# A single external force doing work on a system of particles

Suppose we have a system of particles being acted upon by a single external force ##\mathbf{F}^{e}##. Each individual particle feels a force of ##\mathbf{f}_i = \mathbf{f}_{i}^{int} + \mathbf{f}_{i}^{e}## such that ##\sum_i \mathbf{f}_{i}^{e} = \mathbf{F}^{e}##, and ##\mathbf{f}_{i}^{int}## are internal forces which I pretty much ignore in what follows! I'd like to show that the work done by ##\mathbf{F}^{e}## is equivalent to the sum of the works done by ##\mathbf{f}_{i}^{e}## over all particles.

For context, the system of particles might make up a rigid cylinder and ##\mathbf{F}^{e}## a force applied on the curved face, a bit like how a string exerts a tension force on a yo-yo if you pull it. It's still slightly unclear to me exactly what displacement we use to compute the work done by ##\mathbf{F}^{e}##, though I'd assume this be the displacement of the point on the body where ##\mathbf{F}^{e}## acts which I might call ##\delta \mathbf{r}##.

Consequently, the work done ##W## by ##\mathbf{F}^{e}## will then be ##W = \mathbf{F}^{e} \cdot \delta \mathbf{r}##. Also, the work done on the ##i^{\text{th}}## particle by external forces is ##\mathbf{f}_{i}^{e} \cdot \delta \mathbf{r}_i##, which means that the total work done on all particles by external forces is $$W = \sum_{i} \delta W_{i} = \sum_{i} \mathbf{f}_{i}^{e} \cdot \delta \mathbf{r}_i$$ Had we been considering centre-of-mass work, the external work would fall right out since all ##\delta \mathbf{r}_i## would just be ##\delta \mathbf{r}## (of course assuming we were to also redefine ##\delta \mathbf{r}## as the displacement of the centre of mass). However, things don't appear to be so straightforward for extended bodies, since the ##\delta \mathbf{r}_i## are all different.

So I wondered if there were any way of finishing this off? Even though what I've said should hold true for general systems of particles, the motivation was mainly for rigid bodies. If the total real external work done on a rigid body equals the total change in its kinetic energy (translational, rotational, microscopic, etc.), this should be provable by considering the external work done on a single particle and then summing this up. Thank you!

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Isn't it true that you can write the net force on the ##i##th particle as
$$\vec f_i=\vec f_{e,i}+\sum_{j\neq i}^N \vec f_{ij}$$where the second term is the sum of internal forces? Then $$\sum_i^N\vec f_i=\sum_i^N\vec f_{e,i}+\sum_i^N\sum_{j\neq i}^N \vec f_{ij}.$$The double summation is zero because of Newton's 3rd in which case$$\sum_i^N\vec f_i=\sum_i^N\vec f_{e,i}=\vec F_e.$$Does that answer your question or did I miss what you are trying to do?

• vanhees71 and etotheipi
$$\sum_i^N\vec f_i=\sum_i^N\vec f_{e,i}=\vec F_e.$$

This is certainly true, though I'm concerned about the sum of the works done by the external forces. In a macroscopic sense, you can draw the external forces on your free body diagram and compute the real work done by any given external force by calculating its dot product with the displacement of the material it acts on.

Though if we zoom in, now we've got lots of particles each of which is acted on by external forces, and by definition the sum of these external forces is the macroscopic external force, which was the proof you gave. However, now if we compute the total work done by external forces on each particle, I don't know how it's possible to show that their sum equals the work we computed when we considered the macroscopic external force. Primarily because all of the displacements are different!

