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This has confused me for some time: if one integrates the gravitational force between two radii (let the motion of the particle always point toward the centre of mass of the other mass) from a radius which is further from the centre of mass than the other radius, to the radius which is closer to the centre of mass (in other words, when an object falls), then the work integral yields a negative sign. Nevertheless, the particle may gain kinetic energy in the process.

My main conceptual dearth here is that I do not sufficiently understand the relation between work, kinetic energy and potential energy.

I know that work is defined in terms of an integral of force with respect to displacement, and that a force acting along a displacement yields a positive force. I also understand that said force would also accelerate the particle upon which is does work. Therefore, I understand that positive work is associated with an increase in kinetic energy, because an accelerating object has an increasing velocity. This does not agree, however, with my result from integrating gravitational force (though it would agree, for instance, with the integration of the electrostatic force). I suspect that we would have to treat forces which point radially inwards with a special convention for forces.

If you have time, the clarification which I would actually most appreciate would describe the bases upon which relation between work, potential energy and kinetic energy is founded. The relation is often seen that W=T+U. However, work is otherwise defined in terms of force and displacement. At present, I see that definition to be more powerful because, to the knowledge that I presently have (and perhaps not to the knowledge that I presently lack), T and U are less well-defined in terms of physical intuitions. T is defined as that which integrates with respect to velocity to yield momentum. U is not exactly defined because it varies according to the relevant system.

So, concisely: Given that all concepts more basic than energy (Force, mass, acceleration, displacement, etc.) are all well-defined, can you justify the relation that W=T+U? If so, (and this will likely have been done via definition) can you prove the conservation of T+U using only other well-defined concepts? Thanks.

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If an object is falling freely under gravity, then the force of gravity and the displacement of the object are in the same direction. The value of the integral of force with respect to displacement (what you are calling the "work integral") is therefore positive. Gravity does a positive amount of work $W_g$ on the object and the result is an increase in the kinetic energy $T$ of the object (which we can measure directly). In the absence of drag or other dissipative forces we have

$W_g= \Delta T$

It is conventional to keep track of the work $W_g$ done by gravity by assigning a potential energy $U$ to the object, which depends on its location. Because the location at which $U$ is zero is arbitrary, we cannot assign an absolute value to $U$, but instead we equate the work done by gravity with the negative difference in $U$ i.e.

$W_g = - \Delta U$

So for an object falling freely under gravity (assuming no drag etc.) we have

$\Delta T + \Delta U = \Delta T - W_g = 0$

If we now introduce an external force $F$ that does work $W_F$ on the object (say by lifting it from ground level to the top of a mountain) then the total work done on the object is $W_F + W_g$ and so we have

$W_F + W_g = \Delta T \\ \Rightarrow W_F = \Delta T - W_g = \Delta T + \Delta U$

If the force due to gravity is a constant value $mg$ and the height of the object changes by an amount $\Delta h$ (and noting that $\Delta h$ is in the opposite direction to $mg$) then we have

$W_g = -mg \Delta h \\ \Rightarrow \Delta U = mg \Delta h \\ \Rightarrow W_F = \Delta T + mg \Delta h$

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  • $\begingroup$ Much appreciated. Just noticed the sloppy wording I used. Guess I wrote too much and mathed too little. $\endgroup$ Jul 29 '20 at 17:32
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It is important to understand who is doing the work. In general, the work done by a force is positive, if the displacement of object on which the force is acting is positive with regard to said force.

Take your example of a falling particle of mass $m$ falling toward a larger object of mass $M$. Let $\textbf{r}$ be the radial vector connecting their center of masses, with the origin being at the C.O.M. of the larger mass. The mass $m$ experiences a force $$\textbf{F}_G = -G \frac{mM}{|\textbf{r}|^3} \textbf{r}$$ due to the gravitational field of mass $M$. The work done by this field in taking $m$ from $\textbf{r}_1$ to $\textbf{r}_2$ is given by integrating this force over that interval:

$$\text{W}_G = -G m M\, \int_{\textbf{r}_1}^{\textbf{r}_2}\frac{\textbf{r} \cdot d\textbf{r}}{|\textbf{r}|^3} = -G m M\, \int_{r_1}^{r_2}\frac{1}{r^2}\, dr = Gm M\left(\frac{1}{r_2} - \frac{1}{r_1}\right).$$ Notice that as the object is falling toward $M$, $r_2 < r_1,$ so this work done is positive. This makes sense, since $m$ is moving in the same direction as the field of $M$.

