Why does light not have any mass?

Dimensionless Science and Geometric Maths that provides a completely new perspective on why light does not have any mass.


When the standard model of science suggests that light has a mass of zero, many people will look at the equation E=mc2 and, replacing the value of mass for a zero, see a result that appears to suggest light has no energy. Clearly this cannot be the case, as electromagnetic waves expand at the speed of light (c), which is a universal constant.

Light is not matter

This misunderstanding arises form the fact that electromagnetic waves are not considered to be matter. In fact the term ‘mass’ was derived by Newton, who defined it as a property of matter, which is a rather unambiguous definition. Even today the discussion of what mass actually is continues, however most physicists today agree the light is not matter, and therefore has no mass.

So what is Matter?

You might think that as the concept of mass plays so heavily in the world of science that it would exhibit a clear definition. However, this is not the case. Whilst mass is a property of matter, quite exactly what this property is still lacks any kind of clear definition.

There problem is accentuated what we begin to examine the nature of other massless particles, called bosons. within the atom we find a subtle boundary between particles such as the Proton and Neutron, who mass units are used to establish a vast body of scientific theory, and the quarks that form them, whose existence appears to be mass less.

L.B. Okun from the Institute for Theoretical and Experimental Physics, in Russia explains the problem.

Various facets of the concept of mass are discussed. The masses of elementary particles and the search for higgs. The masses of hadrons. The pedagogical virus of relativistic mass. From “Principia” to Large Hadron Collider (LHC)
The term “mass’ was introduced into mechanics by Newton in 1687 in his Principia”. He defined it as the amount of matter. The generally accepted definition of matter does not exist even today. Some authors of physics text-books do not consider photons – particles of light – as particles of matter, because they are massless. For the same reason they do not consider as matter the electromagnetic field. It is not quite clear whether they consider as matter almost massless neutrinos, which usually move with velocity close to that of light. Of course it is impossible to collect a handful of neutrinos similarly to a handful of coins. But in many other respects both photons and neutrinos behave like classical particles, while the electromagnetic field is the basis of our understanding of the structure of atoms. On the other hand, the so-called weak bosons W +, W , Z0 are often not considered as particles of matter because they are too heavy and too short-lived. Even more unusual are such particles as gluons and quarks. Unlike atoms, nucleons, and leptons, they do not exist in a free state: they are permanently confined inside nucleons and other hadrons. There is no doubt that the problem of mass is one of the key problems of modern physics.

L.B. Okun
State Research Center,

Institute for Theoretical and Experimental Physics,
117218, Moscow,

Russia – February 2, 2008

Putting aside the problem of the definition of mass, we are still able to examine the nature of is property. When energy is divided by the square of the speed of light we get an objects mass, for example. However this is not the whole truth. Whilst E=mc2 is one of the most famous equations in science it is in fact an abbreviation of the full equation:

((p * v) * c)2 + (M * c2)2

When considering an object (or particle) that is completely stationary then the equation E=mc2 can be used to determine the energy equivalence of its mass. But when a object is in motion we need to include momentum (p) and velocity into the equation. So let’s see if we can solve the whole equation.

What is the Velocity of Light?

You may have heard of Velocity high school physics. Velocity can often be confused with speed. However the key difference is that velocity measures the change in speed and position over a period of time.

If an object changes velocity the we can take the average over a period of time. If velocity is consistent then wherever we sample the time interval we will get the same result. Light travels at a continuous speed in a vacuum, at around 300,000 km per second.

This video from Mr Anderson provides a really great explanation on what velocity is.

Constant velocity is Motion that does not change in speed or direction.

Key Points

  • The velocity can be positive or negative which tells us in which direction the object is moving.
  • Constant velocity means the object is in motion, moving in a straight line, and at a constant speed.
  • This line can be represent by the algebraic formula: x=x0+vt, where x0 is the position of an object at t=0, and the slope indicates the object’s speed.

So you might think then that if light travels in a straight line at a constant speed in a vacuum it should be easy to calculate is velocity, right?

Actually no.

For one, light can act a both a wave and a particle, each of which has a different method for calculating velocity.

However, ignoring the various velocities that light can travel at depending upon its structure or wave formation, and going for a simpler photon particle model, we can suggest that light can travel consistently through the vacuum right.

Again not quite. There is a problem with proving the one way speed of light.

Still, let’s ignore that and speculate that light does travel at the same speed through the vacuum. At least then we will be able to move on to the last part of the mass equation. Momentum.

