Physical interpretation of virtual displacement

[Pushoam]

From the above description of Virtual displacement, what I understood is: virtual displacement is another name for $dx_i$ when dt = 0. I didn’t understand its physical interpretation. Could anyone please put some more light upon it?

 

[albertrichardf]

In thermodynamics, when we are trying to carry out a reversible process, we do things very slowly. If we are compressing a piston reversibly, for example, we could compress it by putting one grain of sand at a time on its surface. If we don't do it slowly, then the system will not have time to adjust because heat won't be able to flow, and we end up with an irreversible system. I

A virtual displacement can be thought of as a mechanical analogue to that. Think of a cart on a slope, being held stationary by a pulley system. If we give the cart a quick kick downhill, the cart will start to roll down because the pulley will not have time to adjust to the motion of the cart. But if we do it very slowly, then the pulley has time to adjust and the cart doesn't fall under gravity.

Of course, neither can we physically compress a piston slowly enough to make a process reversible, nor can we cause a displacement so slowly that everything has time to adjust. So we imagine this displacement instant, hence "virtual" displacement. Essentially what we are describing is a system that has every single component move instantaneously. We want heat to flow instantly as we compress the gas, and we want the pulley to adjust instantly as we pull the cart. We thus imagine that the displacement occurs instantly, in zero time. Hence why we say time is frozen for a virtual displacement.

 

[vanhees71]

My advice is to skip over this puzzling question and read on in your book. Everything becomes clear when you get to the most fundamental description of all fundamental laws of nature, which is Hamilton's principle of least action. It becomes immediately clear why the handwaving way to introduce "virtual discplacements" and d'Alembert's principle, because it's just the math of functionals of trajectories in configuration space (Lagrange version of Hamilton's least-action principle) or in phase space (Hamilton's version of the least-action principles). For me, it's among the great mysteries of physics didactics to introduce "virtual displacements" and d'Alembert's principle before introducing the least-action principle. It must be due to the historical approach and laziness of textbook writers to make life for generations of students unnecessarily complicated. It's of the same kind of laziness to use non-relativistic approximations in electrodynamics first and then resolving some apparent paradoxes (mostly in connection with Faraday's law like the homopolar generator), which only occur because the non-relativistic treatment of matter is used outside of its range of applicability.

One great exception is Volume 1 of Landau and Lifshit's Theory Course, where classical mechanics is treated right away using the action principle. The same holds true for Volume 2 concerning classical field theory (i.e., microscopic classical electrodynamics and General Relativity).

 

[Pushoam]

@vanhees71, I take virtual displacement as a name for the term given in the right side of equation 1.9 and go ahead till I get least action principle.

Thanks for pointing it out.@Albertrichardf
Is it needed that reversible process should be slow?

I think we want to keep one grain on the piston so that the disturbance occurred to the system could be observed by us clearly (at least in theory) and hence, by doing things in the reverse order of what we observed, the system could be brought into its original state(at least in theory). So, this process could be termed as a reversible process.

Now, I do not understand why heat flow is related to this.

 

[Pushoam]

A virtual displacement can be thought of as a mechanical analogue to that. Think of a cart on a slope, being held stationary by a pulley system. If we give the cart a quick kick downhill, the cart will start to roll down because the pulley will not have time to adjust to the motion of the cart. But if we do it very slowly, then the pulley has time to adjust and the cart doesn't fall under gravity.

Let's take the case in which the cart is connected to the pulley through the rope. The tension acts as a force of constraint in the sense that it does not let the cart roll down. If I move the cart downwards in such a way that after the movement, when I leave the cart, the cart does not start rolling ( as the tension gets adjusted to not to let the cart go down by this time), i.e. the cart's initial and final acceleration and velocity remains same; then the displacement of the cart is known as virtual displacement. Is it so?

