steevey Posted February 9, 2011 Posted February 9, 2011 (edited) I know there's the uncertainty principal, but why does matter having a large mass decrease its wave packet size? An electron has very low mass, which is why it can encircle and entire atom as one shell, but a proton has a mass some 1800 times larger than an electron, and its wave is concentrated and a very small reigion. But wait, what about quarks? They are small than an individual proton. Whats the deal? Edited February 10, 2011 by steevey
alpha2cen Posted February 15, 2011 Posted February 15, 2011 The problem comes from the beginning of the Universe. It is related to the creation of electron, proton and neutron at the end of the quark-gluon plasma creation. If electron mass were too high, it would not be easy to make present Universe. The reason is like this. 1) Low mass electron can move very fast, so it is easy to make molecules. Electron can easily move around very large area. 2) If electron mass were too high, the electron orbit diameter would be very small, and it would be easy to make fusion reaction with proton or neutron. 3) If electron mass were too high, energy distribution in the Universe would be very heterogeneous, and local energy distribution would be very heterogeneous . As a result , objects would be very unstable. Because heavy electrons have very slow energy transfer rate.
swansont Posted February 15, 2011 Posted February 15, 2011 You say you know there's the uncertainty principle. What's the problem, then? That's the answer.
steevey Posted February 15, 2011 Author Posted February 15, 2011 You say you know there's the uncertainty principle. What's the problem, then? That's the answer. Why does the uncertainty principle act the way it does? It's just a principle, it doesn't actually describe why something more determined causes something else to be less determined.
lemur Posted February 15, 2011 Posted February 15, 2011 Isn't it just simply logical that the less mass a particle has, the less positionally stable it would be? After all, energy has to express itself via momentum in particles with mass and energy doesn't seem to be infinitely divisible into smaller amounts. If the only means you had to move a ping-pong ball around was to use a leaf-blower, would you be surprised if the ping-pong ball always seemed to be everywhere and nowhere at the same time? Sorry if this is a naive analogy.
swansont Posted February 16, 2011 Posted February 16, 2011 Why does the uncertainty principle act the way it does? It's just a principle, it doesn't actually describe why something more determined causes something else to be less determined. Conjugate variables are Fourier transforms of each other; in wave mechanics they are not independent quantities as they are classically. The uncertainty relation is part of the behavior of the Fourier transform. "Just a principle" is not far off from "just a theory." It doesn't mean "guess." 1
steevey Posted February 16, 2011 Author Posted February 16, 2011 (edited) Conjugate variables are Fourier transforms of each other; in wave mechanics they are not independent quantities as they are classically. The uncertainty relation is part of the behavior of the Fourier transform. "Just a principle" is not far off from "just a theory." It doesn't mean "guess." I didn't mean it in that context, I meant it as its just stating the relationship is there, but not explaining why its there. Like E=mc^2, but why does E=mc^2? Edited February 16, 2011 by steevey
swansont Posted February 16, 2011 Posted February 16, 2011 I didn't mean it in that context, I meant it as its just stating the relationship is there, but not explaining why its there. Like E=mc^2, but why does E=mc^2? That's philosophy. Two doors down.
steevey Posted February 16, 2011 Author Posted February 16, 2011 That's philosophy. Two doors down. So in other words, no one knows?
swansont Posted February 16, 2011 Posted February 16, 2011 So in other words, no one knows? It depends how deeply you are asking. E=mc^2 comes about from relativity, and the HUP from details of wave mechanics. But if you want to know why nature behaves in that way, ultimately the answer is no, nobody knows, because you reach a level where all you can do is guess with no way to test your ideas. Which means there's no way to know if you were right or not. (You can define your answer as being right, but then you want religion, which is across the hall from philosophy)
Xerxes Posted February 16, 2011 Posted February 16, 2011 Like E=mc^2, but why does E=mc^2? Like this, as a shameless paraphrase of Einstein. Consider a material body B with energy content [math] E_{\text{initial}}[/math]. Let B emit a "plane wave of light" for some fixed period of time [math] t[/math]. One easily sees that the energy content of B is reduced by [math]E_{\text{initial}} - E_{\text{final}}[/math], which depends only on [math]t[/math]. Let [math]E_{\text{initial}} - E_{\text{final}} = L[/math] i.e.the light energy "withdrawn" from B. Now, says Einstein, consider the situation from the perspective some body, say [math]B'[/math] moving uniformly at velocity [math]v [/math] with respect to B. Then from this perspective, the energy withdrawn from [math]B[/math] is [math]L'[/math] so that, as before, [math]L'[/math] depends only on [math]t'[/math], which is [math] t(1 -\frac{v^2}{c^2})^{-\frac{1}{2}}[/math] by Lorentz time dilation. The difference between [math]L[/math] and [math]L'[/math] is simply [math]L' - L = L[(1 - (\frac{v^2}{c^2})^{-\frac{1}{2}} - 1].[/math]. By expanding [math](1 -\frac{v^2}{c^2})^{-\frac{1}{2}}[/math] as a Taylor series, and dropping terms of order higher than 2 in [math]v/c[/math], he finds that [math]L(1 + \frac{v^2}{2c^2} - 1) = L\frac{v^2}{2c^2} = \frac{1}{2}(\frac{L}{c^2})v^2[/math]. With a flourishing hand-wave Einstein now says something like this: the above is an equation for the differential energy of bodies in relative motion; but so is [math]E = \frac{1}{2}mv^2[/math], the equation for kinetic energy - these can only differ by an irrelevant additive constant, so set [math]\frac{1}{2}(\frac{L}{c^2})v^2 = \frac{1}{2}mv^2 \Rightarrow \frac{L}{c^2} = m[/math] and so [math]L= mc^2[/math]. But, says he, [math]L[/math] is simply a "quantity" of energy, light in this case, that now depends only on [math]m[/math] and [math]c^2[/math] so...... [math]E = mc^2[/math].
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