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Conceptual Question: Light VS ElectroMagnetic Induction 
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#1
Dec1911, 06:05 PM

P: 169

I'm willing to say that I have a fair understanding of Electromagnetism, at least from a classical perspective, but I was never entirely clear about the difference between, say, the timevarying magnetic field created by an inductor and the Electromagnetic field of light in general. Based on what I understand, I should think that they are one in the same, the "Light" from the inductor simply being extremely long wavelength radio waves. Would that be correct? If it is, then why do we observe such incredibly different behavior from a transformer, a microwave, and a lightbulb? Does it really just boil down to the frequency/wavelength of the photons?



#2
Dec1911, 10:20 PM

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Yes you would be correct. Yes it does boil down to wavelengths.
The shorter the wavelength, the more particlelike the behavior. Probably you want Feynman's lectures on WaveParticle duality  youtube  he covers how the "wavelength" of a photon works. How does a photon of EM radiation with a wavelength of several meters not get to be several meters long? 


#3
Dec2011, 10:38 PM

P: 169

Thanks for the reply; perhaps you could help me with one other thing?
I've taken a college level Physics E&M course, but whenever I see maxwell's equations written out they just look like gibberish to me. I'm familiar with things like dot and cross product, but the formulation on Wikipedia, for example, makes use of operators I've never seen before. Could someone clarify that for me? (I've read the related articles on Wikipedia, which seem to me to also be gibberish) 


#4
Dec2011, 11:21 PM

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Conceptual Question: Light VS ElectroMagnetic Induction
Which operators do you have a problem with?
The big upsidedown triangle is a nabla for example. You can get the notation by comparing what they use with the forms you are used to. 


#5
Dec2011, 11:38 PM

P: 169

Well, I'm familiar with your standard integral and derivative, as well as all of the constants and units. I'm a little iffy on surface integrals, as well as integrals in 3 dimensional space. The differential element makes sense to me, however I am unfamiliar with either of the inverted triangle operators (The nabla I guess); It appears that it is/returns a vector, which can then be (and in this case is?) used in a dot or cross product. I'm also unsure what exactly to do with the nested/double integral symbol, especially the one that's supposed to be over a surface.



#6
Dec2111, 01:39 AM

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Well those are all terms you can look up.
It's hard to believe you've done college EMag and still iffy about surface or volume integrals or double integration. It is technically possible to do it without seeing the differential form of maxwell's equations ... but not in the last decade, in the western hemisphere. 


#7
Dec2111, 07:18 AM

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Firstyear calculusbased physics textbooks commonly present Maxwell's equations (Gauss's Law etc.) in integral form. What you're seeing are Maxwell's equations in differential form, using vector derivatives. Physics students usually learn about them in an intermediate or upperlevel E&M course, using a textbook such as Griffiths or Purcell. 


#8
Dec2111, 05:20 PM

P: 169

We also didn't fully cover Maxwell's equations I think. We gained a solid understanding of the first two laws, but the second two were presented in significantly less depth. While we covered Gauss's Law and Gauss's Law for Magnetism from a calculus perspective, the other two were more or less glossed over, so I can only really do simpler things with them. I have, for example, wanted to reproduce the derivation that shows light can be represented as a self propagating EM wave, and while I can see it conceptually, my grasp on the math isn't good enough to actually work it out. 


#9
Dec2111, 07:26 PM

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The nabla notation is used when we don't want to make assumptions about the coordinate system  physics should not depend how we choose to represent space. It's quite involved  you do need to know how differential equations work. For the integrals  you need 1st year calculus, look at volume and surface integrals and learn how to construct them. 


#10
Dec2111, 07:43 PM

P: 169

The biggest problem I have with this stuff is that I never got to work in anything but two dimensional space; as such, I struggle sometimes extending things into that 3rd dimension, most importantly, in integrating equations which involve more than two variables. I can start to picture how one might go about doing that kind of integration, but again, not well enough to actually pull it off. I think the best thing at this point would be if anyone knew of some online resources that would help me put all of the pieces together; I can't help but feel I'm close and just missing a few little things. 