 

[Pushoam]

Of course, neither can we physically compress a piston slowly enough to make a process reversible, nor can we cause a displacement so slowly that everything has time to adjust. So we imagine this displacement instant, hence "virtual" displacement. Essentially what we are describing is a system that has every single component move instantaneously. We want heat to flow instantly as we compress the gas, and we want the pulley to adjust instantly as we pull the cart. We thus imagine that the displacement occurs instantly, in zero time. Hence why we say time is frozen for a virtual displacement.

Does this also mean the following?

if there is a system on which a time depending force f(t) is acting, then giving a virtual displacement $\delta x$ to the system means that I have displaced the system $\delta x$ in such a small time that the force acting on the system before and after the displacement remain same. So, we take the small time to be zero, i.e. time is frozen.

 

[albertrichardf]

Now, I do not understand why heat flow is related to this.

Heat flow is the reason we need to do the process slowly. When we compress the gas, the parts of the gas closer to the piston are hotter than those further away from the piston and heat has to flow throughout the gas. When heat flows, it flows from hot to cold, and not the other way round, which means that any process that involves heat flow is irreversible. There is a rate of heat flow, which means that it takes time for the temperature to equalise throughout the gas, and when we compress the gas quickly, heat does not have enough time to flow during the compression and we end up with gas that is hotter near the piston, and colder away from it. Once we are done compressing the gas, heat will flow through the gas to equalise the temperature, which means that we cannot run the process in reverse. If we do the process very slowly the temperature at any point in the gas is practically the same at any given time during the compression, so there's very little heat flow and the process is more or less reversible. If we keep time frozen and compress the gas, then there's absolutely no heat flow since for heat to flow, time must pass and the process is completely reversible.

Let's take the case in which the cart is connected to the pulley through the rope. The tension acts as a force of constraint in the sense that it does not let the cart roll down. If I move the cart downwards in such a way that after the movement, when I leave the cart, the cart does not start rolling ( as the tension gets adjusted to not to let the cart go down by this time), i.e. the cart's initial and final acceleration and velocity remains same; then the displacement of the cart is known as virtual displacement. Is it so?

Yes.

Does this also mean the following?

if there is a system on which a time depending force f(t) is acting, then giving a virtual displacement $\delta x$ to the system means that I have displaced the system $\delta x$ in such a small time that the force acting on the system before and after the displacement remain same. So, we take the small time to be zero, i.e. time is frozen.

The force throughout the whole displacement must remain the same, otherwise, the acceleration of the system will change. If at any point the force differs during the displacement, it will cause a change in the acceleration, and hence in the velocity.
It's the same principle as for the compression of a gas. You have to make sure that at all times the temperature throughout the gas is the same during the compression, not just before and after. In the same way, the net force should be the same throughout the displacement.

 

[Pushoam]

If we keep time frozen and compress the gas, then there's absolutely no heat flow since for heat to flow, time must pass and the process is completely reversible.

But how can you keep time frozen and compress the gas?
I think as heat flow takes time, so compression also takes time.

You have to make sure that at all times the temperature throughout the gas is the same during the compression, not just before and after.

So, what I understood is:
Compressing the gas leads to heat flow ( an irreversible process) through the gas. As a result, the heat flow makes compression an irreversible process. So, to make compression a reversible process, we have to make heat flow negligible. One way to do it is by compressing slowly. Now, measurement of temperature gives an idea of how much we have succeeded in reducing heat flow.

In the same way, the net force should be the same throughout the displacement.

Should the net force remain same for all kind of virtual displacement or only in this case?

 

[albertrichardf]

But how can you keep time frozen and compress the gas?
I think as heat flow takes time, so compression also takes time.

You're right: physically, we cannot freeze time. Reversible processes exist only in our theories. However, through methods such as compressing gasses slowly, we can approximate reversible processes physically. Therefore, we can imagine freezing time and compressing the gas theoretically, which is approximately the same as compressing the gas very slowly physically. We only imagine freezing time and compressing the gas, or freezing time and causing a displacement (hence the term "virtual" for the displacement). Physically, we approximate it by carrying the process very slowly.

