x(x-y)

That was one of the most hilarious things I've read so far this year - thank you. You still haven't got the idea have you? i apples does not make any sense, neither mathematically nor physically - perhaps it does philosophically, but philosophy is a load of pretentious twoddle most of the time anyway.

You don't really know what infinity looks like - you think you can comprehend its size by applying an object (apples) to it, however (as you said) it is not a real number and one cannot fathom what infinity actually looks like. No, an imaginary unit i of apples doesn't "look" anything like 1 apple as it's just pretty meaningless to label objects such as apples with a non-real quantity. Trying to visualise complex numbers through the use of "apples" is not the correct way to go. This statement requires mathematical justification - simply stating that "you think" i has a property which makes it similar to 1, doesn't really mean anything. Of course one can use complex variables to describe systems (such as quantum systems) - however, as far as I know, when measuring something in a physics experiment the actual quantity (or quantities more likely) that you are measuring has (have) to be real!

Indeed there is a "cubic formula", however it is rather more difficult to derive than the quadratic formula and I wouldn't recommend memorising it.

You seem to be thinking of this in the wrong way - the last question you asked is rather meaningless as it implies that i is a real, measurable number and of course i is purely imaginary. In physics a quantity that you want to measure damn well better be real (i.e. not complex/imaginary) otherwise you simply will not be able to measure it. For example in quantum mechanics a lot of the time the wavefunction ψ will be a complex equation and so it would be erroneous to think of it as an expression representing a quantity of a mechanical wave which can be measured. As to your first question of what i actually is - well, in the most simple and elegant way, i is just the square root of negative 1; it is an imaginary unit, however the arithmetic behind it is purely logical and works perfectly as well as it having a plethora of applications to physics in particular (I am a physics student and so I know and understand how important complex analysis is to so many fields of my subject).

The problem is that you're thinking of the situation just in terms of an ordinary "Cartesian co-ordinate system". Of course, the graph of the function [latex]f(x) = x^2 + 1[/latex] doesn't ever touch the y-axis - however this function does have zeroes in the complex plane (they could be represented on an Argand diagram), and in this case the zeroes are [latex]\pm i[/latex]. It is good to think of complex numbers as a way to "extend" the roots of polynomial equations - so that they all have roots, in the complex plane or real plane.

Thanks, and yeah I suppose I could've, oh well.

I obtain this solution as an indefinite integral: [latex]e^{\frac{ik}{2}}\left(2e^{\frac{1}{2}}-1\right)u + 2\left(1-e^{\frac{1}{2}}\right)ln|u| + C[/latex]

You are assuming that dark energy doesn't exist from the onset - and if that was (were?) the case, then indeed gravity would've slowed the expansion down and we would be seeing a very different picture to the one we see today (i.e. we would see a universe which seemed to be slowing down in expansion or even contracting depending on the time). The need for dark energy is to actually be able to explain how the universe is expanding (and expanding at an accelerating rate as suggested by deep field observations of galaxies and standard candles) - without dark energy we don't really have a beginning of an explanation as to why the universe is expanding. As I understand it too, dark energy may actually appear in Einstein's equations of general relativity in the form of the Cosmological Constant. However, I have very little understanding of Dark Energy as yet (although I have just started a lecture module on "Particle Physics and Cosmology" at university so hopefully I will know more in the coming months) and physicists as a whole don't really know much about it either at the moment.

I was just messing around with integrals to see if I could derive a general solution of this integral: [latex]I = \displaystyle \int \frac{dx}{\sqrt{c+bx+ax^2}}[/latex] and I was wondering if anybody could confirm my result. [latex]c+bx+ax^2 \Rightarrow c + a\left(x^2 + \frac{b}{a}x\right)[/latex] [latex]c+a\left[\left(x+\frac{b}{2a}\right)^2 - \frac{b^2}{4a^2}\right][/latex] [latex]\Rightarrow \left(c-\frac{b^2}{4a}\right)+a\left(x+\frac{b}{2a}\right)^2[/latex] [latex]\Rightarrow \displaystyle \int \frac{dx}{\sqrt{\left(c-\frac{b^2}{4a}\right)+a\left(x+\frac{b}{2a}\right)^2}}[/latex] [latex]\Rightarrow \frac{1}{\sqrt{c-\frac{b^2}{4a}}} \displaystyle \int \frac{dx}{\sqrt{1+\left(\sqrt{\frac{a}{c-\frac{b^2}{4a}}}\left(x+\frac{b}{2a}\right)\right)^2}}[/latex] then let [latex]\sinh t = \sqrt{\frac{a}{c-\frac{b^2}{4a}}}\left(x+\frac{b}{2a}\right)[/latex] [latex]\Rightarrow dx = \sqrt{\frac{c-\frac{b^2}{4a}}{a}} \hspace{2mm} d(\sinh t)[/latex] and we know that [latex]d(\sinh t) = \cosh t \; dt[/latex] So, substituting these terms into the integral gives [latex]\frac{1}{\sqrt{a}} \displaystyle \int \frac{\cosh t \; dt}{\sqrt{1 + \sinh^2 t}} \Rightarrow \frac{1}{\sqrt{a}} \displaystyle \int \frac{\cosh t \; dt}{\cosh t}[/latex] [latex]\Rightarrow \frac{1}{\sqrt{a}}t + C[/latex] [latex]\therefore I = \frac{1}{\sqrt{a}} \sinh^{-1} \left(\sqrt{\frac{a}{c-\frac{b^2}{4a}}}\left(x+\frac{b}{2a}\right)\right)+C[/latex]

As far as I know, negative mass is physically impossible with respect to current established physics. A negative mass implies that something can have a negative kinetic energy which, in turn, (I'd assume) implies a temperature below absolute zero.

