"Energy" means "Capability to do work"

We can now write "light is a form of energy" into "light is a form of capability to do work"

Capability can be thought as a Capacity

Whenever we talk about Capability of doing work we have an object that contains the ability to do work. So in this case who is having the ability to do work? The answer could be electromagnetic wave or something but what it is and how it is having the capability?

Is it necessary to have mass to have Capability to do work?

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    $\begingroup$ In Quantum Mechanics, light can be seen as photons with energy $E = h \nu$, where $h$ is Planck's constant and $\nu$ is the photon's frequency, giving a picture of a "packet of energy". Electromagnetic radiation also has energy associated with it (in a "classical sense" without the need of going at the microscopic level). I recommend you to check chapter 9 of Griffith's book "Introduction to Electrodynamics". $\endgroup$
    – RMC777
    Oct 24, 2021 at 15:17
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    $\begingroup$ This sounds like word salad without any physics questions. $\endgroup$ Oct 24, 2021 at 15:18
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    $\begingroup$ I know this was a silly question. Actually i was going to ask if we say light as energy which is not the case here $\endgroup$ Oct 24, 2021 at 16:07
  • $\begingroup$ If you are asking what kind of energy is light then it is kinetic energy given by $E=pc$. Since photons have no mass, you set $m=0$, and therefore all the energy is due to the photon's movement/momentum. If you wanted to do work with light, then you could read about the solar sail $\endgroup$
    – Tachyon
    Jan 30 at 22:01
  • $\begingroup$ How can we have momentum without mass? Is there any other definition for momentum that I haven't read yet? $\endgroup$
    – user316791
    Feb 1 at 8:35

10 Answers 10


Energy is a physical property of a physical system. There is nothing like pure energy but maybe in Dragon Ball cartoons. Not in our universe. Therefore, the question if the light is a form of energy is not consistent with known physics.

What can be safely said is that light, either described as electromagnetic waves or as photons, is a physical system that has some energy. Then, it may do work.

Having mass is not necessary for the ability to do work. Systems with or without mass do have energy.

  • $\begingroup$ How this system carries the capability to do work or energy in it? Can it be described? How massless thing can do work? Well the most important thing is the force. So how massless things apply force? Or if it is a property of elctromagnetic wave that whenever it interacts with charges, it applies force then, can we say that a given length of field is associated with a fix ed amount of energy? cause the force is applied by a point in field and only displacment is required which can be provided by giving a length of field. Thank you! If yes then light ray is also a line of force. $\endgroup$ Jan 30 at 20:46
  • $\begingroup$ @PredakingAskboss In terms of electromagnetic fields, it should be quite clear how fields can do work. In terms of photons, the most appropriate description is in terms of energy exchanges. However, I have the impression that you should master more physics to formulate your doubts in a more precise way. $\endgroup$
    – GiorgioP
    Jan 30 at 21:44
  • $\begingroup$ @PredakingAskboss, "having" energy does not imply a capability to do work. Work is a transfer of energy, a change in the placement of the different forms of energy in the system. $\endgroup$
    – Steve
    Jan 31 at 12:54

As an easy example, think of solar panels. Inside a solar panel you have atoms, surrounded by electrons. When light hits those electrons they jump to a higher orbit and generate electricity. (Einstein won his Nobel Prize for discovering the Photoelectric effect). This is certainly "work" as you have defined it. It is not necessary to have mass, only energy, to do work.

And of course you can look at nature. Light strikes leaves and performs "work" with photosynthesis.
And you can look at a solar light mill. This is certainly work. Solar Light Mill

