hobz

But isn't the wavelength "along" the ray of the wave? I mean, since EM waves are transversal waves, shouldn't it be their amplitude that "got stuck" in the holes?

I was wondering why the microwaves generated in a micro oven, cannot escape the metal grid that is located just behind the glass in the door of the micro oven?

I see.

Does the response of the electrons depend on the number of electron that respond (are present)?

That is, if I have 10 electrons at the same radius from the antenna, do they experience a force 10 times weaker that if there were only one electron? And is this due to a sort of electromagnetic inductance caused by the 10 electrons?

thedarkshade: Thanks for your reply. swansont answered this. What I meant was: do non-moving charges exist?

swansont: Thank you very much. Your answer suggests that the wave in the electromagnetic field is pure abstraction, and what is "real", is how the wave interacts with the detectors; through the photon.

Is this correct?

If not these detectors were photon detectors, but rather Coulomb force detectors, then as the field stars oscillating, so would the charges inside the detectors (thus allowing me to measure the force of their "movement").

Now all these charges are oscillating around the antenna, and creating their own oscillations in the electromagnetic field. Will these oscillations oppose that of the antenna, and if so, why? Would it be like oscillating a cork in a pool of water, and when the waves from the edge of the pool are reflected, and superpositions with the cork-wave, then the it would become harder to oscillate the cork?

I have questions. Questions that I have been trying to figure out for myself, but now I have realized that I need professional help.

An electron has a charge, and is a source of an electric field.

A moving charge has the property that it is a source of a magnetic field.

What should the charge be moving in relation to, in order to consider it a "moving" charge? Are not all charges moving in some sense, and thus be affected by all other charges however slightly?

I set up a measuring device, at some point in space, that can measure electric force as well as the magnetic force, so that I can measure moving charges (as well as non-moving).

If I shoot one charge in some direction, so that it is moving at a constant velocity, then would my measuring device measure an impulse in the electromagnetic field?

Accelerating a charge gives rise to a constant oscillating electromagnetic wave. Can this be interpreted as a series of continuous impulses (as described above)?

Do electromagnetic waves, created by an accelerating charge, propagate spherically (e.g. from the origin of an radio antenna)?

If I accelerate the charges (e.g. in an antenna) frequently enough, say to make the electromagnetic waves have a wavelength of visible light, would people all around the antenna be able to see this light?

I have read that the human eye can detect photons, which suggest that if the answer to the question above is: yes, then the electromagnetic waves have the property of interaction with the cells in eyes via photons.

But since the charges in my antenna are accelerated using limited energy (as opposed to unlimited energy), then if I crowd the space around the antenna with eyes (or solar cells), then would I not be able to detect photons in every eye/solar cell, since the electromagnetic wave is (assumed by me) omnipresent? And if so, there would be an unlimited number of photons caused by limited amount of energy, which really tells me something is wrong.

I cannot seem to figure out how photons can have a direction, if they have a wave interpretation. Which brings me to my next question.

Do photons have a wave interpretation? I imagine the electromagnetic field to be a large carpet, and a photon to be a beetle crawling (in a precise direction) below that carpet.

Can photons only be created by-- and related to-- quantum leaps in excited atoms?

To me it seems that when life (in form of humans) proclame that it it self is meaningless or purposeless. If all humans could agree upon this, then, as the most advanced known representative of life, this must be true.

Thank you for your explanation! It is very insightful and helped me a lot.

Also thank you for the time you put into explaining this to me.

That is not the error function. The standard definition is

 

[math]\mathrm{erf}(x) = \frac 2 {\surd \pi} \int_0^x e^{-t^2}dt[/math]

 

The Maclaurin series for erf(x) as reported at mathworld uses the defintion I reported, not the function in your original post.

 

BTW, wikipedia has the same definition for erf(x) as mathworld.

Yes I am sorry. I did not mean erf(x).

Thanks. Just what I needed.

Why is C = 0?

With [math]f(x)=\int_0^x e^{-1/2 t^2} dt[/math] Taylor expanded around a=0 leading to the Maclaurin series.

[math]f(0)=\int_0^0 e^{-1/2 t^2} dt[/math] which must be equal to zero because of identical limits I presume.

[math]f'(0)= e^{-1/2\cdot 0^2} = 1[/math]

[math]f''(0)= -1/2 \cdot e^{-1/2\cdot 0^2} = -1/2[/math]

[math]f'''(0)= (-1/2)^2 \cdot e^{-1/2 \cdot 0^2} = (-1/2)^2[/math]

And thus I get the first three Maclaurin terms to be [math] f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x + \frac{f'''(0)}{3!}x = 0 + \frac{1}{1!}x + \frac{(-1/2)}{2!}x^2 + \frac{(-1/2)^2}{1!}x^3 [/math] which clearly differs from the series defined at MathWorld.

I realize that I exchange all [math]t[/math]'s with the value for [math]x=0[/math], but I have no idea what else to do.

Hi. I am trying to figure out what the Maclaurin series of the error function looks like.

I use the reference from Wikipedia to check my results, but I cannot seem to get it right.

with [math] f(x) = \int_0^x e^{-\frac1{2} t^2}~dt[/math]

I would expect [math]f(0) = 0[/math] and [math]f'(0)=-\frac1{2}[/math] and thus the first two terms of the Maclaurin series to be [math]\frac1{1!}x + \left(-\frac1{2}\right) \frac1{2!} x^2[/math]

but this does not seem to be correct.

What am I doing wrong?

Anyone?

I indeed do. I will try that. Thanks for the help! Thank you all.

I have tried the substitution, with [math]u=t^2[/math] I get [math]\frac{du}{dt}=2t \Rightarrow du=2t~dt[/math].

This doesn't really help, since there are no [math]2t~dt[/math] to substitute for [math]du[/math]. Is this correct?

I see. Brought me closer to understanding the nature of the error function.

But I still do not quite understand why I cannot employ the usual techniques for evaluating the integral. What exactly will go wrong, if I try finding the [math]\int e^{\left( -\frac{t^2}{2}\right)}[/math]. And here I would use some product rules, chain rules or what ever, as is taught in calculus introduction courses.

I understand. But it does not clarify why it does not have an antiderivative.

Yes. I meant e^(0.5*t^2). Sorry.

This was my observation also!

 

For [math]e^{-t^{2}}[/math] you will need the error function, which is related to the normal distribution.

But what happens if I try to integrate [math]f(t)=e^{-t^{2}}[/math]? Do I end up with something where I need to integrate the [math]f(t)[/math] again? My calculus is rusty, but I would expect some rules (chain rule if I am not mistaken) to allow an expression of [math]\int f(t)dt[/math] where integration of [math]f(t)[/math] again is needed?

I know that the cdf of the normal distribution can not be evaluated analytically, since e^(-t/2) has no antiderivative. Why is that?