Is Mechanical Energy Conserved for Non-Conservative Forces?

[fog37]

Hello,
A generic force $F$, which may be conservative or not, performs mechanical work which is always equal to $$W=\Delta KE=KE_{final}-KE{initial}$$
i.e. produce a change in the object's kinetic energy $KE$. Work is essentially a way to inject or subtract kinetic energy from a system.
If and only the force $F$ happens to be conservative, then we can define a new type of energy called potential energy $PE$, such that the work done by $F_{conservative}= -\Delta_{PE}= PE_{initial}-PE_{final}$.

This means that, for a conservative force, $$KE_{final}-KE_{initial} = PE_{initial}-PE_{final} $$ or equivalently $$PE_{initial}+KE_{initial} = PE_{final}+KE_{final} $$ $$ME_{initial} = ME_{final} $$ where $ME = PE+KE$

So, it appears to me that the term "mechanical energy", since it encompasses potential energy, is only applicable to objects under the influence of conservative forces. Is that correct? If an object is only subject to nonconservative forces, it will only have kinetic energy and not potential energy. And if a system has potential energy, it has it only in virtue of the fact that conservative forces are acting on it, correct?

Conservative force can never be time-dependent which means that potential energy $U(x,y,z)$is only a position-dependent quantity and is never time-dependent. If conservative forces were time-dependent, the mechanical energy of a system would not be conserved. What is the problem with having a time-varying mechanical energy? Is it not a useful concept? Is it just against the definition of conservative forces, i.e. forces that conserve mechanical energy?

 

[kuruman]

$F_{conservative}= -\Delta_{PE}= PE_{initial}-PE_{final}$

A force is never equal to a change in potential energy. A conservative force can be derived from a scalar potential function $U$, $\vec F_{cons.}=-\vec \nabla U$. Conversely, if a force can be derived from a scalar potential function, it is conservative.

So, it appears to me that the term "mechanical energy", since it encompasses potential energy, is only applicable to objects under the influence of conservative forces. Is that correct?

No it is not. You can have non-conservative forces influencing the motion of an object, yet mechanical energy is conserved. For example, mechanical energy is conserved in the motion of a pendulum (ignoring air resistance), yet the tension that provides the centripetal acceleration cannot be derived from a potential and is non-conservative. It is more correct to say that mechanical energy is conserved when an object is under the influence of conservative forces and non-conservative forces that do no work on the object. In other words, as long as non-conservative forces do not transform energy from mechanical to non-mechanical by doing work on the object, then mechanical energy is conserved. It's kinda obvious when phrased in such terms, isn't it?

If an object is only subject to nonconservative forces, it will only have kinetic energy and not potential energy.

More correctly, an object that is only subject to non-conservative forces may have its kinetic energy changed if the non-conservative forces do work on it. It cannot have its potential energy change because there can be no potential energy change when conservative forces are not there.

Conservative force can never be time-dependent ...

Why not? If the potential is time-dependent, then the conservative forces derived from it will also be time dependent. Mechanical energy conservation is not a sacrosanct principle. "Conservative" does not mean "it is the same at initial time t_i and at final time t_f". It means "it is the same at initial point x_i and at final point x_f". The two ideas are often conflated if one thinks of a moving object that at t_i is at x_i and at t_f is at x_f. You know that the work done by a conservative force is zero if the path that the object takes under the influence of this force is a closed loop. This equivalent definition of conservative force makes no mention of time. So even if the object is taken around the closed loop in zero time, the work will be still be zero. Now imagine the Earth's acceleration of gravity be a function of time $g(t)$. The work done by a gravity around a closed loop in zero time will be zero regardless of the instantaneous value of $g$.

 

[fog37]

Thank you kuruman. I agree with almost everything you said.

Your statement "...It is more correct to say that mechanical energy is conserved when an object is under the influence of conservative forces and non-conservative forces that do no work on the object..." is indeed a better way to say what I was trying to say. But if no conservative forces are around, it does not make sense to talk about mechanical energy but solely about kinetic energy since the work done by non-conservative forces does not play any role in potential energy...

