Perfection
The Great Head.
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Cumulative General Science/Technology Quiz
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Genocidicbunny
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open floor.
It's a triple integral, does that answer your question.
4 because it's a (x,y,z,t) calculation that needs to take into account special and general relativity to give a precise reading. x,y,z technically needs two and some trig, although a third would help as would a forth, time needs 2 and position from 2 and some less than obvious maths. So four.
To put it in laymen's terms the measurement needs four variables, dependant and independent and thus 4 satellites to be accurate.
One for the mathmos and physics bods/engineers.
What is this equation useful for and how is it possible to get a general or "indefinite" integral from it? Or at least an integral within infinite limits.
And I want all working and the sort of physics it's used in.
Should be quite easy for anyone who's done a degree.
What is this equation useful for and how is it possible to get a general or "indefinite" integral from it? Or at least an integral within infinite limits.
And I want all working and the sort of physics it's used in.
Should be quite easy for anyone who's done a degree.
Catharsis
catch u on the flip scythe
It's not an equation because it hasn't been set equal to anything.
It's an integral, it involves relating one value to another, in this case it is equal to the area under the curve of the graph of e^(ax^2). it is also an equation because it is related to something in physics. That's all I can say without giving it away, it's a model thing.
That's incorrect, I presume you either tried substitution or by parts, neither of which will yield an answer. Basically this is something you either know or you don't. If you know it it's obvious.
Mise
isle of lucy
It isn't possible to integrate that function analytically -- it's the normal distribution (or rather the form of the normal distribution), and, IIRC, integrates between -inf and +inf to sqrt(pi/alpha) or something like that.
It's also used in QM, IIRC... Eq'n of a wavefunction or something... don't hold me to that
EDIT: It's something like, if the wavefunction is Y=Y(x), then the probability that the particle lies within the region a to b is integral of Y^2 between a and b. The eq'n of a wavefunction is smth like Y=exp(i.lamda.x), where lamda is the wavelength. Y^2 is therefore exp(-lamda^2 . x^2), or exp(-alpha.x^2), where alpha = lamda^2. Again, don't hold me to that.
EDIT2: No, wait, it's not lamda, cos then the units dont match up, it must be the spatial frequency, i.e. 1/lamda -- the wavenumber or something? God I can't remember... But the symbol is k. So it's Y=exp(i.k.x).
EDIT3: Yes, that definitely rings a bell... exp(i.k.x)... And then alpha = k^2
It's also used in QM, IIRC... Eq'n of a wavefunction or something... don't hold me to that
EDIT: It's something like, if the wavefunction is Y=Y(x), then the probability that the particle lies within the region a to b is integral of Y^2 between a and b. The eq'n of a wavefunction is smth like Y=exp(i.lamda.x), where lamda is the wavelength. Y^2 is therefore exp(-lamda^2 . x^2), or exp(-alpha.x^2), where alpha = lamda^2. Again, don't hold me to that.
EDIT2: No, wait, it's not lamda, cos then the units dont match up, it must be the spatial frequency, i.e. 1/lamda -- the wavenumber or something? God I can't remember... But the symbol is k. So it's Y=exp(i.k.x).
EDIT3: Yes, that definitely rings a bell... exp(i.k.x)... And then alpha = k^2
It isn't possible to integrate that function numerically -- it's the normal distribution (or rather the form of the normal distribution), and, IIRC, integrates between -inf and +inf to sqrt(pi/alpha) or something like that.
It's also used in QM somewhere, IIRC... Eq'n of a wavefunction or something... don't hold me to that
Kind of it is but you have to do a bit of maths manipulation you're in the right area, it's Maxwell's work and it's used in kinetic theory of gases.
If you have I = e^{ax}dx, then you can effectively get rid of the constant, a, from being inside the integral, by simply substituting for, say, t = ax. Differentiating this, we get dt = a dx. Rearranging, therefore dx = dt/a. Substitute in the integral, for ax, and dx, and we now get,
I = (e^{t}dt)/a .We divide I by the factor a, but a has now come outside the integral of (e^{t}dt) . We can do a similar substitution for pretty well any integral.So in general, if I = f(ax)dx, where f is some function, then I=(f(t)dt)/a .
Given that you should be able to solve it with the double integral to polars method for e^(x^2) with the constant outside the integral and it is equal to1/2*sqrt(pi/a).
Close enough, I wouldn't expect anyone to remember that.
And you'll only find it in text books. It's not soluble technically with elementary functions, but you can fenagle a "general" solution. It can do. it's a fairly general equation.
Catharsis
catch u on the flip scythe
Yes, alright, I know what an integral is.
But there's no equals sign in it, so it isn't an equation. It's an expression, I guess.
Perfection
The Great Head.
I think it's the abs|psi|^2 of the ground state for a one dimensional SHO. Could be wrong thoughIt's also used in QM, IIRC... Eq'n of a wavefunction or something... don't hold me to that
EDIT: It's something like, if the wavefunction is Y=Y(x), then the probability that the particle lies within the region a to b is integral of Y^2 between a and b. The eq'n of a wavefunction is smth like Y=exp(i.lamda.x), where lamda is the wavelength. Y^2 is therefore exp(-lamda^2 . x^2), or exp(-alpha.x^2), where alpha = lamda^2. Again, don't hold me to that.
EDIT2: No, wait, it's not lamda, cos then the units dont match up, it must be the spatial frequency, i.e. 1/lamda -- the wavenumber or something? God I can't remember... But the symbol is k. So it's Y=exp(i.k.x).
EDIT3: Yes, that definitely rings a bell... exp(i.k.x)... And then alpha = k^2
Mise
isle of lucy
Yeah, you're right. And for any other energy level "n" you just stick that into the exponent, i.e. Y=exp(i.n.k.x). It pops out of the boundary conditions of the wavefunction (i.e. you have to have a whole number (n) of waves between the boundaries of the potential well).
PS. I'm using Y cos in my head it looks like a psi
@Sidhe: I'm fairly certain that the integral cannot be solved analytically. It can be solved numerically though, and the integral between -inf and +inf comes to a "nice" number. It's basically the normal distribution, for which the solutions and integral between any two arbitrary points are well known.
Yes, alright, I know what an integral is.
It models something it is equal to something, I just couldn't tell you what that was without giving it away.
Like I said Mise it isn't technically soluble but you can get an answer. And that is the answer given by the text book, I know it's not very satisfying, but then technically the solution is a derivation of the error function anyway. That's physics for you though, even if it isn't technically soluble in elementary functions it is wangleable.
Mise
isle of lucy
I guess it's my turn...
What does it mean for an operator to commute?
What does it mean for an operator to commute?
Well I've just done one but that's to do with certain things like charge and other forces in tensors and matrix maths is it neutral or charged, but purely in maths it just means AxB=BxA.
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