Last week, we were introduced to the basic questions that quantum physics theory is needed to answer. This week, we start answering those questions in part two: Curiouser and Curiouser.

- Lesson 1: Classical Thinking: Why Does It Fail?
- Lesson 2: Curiouser and Curiouser
- Lesson 3: Enter Heisenberg, Exit Common Sense
- Lesson 4: Don’t Underestimate the Power of Virtual Particle Exchange
- Lesson 5: Let There Be Quantized Electromagnetic Radiative Energy
- Lesson 6: Quanta, Quanta Everywhere
- Lesson 7: Down the Rabbit Hole
- Lesson 8: One and One and One is Three
- Lesson 9: Like a Record, Baby

inkeye

July 13, 2010@ 11:19 amI have a question: Toward the end, it talks about Carbon-14 dating and says:

We have found that unstable particles can decay when they expe-rience collisions with other particles, or are otherwise given more energy.

Is it then possible that some extraterrestrial form of energy (say an ancient exploding star or perhaps a very large meteorite) could globally accelerate the decay rates?

And if this is true, how is it accommodated for, since no one was around to observe this source of energy (especially if it left no other traces)?

W. Blaine Dowler

July 13, 2010@ 1:06 pmTheoretically, that’s entirely possible. However, other traces would be seen, in that radiation levels significant enough to drive this process would have a devastating effect on organic tissue, and would be seen in fossil records as a mass extinction.

J_W_W

July 14, 2010@ 1:37 pmWouldn’t it also be true that runaway nuclear fission reactions would fit this accelerated decay scenario as well through the process of colliding fissionable material together…?

sinewalker

July 14, 2010@ 4:42 pmYes it is — I think that is part of the reason why C14 cannot be used to date recent items (last few hundred years). Also when I was learning about C14 dating I recall being told that they had to adjust their calculations to take into account the atmospheric nuclear tests conducted in the 1950s. I don’t recall the details (something about radiation levels changing globally, but it also had an impact to the radioctive decay of C14).

jtriley

July 18, 2010@ 5:52 amI find the “inertia” versus “mass” distinction still a bit confusing.

I understood that the two are not the same thing, that in theory they could have different values, but present measurements show them the same in non-energetic systems to at least five significant figures. Furthermore, explaining why they measure the same is a major test for various competing theories of physics.

This lesson seems to say that inertia has to be the one effected by energy because inertia is about the relationship between energy and mass.

Mass is then taken to mean “rest mass” which is about the way matter bends space/time, even at rest.

This seems to be a nomenclature problem that is so embedded in our text books that no one can sort it out.

W. Blaine Dowler

July 18, 2010@ 8:01 amIt’s even more obscure than that. Energy also distorts space/time, so it’s inertia that is used in gravitational equations as well. (It’s the energetic contributions to inertia that are being compared by most formulae.) The “rest mass” is perhaps best thought of as a means to catalogue the total rest masses of the particles that make up a body, be they protons, neutrons, electrons, etc. If you have taken some condensed matter physics, you can think of “rest mass” as the mass the system would have if every particle were in the lowest available energy state, as though the entire body were at absolute zero.

msimkin

July 18, 2010@ 12:31 pmThank you for this. I read a lot about the cosmos, physics, quantum effects etc. but because the math symbols make no sense to me I just have to do my best to understand the text.

In the classic expression E=mc2 isn’t c a constant? Thus c2 a constant? Thus we’re really just saying E=mX where X is an arbitrary value depending on whatever unit of measure we choose for E and m. So why is the speed of light relevant here? Couldn’t you just as easily say E=m times the diameter of the moon, given appropriate units in all three variables?

Chad Cloman

July 19, 2010@ 9:18 amIn the classic expression E=mc2 isn’t c a constant? Thus c2 a constant? Thus we’re really just saying E=mX where X is an arbitrary value depending on whatever unit of measure we choose for E and m.Umm, no. c² also has units, and the conversion factor between E and m will always be the speed of light as expressed in the units needed to convert between E and m (a unit of distance squared divided by a unit of time squared). The diameter of the moon will never have (m/s)² as the units (or mph² or kph² or whatever).

Chad Cloman

July 19, 2010@ 9:18 amWhere you see ‘²’ in my post, just replace it with ‘^2′. Thought I could use HTML entities, but I can’t.

Chad Cloman

July 19, 2010@ 9:20 amIn the classic expression E=mc2 isn’t c a constant? Thus c2 a constant? Thus we’re really just saying E=mX where X is an arbitrary value depending on whatever unit of measure we choose for E and m.Umm, no. c^2 also has units, and the conversion factor between E and m will always be the speed of light squared, as expressed in the units needed to convert between E and m (a unit of distance squared divided by a unit of time squared). The diameter of the moon will never have (m/s)^2 as the units (or mph^2 or kph^2; or whatever).

DGDanforth

July 19, 2010@ 5:05 pmIn the 3rd lecture I slightly disagree with the characterization of the uncertainty principle. Here is why. Assume for the moment that everything is made out of bowling balls: us, measurement devices, elementary particles. Then what do we mean by a ‘measurement’? Well, all we can do is throw bowling balls at bowling balls but how do we get a grip on a single bowling ball? We can’t really. We can try a shake one loose from where it is stuck on another bowling ball but how do we know where it is? We don’t. We can only infer after the fact which is … just another collision between bowling balls whose presence is only made know by an amplification process, a cascade of bowling balls. The cascade structure must be set up before hand to cause the amplification, etc.

So, even without any ‘wavefunctions’ or quantum mechanics it still is not possible to measure where things are precisely, even though (under this bowling ball view) they have a precise location and momentum.

Hence, there is this ‘concrete’ uncertainty but it is also augmented with a ‘wave’ nature of matter.

W. Blaine Dowler

July 19, 2010@ 8:21 pmThe imprecisions in that model come primarily from the fact that the bowling balls have nonzero radii. We’ve already established that this cannot be the case. If the bowling balls had zero volume, we’d still have the imprecision problems dealing with molecules, but an electron-electron collision could have arbitrary accuracy.

DGDanforth

July 20, 2010@ 4:11 pmI am not sure zero radii particles exist. Self energy of fields go to infinity in that case. There are models of electrons with nonzero radii. Either particles do not exist and their seeming existence is just a reflection of wave-wave interaction or they exist with finite radii (or functions of distance like a Bessel function).

W. Blaine Dowler

July 20, 2010@ 4:18 pmSee lesson six.