Minggu, 20 Maret 2011

Fisika untuk Universitas

Fisika untuk Universitas

Ditujukan untuk meningkatkan kualitas proses dan hasil perkuliahan Fisika di tingkat Universitas

Kelistrikan dan Kemagnetan

Topics covered:

Index of Refraction
Poynting Vector
Oscillating Charges
Radiation Pressure
Comet Tails
Polarization (Linear, Elliptical, and Circular)

Instructor/speaker: Prof. Walter Lewin

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I'm going to talk today about energy in electromagnetic waves.

There must be energy in there, because we know that electric fields contain energy, and magnetic fields contain energy.

And you may remember that the electric field energy density is one-half epsilon 0 E squared, this is now in joules per cubic meter, and the magnetic energy field density, which also has come up earlier in the course, is 1/(2 mu 0) times B squared.

Again, joules per cubic meter.

Now, we -- when we deal with traveling waves, in vacuum, at any moment in time, the magnitude of B is E divided by C.

So this is also one divided by 2 mu 0, I can replace this by E squared / C squared.

But C squared is one over epsilon 0 mu 0.

So this is also one-half epsilon 0 E squared.

And when you see this, this is an absolutely amazing result, because what this tells you is that the energy density in the magnetic field of a traveling wave is exactly the same as the energy density in the electric field of an electromagnetic wave.

That's really an amazing thing, the symmetry is absolutely beautiful.

So the total energy density is the sum of the two, so I double this one, so that it's epsilon 0 E squared, joules per cubic meter, and of course, I can also write for that, epsilon 0, just one E, and then the other, I write B times C.

So I write for E, B times C.

So this is, again, in joules per cubic meter.

Yes, I'm happy with that, that's fine.

Now, I want to ask the question, if electromagnetic waves come by me, how much energy passes through one square meter?

It's like an energy flux.

And so I have here one square meter, and this one square meter is perpendicular to the direction that the electromagnetic wave is going, and I want to know how much energy flows through there.

1 square meter every second.

In 1 second, light travels a distance C, which is a horrendous distance, 300000 kilometers.

This side of this box is what light travels in 1 second.

And this is 1 square meter, and I'm going to calculate how energy now goes through this 1 square meter in 1 second.

Of course, I could have chosen this box a billion times smaller, would have gotten the same answer of course, but for convenience, I choose this to be C, and this to be 1 square meter.

So the volume of that box is C cubic meters.

And all the energy of that box is going to come out here in 1 second, because I know that electromagnetic waves move with the speed of light, which is C.

And so therefore, the energy that comes out here per square meter per second is that U total there that I have, which is the amount of energy for every cubic meter, but I have so many cubic meters.

And so I can use this result, now, here, and I can substitute that here, so I get epsilon 0 E B C squared.

I have to multiply it by this C here.

And this, of course, is also E B / by mu 0, because C squared, in vacuum, is one over epsilon 0 mu 0.

And this, now, is joules per square meter, per second.

Because in one second, all that energy comes out, and I have already chosen one square meter area.

Let me see whether I'm happy with that, yes, I'm happy with that.

We call this the Poynting vector.

And we write it, in general as a vector.

We write it, S with a vector, and it's called the Poynting vector with a y, and we write that as E cross B divided by mu 0.

E B divided by mu 0.

You don't need the cross, really, because E and B are always perpendicular to each other in a traveling wave.

The advantage of this notation is that S, which is the energy flux, goes in a certain direction, and the velocity of the wave is always in the direction of E cross B, so it also tells you, then, in which direction the radiation is flowing, whereas here, you -- you lack that information.

And so this, then, to remind you, is in these units, which is watts per square meter.

How many joules per second?

1 square meter through a plane perpendicular to the direction of propagation.

Now, E and B are changing with a frequency omega, cosines omega T or sines omega T.

And so S is changing with the cosine square of omega T.

If you are somewhere in space, and this electromagnetic wave comes by, there are moments that S is 0, namely, when E and B happen to be 0.

And there are moments that it is at maximum, when E and B happen to be at maximum.

And so when we deal with electromagnetic radiation, it's more meaningful to discuss the time-average value.

And the time-average value of the Poynting vector is, first of all, the time-average of the cosine square omega T, or the sine square omega T, whatever the case may be.

And the average value of cosine squared is one-half.

And then I can write now, for E, E0, and for B, I can write B0, which is now the maximum value possible, divided by mu 0.

And if you want to write it differently, if you want to write it only in terms of E 0, then you can write down one-half E0 squared divided by mu 0 times C.

So this is the more practical equation, that gives you a number which is time-averaged over the oscillations.

Let me make sure that there is nowhere, a slip of the pen -- no, there isn't -- I hate to have slips of the pen, because you can't edit them out later on your videotape.

Slips of the tongue, you can edit it.

The slips of the pen, you can't.

OK, this looks good.

So now, we have an average value for the Poynting vector, and so we can calculate, now, how much energy flows through 1 square meter per second.

And I can give you an example, suppose I have a plane electromagnetic wave, and E0 is 100 volts per meter.

That's just what it is, how we get it, that's a different story.

And this could be radio emission, this could infrared, this could be light, I d- I don't specify the frequency.

I don't have to.

Frequency doesn't come up there.

And so the value for -- average value for the Poynting vector, I can pick this equation.

E0 is 100.

I know mu 0, I know C, so I can calculate what it is.

It's 100 squared divided by 2, divided by mu 0, divided by C.

