8.03 - Lect 14 - Accelerated Charges, Poynting Vector, Power, Rayleigh Scattering

[@Originalimoc]

A note: four solid lines in the diagram 20:09 are all at time t + (delta)t

[@nybble]

Time 31:30 ... Now I understand why, in one of Feynman's lectures, when he says if he could shake a charged comb fast enough, one of the first consequences you'd notice is that the effects of oscillation reach out much further than the static effects...

[@maxwellsequation4887]

Yeye

[@meikejune4009]

Your demonstration of romanticism in Physics of sunsets reminds me on a very impressive moonrising at the Bretagne. The sky was dark blue and the hills below dark green. The Atlantic ocean was calm and quiet. The moon was not high above the horizon,but he rises very big and in orange colours. No winds, it was at la Phare de la Vielle. When the night came with the stars, the moon get higher and smaller and white as he used to be. I went back to my house alone. That's la Bretagne. My Physics is more romantic with the moon. The sun is my friend. That's a big difference. .

[@jacobvandijk6525]

Keep enjoying your moments, Meike.

[@jibinkjose1073]

Hi Professor. Great lecture. But I have one confusion: If the acceleration of the charge is constant for a period b/w two points in time, say t1 and t2, then the perpendicular electric field at some point X in space should also be constant during the period the period t1' and t2' right (where t' = t + |X|/c)? My confusion is that this electric field is not varying as a sine wave. So does it still qualify as an EM wave? If yes, how do we determine its frequency/wavelength?

[@eamon_concannon]

55:40 The acceleration vectors for the particles are all in the same plane with the point representing the observer's eye. Hence the perpendicular component of these vectors (a-perp) to the position vector from each of these particles to the observer is responsible for electric field (kink) polarized in this plane and perpendicular to position vector. This electric field kink is anti-parallel to a-perp 59:40 For the light moving upward, the Electric field is parallel to the ground. It excites charged particles causing them to accelerate in a plane parallel to the ground. These accelerated particles emit light. An upright observer's eye in that plane will see light horizontally polarized in that plane

[@beletewube]

The to circle's edge intersects as the inner circle will extend.the question is that in what circumstances fall comparable the two radiuses and determine the travelling wave?

[@sudhir2854]

Is this whole logic (about the production of EM waves by oscillating charges) could be extended to other inverse square fields?For an instance,if we accelerate a point mass in the same fashion to produce kinks in gravitational field produced by that mass,will it lead to production of some kind of 'gravitational waves'?

[@lecturesbywalterlewin.they9259]

google GW

[@Originalimoc]

Brilliant

[@rgudduu]

fantastic thought

[@Skumar-rg4hd]

Sandar , Jabardast , Jindabad....super ... thanks sir..

[@twistedsim]

Wow nice lecture and awesome demonstrations!

[@lecturesbywalterlewin.they9259]

:)

[@theodorei.4278]

Wow just amazing. Professor you give meaning to words like : teaching, learning, academic etc. I'm trying to find the book you are referring in this lecture (at 15:20) but I don't seem to find it. Could you be kind enough to tell me what is it?

[@lecturesbywalterlewin.they9259]

8.01 Physics Hans C. Ohanian 2nd edition W.W. Norton & Company ISBN 0-393-95748-9 8.02 Physics for Scientists & Engineers by Douglas C. Giancoli. Prentice Hall Third Edition ISBN 0-13-021517-18 8.03 Vibrations and Waves by Anthony French CRC Press ISBN 9780748744473 8.03 Electromagnetic Vibrations, Waves and Radiation by Bekefi and Barrett. The MIT Press ISBN 0-262-52047-8

[@theodorei.4278]

Lectures by Walter Lewin. They will make you ♥ Physics. Professor Lewin thank you very much for your prompt response and effort. To answer, to someone you've never met in person in such a short period of time is even more remarkable. One of the best professors I've ever seen in my life even though I don't know you in person. Thank you again.

