Why fundamental quantization of energy is hv?

[Phrak]

That is an imaginative and fascinating thought...and I mean that in a very positive way. If it weren't late and I have to work in the AM, I would kick that can around for awhile. I might want to get back to you on that.

@canon and janakiraman

It's one answer to the question "What would happen if h were different, or if c were different?", where usually the question is pu-poohed, and the questioner is left unsatisfied?

Those who know the least physics have the best questions, in my opinion. Why?

If you have the background, http://en.wikipedia.org/wiki/Gauge_theory should be an interesting place to start, under Classical Gauge Theory.

 

[canoe]

@canon and janakiraman

It's one answer to the question "What would happen if h were different, or if c were different?", where usually the question is pu-poohed, and the questioner is left unsatisfied?

Those who know the least physics have the best questions, in my opinion. Why?

@Phrak

Here it is late again, and I have about 3 minutes...but if h changed as conjectured than uncertainty could likley be causal.

 

[Phrak]

...Here it is late again, and I have about 3 minutes...but if h changed as conjectured than uncertainty could likley be causal.

I'm not sure what you mean, but if h were not everywhere constant, then it would give rise to a field. What sort of field? I don't know. This is the basis of quantum field theory. The mathematic basis qft originated with the connection coefficients found in general relativity.

 

[diazona]

9.8 ms^-2 is not a fundamental value, but I think G - gravitational constant - is, in the same way h is. Both are proportionality constants that we can't calculate, we can only measure them.

Yeah... and on that note (for completeness, since nobody's posted this yet): the gravitational acceleration $g = 9.8 \mathrm{m}/\mathrm{s}^2$ comes from Newton's universal law of gravitation (and second law of motion),

$$F = G \frac{Mm}{R^2} = mg$$

with M as the mass of the Earth and R its radius.

$$g = G\frac{M}{R^2} = \left(6.67\times 10^{-11}\frac{\mathrm{m}^2}{\mathrm{kg}\cdot\mathrm{s}^2}\right)\frac{5.9736\times 10^{24}\mathrm{kg}}{(6371\mathrm{km})^2} = 9.8\frac{\mathrm{m}}{\mathrm{s}^2}$$

In general relativity, the equation is slightly different (I don't remember exactly what the higher-order corrections are) but the procedure is basically the same.