"...In other words, as long as non-conservative forces do not transform energy from mechanical to non-mechanical by doing work on the object, then mechanical energy is conserved..." Well, why do you say that non-conservative forces transform energy from mechanical to non mechanical? We may have a non-conservative force like the thrust from a rocket engine that changes the kinetic energy of the rock by doing mechanical work on the rock. So the work done by thrust changes the $KE$ of the rocket changing, hence the mechanical energy $ME$. I don't see the thrust transforming energy from mechanical to non mechanical in that case. It seems that every time non-conservative forces do nonzero work (transforming energy from mechanical to mechanical or from mechanical to non-mechanical), the mechanical energy changes.

As far as conservative forces being time-dependent, I may need assistance from other forum members to prove that, but I am have read and seen demonstrations that conservative forces cannot be functions of time, only function of position. The work along a closed path (which is zero) may change with time and that is against the definition of conservative forces...Maybe someone on the forum can provide more solid arguments about this...

 

[kuruman]

Well, why do you say that non-conservative forces transform energy from mechanical to non mechanical?

Because they do.

We may have a non-conservative force like the thrust from a rocket engine that changes the kinetic energy of the rock by doing mechanical work on the rock.

That is correct. Non-conservative forces add to and subtract from the mechanical energy of a system. The bigger picture in the rocket case is that you have a heat engine which changes mechanical energy through no-conservative forces. The burning fuel changes the kinetic energy of the rocket (mechanical) but also heats up the rocket and the combustion products. "Heating" means an increase of the internal (non-mechanical) energy. I never said or meant to imply that non-conservative forces only do non-conservative work. Sometimes they do zero work as in the case of the pendulum.

As far as conservative forces being time-dependent, I may need assistance from other forum members to prove that, but I am have read and seen demonstrations that conservative forces cannot be functions of time, only function of position.

Here is how I see it and I may have a blind spot. It is a standard exercise in intermediate Classical Mechanics to list the following five statements
1. A force $\vec F$ is conservative
2. $\vec F=-\vec \nabla U$
3. The line integral $\oint \vec F \cdot d\vec l$ is zero
4. The line integral $\int_a^b \vec F \cdot d\vec l$ is independent of the path from $a$ to $b$
5. $\vec \nabla \times \vec F =0$
and then prove that if any one of these statements is true, the remaining four are also true. Sorry for the math, but that is the underpinning of conservative forces. The derivatives involved with the del operator, $\vec \nabla$, are strictly spatial derivatives that leave time alone. Given a reasonably well-behaved potential $U(x,y,z)$ one can take its gradient and get a conservative force from it (statement 1). Specifically, if you look at statement 4 and you calculate the work done by this force from point $a$ to point $b$, the fact remains that this work is path-independent; the work will be the same whether you follow a straight line or a zigzag path. Of course, if you have a time-dependent potential $U(x,y,z,t)$ the value of the integral will vary from one moment to the next and if you are looking at the time evolution of the system you cannot say that the force is conservative because there is no unique potential from which the force can be derived.

 

[fog37]

In regards to the time-dependence/conservative force concept, are you mentioning that it is possible to define a scalar function $U(x,y,z,t)$ and take its negative gradient to obtain a force $$F=-\nabla U(x,y,z,t)$$
But this force $F$ is not conservative and does not satisfy statements 3 and 4 (the integrals become time dependent and mechanical energy $ME(t)$ is a function of time instead of being constant).
I see how non-conservative forces can either do zero work or non-conservative work. But by nature of their name (non-conservative), why would we call the work they do ever conservative work? Just because the thrust changes the $KE$ of the system does not mean that thrust does conservative work in that case...I think...?