  • $\begingroup$ If it doesn't have mass then it can't have kinetic energy. So how does it store energy in it? Can we say that there is no energy in its motion? $\endgroup$ Oct 24, 2021 at 16:09
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    $\begingroup$ @PredakingAskboss Because a photon has no mass, its kinetic energy is equal to its total energy. Here is an interesting article on why light is pure energy wtamu.edu/~cbaird/sq/2015/01/12/why-is-light-pure-energy $\endgroup$ Oct 24, 2021 at 16:48
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    $\begingroup$ @nasu Einstein discovered the explanation for the photoelectric effect, I’d say. The effect was already known but no one could explain its specific properties until Einstein came along. $\endgroup$ Oct 24, 2021 at 19:36
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    $\begingroup$ -1, because Crookes radiometer has nothing to do with light pressure or photon energy. It's pure thermodynamic effect due to temperature gradient in bulb vacuumed gas. Also look at this nice exploration. $\endgroup$ Jan 30 at 14:31
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    $\begingroup$ @AgniusVasiliauskas +1 because I don't se anything referring to light pressure or photon energy. Are you trying to say that Crookes radiometers are not powered by light energy? $\endgroup$
    – M. Enns
    Jan 30 at 21:08

"Energy" means "Capability to do work"

Many problems with this :

  1. It's circular. If energy is capability to do work, then work is consumption of energy. No way out.

  2. Not all forms of energy can do useful work for us. For example it's estimated that lower bound of false vacuum energy density is about $5~\text{KeV∕cm^3}$, however there's no way to extract these quantum vacuum fluctuations as meaningful work. Of course if to believe that universe accelerated expansion is due to dark energy, then quantum vacuum fluctuations does work actually as per $$\Delta W=P\,\Delta V=wc^{2}\rho~\Delta V$$ where $\rho$ is vacuum energy density, and $w$ is some cosmological dimensionless constant which defines model of universe expansion. So as universe expands in volume,- more and more work is done in each time span. Which means that more and more vacuum energy is spent. Where this energy comes from ? I can't wrap my head still that it comes out of nothing - from virtual particles - which are being generated constantly as each new parcel of empty space is borned, even being a die-hard fan of quantum mechanics. So it's kind of loop - new space produces virtual particles which pushes space apart, producing yet new space, which produces new virtual particles, and so on, so on. If universe will expand forever in accelerated way, then this would mean that there's infinite amount of energy inside it and infinite amount of work will be done too. Kind of perpetual machine, which doesn't make sense.

What do we mean by saying light is a form of energy?

You ask here particularly about light. Light quanta - photon has no rest mass, so based on it my quantum mechanical interpretation would be such that "pure form" of energy is mass-less particle(s) like photon. Given amount of physical rest mass,- how much pure energy you can convert from it, is given by Einstein famous eq: $$ E=mc^2 $$ or if expressed in terms of amount of massless photon particles: $$ n = \frac {mc^2}{h\nu} $$

But again, this explanation is subjective and I doubt if objective one exists in your question.

  • $\begingroup$ Whenever a work is done, Force is applied and displacement occurs. Infact the work is done by force. Is light just a force if you are calling it a pure form? But then a question comes too, for how long(time taken) could that force exist or how long light quanta- photon, as you said, can do work? $\endgroup$ Jan 30 at 20:36
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    $\begingroup$ You have in mind only mechanical work, which does not necessarily the case, it can be also a thermodynamical work or any other. As about how work is related to energy consumption. For example suppose you decreased car kinetic energy by some amount bumping into it, then car has made work towards you $W = \Delta E_k$. Or by using gas internal energy made moving some Stirling engine, which dropped gas temperature, thus gas has produced work $W = \Delta U$ $\endgroup$ Jan 30 at 20:36
  • $\begingroup$ I don't understand the term "pure energy". What do you mean by it? $\endgroup$ Jan 30 at 20:39
  • $\begingroup$ I meant by it massless particles like photon, gluon, graviton (if exists). Most particles which have rest mass, at least hypothetically could be converted into quark-gluon plasma. At least, because universe at some point only existed as quark–gluon plasma, so everything we see came later from it. $\endgroup$ Jan 30 at 20:51
  • $\begingroup$ It seems like a ray of light is actually a line of force. It can apply force over a length when it acts on charge. If it is right then is there a fixed number of rays given off by a light source like a bulb. Otherwise there would be infinite amount of energy that can get released by it. Or the light is beam like thing that has some thickness. And taking length of this beam gives a volume and therefore no infinite energy $\endgroup$ Jan 30 at 21:01

