1
$\begingroup$

I'm trying to implement a simple simulation of a box of photons, to demonstrate mass-energy equivalence, and relativistic time dilation. I want to be able to push the box around interactively. As I understand it, two effects should emerge from this simulation if I implement it properly:

  1. The photons should increase the mass of the box through radiation pressure.
  2. As the box moves faster, the photons should slow down relative to the box, approaching zero as the box approaches c.

But, I'm not sure how to model the box-photon collisions. Looking through Wikipedia, I found classical collision models, and statistical scattering models, but nothing that applies to a single, massless particle.

My best guess is that momentum is conserved:

$$ m\vec{v}_{box} + \vec{p}_{photon} = m\vec{v}'_{box} + \vec{p}'_{photon} $$

and energy is conserved:

$$ {m\over2}|\vec{v}_{box}|^2 + |\vec{p}_{photon}|c = {m\over2}|\vec{v}_{box}'|^2 + |\vec{p}_{photon}'|c $$

and angle is reflected about the normal $\vec{n}$:

$$ \vec{v}'_{photon} = \vec{v}_{photon} - 2(\vec{n}\dot{}\vec{v}_{photon})\vec{n} $$

and solving that system will give me $\vec{v}'_{box}$ and $\vec{p}'_{photon}$.

But I'm skeptical that this is correct, and I'd like to know if this subject is already documented somewhere.

$\endgroup$
4
  • $\begingroup$ Please note that we don't answer homework or worked example type questions. Please read How do I ask homework questions on Physics Stack Exchange? and Are check-my-work questions on-topic? for "check my work" problems. $\endgroup$
    – Yashas
    Apr 11 '17 at 7:56
  • 1
    $\begingroup$ It's not schoolwork. I'm just trying to understand the concepts, and I made my best guess. $\endgroup$
    – jedediah
    Apr 11 '17 at 8:08
  • $\begingroup$ The angle is preserved in the mirror's rest frame. Don't forget to transform. $\endgroup$ Apr 11 '17 at 8:33
  • $\begingroup$ Thanks Jan, that explains how the photons change direction as the box speeds up. Now I'm trying to figure out how to transform the photon velocity in/out of the box frame. The standard velocity addition doesn't work for massless objects. $\endgroup$
    – jedediah
    Apr 13 '17 at 3:12
1
$\begingroup$

Photons always travel with velocity c, by construction of the theory that includes photons. It is only light that can change velocity because it is an emergent phenomenon from a quantum mechanical superposition of innumerable photons . (AFAIK the slowing of light needs a medium) .

Elastic scattering is the same whether one is talking of the photon or other particles .

A perfectly elastic collision is defined as one in which there is no loss of kinetic energy in the collision. An inelastic collision is one in which part of the kinetic energy is changed to some other form of energy in the collision. Any macroscopic collision between objects will convert some of the kinetic energy into internal energy and other forms of energy, so no large scale impacts are perfectly elastic.

Think of a ball bouncing off a wall. The velocity of the wall due to the collision is very close to zero, momentum conservation is easy to achieve .

Your "box-photon collisions." fall into the framework of a macroscopic collision , because the box is a macroscopic object. Because of the large mass of the box the velocity of the box needed to achieve momentum conservation is practically zero, and within measurement errors, which means, even though the photon will lose some of its energy to the box, the energy change is so small that it cannot be measured. Note that energy change in a photon is seen as a frequency change because $E=hν$,not a velocity change.

$\endgroup$
0
$\begingroup$

You have to use elastic scattering, when a photon gives part of its energy to a system, and changes angle.

The photon gives the box part of its energy as kinetic energy, thus slowing it down (compared to light), and to an external observer this will give the box rest mass.

That is why some say an electron or quark is like a box that has energy in it, photons or gluons, that gives them rest mass to an external observer.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.