 

[kuruman]

It looks like we are posting past each other. Consider $\vec F=axu ~\hat x$ where $a$ is a constant and $u$ is a variable that does not depend on $x$. Also consider the line integral over the closed path from $x=0$ to $x=1$ and back. Then $\oint \vec F \cdot d\vec l=\int_0^1 axu~\hat x \cdot d\vec l+\int_1^0 axu~\hat x \cdot d\vec l=au\int_0^1 x dx+au\int_1^0 x dx=au(\frac{1}{2}-\frac{1}{2})=0$. According to the math, the force is conservative for any value of parameter $u$. Now suppose I identify parameter $u$ as the independent variable for time. $u=t$. Mathematically it makes sense to say that the force is conservative because renaming $u$ does not change the derivation shown above. Now suppose I say that a particle of mass $m$ is placed under the influence of this force and that I move it with my hand at constant speed from the origin to $x=1$ and then back to the origin. Question: How much work do I do on the particle over the round trip? Obviously, the amount of work I do is the negative of the work done by force $\vec F$. How much is that? It depends on when I choose to start, how much time the round trip takes, whether I choose to take a nap once I reach $x=1$, etc. etc. In that sense, the force is non-conservative. Physically a finite amount of time is always needed to move a mass from one point in space to another.

But by nature of their name (non-conservative), why would we call the work they do ever conservative work?

I do not use the term "conservative work". I prefer to call it "the negative of the change in potential energy".

Personally, I see no merits in the concept of mechanical energy conservation. I prefer to invoke the work energy theorem of which conservation of mechanical energy is a special case.

 

[DrStupid]

It is a standard exercise in intermediate Classical Mechanics to list the following five statements
1. A force $\vec F$ is conservative
2. $\vec F=-\vec \nabla U$
3. The line integral $\oint \vec F \cdot d\vec l$ is zero
4. The line integral $\int_a^b \vec F \cdot d\vec l$ is independent of the path from $a$ to $b$
5. $\vec \nabla \times \vec F =0$
and then prove that if any one of these statements is true, the remaining four are also true.

How does that look like for the Lorentz force?

 

[vanhees71]

The Lorentz force is "conservative" in @kuruman 's sense if the electromagnetic field is an electrostatic field, because then and only then the electric field is derivable from a potential and the magnetic field vanishes by definition. The force on the particle is then given by
$$\vec{F}=q \vec{E}=-q \vec{\nabla} \phi.$$
If there's also a magnetic field, you have
$$\vec{F}=q (\vec{E}+\vec{v} \times \vec{B}),$$
which is obviously not conservative in the above sense, but it even depends on the velocity of the particle, so that you never have a chance to derive it from a gradient of a scalar field.

If the $\vec{E}$-field is static, however, for the motion of the particle still the conservation of mechanical energy holds, because the magnetic part of the force doesn't do any work, because $\vec{v} \cdot (\vec{v} \times \vec{B})\equiv 0$.

In this sense the expression "conservative force" is a bit confusing, because of course the theorem of energy conservation also holds if the forces are neither "conservative" (as the example of the additional magnetic field above shows) and even for any electromagnetic system (but of course not for the mechanical energy of the moving charges only but also involving electromagnetic field energy).

 

[DrStupid]

If there's also a magnetic field, you have
$$\vec{F}=q (\vec{E}+\vec{v} \times \vec{B}),$$
which is obviously not conservative in the above sense, but it even depends on the velocity of the particle, so that you never have a chance to derive it from a gradient of a scalar field.

If the $\vec{E}$-field is static, however, for the motion of the particle still the conservation of mechanical energy holds, because the magnetic part of the force doesn't do any work, because $\vec{v} \cdot (\vec{v} \times \vec{B})\equiv 0$.

That's why I asked my question. To my understanding 3. and 4. are true in case of the Lorentz force but 2. and 5. are wrong and 1. depends on the definition of conservative forces. Is the mentioned "standard exercise" some kind of Kobayashi-Maru-Test?

 

[kuruman]

That's why I asked my question. To my understanding 3. and 4. are true in case of the Lorentz force but 2. and 5. are wrong and 1. depends on the definition of conservative forces.

3 and 4 are trivially applicable and not a test whether $q\vec v \times \vec B$ is conservative. Sure, $\vec F \cdot d\vec l=0$, because the two vectors are perpendicular so when you do the integral, you are adding a whole bunch of zeroes to get a net of zero. It's always true that when you add a whole bunch of zeroes you get zero and that does not prove or disprove anything. The idea with 4 is that when you integrate from $a$ to $b$ you add a whole bunch of numbers (some positive, some negative, some zero) and the sum you get does not depend on the path that you follow from $a$ to $b$. The idea with 3 is that you go around any closed loop again adding non-zero numbers, the sum of positive numbers matches the sum of negative numbers exactly so that the net is zero. The positive and negative sums may depend on the path, but one is always the negative of the other.

