hobz

Well, I suppose you could argue that the term perpendicular is just as valid in an algebraic sense. For example orthogonal functions, and vectors spaces, need only to satisfy a certain criterium to be called orthogonal, but the criterium has roots in geometry analysis. I mean [math]e^{i\phi}[/math] relates [math]e[/math] to an angle (in the complex plane) [math]\phi[/math] via [math]i[/math], such that e.g. [math]e^{i\pi}=-1[/math]. In essence you've already explained this relation in your previous answer, where [math]e^{i}[/math] was related to trigonometric functions.

Ah yes of course. Thanks for pointing that out. So then, the geometrical property of the imaginary axis being perpendicular to the real axis, can be viewed as a consequence of the angular relation between [math]e[/math] and [math]i[/math]?

Truely magic However this holds for ANY value of [math]i[/math] doesn't it? I mean, the derivation of the fact doesn't involve any characteristic if [math]\sqrt{-1}[/math] does it?

Isn't it a property to have no properties?

I guess of course and have never postulated my guess to be truth. If I had to choose from your list it would be a superposition of the first three

True. Anything can be perceived as something. Even nothing. The "true" nothing must not exibit such categorization, and thus it cannot exist in any form that we can comprehend. I guess.

I am having problems with understanding how the formula can have a geometrical interpretation by having an imaginary axis perpendicular to the real axis. I mean how can an exponent (of e) multiplied by the imaginary unit be related to an angle?

I agree with Gilded. Often when the term "nothing" is used, it refers to the absence of objects of a particular interest. A police investigator replying "nothing" to a question of what he found in a search did obviously not encounter nothing at all, just nothing of interest. If he actually encountered nothing at all, it would be that he was faced with the absence of everything even nothing it self, since nothing, as a concept, word, sound or idea is something. One might argue that the philosophical "nothing" we use in word, sound and idea is just a placeholder for an absolute nothing that is not definable but all people have an intuitive idea about what it means in general. Relating this to a philosophysical meaning, one might consider what a total absence of energy in some region of (timeless) space might be. Is it nothing? Well it is still a region of space, which certainly not is nothing. Does it contain anything? It does not contain any energy, but one might argue that it contains it self, i.e. the empty space we are looking at contains the empty space it self. By shrinking this region of space to zero (multiply all sides of the space by 0). The total volume is now 0. In other words there is no space. This kind of nothing-space might be the closest to an absolute nothing that I can imagine. The nothing-space is however inherently boring, as nothing(!) can exist within it...

I agree. The charge on the area of the terminals will probably be the only current stored in the system, and not that much charge is needed to create such small differences in voltage. Thanks for replies!

I follow. But right when you connect the batteries a current must flow as a reaction to the equilibrium being created?

I guess you're viewing the battery as a diode. But couldn't I then just flip the diode around? I am basically viewing the battery as a charged (to 1.5 V) capacitor, and the series connection as two capacitors in series. By connecting one capacitor to the other, redistribution of charge will occur, such that the plates that are connected obtain equal potential. While this happens, current must flow in the capacitors until the charges are distributed to equilibrium. Is this not true for batteries also?

Good explanations, thanks! I have not made my question clear enough though. The 3 V "wire", was actually not meant to be a wire, but a visual reference to the 3 V. My question is: Why does current not flow (or does it?) from the +/- terminals when the first battery is put ontop of the second battery, without the wire.

When I connect two 1.5 V batteries in series, I get a potential difference of 3 V across the terminals. |+|---- |1| | |-| | | 3 V |+| | |2| | |-|---- (Showing battery 1 and 2 in series). Can anyone explain why the voltage is doubled when voltage sources are connected in series? Why, for instance, is current not flowing from + of battery 2 to - of battery 1 (conventional current flow)?

No I am not familiar. Just did a read up on wikipedia. Interesting stuff. I am guessing that the probability a decay occuring is related to the square of the wavefunction for a given atom. Small probabilities of tunnelling occuring, give a high stability (long half-life). Are all atoms subject to this tunneling effect? If they are, then that would mean that all atoms are unstable.

Exactly what I meant by "some degree" thank you. So "tunnelling" is an event that, when it occurs, can escape the nuclear force?

So what force overcomes the nuclear force to allow the decay to happen?

What is the logic behind this? Shouldn't the strong force hold the nucleons together as it does in heavier stable atoms?

As far as I can tell, an atom's instability is caused by a struggle in the nucleus between the nuclear force and the electromagnetic force. Are neutrons and protons bound by the nuclear force or is it only protons? Iron represents an equilibrium between these two forces. If an atom has more protons (e.g. 92 as is the case with uranium is I'm not mistaken), the electromagnetic force will "win more battles" than the nuclear force, and the atom will be prone to break up into smaller more stable atoms - fission - in which an amount of energy in some form is also released. The energy released is equal to the energy used to bind the original atom together by the nuclear force. If I have a hydrogen atom, and shoot a proton directly into the nucleus, then the two protons fuse together and release some energy. I don't why or how much. I have to shoot my proton rather fast as to get the my proton sufficiently close to the nucleus proton allowing the nuclear force to kick in and overcome the repelling em-force. Are light atoms not also unstable in a way? I mean, could these fusions not occur naturally provided that high energy protons are present. If I fuse iron and a proton, then this configuration will eventually decay into iron again right? Is iron unstable to some degree?

I suppose you theoretically could come infintely close to an acceleration described by a (weighted) delta function. Returning to the main question. Suppose a satellite is told to move from A to B using a minimum amount of fuel. Then why would it be most efficient, in terms of minimum fuel usage, to accelerate the satellite early on in stead of a constant acceleration?

I have come across the statement that it is more efficient to accelerate a mass at the beginning of a total movement, than later on. So having a mass at rest and wanting to accelerate it up to 10 m/s over 10 s, it should be more efficient to accelerate the mass to 10 m/s earlier on. Why is this?

Thank you. A very concise explanation. Much obliged.

Can somebody post (or direct me to) a proof/explanation that an outer product always has rank 1? Thanks!

Greatly appreciated! Is the original paper available on the net?

I have learned that Einstein predicted that it was possible to stimulate and excited atom with an incoming photon, and that the atom would have a high probability of emitting a similar photon having same frequency, phase and direction. (Of course, the incoming photon should match the excited state so as to facilitate an emission in the first place.) My questions are these. How the hell was Einstein able to predict that this phenomenon would occur? As I understand it, no experiment had been done prior to this prediction. Is the probability binary, such that it either emits a similar photon or none at all? Or can it emit a "random" photon, or does stimulated emission overrule spontaneous emission?

You are of course correct. I meant probability, not uncertainty. But the wave packet is an uncertainty (to my understanding). The probability of the photon being "present" is spread out in space, and is not measurable to arbitrary precision at the same time as the momentum? This makes it hard to understand your statement because I thought that HUP prevented this, so to speak.