth. Gravity has a magnitude and a direction at each point in the space, so that would be a "vector field".

Vector fields can be "conservative". The notion does not apply to scalar fields. There are [at least] two ways from here to define what it means for a vector field to be conservative.


The first definition for "conservative" uses a concept known as the "gradient". Suppose you have a scalar field. Like the altitude of every point in a park. From this scalar field, you construct a vector field. At every point in the park you come up with a vector. The direction of the vector is the direction that points downhill. The magnitude of the vector is how steep the hill slopes in that direction. [I can never remember whether the gradient points uphill or down -- it does not really matter until you need to get the sign conventions right].

This notion of "gradient" is very much like the notion of a regular derivative. We're doing multi-variable calculus!

Definition: A vector field is said to be "conservative" if it is the gradient of some scalar field. That scalar field is the "potential" associated with the vector field.


The other [equivalent] definition uses the concept known as a "path integral". Say you have a vector field ##\vec{F(x)}## defined for every point in your space. You add up ##F(x) \cdot ds## for every incremental interval ##ds## along the path. Basically you are computing the work done along the path.

Definition: If the work done around every closed loop is zero then the vector field is said to be conservative.
The third equivalent statement is that the curl of the field is zero.
 
  • #30
Office_Shredder
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I realized there's a better way to demonstrate the issue here. A field is conservative if the energy used to move something is independent of the path.

They have only told you a single path that a particle is taken. How can you possibly know what will happen if a particle takes a different path. Let's not worry about how people normally confirm something is conservative, you only have one path, how can you know what the result would be if you moved in a different path?
 
  • #31
dyn
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I realized there's a better way to demonstrate the issue here. A field is conservative if the energy used to move something is independent of the path.

They have only told you a single path that a particle is taken. How can you possibly know what will happen if a particle takes a different path. Let's not worry about how people normally confirm something is conservative, you only have one path, how can you know what the result would be if you moved in a different path?
If a spring force is given by F = -kx is that force conservative ? We only know the force in one direction.
 
  • #32
Steve4Physics
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If a spring force is given by F = -kx is that force conservative ? We only know the force in one direction.
Hi. Hope no one minds if I chip in.

Yes. 'F=-kx' describes a conservative force.

The simple test is to show that no net work is done moving from some arbitrary extension, x=a, to a different arbitrary extension, x= b, and then returning from x=b to x=a. I.e. show the work done going round any arbitrary 'loop' is zero.

Edted: In physical terms this means (for b>a say) that the work done stretching an ideal spring equals the increase in its (elastic) potential energy. And when restored, this same amount of energy is 'returned' by the spring doing work.

If you really wanted to use ‘curl’ then you could take ##\vec F = -kx\hat{i} + 0\hat{j} + 0\hat{k}## and work out ##curl(\vec F)## in the usual way. (But then you don't get the bonus of deriving the expression for potential energy.)
 
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  • #33
jbriggs444
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If a spring force is given by F = -kx is that force conservative ? We only know the force in one direction.
If we are working in a one-dimensional space where the spring is aligned then, since the force is aligned in that direction, there would seem to be no problem.

We know the force everywhere within the [sub-]space and the force falls within the [sub-]space. It is unambiguously conservative.

By contrast, for a craft in an orbit subject to a radial force, the force does not fall within the [sub-]space and we have a problem. Do we expand the sub-space to handle the force and cry "Not Necessarily Conservative"? Or do we project the force into the sub-space, wind up with a zero force and cry "A Zero Force is always Conservative".
 
  • #34
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If a spring force is given by F = -kx is that force conservative ? We only know the force in one direction.

As a serious question, do you know what the definition of a path is for the purposes of something being path independent? Because this is not really a counterpoint to what I said.
 
  • #35
dyn
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I have been told that force in #1 is not conservative because we only know the force on the circle which can be considered 1-dimensional and not anywhere else.
Yet i am also told that the spring force is conservative despite only knowing the force in 1-dimension and not anywhere else.
I fail to see the difference between the 2 cases
 

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