#11
Dec2111, 08:18 PM

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The concept of a double integral should work if you consider the area between two curves, f and g, in 2D  you know the result is just to do A=∫(fg)dx but I want to demonstrate, formally, how this works: You divide the region between the limits a and b into lots od small squares width dx and height dy. the area of one of them is dA=dxdy  now you have to count up all the little areas. Each column of squares at position x starts at y=g(x) as the lowest point and y=f(x) as the highest. The x position starts at a and ends at b. so we need to sum all the little dA's within those limits  which we write [tex]A = \int_{x}\int_{y} dA = \int_{a}^{b} \int_{g(x)}^{f(x)} dy.dx[/tex]... evaluate them from the inside out  the indefinite integral of dy is just y+c, which evaluates to fg because of the limits. The answer is so easy that we don't normally bother with the double integral. If you put g(x)=0, this will give you the regular integral you were first taught  the area between the function and the x axis. So have I got the right pitch for your level yet? I want to be sure before going into 3D. Hyperphysics can be a useful quickstart, though it tends to be conceptual. Mathinsight has some useful examples. It's difficult to know what to suggest, but after a bit of toandfro we should settle on something. 


#12
Dec2111, 10:37 PM

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#13
Dec2211, 01:20 AM

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OK Cool  so, to make that concrete: if the area is a circle  that would be the set of points so that [itex]x^2+y^2 \leq R^2[/itex] right? So [itex]g(x)=\sqrt{R^2x^2}[/itex] and [itex]f(x)=\sqrt{R^2x^2}[/itex] ... though you can probably think of simpler ways to do this.
But what we can do is exploit the symmetry  if change coordinate system to polar, at position [itex](r,\theta)[/itex] we can put a small area [itex]dA[/itex]  the area between [itex]r[/itex] and [itex]r+rd[/itex] in the [itex]r[/itex] direction, and [itex]\theta[/itex] and [itex]\theta + d\theta[/itex] angular. This [itex]dA[/itex] is roughly square [itex]r.d\theta[/itex] by [itex]dr[/itex] ... the limits will be from [itex]0[/itex] to [itex]R[/itex] in the [itex]r[/itex] direction and [itex]0[/itex] to [itex]2\pi[/itex] in the [itex]\theta[/itex] direction  so [itex]dA=r.dr.d\theta[/itex]. The integral becomes: [tex]A = \int_0^{2\pi} \int_0^R r.dr.d\theta[/tex]... The next step is to do this for surface area. In this case the surface is not going to be flat. Again  concrete: the surface area of a sphere. A sphere surface is the set of points so that [itex]x^2+y^2+z^2 = R^2[/itex]  see how the third dimension pops in there. When you draw this, put the y axis horizontal, and the z axis vertical, then the x axis sticks out of the page. Once again we want to exploit the symmetry so that the math is not too hard. This time I need two angles  I can define them any way I like but the standard is to make one the angle, in the xy plane, from the xaxis and the other the angle from the z axis. A particular position on the surface of the sphere would be at angles [itex](\theta, \phi)[/itex] where [itex]\theta[/itex] is the one from the z axis. The area [itex]dA[/itex] will be the bit between [itex]\theta[/itex] and [itex]\theta+d\theta[/itex] one way and [itex]\phi[/itex] and [itex]\phi + d\phi[/itex] in the other. Since these dangles are really really small, this area is (length times width) [itex]dA=(R.d\theta)(R.d\phi)=R^2d\theta d\phi[/itex]. Since the sphere is for all angles, the limits are [itex]0[/itex] to [itex]2\pi[/itex] for [itex]\phi[/itex] and [itex]0[/itex] to [itex]\pi[/itex] for [itex]\theta[/itex]. Think you can write out the integral? A point of notation  Ive been writing dA for the infinitesimal area  but usually it is written dS  which is often vector because in the math, a plane can be defined by the vector normal to it's surface... we like to pick the outward normal when we have to but if you are careful about how you construct the integral, it won't matter. For instance: E.dS is the amount of E in the direction of dS ... i.e. perpendicular to the surface. But we usually select the surface specifically to make E perpendicular. For instance, for a point charge we use a sphere, for a line of charge we use a cylinder, and a flat sheet of charge requires a box. A volume integral is worked just the same but, but you need to also integrate wrt to a radius component and the volume element, dV = dr.dA(r). That gives you your triple integral. That should cover the integral approaches  practice. Google for "integral form of gauss' law examples". If you feel masochistic, you can try to work out the above integrals in Cartesian coordinates. The differential form otoh requires you to be able to solve differential equations  and that mean assessing the boundary conditions. I'm gonna take a bit of a break now. 


