Why are so many energies represented by $\frac{1}{2} ab^2$? [duplicate]

Why are so many energies in our universe mathematically represented by the equation $$\frac{1}{2}ab^2$$.

For example:

• Kinetic energy $$\frac{1}{2}mv^2$$
• Energy stored in a capacitor $$\frac{1}{2}CV^2$$
• Energy stored in an inductor $$\frac{1}{2}LI^2$$

These are just some I can remember off the top of my head ,I remember that there are more like this. Is this just a coincidence or is there a reason behind this ?

I tried to google it but couldn't find any relevant results.

marked as duplicate by StephenG, Dvij Mankad, John Rennie energy StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 7 at 14:41

• Some can be explained due to potential energy being the integral of the force associated with it: $$F=\frac{dU}{dx} \Leftrightarrow \int F \, dx=U$$ A constant force, such as gravity close to earth $w=mg$, gives a linear potential energy, $U=\int w \, dx=mgh$, while a linear force, such as the spring force $S=kx$, gives a quadratic potential energy, $U=\int S \, dx=\frac12 kx^2$. So, have a look at the underlying forces - at least in the case of potential energies - and you might find some of the explanation. – Steeven Apr 7 at 7:56
• Because physicists like using linear models a lot. – onurcanbektas Apr 7 at 8:56
• Related: physics.stackexchange.com/q/535/2451 , physics.stackexchange.com/q/159021/2451 and links therein. – Qmechanic Apr 7 at 10:06

This is purely because of dimensional reasons. Energy has dimension $$\rm ML^2T^{-2}$$, and you can construct this dimension from different combinations of dimesionful quantities. Note that, the mass $$m$$ and velocity $$v$$ has dimensions $$\rm M$$ and $$\rm LT^{-1}$$ respectively which makes the unique combination $$mv^2$$ to have the dimension $$\rm ML^2T^{-2}$$. Similarly, the capacitance $$\rm C$$ and potential difference (voltage) $$\rm V$$ has dimensions $$\rm M^{-1}L^{-2} T^4 I^2$$ and $$\rm ML^2I^{-1}T^{-3}$$ respectively and so on. For completeness, you can construct a few more. For example, the energy stored in an electric and magnetic field in a volume $$\rm V$$, elastic potential energy in a spring are respectively given by $$\int_{\rm V}\frac{1}{2}\epsilon_0\textbf{E}^2dV$$ and $$\int_{\rm V}\frac{1}{2\mu_0}\textbf{B}^2dV$$, $$\frac{1}{2}kx^2$$ etc.

Now, it remains to explain the factor of $$\frac{1}{2}$$. However, you should note that you can construct many other examples of dimension energy which do not carry the factor of $$1/2$$. For example, the gravitational potential energy of a particle of mass $$m$$ on the surface of earth is given by $$U(R)=-\frac{GMm}{R^2}$$ and so on. Also, energies need not necessarily have a quadratic dependence on some observable. For example, the thermal energy per particle is given by $$\sim k_BT$$.

The factor of $$1/2$$ is simply a matter of definition. For example, you define the kinetic energy to be $$T=\frac{1}{2}mv^2\tag{1}$$ or, alternatively as,

the work done required to change the velocity of a particle from $$0$$ to $$v$$.

When the energy is quadratic in some observable $$\mathscr{z}$$, the origin of this $$1/2$$ factor, if it appears, always come from an integral of the type $$\int \textbf{F}\cdot d\textbf{r}\sim \int \mathscr{z} d\mathscr{z}\sim \frac{1}{2}\mathscr{z}^2.$$ Now, the quantity $$\mathscr{z}$$ will vary from one situation to another. In the example of kinetic energy, $$\mathscr{z}=v$$.

• What is a "dimesioful quantity" ? Googling it has literally zero results. Is it a quantity which can be represented in multiple forms (for ex : energy in a capacitor can be represented as 1/2 CV^2 or 1/2*Q*V or Q^2/(2C)) – Shailendra Sorout Apr 7 at 7:35
• @ShailendraSorout It's a typo. SRS meant to type dimensionful quantity – PM 2Ring Apr 7 at 9:25
• SRS, the factor of 1/2 in the kinetic energy formula is not arbitrary. If you drop an object from rest and calculate its kinetic energy as it falls, you will find that conservation of energy considerations require this factor. – David White Apr 7 at 16:20
• Yes. It's not. @DavidWhite – SRS Apr 7 at 16:51

For completeness, suppose you have any energy expression $$U=U(k)$$ that you are looking at. Energy typically has to be bounded from below, otherwise thermodynamics would send the system to negative-infinite energy to spread that energy everywhere across the universe. Let's say that it attains a minimum at a special value $$k_0$$ and we define $$b=k-k_0$$, then $$U$$ has a Maclaurin expansion,$$U(b) = U_0 + \frac12 U''(0) ~b^2 + \frac16 U'''(0)~b^3+ \frac1{24} U^{(4)}(0)~b^4 +\dots,$$ which must lack a linear term because it is minimized at $$b=0$$.