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You can dot with ##d\vec r_i## to get$$\vec f_i\cdot d\vec r_i=\vec f_{e,i}\cdot d\vec r_i+\sum_{j\neq i}^N \vec f_{ij}\cdot d\vec r_i$$Then sum over ##i## $$\sum_i^N\vec f_i\cdot d\vec r_i=\sum_i^N\vec f_{e,i}\cdot d\vec r_i+\sum_i^N d\vec r_i\cdot\sum_{j\neq i}^N \vec f_{ij}$$The double sum has pairs of terms like $$d\vec r_i\cdot \vec f_{ij} + d\vec r_j\cdot \vec f_{ji}=\vec f_{ij} \cdot(d\vec r_i-d\vec r_j).$$Under what circumstances do you have cancellation?

• • vanhees71 and etotheipi
The double sum has pairs of terms like $$d\vec r_i\cdot \vec f_{ij} + d\vec r_j\cdot \vec f_{ji}=\vec f_{ij} \cdot(d\vec r_i-d\vec r_j).$$Under what circumstances do you have cancellation?

The simple case looks to be when ##\delta \mathbf{r}_i - \delta \mathbf{r}_j= \mathbf{0}##, or ##\delta \mathbf{r}_i = \delta \mathbf{r}_j##. I know that for rigid bodies ##|\mathbf{r}_i - \mathbf{r}_j| = \text{constant}##, which would appear to satisfy that condition in the absence of rotation. I have the feeling that it holds also for rigid bodies undergoing rotation but can't see why yet, so I'd need to play around with it for a little longer!

But if we accept that internal forces can do no work in rigid bodies (i.e. ##\sum_{j\neq i}^N \vec f_{ij}\cdot d\vec r_i = 0##), is it then possible to prove that $$\sum_i \mathbf{f}_i^{e} \cdot \delta \mathbf{r}_i = \mathbf{F}_{i}^{e} \cdot \delta \mathbf{r}$$where ##\delta \mathbf{r}## is the displacement of that macroscopic external force on our free body diagram?

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If you want to include rotations, I would try writing ##\vec r=\vec R+\vec r'## where ##\vec R## is the position of the CM (assumed to be the axis of rotation) and ##\vec r'## is the coordinate of the point relative to the CM.

• etotheipi
If you want to include rotations, I would try writing ##\vec r=\vec R+\vec r'## where ##\vec R## is the position of the CM (assumed to be the axis of rotation) and ##\vec r'## is the coordinate of the point relative to the CM.

In that case I obtain ##\delta (\mathbf{r}_i^{'} - \mathbf{r}_j^{'})=\mathbf{0}##. That is somewhat peculiar, since for a rotation that quantity can certainly be non-zero! I'll keep trying though!

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But if we accept that internal forces can do no work in rigid bodies (i.e. ##\sum_{j\neq i}^N \vec f_{ij}\cdot d\vec r_i = 0##), is it then possible to prove that $$\sum_i \mathbf{f}_i^{e} \cdot \delta \mathbf{r}_i = \mathbf{F}_{i}^{e} \cdot \delta \mathbf{r}$$where ##\delta \mathbf{r}## is the displacement of that macroscopic external force on our free body diagram?
You have to be careful with your statements. When you mention work, you have to specify who does the work (the force) and on whom (the system) the work is done. Internal forces cannot do work on the entire system, here the collection of ##N## particles. However one part of the system (particle ##i##) can do work on another part of the system (particle ##j##). Look at the equation you want to prove. I think you meant to write $$\sum_i \mathbf{f}_i^{e} \cdot \delta \mathbf{r}_i = \mathbf{F}^{e} \cdot \delta \mathbf{r}$$How do you define ##\delta \mathbf{r}##?

• etotheipi
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In that case I obtain ##\delta (\mathbf{r}_i^{'} - \mathbf{r}_j^{'})=\mathbf{0}##. That is somewhat peculiar, since for a rotation that quantity can certainly be non-zero! I'll keep trying though!
Why "certainly" or have we stopped considering a rigid body with axes painted on it and the origin at the CM?

• etotheipi
Look at the equation you want to prove. I think you meant to write $$\sum_i \mathbf{f}_i^{e} \cdot \delta \mathbf{r}_i = \mathbf{F}^{e} \cdot \delta \mathbf{r}$$How do you define ##\delta \mathbf{r}##?