Here is a classical explanation, (not involving QM), of the effect of a EM wave in a conductor like an antenna:

Any charge $q$, under the joined effect of the electric and magnetic field of a plane wave, is under Lorentz force:

$\mathbf F = q(\mathbf E + \mathbf v\times \mathbf B)$

and the wave does work on the charge due to this force. For a small displacement $\Delta x$:

$\Delta W = q(\mathbf E + \mathbf v \times \mathbf B)\boldsymbol{.Δx}$

Supposing propagation in the $x$ direction, there are no field components in this direction:

$\Delta W = q((0 ,E_y ,E_z ) + (v_y B_z - v_z B_y , -v_x B_z , v_x B_y )).(\Delta x,\Delta y,\Delta z)$

$\Delta W = q((v_y B_z - v_z B_y )\Delta x + (E_y – v_x B_z )\Delta y + (E_z + v_x B_y )\Delta z)$

Dividing both sides by the volume, and by a small time interval we get the energy per time in a unit of volume. Taking the limit when the $\Delta$'s go to zero:

$$\frac{\partial W_v}{\partial t} = \rho(v_x v_y B_z - v_x v_z B_y + E_y v_y – v_x v_y B_z + E_z v_z + v_x v_z B_y )$$

where $\rho$ is the density of charges. The components of $\mathbf B$ vanishes, and only $\mathbf E$ does work on the charges.

$$\frac{\partial W_v}{\partial t} = \rho(E_y v_y + E_z v_z)$$

If we call the product $\rho \mathbf v = \mathbf j$ as density of current: $$\frac{\partial W_v}{\partial t} = \mathbf {j.E}$$

That is the Joule effect on the conductor, leading to an increase of its temperature, what means an increase of its internal energy.

  • $\begingroup$ Is there a continuous beam of light coming towards me or is it quantised? $\endgroup$ Jan 30 at 21:03
  • $\begingroup$ For this type of effect it is not necessary to deal with quanta. As to design a hydroelectric plant it is not necessary to deal with the molecular nature of water. But in other cases, as for the power generated by a fotovoltaic panel, QM have to be used. $\endgroup$ Jan 30 at 23:14

Capability can be thought as a Capacity

Not necessarily. Although energy is a capability that represents a capacity, there are also capabilities that are not capacitive in the sense that they are filled up by something that is drawn from another resource. Just to take a physical example: you have the capability to destroy a precious china vase by hitting it strongly with a hammer. And although there is also energy involved in this process, you could also have used this same amount of energy to tinily tickle the vase a billion times without destroying it, or to lift up the vase several meters or so. The actual physical quantity that makes the difference between destruction and integrity is called "entropy", and in contrast to energy, the entropy of a system is not conserved, but rather grows all the time. And because it grows, you cannot make a broken vase whole again, at least not without working about as hard as someone did when it was first manufactured. So, the capability to destroy (or disorder, or produce entropy) is not a capacity, but rather depends on how you do something, and hence, is related to information. Contrary to that, energy is conserved, and that justifies the special role it has in physics. Any force (e.g. friction) can "do something", but only energy (and some others, like momentum and certain elementary particle numbers) can do it in a way that involves taking a specific amount from A and putting it into B.

Whenever we talk about Capability of doing work we have an object that contains the ability to do work. So in this case who is having the ability to do work?

An "object" is a philosophic concept. Is a hammer an object? In your mind, it is, because you can think about it. In technology it is, because somebody manufactured it. In physics, it depends on the level of abstraction.

If you apply classical mechanics, a hammer is an object with a mass, a center of mass, a position, and a velocity, probably also an inertia tensor and angular orientation and velocity. In continuum mechanics and/or thermodynamics, a hammer is a completely different object, namely a field of densities, displacements from equilibrium, velocities, maybe temperatures, chemical potentials etc. In microscopic physics, the hammer is a bunch of atoms, hopefully stuck together strongly enough to support the function of the hammer.

So is the rigid body the "object" hammer? Is the field in continuum mechanics "the" object? Is it the configuration of atoms that makes up the hammer? No, because "object" is not a term that is well-defined independently of context.