#14
Dec2311, 05:02 PM

P: 169

Alright, so that was pretty easy; I think I've done most of this before, though the notation must have been different.
I put the numbers you gave me into an integral, though it didn't come out quite right: assuming dA = R^{2} * dθ * d[itex]\varphi[/itex], one gets ∫[0,2∏]∫[0,∏]R^{2}*dθ*d[itex]\varphi[/itex] evaluating to 2(∏R)^{2} Which isn't quite the right answer. Checking my work, it appears that I'm missing a sine/cosine in the initial formula, which makes sense: If you consider the first integration as determining the arc length along a half circle, then your second integration is not against the radius, but rather against the distance from the half circle to a line connecting perpendicular to the line between it's endpoints. I can barely follow that when I write it out, but instead of (R . dθ)(R . d[itex]\varphi[/itex]), one gets (R * dθ)(R * sin(θ) d[itex]\varphi[/itex]) and integrating with the same limits, one gets 4∏R^{2}, as they should. I'll try a few more, though this is making good sense in these simple cases. I have always wondered how one would go about representing an arbitrary surface in 3 dimensional space, and working with it in terms of things like surface integrals. Typical computer modeling seems inadequate to me. PS: Sorry for the lengthy time to respond, I've been afk for a rather long time. 


#15
Dec2311, 05:47 PM

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Technically that integral goes:
[tex]\int_0^{2\pi}\int_0^\pi d\theta d\phi = \int_0^{2\pi}\pi R^2\ d\phi = 2\pi^2 R^2[/tex] The indefinite integral is: [itex]\int \int d\theta d\phi = \int (\theta + c_1)d\phi = (\theta+c_1)\phi + c_2[/itex]... where c_{1} and c_{2} are constants of integration. I skrewed up didn't I? Did you have a go figuring out what I missed? If you draw out the coordinate system, and do the conversion, you'll see I forgot a [itex]\sin(\theta)[/itex] in there. (the arc due to [itex]d\phi[/itex] is actually [itex]R\sin(\theta)d\phi[/itex]  remember how you get arclength?). The integral should be: [tex]\int_0^{2\pi}\int_0^{\pi} R^2 \sin(\theta) d\theta d\phi[/tex] If you look up "area of a sphere" in Wikipedia, they have another derivation using the volume integral. But if you start here you get the volume of a spherical shell of thickness dr and radius r, is 4πr^{2}.dr ... integrate from 0 to R to get the volume of the sphere and you'll see where that 1/3 comes from. It is a good idea to practice constructing integrals like this ... try deriving surface area and volumes for oblate and prolate spheroids. It is very likely you have met this before  either different notation or you/your teachers just got into all the tricks too fast to take all this in (often happens). For instance  you'd have done the volume of a sphere either as a series of disks or a solid of revolution  turning it into a single integral. It is uncommon for teachers to spellout the integrals like this. From these concepts and the definition of nabla, see if you can figure out div.E=\rho for an infinite sheet of uniform charge density, thickness t. (You have to set up your coordinate system, figure which form nabla should take, are resolve the boundary conditions.) It's xmas eve here, so I'll be gone for a bit. Have fun. 


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