It makes me wonder about how much we might have achieved by now if, for the last 10 years, the military spending and space spending budgets were swapped around for all countries around the globe - so that the space spending was astronomical, excuse the pun, instead of the military spending.

Greetings. I think I may have joined this forum quite some time ago, however I cannot remember the email address I used nor the username or password - so I just joined again (or maybe not?) with this account, using a username that I have on 2 other science forums. Anyway, I am 18 (nearly 19) years old and am studying Physics at degree level at the University of Birmingham (UK - Russell Group and Red Brick Uni) which I am very much enjoying - I'm in my first year. My main interests within Physics are Quantum Mechanics, Relativity, Cosmology, Particle Physics to name a few (QM being my favourite I think). And my interests within Mathematics are Calculus, Vectors, Complex Analysis, Group Theory again to name a few. I recognise a few members here (such as iNow, Ophiolite and DrRocket) from other science forums, and so I hope that I shall enjoy my time on here.

Think about the quantities they you have been given - How might you compute energy from voltage, current and time? - Gravitational potential energy is approximately = mgh, so what will Δ(mgh) be? And for the last part - think about energy losses and where they might occur in such a system.

Hmmm... Yes, this is a bit beyond my current level of understanding - I suspect that I will have to wait until the Quantum Mechanics 2 course of the second year of my Physics degree to begin to gain an understanding of these more advanced processes. I very much look forward to it - quantum mechanics is one of my favourite topic of physics, it's so profound and weird; yet incredibly interesting and exciting to study!

It sounds to me that you are just very bad at mathematics and therefore are bitter and try to make out that mathematics is to blame when you yourself just don't understand it.

Just because you don't understand it does not mean it is incorrect - perhaps you should study mathematics and physics in more detail before making such wild, nonsensical statements. Mathematics is beautiful for the very fact that is the most rigorous, logical and definite of all subjects - if one proves something in mathematics, then nothing can take that away from you; it is a definitive proof. Your statement of "mathematicians and scientists being the same thing these days" seems to hold no merit and has no substantiation behind it - why did you say such a thing? If (pure) mathematics - where no assumptions have been made and everything has been derived, either directly or indirectly, from first principles - says that "something can't" (as you said it - quoted above) then that means that it absolutely can't. I hope you're joking, that's utter nonsense.

So your talking about the conservation of symmetry essentially, aren't you? And, whilst it is true that symmetry forms a large part of physics (for example the inherent symmetry in much of physics is what inspired the likes of De Broglie to begin searching for further evidence for wave-particle duality - which is know a central concept in QM), it is also found to be broken when studying the matter-antimatter asymmetry (CP violation etc) as just one example. Also, I must say that your post seems to be mostly philosophical rambling with the occasional buzzword - just pointing that out.

Ah, thank you - I suspected that this wavefunction can't be normalised in a way that I know of (as shown by the link in the OP). Anyway, I'm afraid that link are your explanation seem a bit too advanced for my current level; however, I love to learn and shall take a closer look later on. Thanks!

So, I was just messing around with some practice questions in my university physics textbook and I came across a particular question which was fairly easy to answer but which I would like to take further - however, it may be too advanced for me at the moment being a first year university physics student. So, the question basically was: "The wavefunction [latex]\psi (x,t) = Ae^{i(k_1 x - \omega_1 t)} + Ae^{i(k_2 x - \omega_2 t)}[/latex] is a superposition of 2 free-particle wavefunctions, both k1 and k2 are positive. a) Show that this wavefunction satisfies the Schrödinger equation for a free particle of mass m. b) Find the probability distribution function for ψ(x,t)." So, the first question (a) was simple of course - just taking partial derivatives then a bit of algebra. And the next part (b) gave me an answer of [latex]|\psi (x,t)|^2 = 2A^2(1 + cos[(k_2 - k_1)x + (\omega_1 - \omega_2)t])[/latex] using the Born interpretation procedure (complex multiplied by complex conjugate gives mod. complex squared which is the probability density - or "distribution function"). So, I was wondering how one could use this result for the probability distribution function to normalise the wavefunction - as I have done for simpler complex wavefunctions before. Now I know that this result should be true: [latex]\displaystyle \int \limits_{-\infty}^{\infty} |\psi (x,t)|^2 \, dx = 1[/latex] as the particles must be somewhere in this one dimensional space. However, when working through this I found that A = 0; which is nonsensical and obviously incorrect. I have also tried to compute the problem using the identity: [latex]\frac{d}{dt} \displaystyle \int \limits_{-\infty}^{\infty} |\psi (x,t)|^2 \, dx = 0[/latex] and then using Schrödinger's 1D time dependent equation, one arrives at the result: [latex]\frac{d}{dt} \displaystyle \int \limits_{-\infty}^{\infty} |\psi (x,t)|^2 \, dx = \frac{i\hbar}{2m}\left[\psi (x,t)^{\ast}\frac{\partial \psi (x,t)}{\partial x} - \psi (x,t)\frac{\partial \psi (x,t)^{\ast}}{\partial x}\right]_{-\infty}^{\infty} = 0[/latex] However, yet again when working through this I found that A = 0 again! After reviewing page 38 of "Quantum Mechanics by E.Merzbacher, 2nd Edition, 1970" I found that the above mechanism only works if the expression below is true: [latex]\displaystyle \lim_{x \to \infty} \psi (x,t) = 0[/latex] which the wavefunction in my case doesn't obey. So, I was wondering how one may go about computing such a situation - it may be too advanced for me at the moment, but I love to learn! By the way, no this is not a "homework" question - I am not the kind of person who cheats in such a way for homework answers; however, evidently I cannot prove this so believe this is a homework question if you want to, it makes no difference to me as I know it isn't! Thanks, x(x-y) P.S: Here is a more complete review of what I've done for this problem so far on another forum