[@lecturesbywalterlewin.they9259]

:)

[@jainecruz7470]

Hello Professor Lewin! Love your lessons. Could you say a biography where I can find the content you gave in this video? I tried read Griffhts but it was too much dense for me to understand. I need some reference to use in my project. Thank you!

[@lecturesbywalterlewin.they9259]

Watch all my 94 MIT course lectures. Start with 8.01, then 8.02, then 8.03. Do all the homework and take all my exams. *I guarantee you that you will then do very well on the Physics portion of any freshman college or JEE exam* You will find all information you need on this channel (notice the three playlists "Homework & Solutions"). 8.01 & 8.02 will each take about 200 hours, 8.03 about 250 hours.

[@bs694]

Thanks for a great lecture course. The Poynting vector ExB goes to zero twice a period. So where does the energy disappear too? Energy is conserved surely?

[@lecturesbywalterlewin.they9259]

how many minutes into the lecture?

[@AntarcticaNZ]

About 5 to 6 minutes. Obviously you are assuming x=0

[@aeroscience9834]

+B S at each point the Poynting vector oscillates between zero and a maximum, however, at different points in space the Poynting vector hits zero at different times (it depends where the crests and nodes of the wave is at any time) such that the total energy given by integrating over all points in space is constant.

[@Originalimoc]

Aeroscience makes sense

[@金至超]

sir, thank you very much for your enthusiastic tutoring. But why at 19:30, when charge moves from O' to O'' constantly, the field line is radially out from the charge. For example, when the charge arrives at O'', I think only the E field at this point is radially out, but the place beside it haven't known charge have reached O''. Therefore, I think the field line is curved, but not parallel to r vector.

[@ahmadsalameh7959]

I have this same problem with the derivation.

[@rgudduu]

why don't we consider electrostatic field to be a propagating wave? It is a disturbance alright which propagates at speed c to distant points and can do some work, because a charge placed at the distant point feels a force and may possibly move by it. So why is it not a propagating wave energy?

[@lecturesbywalterlewin.they9259]

it is "static"

[@rgudduu]

@@lecturesbywalterlewin.they9259 , sir, by 'static' i meant the location of the charge is fixed, not necessarily the value of the charge. What if the charge is at rest, but it is Q(t). Then the effect of change in Q would reflect at a distant point after a time r/c. E.g. we introduce some charge Q at a previously charge-free region. Then new E field lines will reach distant points, propagating at speed c. Plus, these field lines can do some work. Is it not a wave energy?

[@carultch]

@@rgudduu There is energy in the electric field of stationary charges, but it isn't propagating energy. It is energy stored in the space itself, in the electric field specifically. Per unit volume, it is equal to 1/2*epsilon0*E^2. Or in a field in a dielectric material, epsilon0 gets replaced with a material-specific epsilon. At one point it did propagate to set up the electric field, and there were electromagnetic waves involved in doing so. But once it settles to an equilibrium of a static electric field, the waves are no longer propagating.

[@sundaranarasimhan58]

Sir as you have mentioned it would be naive to associate 1000V/m to sun's rays ,can I but tell it's time average of the amplitude is around 1000V/m ???

[@lecturesbywalterlewin.they9259]

how many minutes into the lecture did I say that?

[@Originalimoc]

No it's more like many 1V/m with different wavelength.

[@stargazer7644]

In the Rayleigh scattering experiments, if the incoming light were horizontally polarized, would there still be vertically polarized light being scattered at 90 degrees? Or would there be no light emitted at 90 degrees?

[@lecturesbywalterlewin.they9259]

find your answer in my 8.03 lecture about the Fresnel equations

[@Jirayu.Kaewprateep]

Accelearted charge create of radious velocity but they can reduce of shapes such one tail spherical curcuits. If target has velocity can they more effective efficiency⁉️Do solar cells coat create different ⁉️ Can we breaths without water or movements is that the aerogel for⁉️

[@munguis]

Professor Walter, In one of your previous lectures you mentioned that when a traveling wave encounters a different medium there will be a reflected and a transmitted wave. I recently read that in the case of EM waves when an incident EM wave encounters a different medium it can either get reflected or absorbed. Is the word "absorbed" being used as a synonym of "transmitted" in this case? Or, is it a totally different situation?