Momentum of Light

Have you ever tried to push a car? At first it feels really heavy, but then as the wheels start moving suddenly it gets easier to push, unless to are trying to push up hill or course. So what is happening? Is the car getting lighter? Or course not. The car has the same weight, but it is acquiring velocity (v), a change in position and speed, which gives it momentum (p).

When we examine the equation of p (momentum) we find that:

p = m * v

So that seem simple enough, as long as we ignore the fact that mass hold no true definition, and a clear value for velocity has never been established for light. Unfortunately no.

The problem is that velocity comes in different forms. Linear momentum is different to angular momentum that describes a object circling a central axis.

Still if ignore the fact that light bends around gravitational fields, we assume an ideal condition of space. So let’s assume that light travels at a constant velocity, then we should be able to solve the mass-energy equation?

Not quite. There is one more factor that is often forgotten about, and is even more often misunderstood.

Lorenz Gamma and the hidden square

Whilst the equation for momentum is produced from a factor of Mass multiplied by velocity, this is only appropriate for non-relativistic equations. For objects moving at relativistic speeds, close to the speed of light, the concept of γ (gamma) needs to be introduced. Therefore the correct equation for momentum should be written as:

p = γ * m * v

The reason gamma is not included in non-relativistic equations is because its value is extremely close to the number 1, and so it is often ignored. However, as an object with mass is accelerated towards the speed of light, gamma begins to increase exponentially. The equation for gamma is:

γ = 1 / (1-(v/c)2)

The video from Fermilab’s Dr. Don Lincoln explain the concept in more detail.

So now that we understand Mass, Velocity, Momentum and Gamma, we finally solve the full Einstein equation and understand why photon has to be a massless particle. And to provide the solution we can use the system of dimensionless science.

Dimensionless Mass, Light, and Energy

To begin we are able to simplify the speed of light to the number 3, which is derived from a unit of time 1 VS a unit of space 3. The speed of light come out of distance over time. So 3 divided by 1 is 3, the speed of light. In dimensionless science we do not apply the concept of a second or the meter, just a ratio. In this case the Si unit equivalent would be 299,792,458 m / s, so we are operating at a 100,000m scale.

When the speed of light is set to 3, then this provides a very simple relationship for the conversion of mass to energy, whereby for each unit of mass (1) we can multiply by 32 to ascertain its energy equivalence:

m = 1 1 * 32 = 9

m = 2 2 * 32 = 18

m = 3 3 * 32 = 27

m = 4 4 * 32 = 36

Notice then that for a non-moving particle, the relationship of mass to energy is 9. Using dimensionless science we have understood the nature of this relationship, as a squaring function that takes mass as a variable operator.

OK so a little brain teaser for you.

What happens if we put a zero value in for the speed of light?

m = 0 0 * 32 = 02

We get Zero2

Those who are familiar with geometric mathematics will have an understanding of the concept that 02 produces a cross dividing infinite space into 4 quadrants, positive/negative, and whole/reciprocal.

With the first half of this equation solved, and resolved to 0 let’s see if we can do solves the second half of the equation, to produce a more tangible result.

Dimensionless Momentum

Let us return to a dimensionless mass value of 1 to help us grasp the finer details of this more complicated equation. Now we need to find a value for the velocity of light. If we define light as travelling in one direction at a constant speed, the we can produce a simple graph. Which a time unit of 1 and a distance of 3, the velocity of light is 3. So mow lets solve for momentum.

p = m * v = 1 * 3 = 3

Therefore p = 3.

However we have another problem. If light has no mass then guess what. Yep we get zero again. This ‘zero mass’ is really proving to be a problem. But don’t worry. We have yet to introduce gamma. Perhaps that might reveal some missing detail.

A particle travelling at a constant velocity is determined by the formula:
v=x(t2)−x(t1)t2−t1, where x(ti) is the position at time at the starts and end. In the case of the speed of light, from a position of zero (t1) to it position one second (t2) later is 3. therefore: v=3*(1) − 3* (0) * 3−0 = 3.

Dimensionless Gamma

The concept of gamma is velocity (v) and the speed of light (c) and also involves finding a reciprocal square root of a value that is squared.

