We consider a thought experiment. Let weightless cylinder is located in the reference frame $K'$ (axis X, Y, Z). The height of the cylinder is equal to $h$ (Fig.1). The top cover of the cylinder is denoted by the letter $S_2$, and the bottom cover is denoted by the letter $S_1$.

(source: liveinternet.ru)

Let the system of reference $K'$ (cylinder) is moving with uniform acceleration in the direction of positive values of $Z$ (acceleration $\gamma$). Let from $S_2$ to $S_1$ emitted quantum of light - a photon with energy $E_0$. We consider this process in the system $K_0$, which has no acceleration. Assume that at the moment when the radiation energy $E_0$ is transferred from $S_2$ to $S_1$, $K'$ system has a speed equal to zero (with respect to system $K_0$). Light quantum will appear in $S_1$ after time $h/c$ (in first approximation), where $c$ is the velocity of light. At this time, the bottom of the cylinder $S_1$ has a speed $v=\gamma h/c$. Therefore, according to special relativity, reaching $S_1$ radiation has an energy $E_1$, which is equal to $$ E_{\,1}\approx E_{\,0}\,(1+v/c)=E_{\,0}\,(1+\gamma h/c^2) \tag{1}$$ Momentum is $$P_{\,1}=E\,_1/c=E_{\,0}\,(1+\gamma h/c^2)/c \tag{2}$$ Let the light quantum with the same energy $E_0$ is emitted from $S_1$ in the direction $S_2$. Then the energy of the radiation reaching the wall $S_2$ and momentum are of the form $$E_{\,2}\approx E_{\,0}\,(1-v/c)=E_{\,0}\,(1-\gamma h/c^2) \tag{3}$$ $$P_{\,2}=E\,_2/c=E_{\,0}\,(1-\gamma h/c^2)/c \tag{4}$$ If we simultaneously send two light quanta of equal energy - one in the direction of $S_1$ and the second in the direction $S_2$, the recoil momenta mutually balanced by, and will play a major role (2) and (4). Then we get $$\Delta P=P_1-P_2=(2E_0/c^2)(\gamma h/c)=2\,m\,\Delta v \tag{5}$$ where $2\,m=2E_0/c^2$ is inert mass; coefficient $2$ corresponds to two photons.

When we accelerate the cylinder, we break the symmetry of the transmitted impulses.

Weightless cylinder in which there is radiation, as a result of the acceleration behaves as if it has an inertial mass $2\,m$, and the momentum $\Delta P$ this inert mass, as is easily seen from Fig.1, is directed in the direction opposite the acceleration vector $\gamma$. Cylinder with photons within it resists an accelerating force. It is one of the characteristic manifestations of the physical property, which is called "mass".

We also get the gravitational mass in accordance with the principle of equivalence of $\gamma h=\varphi$.

Question: How can the Higgs boson explain the inertia of the body?


1 Answer 1


Force can be defined using Newton's second law, $F=ma$, but it can also be defined as the rate of change of momentum. These are not necessarily equivalent because particles can carry momentum even when they have no mass. In your example the resistance to acceleration is due to the change in momentum of the (massless) photons, where the momentum of a photon is given by:

$$ p = \frac{h}{\lambda} $$

So the Higgs mechanism is entirely irrelevant to the resistance to acceleration you describe.

For completeness I should add that even for massive bodies the Higgs mechanism makes a very small contribution to the mass. The majority of the mass is due to the QCD interaction energy in the nucleons.

  • $\begingroup$ @ John Rennie thanks. I think that the essence of inertia must be the same for all particles. $\endgroup$ Apr 3, 2016 at 9:02
  • $\begingroup$ @AleksanderKlimets: that is not the case. $\endgroup$ Apr 3, 2016 at 10:52

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