# Falling charged objects: energy conservation paradox?

Imagine that we start with two oppositely charged objects on the ground, separated by a distance $d$, with charges $+q$, $-q$ and masses $m$.

We raise them both up to a height $h$.

In doing so we only move them at constant velocity so that neither charge produces a radiative electromagnetic field in the vicinity of the other.

Also assume that the objects move simultaneously on frictionless poles so that their Coulomb attraction is always horizontal and thus plays no part in the experiment.

The energy that we have put into the system is simply:

$$E_{grav} = 2mgh$$

Now we simultaneously release the two objects and they fall back to the ground.

The force of gravity or weight, $mg$, acts on each object over the distance $h$ so that we end up with the final kinetic energy of the objects, assuming the effect of gravity alone, given by:

$$KE_{grav} = 2mgh$$

Therefore we seem to have an energy balance as expected.

But this is not the end of the story. As the charged objects accelerate to the ground they each produce a radiative electric field at a distance $d$ given by:

$$\mathbf{E_{rad}} = -\frac{\pm q}{4 \pi \epsilon_0c^2d}\mathbf{g}$$

Therefore there is an extra downward force on each object given by:

$$\mathbf{F_{em}} =\frac{q^2}{4 \pi \epsilon_0c^2d}\mathbf{g}$$

As this force acts on each object over a distance $h$ then the extra kinetic energy of the objects when they reach the ground due to the mutual effect of their radiative electric fields is:

$$KE_{em} = \frac{2 q^2 g h}{4 \pi \epsilon_0c^2d}$$

We seem to be getting more energy out than we have put into the system.

What's wrong with this calculation?

Postscript

Mark Mitchison has pointed out that I need to find the acceleration for the case where both the electromagnetic force and the gravitational force are acting. I can't assume that it is just the gravitational acceleration $\mathbf{g}$.

Let me try to work out the equation of motion for one of the objects, where both objects have a downward acceleration $a$.

We have:

$$F = m a$$

The total force $F$ on one of the objects is the sum of the gravitational and electromagnetic components:

$$F = m g + \frac{q^2a}{4 \pi \epsilon_0 c^2 d}$$

If we define:

$$m_{e} = \frac{q^2}{4 \pi \epsilon_0 c^2 d}$$

then the total force on an object is given by:

$$F = m g + m_{e} a$$

The equation of motion is then:

$$m a = m g + m_{e} a$$

Thus the acceleration $a$ is given by:

$$a = \frac{mg}{m - m_{e}}$$

The total energy gained by an object under the influence of both the gravitational force and the electromagnetic force as it falls a distance $h$ is then:

$$E = F h$$

$$E = \left(mg + m_{e} \cdot \frac{mg}{m - m_{e}}\right)h$$

$$E = mgh \cdot \frac{m}{m-m_{e}}$$

This is more energy than the $mgh$ that we put into the object in the first place.

P.S. If $m_e$ is small then we have:

$$E \approx mgh + m_{e}gh$$

which is consistent with my original calculation above where I took the acceleration to be simply $g$.

• The objects move on poles? What are the charges of those poles? – Sofia Jan 30 '15 at 11:58
• Your statement that $KE = 2mgh$ is the kinetic energy gained is only true if are no forces acting on the charges as they fall. You have argued that this is not true here. At the end of the experiment, the total kinetic energy + energy in the field should be equal to $2mgh$. This follows just from energy conservation. – Mark Mitchison Jan 30 '15 at 12:02
• The poles are just neutral frictionless poles to force the objects to only move vertically. – John Eastmond Jan 30 '15 at 12:02
• I work out the kinetic energy of the objects due only to gravity first to show that there is an energy balance if only gravity is taken into account. When the forces due to the electric fields is added we end up with more energy out than we put in. – John Eastmond Jan 30 '15 at 12:06
• @JohnEastmond You didn't really "work out" the kinetic energy. You just wrote down an (incorrect) expression and assumed it to be true. It is true that $KE = 2mgh$ if the objects are uncharged, or more precisely if there are no other forces acting on them apart from gravity. If they are charged then you need to include any additional acceleration due to the presence of electric fields. Then you will find that $KE\neq 2mgh$. You will also find that energy is conserved when the energy of the field is calculated properly and included. – Mark Mitchison Jan 30 '15 at 12:09

Your thought experiment stumbles upon an important idea in electrodynamics which is quite counter-intuitive.The EM field produced as radiation due to the charge in fact produces a reaction force on the charge itself.

This is known as the Abraham–Lorentz force which is proportional to rate of change of acceleration of the charge.

In SI units it is given by,

$$F_{rad} = \frac{q^2}{6 \pi \epsilon_{0} c^3} \dot{\textbf{a}}$$

If this force is included as well in your energy conservation,it will hold.

References

Abraham Lorentz Force

• But the acceleration due to gravity g is constant so the charged objects should not feel any Abraham-Lorentz reaction force. – John Eastmond Jan 30 '15 at 12:40
• You can't just assume it to be constant.Include Abharam force in your force equation and we get an ODE for $a$ which does not imply $a$ being constant. – Sandesh Kalantre Jan 30 '15 at 14:17
• You get an exponential dependence of $a$ with time in such a way that the $\dot{a}$ is negative and hence the Abraham forces does negative work nulling the positive work done by the $m_e a$ force. – Sandesh Kalantre Jan 30 '15 at 14:39
• If we let $\dot{a}$ be non-zero initially then we get an unphysical runaway solution that increases exponentially. – John Eastmond Feb 6 '15 at 11:38