[@lecturesbywalterlewin.they9259]

Depending on the wavelength and on the geometry it can be partly or 100% reflected, it can be partly reflected and partly entering "transmitted into" the medium in which case it can be partly (or fully) absorbed.

[@munguis]

Thanks for the clarification!

[@sujitmohanty1]

How lucky these students are!!

[@Skyrime007]

Sir maybe in wrong but i think the discussion around 20:00 is not correct it doesn't making sense to me ..( please comment me if someone thinks as I do . Or correct me if I'm wrong)

[@lecturesbywalterlewin.they9259]

Einstein's theory of General Relativity will also make no sense to you. But that does not mean that it is wrong.

[@Skyrime007]

@@lecturesbywalterlewin.they9259 you are right ..with your confirmation i should expel my suspicions and follow the content with a greater concentration and time. 🙏

[@amey22258]

Hello Sir. Great lecture and Thanks for it. I have one doubt which I am posting here as I did not get your email i.d. . In the diagram you draw Charge move from O to O- - in T+delta t seconds. So charge just come to the position of O- - at t+delta t and hence electrical lines of force from charge located at O - - will need a additional time of r/c seconds to intersect the circle . So total time required for Electrical lines of force emitted by charge from location O - - which intersects the circle is t+delta t (for the charge to come to O - -) + r/c ( For electrical lines of force to reach the inner circle) ( r>>>distance moved by charge). with this additional time I see the world which does not know about the change that happened has got much further and hence distance between the 2 circles should be (delta t+r/c) . Am I missing something.

[@abhishekbhardwaj1603]

Professor, can a charge moving with uniform velocity radiate E.M waves? If we have a non-zero initial velocity at Point O (22:45) will it still radiate E.M waves if it does not accelerate ? or we need the small push to generate E.M waves?

[@lecturesbywalterlewin.they9259]

watch this lecture again!

[@bilalhussein9730]

Think about the kink in the field in the region of width c delta t. Could it be present if the charge was always moving with uniform velocity? If you think about for a bit you'll know the answer to your question must be no.

[@tareklule9249]

Dear Prof. Lewin. I do not see the logic why you draw the electric field line from O'' to the sphere around O' with r = c t, as being the reference Electric field line at the end of acceleration. You argument that the observer at sphere r = c(t + Dt) around O does not yet know at time t+Dt, that the charge has accelerated away from O, and sees the E field sticking out of O.By the same argument the observer at sphere around O' with radius r = c t does not yet know at time t+Dt, that the charge has continued to move to O''. Instead, the observer should see an Electric field line coming out of O', not out of O''. What do I miss?

[@lecturesbywalterlewin.they9259]

>>> I do not see the logic>>> *watch my lecture again. I cannot add to the clarity fo this lecture.*

[@tareklule9249]

Thanks for the fast answer, though not what I expected, as I may not have expressed myself clear enough. What I meant to say is that I see a major flaw in the argumentation around time 20:00. The observer at the circle around O', with radius r=ct can only see Electric field line originating from O'' if that Electric field line travels with infinite speed. At the same time you claim the observer a tiny bit (c dt) further away still perceives the Electric field originating from O. You give no reason why the one electric field travels with speed c, while the other with infinite speed. Can you please try to give a physical explanation? Thank you. BTW, otherwise a great lecture! Wish my physics studies had been as well animated :)

[@fernandochiaramello4349]

@@lecturesbywalterlewin.they9259 Dear Dr Lewin, the question is very clear to me. I've searched into the comments because i have exactly the same question. I saw this video several times and i still don't get why the direction of the electric field in the sphere around O' with r = c t is radial towards O''. Can you (or someone else) plaease explain to us what is wrong with the question or what we are not understanding? Best regards!

