# A photon cannot exist if it travels at speed lower than $\mathrm{c}$ in a vacuum, why?

This is known in physics that photons travel only at fixed speed $$\mathrm{c}$$ in vacuum but also inside a medium going from one atom to the next and taking advantage of the vacuum that exists between atoms in a material medium.

Any other speed lower than $$c$$ and the photons would cease to exist and get "destroyed".

Why is that?

This question is fundamental and I have a hard time to derive an analysis or physical explanation.

Why a photon is possible only at $$\mathrm{c}$$ propagation speed independent if there is a medium or not?

There must a deeper connection to vacuum spacetime.

• "but also inside a medium going from one atom to the next and taking advantage of the vacuum that exists between atoms in a material medium." Reference? I don't think this is right. Nov 16 '21 at 10:59
• "Any other speed lower than c and the photons would cease to exist and get "destroyed"." -reference? Nov 16 '21 at 11:06
• The light wave of photons propagating through a medium is slowed down due the absorption-emission mechanism of atoms causing a delay of the light wave (phase shift) from the atoms but photons propagate from one atom to the other only at speed c. Nov 16 '21 at 12:24
• As anna v pointed out massless particles always travel at speed c. Nov 16 '21 at 12:46
• @Markoul11 That's not quite right. If this was the case we would see differences in the time it takes for light to propagate through a medium as well as fluctuations in the light intensity of light leaving the medium, as different photons would take different amounts of time in the medium depending on the path they take. This is not tha case. Please give your references Nov 16 '21 at 15:18

Photons are part of the elementary particles in the standard model of particle physics, which is modeled with quantum field theory (QFT). The model describes successfully the plethora of data in elementary particles and astrophysical observations and is also very predictive. In this QFT all zero mass particles travel with velocity c, because it is a result of Lorentz covariance of the theory. As the theory fits the data , it is a rule when discussing zero mass particles that they go with velocity c.

It is not a matter of "being destroyed", but of not being able to be created with non-zero mass .

• "...with zero mass" I guess you mean with non-zero mass? Nov 16 '21 at 12:07
• @Markoul11 yes, thanks Nov 16 '21 at 12:12
• I have a question about the speed of light and black holes: physics.stackexchange.com/questions/677950/… Nov 22 '21 at 16:30

The property of the photon moving with speed of light can be derived through the relationship between the Lagrangian and the Hamiltonian of a free particle (with mass or without mass):

$$H = p\cdot \dot{q} -\cal{L} \equiv p\cdot v -\cal{L}$$

where $$v$$ is the velocity as time derivative of the canonical coordinate $$q$$ of the free particle. $$p$$ is the momentum of the free particle.

Now in relativistic mechanics the Lagrangian of the free particle is

$$\cal{L} = const \int ds$$

the constant term const is only necessary to keep up with the units and $$ds$$ is the invariant line element:

$$ds = \sqrt{c^2dt^2 -dx^2 -dy^2 -dz^2}$$

So for photons the line element ds vanishes: $$ds=0$$, so the Lagrangian is zero. So the Hamiltonian is simply:

$$H = p\cdot v$$

Furthermore we know from relativistic mechanics that the energy of a free particle is :

$$E =\sqrt{ (pc)^2 + m^2 c^4}$$

For a massless photon m=0 we get:

$$E =p\cdot c$$

Knowing that the Hamiltonian of a particle corresponds to its energy we get:

$$p\cdot v = H \equiv E = pc$$

which finally provides us with the answer: $$v =c$$.

The main reason why photons move with speed of light is that they are massless $$m=0$$.

But one could also just "define" photons as free particles (possibly adding another property to distinguish them from other massless particles -- gravitons for instance) whose line element is zero --- since this property is essential for the proof.

• Great answer! Thanks. Nov 16 '21 at 14:32
• @Markoul11 nice that you like my answer. What I already imply in my last paragraph is that the assumption of $ds=0$ already implies $v_{\gamma}=c$. So it quasi boils down to the definition that particles that fulfill $ds=0$ move a speed of light. If they did not, the whole relativity theory would be wrong. Possible maybe in a parallel universe, but in our universe it would simply contradict experience. Nov 16 '21 at 14:47