The force represented by $q\vec v \times \vec B$ is non-conservative. It does zero work on a particle moving in a uniform magnetic field just like the tension does zero work on a bob swinging in a gravitational field.

Is the mentioned "standard exercise" some kind of Kobayashi-Maru-Test?

Nope. Anyone who has taken a decent intermediate course in Classical Mechanics or Vector Calculus is sure to have encountered it (unless absent or totally dazed) and perhaps learned and understood it.

 

[DrStupid]

3 and 4 are trivially applicable and not a test whether $q\vec v \times \vec B$ is conservative

That depends on the definition of conservative forces and it is relevant for condition 1 only.

The idea with 4 is that when you integrate from $a$ to $b$ you add a whole bunch of numbers (some positive, some negative, some zero) and the sum you get does not depend on the path that you follow from $a$ to $b$. The idea with 3 is that you go around any closed loop again adding non-zero numbers, the sum of positive numbers matches the sum of negative numbers exactly so that the net is zero. The positive and negative sums may depend on the path, but one is always the negative of the other.

Where do the additional conditions "some positive, some negative" and "non-zero numbers" come from? In cannot find them in your "standard exercise" above.

The force represented by $q\vec v \times \vec B$ is non-conservative.

That depends on the definition of non-conservative forces and it is relevant for condition 1 only.

Anyone who has taken a decent intermediate course in Classical Mechanics or Vector Calculus is sure to have encountered it (unless absent or totally dazed) and perhaps learned and understood it.

No course in Classical Mechanics or Vector Calculus can help you proving a wrong statement.

 

[kuruman]

That depends on the definition of non-conservative forces and it is relevant for condition 1 only.

It is not relevant to 1 only. If a force is known to be conservative (1 is true), it must satisfy 2-5.

Where do the additional conditions "some positive, some negative" and "non-zero numbers" come from? In cannot find them in your "standard exercise" above.

They are not conditions, they are part of what you have to do when you compute an integral. When you are called upon to integrate, your task is to add a whole bunch of numbers, no? That funny looking symbol ∫ stands for "sum". I'm sure you know that. And what is it that your adding? Whatever the value of the expression to the right of the ∫ has at each value of the integration variable on the interval between the integration limits. Thus, when you integrate, you are adding (continuously) numbers that could be positive, negative or zero depending on where you are along the integration path.

That depends on the definition of non-conservative forces and it is relevant for condition 1 only.

The definition of non-conservative is clear from the name, " A force is non-conservative if any of the statements 1 - 5 is false."

No course in Classical Mechanics or Vector Calculus can help you proving a wrong statement.

Maybe not, but it can help you understand what line integrals are all about and how they are used to test whether a force is conservative. I can see why you might think that $q\vec v \times \vec B$ passes tests 3 and 4. There is a hidden understanding that 3 and 4 are not tests whether the force is conservative in the trivial case when $\vec F \cdot d\vec l=0$ either because $\vec F=0$ or because $d \vec l=0$ or because $\vec F \perp d\vec l$.

 

[DrStupid]

When you are called upon to integrate, your task is to add a whole bunch of numbers, no?

These numbers don't need to be non-zero.

If a force is known to be conservative (1 is true), it must satisfy 2-5.
[...]
The definition of non-conservative is clear from the name, " A force is non-conservative if any of the statements 1 - 5 is false."

That means Lonrentz force is neither conservative nor non-conservative.