[@fernandochiaramello4349]

I think i have the answer to your question: If you have a charge moving at constant speed, at any point, at time t, the electric field MUST have the direccion radial to the position of the charge at time t, not the position at time t' (where t'=t-r/c, and r the distance between the charge and the point), because if you take a second reference system solidary to the charge, let it be S', the electric field will be radial on S', and since both systems are inertial to each other, the electric field must be the same on both systems. In other words, if you have a charge moving at constant speed since ever, at any point of space the world knows that the charge is moving at that speed and knows where to point the electric field at any point at any time. Is the variation on speed that must travel to light speed to tell the world. A time t + Dt, on the whole sphere around O' and radius c*t, the last information about the charge, is that the charge was traveling at constant speed (a*Dt), so, if the charge acelerate again, in that sphere "would think" that the charge is still moving at constant speed, so the electric field will still pointing toward O'' at time t+Dt, regardless the charge is there or not. Do i express myself clearly? English is not my lenguage. Best regards!

[@gopalsolanki8651]

I am also puzzled by the same question. At time t+dt, the charge just arrived at point o’’. How come the field line has already travelled ct distance from o” instantaneously.

[@subramaniannk3650]

Prof, It's to my surprise, I've come to learn that accelerated charges have radiating/travelling electric field as well as static electric field. It's possible for me to imagine travelling electric field as disturbance traveling thru field lines emanating from the charge. But, I can't account for the static Electric field. I thought static Electric field transformed into radiating electric field when charges are accelerated. Now, I know I am wrong in that static and radiating field are different. Would I be correct in thinking static electric field are due to persistence of field lines of the charge at previous instants of time?

[@lecturesbywalterlewin.they9259]

If that helps you, then think that way.

[@yaoshendeng5250]

Dear professor Walter Lewin, in one of your derivations (around 48:20), you used x as the direction of motion. But shouldn't the charge q move up and down (along the direction of E field)? Thanks for the clarification

[@lecturesbywalterlewin.they9259]

+yao shen deng I call the up and down direction x. I could have called it anything.

[@josedanielbernalrodriguez7394]

profesor, in 19:39 why you say that inside that sphere the field comes from O"? i have problems with that because the particle arives 0" at that very instant, so NO ONE know that the field comes from O", information has not traveled yet, thanks

[@lecturesbywalterlewin.they9259]

I cannot add to the clarity of this lecture. Plse watch it again

[@sunilkukreja2919]

Hats off sir, you are the best

[@jacobvandijk6525]

@ 16:31 The part between O' and O'' is totally irrelevant. Just draw a field line from O' (parallel to O) and analyze the kink.

[@eamon_concannon]

52:30 You show that the blue light that is scattered has about 5 times more energy than red light that is scattered. According to E=hf , blue light has intrinsically more energy than red light. I see that the energy of a blue photon is 1.5 times energy of a red photon. After scattering, the energy of the blue photons is 5 times greater than the red photons, so there must be many more blue photons scattered than red photons. Perhaps you could explain a bit more why the probability of a blue photon being scattered is 5 times that of a red photon. While formulating this comment, I think that I am becoming better illuminated on this point! Thanks a lot

[@lecturesbywalterlewin.they9259]

the intensity of rayleigh scattered light is proportinal to 1/lambda^4 (1.5)^4 is about 5. lambda blue is about 1.5 times smaller than lambda red.

[@eamon_concannon]

@@lecturesbywalterlewin.they9259 Thanks for quick reply. I often like to post post up comments as I try to improve my own understanding. I will edit out any mistakes to make comments more useful. It is fantastic that you continue to be of service to informal learners after creating these amazing videos.

[@eamon_concannon]

I would like to know how the number of photons in white sunlight is distributed amongst the different colours.

[@lecturesbywalterlewin.they9259]

that depends on the light source. Its very different for your lights at home. use google for the sun and for white LEDs.

