In my Physics class I studied that work done by an internal conservative force is equal to the negative of the change in potential energy of the system. That is,

$$dW_{internal conservative force} = dU_{system}.$$

But I have a question. Why is the work done by only 'internal' conservative force accounted for? There may also be external conservative forces which behave no differently from internal conservative forces. I mean they also have the capability of storing energy in the system in the form of potential energy, right? Then why don't we account for external conservative forces. Shouldn't the equation actually be :

$$dW_{internal conservative force} + dW_{external conservative force} = dU_{system}.$$

Why work done by only 'internal' conservative force is equal to the negative of the change in potential energy of the system? Shouldn't it be : Work done by all the conservative forces including internal conservative forces and external conservative forces is equal to the negative of the change in potential energy of the system. Why do we consider only internal conservative forces and not external conservative forces? Can someone please explain?


2 Answers 2


Well there are different notions of potential energy. One is "potential energy of an isolated system" which does not depend on position or state of bodies outside the system, and this is associated with internal forces only.

And the other one is "potential energy in a field of external force" which does depend on position/state of other bodies, and this is associated with external forces.

Consider two balls of masses $m_1,m_2$ are connected by a spring with stiffness $k$. If considered as isolated system, potential energy of this system is

$$ E_p = \frac{1}{2}k(d-d_0)^2 $$ where $d$ is distance between the balls and $d_0$ is the equilibrium distance for which the spring is not stretched nor compressed. This potential energy is due to internal forces only (forces between the balls and the spring). If the system is isolated, then sum of kinetic energy and this potential energy is constant in time:

$$ \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 + \frac{1}{2}k(d-d_0)^2 = const. $$

However, if this system is also placed in field of gravitational force at some height $h$ above ground in gravitational field with intensity $g$, and this height can change, then the above equation does not hold, because kinetic or potential energy of the spring can change due gravitational forces. So the first concept of the potential energy is not sufficient (to formulate conservation of energy).

In Earth's reference frame, conservation of energy can be recovered if we use the other concept of total potential energy, which includes the above potential energy, and also potential energy due to presence in gravitational field. This potential energy is

$$ E_p = \frac{1}{2}k(d-d_0)^2 + mgh. $$ Notice the second term depends on mutual distance between the system and Earth's ground. So this potential energy takes into account existence of both internal forces (due to spring) and external forces (due to gravity).

This second concept of potential energy has the advantage that now total energy is conserved: $$ \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 + \frac{1}{2}k(d-d_0)^2 + mgh = const. $$

So whether potential energy is due to internal forces only, or due to external forces, depends on which kind of potential energy we are talking about.


In short, because:

Conservation of energy theorem =>

$T + V = Constant$

So: $dT + dV = 0$

So: $dT = - dV$


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