By the way: The name "conservative" comes from conservation (of energy). If the definition would be clearly from the name, than Lorentz force would be clearly conservative.

but it can help you understand what line integrals are all about

Until you start doing that, let me tell you what I actually know:

If a curve C (e.g. the circular path of an electron in a homogeneous magnetic field) is given by the scalar functions $x\left( c \right)$ , $y\left( c \right)$ and $z\left( c \right)$ of a parameter c (e.g. the arc length) and a continuous scalar function $f_x \left( {x,y,z} \right)$ (e.g. the x-component of the Lorentz force acting on the electron along the path C) is defined over the path C, than the line integral of this scalar function is given by

$\int\limits_C^{} {f_x \left( {x\left( s \right),y\left( s \right),z\left( s \right)} \right) \cdot ds} = \mathop {\lim }\limits_{\Delta s_i \to 0,n \to \infty } \sum\limits_{i = 1}^n {f_x \left( {x\left( s \right),y\left( s \right),z\left( s \right)} \right) \cdot \Delta s_i }$

if the limit is independent from the individual steps $\Delta s_i$.

Are $f_y \left( {x,y,z} \right)$ and $f_z \left( {x,y,z} \right)$ additional functions defined on C (e.g. the y- and z-component of the Lorentz force acting on the electron along the path C) than the sum

$\int\limits_C^{} {f_x \left( {x\left( s \right),y\left( s \right),z\left( s \right)} \right) \cdot x' \cdot ds} + \int\limits_C^{} {f_y \left( {x\left( s \right),y\left( s \right),z\left( s \right)} \right) \cdot y' \cdot ds} + \int\limits_C^{} {f_z \left( {x\left( s \right),y\left( s \right),z\left( s \right)} \right) \cdot z' \cdot ds}$

is the line integral of a vector function. That is not obvious in this notation, but with $\vec F\left( {x,y,z} \right) = \left( {f_x \left( {x,y,z} \right),f_y \left( {x,y,z} \right),f_z \left( {x,y,z} \right)} \right)^T$ and $d\vec l = \left( {x',y',z'} \right)^T \cdot ds$ it can be written as

$\int\limits_C^{} {\vec F \cdot d\vec l}$

If you do that with the Lorentz force acting on the electron on the circular path you will get zero. You will also get zero for any other path meeting the requirements for the line integral. The only restriction is the continuous functions defined on C. That means C must not intersect itself in a way that results in different values for $\vec F\left( {x,y,z} \right)$ at the same location. Such a path would need to be separated into intervals between these intersections. But that doesn’t matter if $\vec F \cdot d\vec l$ is always zero.

And now I am waiting for your “hidden understanding”.

 

[kuruman]

I have already said all that I have to say on this. Perhaps someone else may wish to add or subtract from it.

 

[vanhees71]

3 and 4 are trivially applicable and not a test whether $q\vec v \times \vec B$ is conservative. Sure, $\vec F \cdot d\vec l=0$, because the two vectors are perpendicular so when you do the integral, you are adding a whole bunch of zeroes to get a net of zero. It's always true that when you add a whole bunch of zeroes you get zero and that does not prove or disprove anything. The idea with 4 is that when you integrate from $a$ to $b$ you add a whole bunch of numbers (some positive, some negative, some zero) and the sum you get does not depend on the path that you follow from $a$ to $b$. The idea with 3 is that you go around any closed loop again adding non-zero numbers, the sum of positive numbers matches the sum of negative numbers exactly so that the net is zero. The positive and negative sums may depend on the path, but one is always the negative of the other.

The force represented by $q\vec v \times \vec B$ is non-conservative. It does zero work on a particle moving in a uniform magnetic field just like the tension does zero work on a bob swinging in a gravitational field.

Nope. Anyone who has taken a decent intermediate course in Classical Mechanics or Vector Calculus is sure to have encountered it (unless absent or totally dazed) and perhaps learned and understood it.

Well, I must say, I never understood, why you call forces with a scalar potential "conservative", as if these were the only forces for which energy conservation holds. In fact energy conservation holds within the most recent fundamental theories as long as no gravity is involved. That's because Minkowski spacetime is time-translation invariant, and by definition energy is the related Noether charge of this symmetry and thus conserved. I'd simply say, a force is the gradient of a scalar field (or equivalently line integrals along any closed path of the force vanish or the force field is curl free, and defined in a simply connected region of space). A force, having a (time-independent!) scalar potential is a sufficient condition for the validity of the energy-conservation law, but it's not necessary, as the example of the motion in a magnetic field shows.