[@eamon_concannon]

@@lecturesbywalterlewin.they9259 I had tried googling, but I think now that i've resolved the issue I had about the meaning of light intensity. Forgetting about the idea of light as photons, I think you mean that for red and blue light of equal intensity being scattered in a region of the atmosphere, the intensity of the scattered blue light will be 5 times the intensity of the emitted red light. In terms of light as photons and the probability of a scattered photon being red or blue, I am not sure how we calculate these as blue and red light of the same intensity have photons of different energies according to E=hf. I guess this involves quantum mechanics.

[@carultch]

@@eamon_concannon You cannot measure frequency of light directly (at frequencies approximately above 1 Terahertz), so you often find this information reported as a function of wavelength, rather than frequency. You then have to infer the frequency by dividing wavelength by it. Even though frequency identifies light color, rather than wavelength, because wavelength reduces in optically dense transparent media. You can look up the sun spectrum which will give you the Watts/meter^2 per nanometer, throughout the color range of visible light. The sunlight peaks at green light, at the center of the roughly 1 octave of our visible band. And it falls off slowly, such that it is relatively close to uniform in most of the visible spectrum.

[@santoshkamar3082]

Sir at 18:07radius of the sphere is c×t Wht there is c

[@lecturesbywalterlewin.they9259]

c is the speed of light

[@tanviruddin5046]

sir i couldn't download the recommended book for 8.02.

[@lecturesbywalterlewin.they9259]

that is not possible

[@tanviruddin5046]

I downloaded the rest of two.

[@luchomolinari]

i loved when you smoked the brown weed! LOL

[@namesurn8590]

at 23:00 , why need to round corners of connection through thin layer? otherwise it means infinite acceleration?

[@lecturesbywalterlewin.they9259]

changes do occur in zero time

[@shlokvaibhav]

Acceleration in reality is a smooth function of time , it starts from 0 , rises to a and falls to 0, this fact necessitates the smoothening of corners

[@santoshkamar3082]

Professor lewin u draw electric field line at 0'' smaller than the field line at 0 What is the means

[@lecturesbywalterlewin.they9259]

how many minutes into the lecture?

[@santoshkamar3082]

1 hour 17 minutes Professor Lewin at 19:36 you draw electric field at 0" smaller than at 0 I am very confused why you draw such lines

[@mastipathsala912]

Sir how to define the value of Eo/Bo?

[@lecturesbywalterlewin.they9259]

kzbin.infoesyw1xihCWk

[@dylanbeard9770]

Why was the transmitter demonstration creating a hum in the audio?

[@lecturesbywalterlewin.they9259]

how many minutes into the lecture?

[@dylanbeard9770]

41:45. When it's pointed at the camera there is quite a bit of humming and i'm curious as to why physically.

[@lecturesbywalterlewin.they9259]

vry funny. must be interference of some kind between the antenna and my microphone.

[@twistedsim]

Might be that the microphone coil/capacitor get excited with the varying EM field.

[@augustinemmuogbana3382]

Why was I not taught physics like this for crying out loud!

[@subramaniannk3650]

Prof, in this lecture at 45:00 you've said that in order to accelerate a charge, we need to subject it to EM radiation. Are charges not accelerated in an AC current, which is obtained from AC voltage? Or, is it not possible to simply move mechanically a charged particle? Is it in the above two cases, we are kind of subjecting to EM radiation?

[@lecturesbywalterlewin.they9259]

You can accelerate charges in an E-field. F=qE. Thus by using a battery (DC) or AC. You can also accelerate them with EM radiation as they carry an E-field

[@subramaniannk3650]

Lectures by Walter Lewin. They will make you ♥ Physics. Thank you professor.... I was just watching your video on who would take over your channel and it kind of evoked (philosophical) sadness.

[@luizcastro7039]

great class, thank tou!

