Leo Liu
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Can anyone please tell me why time varying force F is not conservative? That is, what makes a force not depending on the position nonconservative?
Why is a time varying force nonconservative?
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Can anyone please tell me why time varying force F is not conservative? That is, what makes a force not depending on the position nonconservative?
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Can you see why (i) might be considered a consequence of (ii)?Leo Liu said:
Physics4Funn
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If a force depends on time, then the work done in moving between two points can be different, especially if that path is taken at different times.
If a force depends on velocity, similarly, take two different paths at two different speeds, then the work is different.
From the history of the concept of work and energy, this very strict definition helped tell the difference between a fundamental force, electric, gravity, from a force that they had no way to precisely describe that force. E.G. friction, viscosity. Non-conservative forces come from complex, non-linear interactions between objects. Friction comes from: applied normal force, surface roughness, chemical composition of objects. Viscosity comes from highly non-linear Navier-Stokes fluid equations.
Our understanding of the universe would be very different without this definition.
If a force depends on velocity, similarly, take two different paths at two different speeds, then the work is different.
From the history of the concept of work and energy, this very strict definition helped tell the difference between a fundamental force, electric, gravity, from a force that they had no way to precisely describe that force. E.G. friction, viscosity. Non-conservative forces come from complex, non-linear interactions between objects. Friction comes from: applied normal force, surface roughness, chemical composition of objects. Viscosity comes from highly non-linear Navier-Stokes fluid equations.
Our understanding of the universe would be very different without this definition.
Physics4Funn
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1) and 2), time and velocity are definitely related: v=distance/time. But, especially at the time of creation of the concept of conservative force, time and space were considered completely separate and different. Now that we know that time and space are related, then whole subject can become more complicated.PeroK said:
I always keep specific examples of phenomenon in mind to keep my understanding specific.
etotheipi
For a field ##\vec{F}## to be conservative, you just need that ##\nabla \times \vec{F} = \vec{0}##. This can still be satisfied with a time-dependent field, however it would be strange to call such a field 'conservative' in a physics sense.
Mainly because, suppose a particle moves under a force ##\vec{F} = \vec{F}(\vec{r}, t)##, which satisfies ##\nabla \times \vec{F}(\vec{r}, t) = \vec{0} \implies \vec{F}(\vec{r}, t) = - \nabla \phi(\vec{r}, t)##. Then$$\vec{F} = -\nabla \phi = m\ddot{\vec{r}}$$ $$m\ddot{\vec{r}} \cdot \dot{\vec{r}} + \nabla \phi \cdot \dot{\vec{r}} = \vec{0}$$Chain rule tells you that$$\frac{d\phi}{dt} = \nabla \phi \cdot \dot{\vec{r}} + \frac{\partial \phi}{\partial t}$$Substitute for ##\nabla \phi \cdot \dot{\vec{r}}## in the previous expression (and noting that ##v^2 = \dot{\vec{r}} \cdot \dot{\vec{r}}##),$$\frac{d}{dt} \left (\frac{1}{2}mv^2 + \phi \right) = \frac{\partial \phi}{\partial t}$$You can see that the energy measure ##E = \frac{1}{2}mv^2 + \phi## is not conserved if the potential is time-dependent.
Mainly because, suppose a particle moves under a force ##\vec{F} = \vec{F}(\vec{r}, t)##, which satisfies ##\nabla \times \vec{F}(\vec{r}, t) = \vec{0} \implies \vec{F}(\vec{r}, t) = - \nabla \phi(\vec{r}, t)##. Then$$\vec{F} = -\nabla \phi = m\ddot{\vec{r}}$$ $$m\ddot{\vec{r}} \cdot \dot{\vec{r}} + \nabla \phi \cdot \dot{\vec{r}} = \vec{0}$$Chain rule tells you that$$\frac{d\phi}{dt} = \nabla \phi \cdot \dot{\vec{r}} + \frac{\partial \phi}{\partial t}$$Substitute for ##\nabla \phi \cdot \dot{\vec{r}}## in the previous expression (and noting that ##v^2 = \dot{\vec{r}} \cdot \dot{\vec{r}}##),$$\frac{d}{dt} \left (\frac{1}{2}mv^2 + \phi \right) = \frac{\partial \phi}{\partial t}$$You can see that the energy measure ##E = \frac{1}{2}mv^2 + \phi## is not conserved if the potential is time-dependent.
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If you do work against a force, and then the force disappears, how could you recover the energy you put in?Leo Liu said:
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Thanks for that, but my question was addressed to the OP.Physics4Funn said:
Leo Liu
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Yet how could you make a pair of forces disappear?A.T. said:
Also, I would like to know how you can recover the energy after counteracting a force to move an object to a region in which the force and its paired force disappear.
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Make it time dependent.Leo Liu said:
Move it back where there is a force, then let the force do the work.Leo Liu said:
Physics4Funn
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An example: a book on the floor. No net force: a pair of forces that seem to disappear. Gravity pulling down. Floor pushing up. Lift the book to the table: work is done on the book. On the table, again no net force: gravity pulling down, table pushing up. Since gravity is conservative, then extract that work by pushing the book off the table and gravity pulls it down returning that work of lifting as kinetic energy.Leo Liu said: