I am something of a dilettante in physics, so please forgive me if the answer to this question is painfully obvious. The question is simple, can something that theoretically has no mass exert a force. I have been tossing around this and other similar questions in my head for a while now and have not really found any concrete answers to my inquiry. I am thinking about how light seems to be able to push objects but yet has no mass, however I expanded the question to be more encompassing in hopes of further learning.

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    $\begingroup$ It looks that you already know the answer to your question: yes, light is massless yet it can exert a force. Maybe your question should be "how is this possible?", right? $\endgroup$ Commented Apr 26, 2016 at 6:05
  • $\begingroup$ Think of electric charges. An ideal point charge does indeed exert large forces on other charges nearby. $\endgroup$
    – Steeven
    Commented Apr 26, 2016 at 7:03
  • $\begingroup$ I feel all the answers here are simply bad. (Sorry!) All answers here are ontological assertions which quickly degenerated to core arguments on the topic. $\endgroup$
    – Fattie
    Commented Apr 26, 2016 at 12:57
  • $\begingroup$ Related: physics.stackexchange.com/q/2229/2451 $\endgroup$
    – Qmechanic
    Commented Apr 27, 2016 at 10:18

3 Answers 3


Yes, photons can. See https://en.wikipedia.org/wiki/Radiation_pressure (and photons are certainly massless).

PS In fact, any massless particle has momentum(*) and if it is scattered on a body, it changes its own and the body's momentum, which is what a force does.

(*) $p = \hbar k = E/c$ where $E$ is its energy and $c$ is speed of light

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    $\begingroup$ Might be helpful to elaborate on "photons are surely massless". They don't have a rest mass, but have momentum, and are deflected by gravity and do gain energy falling towards a large mass. I try not to say one way or the other! Look here in your linked article and you'll see "Although photons are zero-rest mass particles, they have the properties of energy and momentum, thus exhibit the property of mass as they travel at light speed." in your linked article. $\endgroup$
    – uhoh
    Commented Apr 26, 2016 at 6:06
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    $\begingroup$ @uhoh That isn't very popular way of putting it anymore. The concept of rest mass is kind of a confusion of relativity (kind of like the Schröedinger's cat is a ridicule of a confusion of quantum physics). The reason "massive" objects have mass is still the same reason why "massless" objects have mass, be it ultimately sourced from the electro-magnetic force or the nuclear forces or whatever else you fancy. $\endgroup$
    – Luaan
    Commented Apr 26, 2016 at 8:06
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    $\begingroup$ One should add that F=dp/dt for the argument to be complete $\endgroup$
    – anna v
    Commented Apr 26, 2016 at 8:26
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    $\begingroup$ @uhoh Depends on what context you're speaking in, I recon. Within relativity, there's no meaning to an object that has literally zero mass - if it doesn't have mass, it doesn't have energy, which means it doesn't interact, which means it doesn't exist. So "massless" can not be misinterpreted as meaning "literally zero mass", and there's little harm in using it to describe what "particles with zero rest mass" used to mean. Assuming the Higgs theory is correct, the distinguishing feature is the Higgs interaction - massless particles don't interact Higgsy and move at the speed of light in vacuum. $\endgroup$
    – Luaan
    Commented Apr 26, 2016 at 8:43
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    $\begingroup$ @uhoh although your note is relevant as we should make this clear for non-professionals, the modern (professional) view on this is rather simple and established: there's only one mass (which is the same as an obsolette term "rest mass") and it is conserved as long as inner structure of an object is conserved. This is discussed in detail in DOI: 10.3367/UFNr.0158.198907f.0511 which has a translation DOI: 10.1070/PU1989v032n07ABEH002739 but unfortunately the english version seems to be behind a paywall iopscience.iop.org/article/10.1070/PU1989v032n07ABEH002739/meta $\endgroup$
    – YakovL
    Commented Apr 26, 2016 at 21:13

Newton's 2nd Law of Motion gives the impressed force as $F=dp/dt$, so a physical theory for a massless particle exerting a force requires that the particle have momentum, $p$.

First we will discuss mass, momentum, the force law, and Special Relativity.

In Newtonian physics mass is identified in two ways: by it's inertia, or as the quantity of matter. The ordinary measurement is by comparison, with a known force, or a balance. Early experiments, 1905-06 with charged particles accelerated through a controlled voltage found that the inertial mass varied with the change in kinetic energy acquired, confirming earlier predictions of Lorentz, 1904, and Einstein, 1905.

The term rest mass, denoted $m_0$, entered the physics lexicon, along with the longitudinal and transverse mass; these two additional terms were required because the measurements vary depending on where you are. For some purposes these are still useful, but it turns out that the rest mass is a relativistic invariant, remaining unchanged under a Lorentz boost. So in modern terminology mass means rest mass, and is denoted by $m$, or for the old fashioned, occasionally $m_0$.

The Lorentz factor, $$\gamma=1/\sqrt{1-v^2/c^2},$$ provides the relativistic correction required for momentum, $p=\gamma mv$ replacing the Newtonian $p=mv$ for a particle with mass. So Newton's Second Law of Motion remains $F=dp/dt$.

Now we will discuss light, and how it carries momentum, and the concept of being massless. For the Physics FAQ summary, see here.

Light, as first noted by Maxwell, travels with the same speed in the vacuum, independent of the observer's inertial reference frame; for the calculation see Deriving the speed of the propagation of a change in the Electromagnetic Field from Maxwell's Equations. For the history, see here.

The relation between energy and momentum of light can be found from the Poynting vector, as derived from the field equations in vacuum. The result is that $p=E/c$, which follows from the radiation pressure.

The relativistic equation for total energy includes momentum, and is $E^2=(pc)^2+(mc^2)^2$; when $p=0$ this simplifies to the iconic $E=mc^2$. For the case with momentum, but no rest mass, we get $E=pc$, which provides the relativistic expression for the momentum of light: $p=E/c$, which is consistent with Maxwell's equations.

So for a self-consistent relativistic theory, if we start with Maxwell's equations, we end up with freely moving light having momentum, but no mass. If the light is trapped in a stationary box it will contribute to the weight of the box in proportion to the energy of the light, $m_L=E_L/c^2$ to the mass of the box without the trapped light.

At this point we have shown the road map for (a) relativistic force law, $F=dp/dt$, and (b) that light has momentum, $p=E/c$, and (c) that this momentum implies that light has no mass. So we now introduce the photon, a particle of light.

Historically, Planck introduced the hypothesis that the energy of light may be quantized, $E=hf$, where the energy of each quanta is determined by it's frequency. Einstein applied this concept to the photo-electric effect, and de Broglie, using Einstein'g Special Relativity plus the Planck relation proposed a complementary relation for the wavelength of a mass with momentum, $p=h/\lambda$. This expression is equivalent to the Planck relation when $p=E/c$ is inserted on the left hand side, because $\lambda f=c$.

Planck and de Broglie, together, provide the foundation for wave mechanics; the term "photon" for this massless quanta, or particle of light first appeared in the literature in 1926.

So in conclusion, yes, something without mass, the photon, can apply a force; this is done through it's momentum.

Experimental verification must be done carefully, for a force may be applied by absorption, or reflection. In the case of absorption the change of momentum is $|p|$, while for reflection it is doubled, as the momentum acts both coming and going.

For absorption the demonstration can be made with Crooke's light mill, often referred to as a radiometer; this clearly responds to light, but the analysis is complex, and does not directly show the pressure due to light.

The direct detectionof light pressure due to reflection requires a fine torsion balance mirror in a vacuum, first performed successfully in 1901; today it can be performed in an undergraduate advanced physics lab.

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    $\begingroup$ radiometers do not demonstrate light pressure; if they did, they would not spin backwards in freezers. $\endgroup$ Commented Apr 27, 2016 at 2:46
  • $\begingroup$ @RossPresser: Clarified that the radiometer does not directly detect light pressure. $\endgroup$ Commented Apr 27, 2016 at 14:33

According to the relation

$$\mathrm{Force = Mass \times Acceleration}$$

force is defined as the phenomenon, which creates acceleration in a body with mass. The mass is for the body on which it acts and not the body which acts. Theoretically it is possible for light to exert force since it has energy and momentum.

Hope your doubt is clear

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    $\begingroup$ The equation stated is not accurate. A force does not imply acceleration of a mass - only the sum of forces $\sum F$ does. $\endgroup$
    – Steeven
    Commented Apr 26, 2016 at 7:07
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    $\begingroup$ No, $F = \dot{p}$ $\endgroup$ Commented Apr 26, 2016 at 14:18
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    $\begingroup$ This answer just shows one law that doesn't forbid massless things exerting a force. However, it doesn't imply that there isn't another law that forbids it. $\endgroup$
    – JiK
    Commented Apr 27, 2016 at 12:16

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