So, what I understood is:
Compressing the gas leads to heat flow ( an irreversible process) through the gas. As a result, the heat flow makes compression an irreversible process. So, to make compression a reversible process, we have to make heat flow negligible. One way to do it is by compressing slowly. Now, measurement of temperature gives an idea of how much we have succeeded in reducing heat flow.

Not necessarily. I think I was a bit unclear, sorry. When I said that the temperature has to remain the same throughout the compression, I meant that the temperature of the entire gas must be the same. What I meant by that is that you could have a gas at 2 K initially, and after compressing it, the gas could be at 20 K, and it would still be reversible as long as every single part of the gas is at 20 K. A measure of how successful you were in compressing the gas reversibly would be the temperature gradient of the gas. The closer it is to zero, the more uniform the temperature distribution of the gas, and thus the more successful you were in compressing the gas reversibly. It's a bit like those Christmas lights that are connected in series: either you can have them all on, or all off, but you can't have some lights on and some off. As long as everywhere in the gas is at the same temperature, even if that temperature changes, the process will still be reversible.

Should the net force remain same for all kind of virtual displacement or only in this case?

The net force should always be the same.

 

[lightarrow]

From the above description of Virtual displacement, what I understood is: virtual displacement is another name for $dx_i$ when dt = 0. I didn’t understand its physical interpretation. Could anyone please put some more light upon it?

Here I only want to give you a simple example on which to think about. Sometimes it is useful to grasp what the theory is trying to say.

Take a table, and a body constrained to move on it bi-dimensionally, e. g. without friction.
Let's say, once drawn orthogonal X, Y axis, that it moves, at constant speed, along the direction of the vector v = (v_x, v_y).

What if the same table can be moved, e. g. up, by some external mechanism?
Let's say it moves up at constant speed u_z.

The real displacement of the body will then be in 3 dimensions, along a direction U = (v_x, v_y, u_z).

But even in this scenario, the virtual displacement is still along v = (v_x, v_y).

In other words: here to understand which is the virtual displacement we "freeze" the movement of the constraint itself.
If the constraints are already stationary, there are no differencies between virtual and real displacement.

Another similar example: a little ball is constrained to move mono-dimensionally inside a slot made on a disk, without friction. If the disk can be made to rotate around its (principal) axis by an external mechanism (a motor) then the virtual displacement of the ball is still the monodimensional one inside the slot. In other words: the ball has just one degree of freedom, despite the fact that its real motion actually is bidimensional.

That's the interest in virtual displacements: they are used to define the system's degrees of freedom.

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[Matt Smith]

It is simple.Because work W=∑Fi*(the corresponding virtual dispalcement).Fi is generalized force.If W=0 under a virtual displacement,it means the total force is 0.Very simple.And if Fi is conserved force,V=-W.If W=0,potential energy don't change,but it is a function of the position.What is the meaning?It means at this point,the potential energy minimize.

 

[Pushoam]

What is meant by the phrases " freeze time"?
Does it mean that we displace the system so fast that nothing about the system gets changed except its displacement and position related forces?
What is meant by the phrases "freeze the movement of the constraint"?
Does it mean that we move the system in such a way that the constraint force (or net constraint force?) on the system remains same?

Thanks for replying.

 

[lightarrow]

What is meant by the phrases " freeze time"?
Does it mean that we displace the system so fast that nothing about the system gets changed except its displacement and position related forces?

No, it means that the position of every point in the system does not depend explicitly on the time, but it can depend on the time implicitly through other coordinates.

What is meant by the phrases "freeze the movement of the constraint"?
Does it mean that we move the system in such a way that the constraint force (or net constraint force?) on the system remains same?

No, it means to stop all the movements which is imposed from outside, for example imposed by an external engine on the system or part of it.

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