Likewise in electromagnetism. All you need to be able to do is measure something that contributes to the evolution of the system. In this case you could say, the vaccuum itself "possesses" the electromagnetic field and hence the energy. But the vacuum is already an abstraction, probably only one that serves mnemonic purposes, the vacuum does not exist in the strict sense (because there is almost always something in it, an empty universe would be pretty boring). All you have is the measurement and how you got it. And if you look deeper, you might find something more complicated.

On the most fundamental level of contemporary physics, the energy of the electromagnetic field is ascribed to an "object" (if that is what you desire to call it) called photon, which is present in certain numbers and probabilities (and each species of photon, of course, has different polarization and momentum/wavelength).


Well actually I don't think we do say "light is a form of energy". Not in science discussions anyway. I would be interested to know if that phrase can be found in any physics textbook. It probably either cannot, or else it might be used in a loose way of speaking when talking about energy conservation or something like that. But no good physicist would use such a phrase in order to say what light is. Anyone expressing physics with a correct grasp of the meaning of the words would say "light is an oscillation of the electromagnetic field" or "light is the name we give to the electromagnetic field oscillations in the frequency range ..." or "light is an excitation of the electromagnetic field" or some such phrase. They might go on to add, "such oscillations have energy and momentum. So if something emits light, it has to provide this energy and momentum, and if something absorbs light, it will absorb the energy and momentum of the light".

In physics there is no such thing as "pure energy"; rather, energy is always an attribute of some physical system.

There are various ways to get mechanical work as an end result after absorbing light. One is via the photo-electric effect. In this effect, as a result of absorbing light, electrons are transferred from one place to another place at higher electric potential (voltage). This provides an energy source which can be used to power an electric motor for example.

The energy in radio waves can similarly be used to drive an electrical circuit by the use of an antenna.


Light is a form of energy just like mass. At quantum scales energy transactions happen by absorption or emmision of photons(meaning light). We also know that light contains energy which is determined by the frequency of photon.E=hf tells about the energy of a photon. For e.g If a electron wants to escape a atom it needs some specific energy to escape. That energy is provided by light of sufficient energy so that it escapes the atom. An electron can also gain kinetic energy by absorption of a photon. I hope you understand why light is called form of energy.


Energy means capability to do work. It's very true, both in principle as well as practice. From Einstein equation, we have $$E^2 = (mc)^2 + (pc)^2$$ Now, as you want, for a photon, mass should be equated to zero. Let's do it. $$E^2 = ( pc)^2$$ Hence, the work is provided by the momentum .


The best proof that light is energy is... Earth.

Except from volcanism and earthquakes that come from the inside, and tides due to its rotation around its own axis,that is not synchronized with the moon, all the energy on Earth got here... by way of light from the sun.

Weather is driven by differences in temperature of regions that get more or less light from the sun, of the amount of water vapor that is evaporated from the seas by the heat generated by light from the sun.

Except for deep-ocean volcanism that allows for pockets of life, almost all life on Earth is supported by photosynthesis that uses light from the sun.

Coal and oil come from plants buried long ago but that also grew thanks to photosynthesis.

Energy that arrived in form of light from the sun is everywhere.


Here I will answer more precisely to a comment by the author of the question, namely Predaking Askboss, just under his original question.

If it doesn't have mass then it can't have kinetic energy. So how does it store energy in it? Can we say that there is no energy in its motion?

It is a very frequent fallacy to believe that light, or a photon, has no mass.

Of course it has mass. What is zero is its rest mass. Since light, that is photons, can never be at rest, one cannot measure directly rest mass. But we know it is zero as I'll explain below.

Particles (or objects) that have a nonzero rest mass can be at rest and have their rest mass ($m_0$) measured. When in motion, their total mass in motion $m$ is higher than their rest mass by their kinetic energy divide by $c^2$


For a photon, what is zero is the rest mass $m_0$. The total mass in motion of a photon is just


where $E_k=\hbar \omega$ is all its energy. And this is how we know it has zero rest mass.

