Electromagnetic waves from Maxwell's equations

  Рет қаралды 15,149

Dr Ben Yelverton

Dr Ben Yelverton

Күн бұрын

Using Maxwell's equations in free space to demonstrate the existence of electromagnetic wave solutions, and investigating the properties of the oscillating electric and magnetic fields. For example, we show that the fields oscillate in phase with each other, but are perpendicular to each other and to the wavevector.
About me: I studied Physics at the University of Cambridge, then stayed on to get a PhD in Astronomy. During my PhD, I also spent four years teaching Physics undergraduates at the university. Now, I'm working as a private tutor, teaching Physics & Maths up to A Level standard.
My website: benyelverton.com/
#physics #mathematics #electromagnetism #waves #electromagneticwaves #wavespeed #light #maxwellequations #vectors #vectorcalculus #differentialequation #waveequation #partialdifferentialequation #planewave #wavevector #dispersion #crossproduct #vectorproduct #dotproduct #scalarproduct #calculus #differentiation #integration #emwaves #maths #math #science #education

Пікірлер: 29
@richardwilliams3215
@richardwilliams3215 Ай бұрын
Very well presented mathematics, thanks.
@DrBenYelverton
@DrBenYelverton Ай бұрын
Thanks, happy to hear that!
@abdurrahmanlabib916
@abdurrahmanlabib916 9 ай бұрын
Great video!!. It would be really helpful if you could make a video for reflection and transmission coefficients. Thanks
@DrBenYelverton
@DrBenYelverton 9 ай бұрын
Thanks for the suggestion!
@MissPiggyM976
@MissPiggyM976 Жыл бұрын
Very useful, thanks !
@DrBenYelverton
@DrBenYelverton Жыл бұрын
Good to hear! Thanks for watching.
@JOVENSTEVEADAPTAR
@JOVENSTEVEADAPTAR 10 ай бұрын
Thankyou for this !
@mrslave41
@mrslave41 3 ай бұрын
❤❤❤❤amazing stuff.
@DrBenYelverton
@DrBenYelverton 3 ай бұрын
Thank you!
@milessodejana2754
@milessodejana2754 5 ай бұрын
isnt the first maxwell equation the gauss law and is not equal to 0? And the reason why del(del \cdot E) is 0 is because del \cdot E is constant?
@hutruth5383
@hutruth5383 5 ай бұрын
in a vacuum the charge density is 0
@DrBenYelverton
@DrBenYelverton 5 ай бұрын
We're considering waves in a vacuum here so ρ = 0. That ∇(∇·E) term may or may not be zero in general, depending on how ρ varies in space.
@milessodejana2754
@milessodejana2754 5 ай бұрын
Oh, okay. Thank you!
@RelaxedSun.
@RelaxedSun. 3 ай бұрын
@@DrBenYelvertonfor simulation purposes is it fine to use these when the travel medium is air since it’s closely related to vacuum?
@DrBenYelverton
@DrBenYelverton 3 ай бұрын
@@RelaxedSun. Yes, unless you need extreme accuracy! There are no free charges or currents in air, so the source terms are still zero. And the relative permeability and permittivity of air are so close to 1 that we can treat them as exactly 1 for most practical purposes.
@sonusangeeths
@sonusangeeths Жыл бұрын
sir one doubt we know that E0/B0 = C ------- (1) Meaning that magnitude of B0 is very much smaller compared to E0. So when we are taking (K x E0)/w = B0 -----(2) By comparing (1) wrt (2) to make B0 a small value we should take a higher value of omega(w). w >> (K x E0). if w(2πf) is greater meaning we have a higher frequency. So no of oscilations of magnetic field at a time interval ΔT will be higher than EF. ?
@DrBenYelverton
@DrBenYelverton Жыл бұрын
They both oscillate with exactly the same frequency - you can see this from the fact that E₀ and B₀ both have the same ω in their exponent. Your equations (1) and (2) are correct, but comparing them just tells us that c = ω/k, which we indeed know to be true. We can't really use your reasoning because it's not valid to say that B₀ is much smaller than E₀, as this depends on the unit system being used. For example, consider using natural units in which c = 1, then the two fields have the same magnitude.
@SurprisedDivingBoard-vu9rz
@SurprisedDivingBoard-vu9rz 4 ай бұрын
Why we use cross product dot products. Because of trigonometry of material structure. The shape of an electron is a tetrahedron structure with a packing effect. I mean if you write the field equations for electrons. Probability and statistics is about study of structures.
@sonusangeeths
@sonusangeeths Жыл бұрын
is E0 a constant vector ?
@DrBenYelverton
@DrBenYelverton Жыл бұрын
Indeed, it's the vector amplitude of the electric field oscillations.
@philoso377
@philoso377 8 ай бұрын
