When using De Broglie’s formulas to calculate the wavelengths of different objects, we typically use the speed of that object relative to an observer. But as we know celestial bodies are also moving through space with some velocity, and so every object on Earth is moving with the speed of the Earth +- the velocity of that object. Why do we not consider this when applying these formulas?
In: Physics
Perhaps because the relative speed given is always assumed without the delta as a sort of standardization. That way speed is always isolated regardless of location, but ambient is not always constant between celestial bodies.
We do. But you see in physics when you get to the scales particularly the assumption that there is no absolute velocity is universal.
The peculiarity in the system is that every observation is taken from the center of the universe because no matter where you are in the universe everything is moving towards or away from you by the same set of ratios.
So it becomes unnecessary to even mention the fact that we are considering the earth as the still center for these measurements of how fast things are moving.
From then on it’s all about triangles. I know how fast suck star A is moving I know how fast Star B is moving I know the directions they’re both moving so I can discuss their movement relative to each other. There’s no point in adding “relative to the earth” for the same reason that when you’re driving down the road and you say I’m going 55 mph you don’t say relative to the Earth in that case either.
That’s also why we know about red shift and blue shift and all that stuff. We know the Spectra lines for things like hydrogen and helium, we find the spectral lines for hydrogen helium in star A. No we can calibrate everything we need to discuss about Star A by measuring its red/blue shift of that hydrogen line and then all measurements are corrected by that same value.
And so on.
Indeed most of what we know about the cosmos is something we wouldn’t know at all if we weren’t correcting for red and blue shift by understanding the relative motions and their effects on the waves being produced.
Yes, the structure of the wavefunction looks different to observers in different reference frames. It turns out that observers in different frames looking at the same interaction will make the same predictions, even though they observe different wavefunctions, so the theory is consistent and the choice of frame doesn’t matter. In fact choosing a frame in which the problem is easier is a good problem solving strategy.
Note that it’s not consistent under special relativistic transformations, which is one of the issues quantum field theory solves.
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