[@santoshkamar3082]

Sir at 18:07 radius of circle is c×t Why there is c

[@lecturesbywalterlewin.they9259]

c is speed of light

[@santoshkamar3082]

But sir i want to ask, why there is 'c' there could be any velocity also

[@lecturesbywalterlewin.they9259]

@@santoshkamar3082 the kink in the E-field moes with the speed c

[@sandrocavali9810]

Thank you

[@lecturesbywalterlewin.they9259]

You're welcome

[@JoinTheTechnicians]

Just WOW. This lecture is the most practical presentation of field propagation physics I have ever seen. You are a legend Walter Lewis, and you have earned my respect. I hope to converse with you someday in the near future, I'm seeking to answer a burning question I have on magnetic field propagation and Lenz law. Specifically, what happens in that short time-space between the secondary generation of opposing magnetic fields from a loaded transformer coil, and the load the primary coil "sees" prior to the secondary's opposing-field propagation reaching it... I have a detailed practical example presented in document form; which clearly illuminates the issue and entirely clarifies my question. I would love to send you the document for your review if you would be willing to take a look. It will be a great exercise for your already sharp mind. Please let me know; if possible via my conduit (Facebook/JoinTheTechnicians.) In any case, please keep doing what your doing! Yours is among the finest work of it's kind publicly available.

[@lecturesbywalterlewin.they9259]

:)

[@santoshkamar3082]

Sir at18:05 you write r=ct Sir why can't any other velocity replace 'c'

[@lecturesbywalterlewin.they9259]

the kink in the E-field moves with speed c

[@santoshkamar3082]

But sir how could I know that it's moving with speed c I am just accelerating this charge What is the factor which clear- cut that kink is moving with speed c

[@lecturesbywalterlewin.they9259]

@@santoshkamar3082 according to Maxwell's eq EM waves (in vaccum) move with speed c. the kink that is caused by the acc of the charge produces EM radiation. The associated E field thus moves with speed c. speed of light in air on Earth is very very very close to c as index of refraction is near 1. .

[@santoshkamar3082]

Thanks for your precious time sir

[@arijitchattopadhyay2045]

Sir why it is that when the size of the particles go way beyond the wavelength of light , raylleigh's scattering does not hold ?

[@lecturesbywalterlewin.they9259]

watch my 8.03 lecture in which I derive Rayleigh scattering. Also read up in Feynman's comments and use google.

[@whitesheatingairappliancer7101]

I thought "C" is colombs? Never mind C is the speed of light.

[@DarkRedZane]

Big C is Coulombs, small c is speed of light

[@alirezaparsai650]

Interesting

[@padraiggluck2980]

Bekefi & Barrett

[@maunster3414]

Sometimes the good Professor sounds like Bulwinkle, the moose.

[@physicsunique7877]

Professor you started the derivation to prove that the accelerated charge partical generates em waves but then you have taken account the perpendicular electric field perpendicular to your chosen r .....and then you had mentioned in the lecture that you are interested about that component because it is perpendicular to the direction of propagation of em waves but your goal was to prove that and secondly what is the garenty taht the direction of propagation of em wave mathches with your chosen r vector because at first you have mentioned that you have chosen the r vector arbitrarily and the perpendicular e field which is perpendicular to r vector may not match the e field that is associated with the direction of propagation of em wave due to the accelerated charge q .....

[@lecturesbywalterlewin.they9259]

how many minutes into the lecture?

[@physicsunique7877]

@@lecturesbywalterlewin.they9259 15 mins to 30 min in the lecture

[@lecturesbywalterlewin.they9259]

@@physicsunique7877 I have no time to watch 15 min of my lecture. be more precise what your problem is.

[@physicsunique7877]

@@lecturesbywalterlewin.they9259 problem is that how can u be sure that your arbitrarily chosen r vector is along the the direction of the electromagnetic wave that is generated due to the acceleration of the charged partical in the derivation .....started at 15 mins of your lecture 14

[@lecturesbywalterlewin.they9259]

I watched from 13 - 17 min into my lecture. I cannot spend more time on this. You should have nailed your problem to an accuracy in time of about 15 sec. I suggest you check Bekefi and Barrot whose classic derivation I used.

[@grahamconquer4934]

Graham conquer