Now, consider an external force $\textbf{F}_{\text{ext}}$ acting on $m$ such that it is always equal and opposite to $\textbf{F}_G$: $$\textbf{F}_{\text{ext}} = -\textbf{F}_G$$ In this case, $m$ will move toward $M$ without acceleration. If you calculated the work done by this external force on $m$, you would get $\text{W}_{\text{ext}} = - \text{W}_G$; i.e. the work done is negative. This also makes sense, since $m$ is moving opposite in relation to the external force.

This is incidentally also how the potential of a field is defined: It is the work done by an external agency in moving a unit mass from infinity to a point. For a gravitational field, this potential is thus negative, as our example showed. The $net$ work done on the object, that is by both forces, is zero in this example, since both the field as well as the external force did equal amounts of work opposite to each other.

As for the work-energy relation you mention, it follows readily if you accept the definition of energy as the capacity for doing work, that is to say the work done on an object gets converted into energy: $$\Delta W = \Delta T + \Delta U.$$ This relation holds for all agents doing path-independent work on the object once you make sure that you use the $\Delta U$ corresponding to the force under consideration. I invite you to verify this relationship for the examples above.

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  • $\begingroup$ Thanks. Only, when you integrate work, the dr used points from the larger mass to the less large mass. For when you reduce r dot dr divided by the magnitude of r cubed you reduce it to dr rather than - dr (divided by 1 over r squared). However, the actual path travelled by the smaller mass is from the smaller mass to the larger mass. It appears that your integral is the integral of the gravitational force from A to B with respect to a displacement vector from B to A. This, to me, is confusing. Could you explain? $\endgroup$ Jul 30 '20 at 16:00
  • $\begingroup$ Nevermind, I suppose that the position of the less massive particle is defined as relative to the position of the more massive particle. Hence, the work done on the less massive particle in moving closer to the more massive particle is from r1 multiplied by a unit vector pointing from M to M, to r2 multiplied by a unit vector from M to m. Important take away for me: gravitational work integrals use reverse vectors even for displacement because a particle's position is defined as FROM another particle. Again, thanks. $\endgroup$ Jul 30 '20 at 16:21
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In Newtonian mechanics, energy arises from the line integral of the second law along the object's path:

$\Sigma_i\overrightarrow F_i = m\overrightarrow a = m\frac{d\overrightarrow v}{dt}\\ \int (\Sigma_i\overrightarrow F_i) \cdot d\overrightarrow r = \int m \frac{d\overrightarrow v}{dt} \cdot d\overrightarrow r\\ \Sigma_i \int \overrightarrow F_i \cdot d\overrightarrow r = m\int d\overrightarrow v \cdot \frac{d\overrightarrow r}{dt}$

That last step on the right-hand side is a key concept: we can move the $dt$ from $d\overrightarrow v$ to $d\overrightarrow r$, because the change in velocity and change in position happen during the same time interval -- all three differentials correspond to the same little piece of the path. Also, on the left-hand side, we simply define the work done by a force as $\int\overrightarrow F\cdot d\overrightarrow r$, so we have:

$\Sigma_i W_i = m\int d\overrightarrow v\cdot \overrightarrow v$

Now the other key point: $d\overrightarrow v \cdot \overrightarrow v$ is the differential of $\frac12 \overrightarrow v \cdot \overrightarrow v = \frac12 v^2$. So

$\Sigma_i W_i = m\Delta(\frac12 v^2)$

Likewise, we define kinetic energy as $K = \frac12 mv^2$. That makes intuitive sense: the more massive an object is and the faster it's moving, the more motion energy it has.

$\Sigma_i W_i = \Delta K$

That's the work-energy theorem: the sum of the works done by all forces gives the change in kinetic energy.

Now, for a given force, that line integral will in general depend on the path the object takes. But oddly, it turns out that the fundamental forces of nature, such as gravity and the electric force, do the same amount of work for any path between two given points. So we don't need to do the integral every time. We just need a formula for the work they will do, in terms of the two endpoints.