That is correct, apologies for the typo! I haven't been able to come up with a rigorous definition for ##\delta \mathbf{r}##, which poses a slight problem! The only hold I have on it is from thinking about how we use free body diagrams.

Say I have a box floating in a vacuum, and I apply a force ##\mathbf{F}## to a corner of the box. We're quite happy to speak about these sorts of macroscopic forces acting on certain parts of rigid bodies, but in reality this ##\mathbf{F}## is divided over many different particles which all feel only parts of that force, as in ##\sum \mathbf{f}_i = \mathbf{F}##.

So the best definition of ##\delta \mathbf{r}## I can think of is the displacement of the material directly at the point of application of the macroscopic force on the free body diagram.

This is I think the main problem with the proof I'm proposing, since it's ill-defined from the outset. We know from experience that it must be true, since we can only ever work with macroscopic descriptions of forces when solving most mechanics problems. But linking it with the particle description seems a little more problematic!

a rigid body with axes painted on it and the origin at the CM?

Ah, you were referring to a body fixed frame. In that case, yes, the result about internal forces follows directly.

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$$\delta\vec r_i=\delta\vec R+\delta\vec r'_i.$$
$$\delta W_i=\vec f_{e,i}\cdot \delta\vec r_i+\sum_{j\neq i}^N \vec f_{ij}\cdot \delta\vec r_i=\vec f_{e,i}\cdot (\delta\vec R+\delta\vec r'_i)+\sum_{j\neq i}^N \vec f_{ij}\cdot (\delta\vec R+\delta\vec r'_i)$$
$$W=\sum_i^N \delta W_i=\delta \vec R \cdot\sum_i^N\vec f_{e,i}+\sum_i^N\vec f_{e,i}\cdot\delta\vec r'_i+\delta\vec R\cdot\sum_i^N\sum_{j\neq i}^N \vec f_{ij}+\sum_i^N\delta\vec r'_i\cdot\sum_{j\neq i}^N \vec f_{ij}$$The first term is the work done by the external force on the CM, ##W_e=\vec F_e\cdot \Delta \vec R_{cm}##. The third term is zero because of Newton's third law. This can be consolidated to$$W=\vec F_e\cdot \Delta \vec R_{cm}+\sum_i^N\delta\vec r'_i\cdot\sum_{j\neq i}^N (\vec f_{e,i}+\vec f_{ij}).$$I don't know if you can do much more with this. You might consider treating a specific example to focus your thinking. Say you have particles ##m_1## and ##m_2## with charges ##+q_1## and ##+q_2## placed in an external electric field and constrained to move in a straight line. Calculate all the works and displacements if the particles are initially at rest separated by ##d## and then let go until they reach final separation ##2d##.

• etotheipi
This can be consolidated to$$W=\vec F_e\cdot \Delta \vec R_{cm}+\sum_i^N\delta\vec r'_i\cdot\sum_{j\neq i}^N (\vec f_{e,i}+\vec f_{ij})$$

Thank you for this treatment, it's really clear! And furthermore in the case of a rigid body (I 'found' another proof of this fact in the answer to the question here) this just becomes $$W=\vec F_e\cdot \Delta \vec R_{cm}+\sum_i^N\delta\vec r'_i\cdot\vec f_{e,i}.$$And we know that this has to be equal to ##\vec{F}_e \cdot \delta \vec{r}##. So really what we're saying is that the term ##\sum_i^N\delta\vec r'_i\cdot\vec f_{e,i}## makes up the difference ##\vec{F}_e \cdot(\delta \vec{r} - \Delta \vec{R}_{cm})##. This makes sense, since if the point of application of the force really is through the centre of mass, then the real work should equal the centre-of-mass work.

Thanks for your patience with this, I'll give the examples you suggested a go and see if the results are consistent with this!

• kuruman