A cyclist free-wheels along a hilly road near Shimla without pedalling. On a steep downhill stretch the bicycle is pushed hard along the road. On a gentle slope the push is weak. On a flat stretch there is no push at all, and uphill the push turns backward. The cyclist feels the force change with the steepness of the road, not with its height.
From potential energy back to force
So far we went from a force to its potential energy: \(\Delta U = -W\). Now go the other way. Let a conservative force \(F\) act along the \(x\) axis. Over a small step \(\Delta x\), the force hardly changes, so its work is \(F\,\Delta x\). The potential energy therefore changes by \(\Delta U = -F\,\Delta x\).
Divide by \(\Delta x\) and let the step shrink to zero. The ratio \(\Delta U/\Delta x\) becomes the derivative \(dU/dx\), the slope of the graph of \(U\) against \(x\). So the force is minus the slope.
The minus sign carries the meaning. Where \(U\) rises to the right, the slope is positive, so \(F\) is negative: the force points left. Where \(U\) falls to the right, the force points right. In both cases the force points the way \(U\) decreases, downhill on the curve. The steeper the curve, the larger the force.
Move the slider across the sim. At \(x = 1\,\text{m}\) the curve drops steeply, with a slope of \(-22.5\,\text{J/m}\), so the force is \(22.5\,\text{N}\) to the right. At \(x = 5\,\text{m}\) the curve rises with a slope of \(4.5\,\text{J/m}\), so the force is \(4.5\,\text{N}\) to the left, back towards the well. At \(x = 3\,\text{m}\) and \(x = 7\,\text{m}\) the tangent is flat and the force is zero.
At a point \(P\) the graph of \(U(x)\) rises steeply to the right. What is the force on a particle at \(P\)?
A flat stretch means no force
If \(U\) is the same everywhere in a region, its slope there is zero, and so is the force. A puck on a level, smooth table is the plainest example. Its gravitational energy is the same at every point of the table, and no horizontal force acts on it. The force is also zero at a single flat point, such as the bottom of the well or the top of the hump. Those points are equilibria, the subject of the next lesson.
Now add a constant \(C\) to \(U\) everywhere. The whole curve moves up by \(C\), but every slope stays the same, so every force stays the same. This is the free choice of zero from lesson 1, seen from the other side. The force knows only the slope of \(U\), never its value.
Over a region of space the potential energy of a particle is constant at \(8\,\text{J}\). What is the force on the particle there?
Hooke's law and the weight, recovered
Apply the rule to the spring. Its potential energy is \(U = \tfrac12 kx^2\). The slope is \(dU/dx = kx\), so the force is \(F = -kx\). That is Hooke's law, with its minus sign: the force always points back towards \(x = 0\), where \(U\) is least. Apply it to gravity, with \(y\) measured upward: \(U = mgy\) gives \(F = -mg\), the weight pointing down.
\[ U = \tfrac12 kx^2 \]
Why. the energy stored in a spring stretched by \(x\), from lesson 1.
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Your turn. differentiate both sides with respect to \(x\). Which line is it?
\[ F = -\dfrac{dU}{dx} = -kx \]
Read it. the force is minus the slope of \(U\).
The potential energy of a particle moving along the \(x\) axis is \(U = 5x^2 - 10x\), with \(U\) in joules and \(x\) in metres. Find the force on it at \(x = 2\,\text{m}\), and the point where the force is zero.
The slope is \(dU/dx = 10x - 10\), so \(F = -(10x - 10) = 10 - 10x\) newtons.
At \(x = 2\,\text{m}\): \(F = 10 - 20 = -10\,\text{N}\), that is \(10\,\text{N}\) in the negative \(x\) direction.
The force is zero where \(10 - 10x = 0\), at \(x = 1\,\text{m}\). There \(U\) has its least value, \(-5\,\text{J}\).
A particle has potential energy \(U = 4x^3 - 6x^2\), with \(U\) in joules and \(x\) in metres. What is the force on it at \(x = 2\,\text{m}\), in newtons? Give the sign.
In three dimensions: the gradient
Now let \(U\) depend on \(x\), \(y\) and \(z\). Move a small step along \(x\) only, keeping \(y\) and \(z\) fixed. The force's \(x\) component does all the work, so \(F_x = -\partial U/\partial x\). The symbol \(\partial U/\partial x\) is a partial derivative: the slope of \(U\) along \(x\) with the other coordinates held fixed. The same argument along \(y\) and \(z\) gives the other two components.
The three slopes together form a vector, the gradient of \(U\), written \(\nabla U\). It points in the direction in which \(U\) rises fastest. The force is its negative, \(\vec F = -\nabla U\), so the force points the way \(U\) falls fastest: straight downhill.
A contour is a line of equal \(U\), like a contour line of height on a map. Moving along a contour changes no potential energy, so the force does no work there. The force must therefore be perpendicular to every contour, as Fig 5.1 shows. Where the contours crowd together, \(U\) changes quickly with distance and the force is strong. For the puck, \(\vec F = -(4x\,\hat i + 4y\,\hat j)\), which points straight at the centre with size \(4r\).
A particle moves in the \(xy\) plane with potential energy \(U = 2x^2 + 3y\), with \(U\) in joules and lengths in metres. Find the force on it at the point \((1\,\text{m}, 2\,\text{m})\).
\(\partial U/\partial x = 4x\) and \(\partial U/\partial y = 3\), so \(\vec F = -4x\,\hat i - 3\,\hat j\) newtons.
At \((1, 2)\): \(\vec F = -4\,\hat i - 3\,\hat j\) N. Its size is \(\sqrt{16 + 9} = 5\,\text{N}\).
The \(y\) coordinate of the point did not matter, because \(U\) rises with \(y\) at the same rate everywhere.
A particle in a plane has potential energy \(U = 3x^2 + 4y^2\), with \(U\) in joules and lengths in metres. What is the size of the force on it at the point \((1\,\text{m}, 1\,\text{m})\), in newtons?
Not every force field is minus a gradient. If \(\vec F = -\nabla U\), then \(\partial F_x/\partial y = \partial F_y/\partial x\), because both equal \(-\partial^2 U/\partial x\,\partial y\). A field that fails this test, such as \(\vec F = y\,\hat i - x\,\hat j\), which swirls around the origin, does nonzero work around a circle. It is non-conservative and has no potential energy.
Now think of the cyclist near Shimla again. Along the road, the potential energy is \(mgh\), where \(h\) is the height at each point. The push along the road is minus the slope of \(mgh\) with distance: large on a steep stretch, zero on a flat one, backward uphill. The cyclist was reading \(F = -dU/dx\) from the shape of the road.
George Green ran his father's windmill in Nottingham and had about one year of school. He taught himself mathematics from library books. In 1828, at the age of thirty-five, he had a small essay printed, paid for by a few subscribers. In it he introduced a single function, which he called the potential, whose slopes give the electric and magnetic forces. Few people read it. Years later William Thomson, the future Lord Kelvin, found a copy and had it reprinted, and Green's potential became a basic tool of physics.
- Dropping the minus sign in \(F = -dU/dx\).The force points the way \(U\) decreases, down the curve.
- Reading the force from the height of the U-x curve.Read the slope. A high flat point has no force; a low steep point has a large force.
- Differentiating with respect to the wrong variable in \(\vec F = -\nabla U\).For \(F_x\) differentiate with respect to \(x\) alone, holding \(y\) and \(z\) fixed.