$W = W(\overrightarrow r_i, \overrightarrow r_f)$ for a path-independent force

But we can also think of that work as the kinetic energy the object stands to gain by moving from $\overrightarrow r_i$ to $\overrightarrow r_f$. So we call that the potential energy of $\overrightarrow r_i$ relative to $\overrightarrow r_f$:

$U(\overrightarrow r_i) - U(\overrightarrow r_f) = W(\overrightarrow r_i, \overrightarrow r_f)\\ U(\overrightarrow r_f) - U(\overrightarrow r_i) = -W(\overrightarrow r_i, \overrightarrow r_f)\\ \Delta U = -W$

So the potential energy difference has the opposite of the kinetic energy difference: for a path-independent force, the object loses the same amount of potential as it gains kinetic, by virtue of our definition. Which means that if you add the potential and kinetic just due to that force, they remain the same.

Therefore, you can take the work-energy theorem above, and replace the work terms in the sum by potential energy changes - but only for those forces that have potential energy, ie. path-independent forces:

$\Sigma_i W_i = \Delta K\\ \Sigma_{path-independent}(-\Delta U_i) + \Sigma_{path-dependent}W_i = \Delta K$

Now, if all forces do path-independent work, we have:

$\Sigma_i(-\Delta U_i) = \Delta K\\ -\Delta U_{total} = \Delta K\\ \Delta U + \Delta K = 0\\ \Delta(U + K) = 0$

That is, total mechanical energy is conserved. Therefore, we call those path-independent forces conservative forces.

So that's mechanical energy in a nutshell. But as to your problem, it actually goes back to that definition of work, because it's the integral of the dot product. At all points along your path, gravity is pointing toward the second object's COM, and so is the displacement of the first object. Since those two vectors have the same direction, their dot product is positive, not negative. So your work will come out positive.

The intuition behind the work formula is that, when you push an object and it moves the direction you're pushing it, you are contributing to the accomplished motion, so you're doing positive work. But if someone else pushes harder, and it moves against your push, you're hindering the process, hence doing negative work. The harder you pushed and the further it moved, the more work you did.

So, like you said, mathematically, energy is just a set of definitions on top of the second law - but those definitions match the intuitive understanding of the terms, and the math gives us shortcuts to solving problems.

EDIT: Note I said it was "odd" that the fundamental forces of nature are path-independent, ie. conservative. But, in retrospect, that fact suggests that energy is actually a very natural concept in physics, perhaps more so than force. The derivation from force is slightly awkward, but energy conservation is a much more universal principle (even if it's not quite as strict in modern physics).

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  • $\begingroup$ A further question: is G positive or negative? This is highly important. For if G is positive then negative the gravitational force is positive, meaning that a force vector from A to B would have the value of negative GMm over r - squared, right? This would explain a lot of my confusion $\endgroup$ Jul 29 '20 at 20:04
  • $\begingroup$ Take a look at the $\mathbf F_G$ formula in Yejus' answer. $G$ is positive, but there is a negative sign in front of it. Also, $\mathbf r$ by definition points from the object that's pulling to the object being pulled. So you have an overall negative scalar, multiplied by a vector that points to the object being pulled, giving you a resulting force that points toward the object that's pulling. The displacement also points in that direction. Then you can define whatever coordinate system you want, but $\mathbf F_G$ and $d\mathbf r$ will have the same sign, so a positive dot product. $\endgroup$ Jul 29 '20 at 20:30
  • $\begingroup$ Thank you so much. I was getting weird propositions after trying negative G. I am sure past physicists have gone through what I have, and found that the best solution was, as you god-sendingly recommended: to reverse the direction of r (ingeniously). Just to make sure, is it so that this solves the problem because when you integrate the negative gravitational force (which is the right integration to obtain the work) the vector direction disappears? I integrate F= - GMm/(r^2) multiplied by unit vector r from b to a (i.e. reverse). Does the unit vector disappear in the integration? Agradecido! $\endgroup$ Jul 29 '20 at 21:12
  • $\begingroup$ Take a look at the dot product. Say you want the work done on yourself by Earth. Then by definition, $\mathbf r$ points away from Earth. If you use polar coordinates, then the coordinate vector $\hat r$ also points away from Earth. So $\mathbf r = r\hat r$. But your displacement is toward the Earth, $d\mathbf r = -dr\hat r$. So your integrand is $(-G\frac{mM}{r^2}\hat r)\cdot(-dr\hat r)$. The negatives cancel, and as you say, $\hat r \cdot \hat r$ disappears, because its magnitude is 1 and its angle with itself is 0. $\endgroup$ Jul 29 '20 at 21:58

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