And light, or photon also has momentum. The momentum $p$ of a photon is its total mass in motion multiplied by its velocity, which is $c$

$$p=mc=E_k/c=\hbar \omega/c$$

So you can actually use the momentum of light to push a lightsail in space, for instance. This is direct use of the radiation pressure of light, which is the incoming flux of energy by unit surface divided by $c$. This however, is a very inefficient way to use this energy.

Consider a car at rest on a very flat road. If you have a squirt gun, you can use 10 gallons of gasoline to move the car : fill your squirt gun with gasoline and aim the jet at the car. Over and over and over till the 10 gallons are used. You will probably manage to displace the car by a few centimeters. Use the same amount of gasoline in the engine, you drive many kilometers.

Using the momentum of light on a lightsail may be practical in deep space, not very much on Earth. What you really want to do is turn the energy $\hbar \omega$ of light into heat, or even better, into chemical energy and use it later.

Incidentally the solar light mill, or Crookes radiometer presented by foolishmuse does not really use the momentum of light. If it were the case, it should turn black face forwards, because the dark face absorbs the photons an receive their momentum, but the silver face reflects them and thus receive twice their incoming momentum, since when they are reflected they carry away the same momentum in the opposite direction. That would indeed be the case in perfect vacuum.

But even though there is a partial vacuum in the glass sphere there is much too much air left to work this way. In fact you can see on the wiki page that it turns silver face in front.

The silver face reflects the light and stays cool. The black face absorbs then and heats up. This heats up the small amount of air in the immediate neighbourhood of the black face. This air expands by heating and pushes the black face away. Hence the rotation silver face in front.

So even this solar light mill is an application of the heat transported as energy by the light, not its momentum.

But energy, momentum and total mass at rest are all nonzero for photons.

  • $\begingroup$ But that was not the question!🥶 Question was to describe that form of energy. Just like a mass with kinetic energy has the capability to apply a force when it decelerates and can do some displacement too on any mass. $\endgroup$ Jan 31 at 21:43
  • $\begingroup$ @PredakingAskboss What do you mean "describe that form of energy" ? When one photon is absorbed by a chlorophyll molecule, it can displace an electron from a low level to a higher level in that molecule. The activated chlorophyll molecule is able to initiate a series of chemical reactions leading to extract a carbon atom from a CO$_2$ molecule. All that from the ability of this photon to displace an electron. Is this the answer to your question ? This is a QM form of energy that has no classical explanation. $\endgroup$
    – Alfred
    Jan 31 at 22:32
  • $\begingroup$ Claudio Saspinski gave you an example that makes sense in a classical way, but can also have a QM interpretation. As for the mass you miss the distinction between "rest mass" and "total mass in motion". If the rest mass is nonzero, the "total mass in motion" is the rest mass plus kinetic energy divided by $c^2$. For a "massless" particle like the photon, which means zero rest mass the total mass in motion (and in this case it must be motion at light speed) is just energy divided by $c^2$. And the momentum is just energy divided by $c$, i.e. total mass in motion times $c$. $\endgroup$
    – Alfred
    Jan 31 at 22:42
  • $\begingroup$ Describe meant to explain how light does some work whenever it gives energy to something. $\endgroup$
    – user316791
    Feb 1 at 8:32

Let us consider a charged rod. If said rod moves at speed 0.86 c, it length-contracts to half of its rest-length, Which doubles the energy of its electric field.

Now the rod may move past a resistor uncharging through said resistor, in which case the resistor heats up gaining heat energy E.

Or the rod may uncharge through a resistor that is moving with the rod, in which case the resistor heats up gaining heat energy E/2. The heat energy is moving at speed 0.86c so it has kinetic energy E/2.

What we have learned from this thought experiment is that a moving electric field has kinetic energy and field energy, and said kinetic energy and said field energy are not separate things.

Now as we may perhaps understand light as bunch of moving electric fields, we can understand that those moving electric fields can do work the same way as electric fields do work: by accelerating electrons.

Or those moving electric fields can do work the same way as moving things do work: by colliding with a piston for example.


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