Nice video and presentation. Vacuum is empty of anything? Yes to mechanical engineer and some mathematicians, and no to RF radio engineers. This video just confirms that light speed in vacuum is dictated by e0 and u0 in vacuum. Also that vacuum is a medium of light. The logic and mathematics in the Michelson and Morley Experiment are correct except the concept. Concept? To proof that Aether, a light medium, is like wind can blow through all matter and not dragged by it. If that is correct, we’d expect that electromagnetic and light waves riding on a medium of Aether can also blow through any matter, and as a result, light becomes undetectable.
@alirexasedghi8378
@alirexasedghi8378 4 ай бұрын
aether has been debunked.
@philoso377
@philoso377 4 ай бұрын
@@alirexasedghi8378 to the shallow minded disciples of celebrity theories Aether is absent. No sound in absence of air and sea. No ripples in in absence of water. No light in absence of a medium Aether. The M&M experiment tell us two but one things. That is either 1) Aether wind track with the interferometer or 2) Aether is absent. On the other hand, No e field in space in absence of e0. No b field in space in absence of u0. Both e0 and u0 are attributes of Aether. No light or internet without Aether.
@philoso377
@philoso377 4 ай бұрын
@@alirexasedghi8378 yes to those who chase after rainbows, celebrated brand name products and theories. Those who learn very well and not think deep enough.
@alirexasedghi8378
@alirexasedghi8378 4 ай бұрын
@@philoso377 this is not a counterargument, it's irrelevant.
@philoso377
@philoso377 4 ай бұрын
Debunked? That in the eye of the beholder. If you believe it that become real to you. If we are unable to scrutinize a theory independently and refute it we become a follower. If we with IQ120 and follow a theory offered by someone who claimed to be IQ200 our IQ become zero. Aether is measurable by e0 and u0 exists in vacuum where there is empty of matter. Just because we can’t detect it doesn’t make it absent. The M&M experiment detects no Aether wind because scientists at the time falsely define what Aether is and chased something false, in the end found nothing. Don’t let me interrupt your dream of no Aether.
@williamwalker39
@williamwalker39 3 ай бұрын
What is presented here is a very simple approximation. Setting the wave equation to zero is a boundary condition valid at infinity, which does not exist in reality. The correct way to analyse this problem is to set the wave equation to a source and then analyse the propagation of the resultant electromagnetic fields. When this is done a lot of detail overlooked here is revealed. The fields propagate instantaneously when they are created by the source in the nearfield, and they reduce to about the speed of light at about one wavelength from the source, and never becomes exactly c even at extreme astronomical distances from the source. This corresponds to the phase speed, group speed, and information speed. Another important detail that is revealed is that there is a circulation of the electric field in the nearfield. The longitudinal electric field that is launched from the source, curves toward the opposite pole and creates the transverse electric field at about 1/4 wavelength, and launches transverse electric fields back toward the source and also away from the source. These propagating fields in the nearfield completely violates Relativity, which says nothing goes faster than speed c. So the speed of light is not a constant as once thought. Inserting c=infinity into the Lorentz Transformation yields Galilean transformations, where time and space are absolute. These results indicate that if a moving body is observed with nearfield light, then no Relativistic effects will be observed, but if farfield light is used then Relativistic effects will be observed. Since time and space can not depend on the frequency of the light used to observe effects on time and space, then one must conclude the effects of Relativity are just an optical illusion. Any theory based on Relativity such as General Relativity, and Quantum electrodynamics are wrong. See the following short KZbin presentation for details, as well as the paper it is based on: *KZbin presentation of above arguments: kzbin.info/www/bejne/qZazlX1tq7iErLM *More extensive paper for the above arguments: William D. Walker and Dag Stranneby, A New Interpretation of Relativity,2023: vixra.org/abs/2309.0145 *Electromagnetic pulse experiment paper: www.techrxiv.org/doi/full/10.36227/techrxiv.170862178.82175798/v1
@l1mbo69
@l1mbo69 Жыл бұрын
Gosh there’s 5 ads in 9mins
@JaganJagan-t8r
@JaganJagan-t8r 8 ай бұрын
I wana mathematics proof of coulomb law in classical mechanics and how epsilon=8.85×10^-12 and not use proportional term
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