γ = 1 / (1-(v/c)2)

Let’s go dimensionless and set Light to 3 and velocity to 3

γ = 1 / (1-(3/3)2) = 1 / (1-12) = 1 / (02) = 1 / 0 = 0

OK. so still zero right? So how does that add up? Well whilst this equation makes no sense from the view of conversional maths, geometric maths does provide an interesting solution. Geometrically the squaring function produces a square (or a triangle) in 2D space. The number 1 represents a single line in space. It we subtract a 1D line from a 2D square, does the square disappear?


Think about it.

Only if we are left with a numerical object does the concept of a square root find an application. A root number is express on a 2D plane, therefore, the ‘square’ minus 1 must still ‘exist’.


Therefore, whilst in conversional maths we perceive the value of zero as producing a null result, in geometric maths our mind remains open to an exploration of numerical space from a wider multi-dimensional perspective.

Confused? OK. Let’s try a quick thought experiment.

Imagine you are standing in the middle of a box. Turning to one if its faces, you can see 5 of its six sides. Can you be sure that the 6th side exists if you are not observing it?

The same can be said of our square in the above example. One of the sides (-1) has been removed. But the square still exists, just as the inside of the box exists. However, in the unobserved space, there is no way of determining where is exists or not. Like a cat in a box. Or similar.

Dimensionless Momentum of light

So let’s see if we can solve for momentum (p)

p = γ * m * v = (1-12) * 0 * 3 = 0

OK still zero but lets look at the process. The square with its side removed is multiplied by velocity 3. As we have defined light as travelling with a continuous velocity, light can be seen to move forward 3 units within the box even though is has a mass of zero. Therefore the dimensionless speed of light is still 3.

Zero Mass Equations

Let’s remind ourselves about the second part of mass equation we are trying to solve.

((p * v) * c)2

The nature of momentum we have already explored, revealing how geometrically speed can be maintained even though its appears as if we are obtaining a zero result. With this in mind we can complete complete the equation using our dimensionless values.

((p * 1) * 3)2 = ((0 * 3) * 3)2 = 02

So we get the same result at the famous E=mc2 equation, Zero2. However, it makes more sense when we realise that the reason the results are both square value. Each side of the equation is part of an even more famous mathematical equation, one that stretches back into Ancient Greece, and possible before.

Light, mass and Pythagorean theorem

The problem that both sides of the mass equation result in a nulled zero might at first seem to hold no reason. After all Zero is nothing right. In geometric maths we are able to provide zero with qualities. For example 02 we define as a cross and 0 the internal geometry of an octahedron. We use the combination of two squared values to work with higher dimensional 4D space; for example x2 * x2,.

Yet two square numbers have also been profoundly embedded in the famous Pythagorean theorem, that is used to define the hypotenuse of a triangle.

A2 + B2 = C2

And it is this same formula that lies at the heart of the conventional reasoning at to why an object with mass cannot we accelerated to the speed of light. But that does not explain why light that has zero mass should be travelling in the first place. Yet dimensionless science and geometric maths can provide a solution.

Let’s look at a simple example. If we take a right angled triangle, with a base and adjacent of ONE, the hypotenuse will be 2. if we double the sides then the triangle gets bigger but the relationship as a ratio will always be 1 : 2. This will apply to all numbers whose values form an equal value. So in the example of the mass equation, both sides have an equal value of zero, which is squared. Therefore, energy (e) that is found on the hypotenuse will be zero multiplied by 2.

At first this idea might seem strange, especially if you are used to performing linear mathematics, but as the ratio remains intact there is no reason to assume that the mathematical expression is not valid. For example, 1 forms a square, but in linear maths the true result remains obscured.


Whenever the momentum side of the mass equation is exactly equal to the stationary value, the value for energy should be an increase in the values for energy be a factor of 1 : √2.

Furthermore, Geometric Maths, which can describe zero suggests that any object who has a zero mass can still contain energy that is √2 greater in value. With a dimensionless value mass 1, and 3 the speed of light contain (c) that appears on both sides of the equation, guarantees that the minimum value for energy (e) will be 9, whenever the two sides fall into equilibrium.

At first this might seem rather absurd, however, is also makes perfect sense of when we examine the nature of electromagnetic waves, that are also offset in equilibrium by 90°.

Light does not exhibit any value of mass and it is not considered to be mass, just like a wave does not change the mass of a volume of water. However through dimensionless science and geometric maths we are able to explore the function of light, without having to perform any kind of magical maths. Presently the convention is to just remove the zero when mass energy calculation are performed with light.

For more perspectives on the reason as to why light does not have or need mass you can check out this video.

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