Is the photon an exception to Special Relativity?

Question

This may have been asked before, but nonetheless I am unable to find it if so. We know Einstein's theory of relativity is confirmed experimentally, and so any postulations made in theory must be true.

I am having difficulty understanding why nothing can travel at the speed of light and yet photons can travel at the speed of light. From $E=mc^2$ we know that energy and mass are basically equivalent, and that the faster something travels the more mass it gains, thus at 99.999% the speed of light, it's mass is increased by a factor of 224, and at 99.9999999999% it's mass is increased by a factor more then 70,000, and at the speed of light it's mass is increased by a factor of infinity, and therefore would require an infinite amount of energy to push anything to the speed of light, thus making it impossible for anything to travel at the speed of light.

Now I know a photon is massless and this is why photons and gravitons (if they exist) are the exception to this rule, but it still carries energy, but if energy is equivalent to mass then how is it that a photon can travel at the speed of light? Does this not violate the very laws set forth in Special Relativity, or more likely, have I missed something very fundamental in the theory itself?

Answer 1

As @Gautampk specified in the comment, it is the problem of the formula.

it still carries energy, but if energy is equivalent to mass then how is it that a photon can travel at the speed of light?

Photons still carry energy, as you have pointed out. So where is it stored, if not in form of their rest mass? Their kinetic energy - and therefore momentum. It turns out that because of publicity, people are so familiar with $$E=mc^2$$ that they forget that this is the rest energy - energy of a particle in its rest frame, related to its invariant mass. When a particle is moving, there is an additional momentum term to it. In other words, energy is equivalent to mass only in the rest frame of an object - but photons have no rest frame by the postulates of SR. They move at $c$ in all frames.

The total energy of a particle comes from energy-momentum fourvector norm invariance under Lorentz transformation. It reads $$E^2=(pc)^2+(mc^2)^2.$$ Note that in a particle rest frame $p=0$ so we are only left with good ye olde familiar E is m c squared. Now, photons have zero rest mass, and so have zero rest energy. All of it is associated with their momentum: $$E=pc$$

On a side note, a more interesting bit:

Back to the original question, are photons an exception? Well, they are, but only in the sense that it is their speed that is selected to be invariant. Eistein could have equally valid say that gravitons travel at the speed of gravity and therefore photons travel at the speed of gravity. But this is only human perspective. The mathematics of SR is the same to them.

In Rindler's Introduction to Special Relativity, you can find an indeed very profound remark: if suddenly all the electromagnetism was just wiped out of the world like it's never been there, SR would be still valid as long as there are things that propagate through vacuum.

Answer 2

The mass energy equivalence formula $E=mc^2$ does apply to photons if $m$ is the so-called relativistic mass. However, if one uses the invariant rest mass $m_0$, more natural in SR, an additional term is needed: $$E^2 = (m_0 c^2)^2 + p^2 c^2.$$ The formula relating the relativistic mass to the rest mass, which follows from equating them, $$m = \frac{m_o}{\sqrt{1 - v^2/c^2}}$$ applies to photons, in a way, but it gives the indeterminacy of $0/0$. The relativistic mass $m$ resolves this indeterminacy. Using de Broglie's energy relations one can even express it in terms of photon's wavelength $\lambda$: $$m = E / c^2 = h / \lambda c,$$ see If photons have no mass, how can they have momentum?

So photons (and zero rest mass particles generally) are not "exceptions" to SR, they are a borderline case described by it. This description implies that their rest mass can only be $0$, and their speed can only be $c$. The infinite acceleration energy paradox is therefore resolved because photons must retain constant speed, and can neither be accelerated to nor decelerated from it. The constancy of the speed of light in all frames was one of the two postulates in the Einstein's original formulation of SR.

Answer 3

but if energy is equivalent to mass

Is it? The relativistic Energy-momentum relation for a particle is

$$E^2 = p^2 + m^2,\qquad c = 1$$

where $m$ is the particle's invariant mass (also called rest mass) and $p$ is the particle's momentum.

This relation is a fundamental result from SR and so, simply put, energy isn't equivalent to (invariant) mass.

For a particle with zero invariant mass, we have

$$E = |p|$$

That is, for a zero invariant mass particle, the particle's energy is proportional to the particle's momentum.

Answer 4

The postulates of special relativity:

1. The Principle of Relativity The laws of physics are the same in all inertial frames of reference.

2. The Constancy of Speed of Light in Vacuum The speed of light in vacuum has the same value c in all inertial frames of reference.

Note, if one has only the first postulate, one is back in the classical Newtonian frame . The Galilean transformations describe kinematics with vector algebra , where three vectors defined in space $(p_x,p_y,p_z)$ retain their length, which is the scalar(dot) product, in all frames .

The special relativity framework enlarges the dimensions to four by including time with space coordinates , and energy with momentum coordinates, and changing the transformation from Galilean ( where the energy is not entangled with the momentum and time is a parameter), to Lorenz transformations.

Lorenz transformations follow a four vector algebra, with a dot product and they are the mathematical result of the two postulates above.

For the four momentum vectors :

$$ \begin{matrix} \vec{P_a}=\begin{bmatrix}E_a \\ \vec{p_a}c \end{bmatrix} & \vec{P_b}=\begin{bmatrix}E_b \\ \vec{p_b}c \end{bmatrix} & \vec{P_a}\cdot \vec{P_b} = E_aE_b - \vec{p_a}\cdot\vec{p_b}c^2 \end{matrix} $$

The "length" of this four vector, invariant to Lorenz transformations by construction of the algebra, is the invariant mass squared of the particle characterized by this four vector.

$$\sqrt{P\cdot P}=\sqrt{E^2-(pc)^2}=m_0c^2$$

The length of this 4-vector is the rest energy of the particle. The invariance is associated with the fact that the rest mass is the same in any inertial frame of reference.

This mass is called "invariant mass" and characterizes all elementary particles and the complex additive four vectors of all matter. It can be zero as is seen in the elementary particle table of the standard model of particle physics.

Because of the minus sign in of the contribution of the fourth component to the length of the four vector, in the mathematics of the Lorenz transformations, complex numbers describe the lorenz four vectors simply.

As the other answers state, the E=mc^2, called the relativistic mass, though useful in calculating trajectories with special relativity, is confusing on the basic issues, as it coincides with the invariant mass of systems only in the rest frame. It is a measure of how the inertial mass changes at high velocities, but is not used by particle physicists because of the confusing way it is mentioned in popularization of science. The invariant mass is the concept to use, in the same way one characterizes objects by their constant volume in space even though they may be moving.

See also this answer of mine on a similar question.

Answer 5

Photons aren't the exception, photons are the cause - they are the agents of information/energy transfer via the EM field.

Whatever speed it is photons travel at, that is the speed limit of the universe.

It's a bit like saying a sailboat can't sail faster than the wind; the wind speed sets the speed limit of the boat (ignoring tacking).