The parameter $$U_0$$ turns out to never affect the dynamical behavior of the system and thus it can always be artificially forced to $$U_0 = 0$$ and then nearly every system for which $$U''(0)\ne 0$$ will have a domain of valid $$b$$ where, \begin{align} b &\ll 3~ U''(0)/U'''(0),\\ b &\ll \sqrt{ 12~ U''(0)/U^{(4)}(0) },\\ b &\ll \sqrt[3]{60~ U''(0)/U^{(5)}(0)}\dots \end{align} and there exists a valid approximation $$U \approx \frac12 a b^2.$$ Note that once you have this, it is natural to simply define $$a$$ as this $$U''(0)$$ parameter as the system will then have a “restoring force” where any change by $$\delta b$$ changes the energy by some amount $$\delta U$$ such that $$\delta U/\delta b = a~b$$. So the “force” is proportional to the “displacement” and this parameter $$U''(0)$$ is the proportionality factor, so it is usually a very natural thing to just give a specific name to like “capacitance” or “inductance” or “mass”.

All of your examples have this form. For example we discovered at some point that the above formula for kinetic energy was wrong and that the appropriate expression instead looks like,$$E(v) = \frac{mc^2}{\sqrt{1 - (v/c)^2}}.$$Does this mean that everything we have been doing before this formula was completely backwards? No, in fact we can expand this as,$$E(v) = m c^2 + \frac12 m v^2 + \frac38 \frac{mv^4}{c^2} + \dots.$$But for $$v \ll c$$ we can just ignore the terms on the right and the term on the left does not affect the dynamics so we can artificially replace it with $$0$$ and simply have $$E = \frac12 m v^2.$$

Similarly for your other examples; for example your capacitor equation says that the change in energy due to adding a little charge $$\delta Q$$ to an existing charge $$Q$$ is $$\delta E = V~\delta Q = V~C~\delta V.$$ But this sort of behavior is an idealization and one might expect nonlinearities for any real capacitor.

• you must become a writer and write an introductory physics book – user225529 Apr 7 at 15:27
• agree. CR, this is most useful and clear. – niels nielsen Apr 7 at 18:05

In the time before the concept of kinetic energy was clearly recognized there were two distinct concepts of 'quantity of motion' in circulation. For instance, Leibniz favored a quantity $$mv^2$$ as representing 'quantity of motion'. In elastic collisions the quantity of motion is conserved.

So that precursor to the concept of kinetic energy didn't have that factor $$\frac{1}{2}$$

$$mv^2$$ has the following property: if you look at a perfectly elastic collision between two particles then $$mv^2$$ for the two added together is the same before and after the collision. This holds good for any frame of reference. That is, you can look at the motion relative to the coordinate system in which the common center of mass is stationary, and obviously the quantity of motion is the same before and after the collision then, but it also holds good for any inertial coordinate system.

Later on physicists noticed more and more that in all kinds of energy conversions the total amount seems to be conserved.

For example, James Prescott Joule did experiments where a paddle wheel would churn water, driven by a string that ran over a pulley, being pulled down by a weight. The water was inside a calorimeter. Joule found that within the margin of achievable measuring accuracy the rise in temperature of the water was proportional to the mass of the weight and the height of the drop.

So that line of experiments was strongly suggestive that there is such a thing as gravitational potential energy, and that the apparatus converted gravitational potential energy to heat.

But you can also see that there is an intermediate energy form: kinetic energy of the motion of the weight as it moves down.

The finding that temperature rise was proportional to heigh of the drop was corroborating evidence for a notion that it is meaningful to define a concept of work. Work being done is proportional to the force and proportional to the distance over which the force has been doing that work. In barebones notation: $$W=Fs$$

If you take a concept of gravitational potential energy as granted, and global conservation of energy, then you arrive at the following expression for kinetic energy: $$\frac{1}{2}mv^2$$

Generalizing: presumably all expressions for types of energy that have that form with a factor $$\frac{1}{2}$$ and a squared quantity arise in cases where some force is doing some form of work over some form of distance, so that it can be expressed as $$W=Fs$$

As pointed out in the earlier answer, the factor $$\frac{1}{2}$$ can be seen as arising from an integration step.

Conclusions:
When you move to a global concept of conservation of energy the demand of self-consistency narrows down the expression for kinetic energy to: $$\frac{1}{2}mv^2$$

Before that, before there was any notion of conversion of kinetic energy to some other form of energy the definition of $$mv^2$$ was good, as the only constraint was that the total amount of it is conserved.