If an observer traveling towards a microwave oven at almost the speed of light blue shifts the microwaves enough to be visible light, how can the mesh on the oven door still stop to waves from escaping the oven?
And conversely, if an observer traveling at almost the speed of light away from a microwave oven red shifts the visible light enough to be microwaves, how can the mesh on the oven door still allow the waves to escape the oven?
Best Answer
First of all, let's get a better picture of why microwave doors keep waves inside the oven in the first place. Using a generous amount of hand-waving: imagine that an electromagnetic wave is incident on a circular hole in the microwave oven door. Furthermore imagine that at some moment in time, the electric field is pointed towards the right side of the hole. Then electrons will move towards the left side, creating a new wave in the electric field. However, it takes some time for the wave to travel all the way around the edge of the hole. If the distance around the hole is roughly the same size as the wavelength of the incident wave [1], then the new wave is exactly out of phase with the incident wave. Thus, outside of the box, the incident wave gets canceled out.
All right, how does this look in the fast-moving observer's frame of reference? Well, the incident and response waves are propagating in the same direction, so the Doppler effect is exactly the same for both. [2] Therefore, they still cancel each other out in the observer's frame.
Notes:
If the hole circumference is smaller than the wavelength, there is still a path inside the metal with the same length as the incident wavelength, so the same thing happens. Waves created along paths longer and shorter than the wavelength cancel each other out. A small amount of the incident wave gets through because this cancellation process doesn't work perfectly for nonzero hole sizes. Getting a more detailed picture of the interaction would require a full numerical simulation, but this model is accurate enough for the question.
We can be a bit more precise about why the Doppler effect is the same for both the incident and response waves. The relativistic Doppler effect has two components: time dilation, which shows up as $\gamma$, and the ordinary Doppler effect (due to the finite speed of light and the distance between wavefronts), which appears as $1-\beta$. From the observer's perspective, the ordinary Doppler component is the same for both the incident and response waves, while time dilation shows up as the electrons in the door appearing to respond much more quickly than they do in the microwave oven frame.