| I studied wave mechanics in college, but the origin of mass didn't click for me until several years later (and in fact I don't believe it was every brought up in the context of wave mechanics, which seems like a problem in retrospect). The conceptualization that worked for me is this: The normal wave equation is (ignoring constant factors like mass and propagation velocity): d^2/dt^2 f(x,t) = d^2/dx^2 f(x,t) <acceleration> = <pulled towards neighbors> This says "if a point in the field is lower than its neighbors, it will be accelerated upwards. If a point in the field is higher than its neighbors, it will be accelerated downwards." This equation is the lowest-order description of most wave phenomena like sound waves, water surface waves, EM waves, etc. and it's usually pretty accurate. If you look for solutions to this differential equation, you can get f(x,t) = exp(i * w * (x±t)) w is the frequency of the wave This tells you that the frequency and wavenumber of waves is determined by the same parameter (w), so they are proportional to each other Now, what if we add a restoring force to this equation? This is a force that pulls the value of the field towards zero. d^2/dt^2 f(x,t) = d^2/dx^2 f(x,t) - M^2 f(x,t) M is just a parameter that tells you the strength of the restoring force. The force increases as the field gets farther from zero, like a spring. Now, solutions to the equation look instead like f(x,t) = exp(i*k*x ± i*w*t) Where w^2 = k^2 + M^2 (or something like that, I need to re-derive this on paper, just going off memory, but I think if you plug it in it should work) Notice that now, if you have a spacial frequency k, your temporal frequency is actually higher. In fact, if your spacial frequency k is 0 (corresponding to a stationary wave), your temporal frequency is still M! This is what mass is. Having a non-zero frequency even if the wave is the same everywhere in space (which corresponds to no movement) A field with no restoring force is e.g. the EM field, so photons are massless. The rate at which they oscillate in time is the same rate at which they oscillate in space. A massive particle has a restoring force, so its temporal frequency is higher than its spacial frequency. In physics, this equation is often reordered like this: d^2/dt^2 f(x,t) - d^2/dx^2 f(x,t) = - M^2 f(x,t) (d^2/dt^2 - d^2/dx^2) f(x,t) = - M^2 f(x,t) (d^2/dt^2 - d^2/dx^2) f(x,t) + M^2 f(x,t) = 0 ◻ f(x,t) + M^2 f(x,t) = 0 (the d'alembert operator) (◻ + M^2) f(x,t) = 0 Again, this is ignoring constant factors like c, h, etc. The above equation is nice because it's relativistically invariant. The d'alembert operator is the contraction of the 4-momentup operator with itself, p^u p_u. This is a concept worth studying - tells you a lot about what mass, energy, velocity, and momentum actually are in a general sense |
Wouldn't it be the opposite, that they do not oscillate in time at all so that they oscillate in space as rapidly as possible (since, as we know, time doesn't pass for photons)? And stationary particles don't oscillate in space, so they oscillate in time as rapidly as possible. Or are you using "oscillate" in a different sense here?