And when I do that, I find that this is 13 watts per square meter.

So imagine that you would stand in this electromagnetic wave coming to you, we take all our clothes off and we let it hit us, and suppose it absorbed us.

Suppose it is radiation that absorbs us -- some radiation may go through you.

Gamma rays may go straight through you -- they are electromagnetic radiation.

But certainly, light will not go through you.

And radio waves, some of them will not go through you.

So you observe them with your body.

Would you notice that?

I doubt it.

You probably have a surface area that's close to 1 square meter.

13 watts, 13 joules per second, not very noticeable.

You radiate, yourself, 100 watts, 100 joules per second, so I don't think you will notice that.

But imagine, now, that we increase the value of E0, and we make it 1000 volts per meter.

Now, this goes up by a factor of a 100, because if E goes up by a factor of 10, automatically, B goes up by a factor of 10 -- remember, in electromagnetic waves, they're always coupled.

So the Poynting vector, which is the product of the two, goes up by a factor of 100, so now, you're talking about 1.3 kilowatts per square meter.

And if you absorb that on your body, believe me, uh, it may fry you.

Certainly if you do that out on the beach, you get a -- a deep suntan.

And if you do it long enough, then you can hurt yourself very badly.

So now comes the question, does a light bulb emit plane waves?

Well, not really.

Plane waves have no beginning and they have no end, they exist at all time and all space.

Look at my plane wave solutions from last lecture.

You can substitute in there any value for X, Y, and Z, and any moment in time, the year 5000 B.C., you get an answer.

Doesn't specify when, doesn't specify where.

And of course that's not very realistic.

In the real world, there is a beginning and there is an end to the electromagnetic radiation, and therefore, they have also a finite length.

Remember the quarter-nanosecond pulses that we sent to the moon that we discussed last time.

They were only 7 centimeters long.

That's not very much like a plane wave.

So I want to discuss with you a little further what these waves look like, and I want to take a closer look at how electromagnetic waves are produced by charges that we begin to shake, that we are accelerating.

Key in the whole process is that you accelerate a charge.

If a charge is just moving at a constant velocity, it will not produce electromagnetic radiation.

But I want to give you some feeling, at least some classical physical feeling on how these electromagnetic waves are produced.

It is a picture that has its own limitations, but it's still useful.

It's not a quantum mechanical treatment, but it is something that you and I can see and therefore, perhaps, appreciate.

Suppose I have here a charge which is not moving.

Just sitting there.

These are field lines, I just draw only a few field lines.

And if it's a positive charge, the arrows are outwards, if it's a negative charge, the arrows are inwards.

And I'm going to accelerate this for a time delta T.

Let's accelerate it in this direction.

And then we bring it, again, to a halt, say.

I redraw this.

This point is the same as this point here.

I accelerate it, and delta T seconds later, it happens to be here.

So this is T0, and this is T equals delta T.

And now I'm going to draw a circle -- it should actually be a sphere, three-dimensionally -- about this point.

And here is that sphere.

And this sphere has radius C delta T.

This field line, which was the field line that goes with this charge, is here, and this field line is here, and this one is here, I only draw three.

This one, this one, and this one.

And the message that I accelerated this charge could not possibly have reached this location in space, because that message can only travel with the speed of light.

So the electric field here is exactly the same as it was when it was still here.

So when the object is here, the electric field must still be like this here, and it must be like this here, and it must be like this there, because the message hasn't reached that point.

But now, look at this charge, which is now here at time d delta T.

Now, the electric field is like so, is like so, and is like so.

So this field line must somehow meet up with this one, it's one and the same field line.

And what does that mean?

That somewhere there, there must be a kink in the electric field.

And there must be a kink here.

Notice there is no kink here, which is interesting.

And so it is the collection of these kinks that propagate outwards, with the speed of light, and they produce an electromagnetic disturbance, a change.

If you were out in space here, and if, for instance, I were to oscillate this charge back and forth, you would see these kinks go by all the time, these breaks in the electric field, and you would experience that as an electromagnetic wave, if there is a changing electric field, according to Maxwell's equations, there has to be also a changing magnetic field.

But the interesting thing is, even though this is an extremely simple picture, notice that in this direction, if you were here, you would not see any kinks.

So there's no electromagnetic radiation going in this direction, nor is there any going in this direction.

And the maximum is going in this direction, and something in between is going in that direction.

It's not much of a plane wave, for that matter.

I mean, if anything, it's more like a spherical wave.

But it's a very special spherical wave, not the same strength in all directions.

And so even though this is a rather classical picture, it helps me, at least, to see how these changing electric fields -- and therefore, associated B fields -- are formed by charges that we accelerate.

And I have a two-minute movie that shows that, also, in a slightly more detailed way than I was able to do.

And so let's look at that movie, Marcos, you ready for that?

OK, then you can start that.

Pengembangan Perkuliahan

1. Buatlah sebuah Esai mengenai materi perkuliahan ini

2. Buatlah sebuah kelompok berjumlah 5 orang untuk menganalisis materi perkuliahan ini

3. Lakukan Penelitian Sederhana dengan kelompok tersebut

4. Hasilkan sebuah produk yang dapat digunakan oleh masyarakat

5. Kembangkan produk tersebut dengan senantiasa meningkatkan kualitasnya

Ucapan Terima Kasih Kepada:

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

b. Prof. Bernd Surrow, Ph.D.


Prof. John Belcher

Dr. Peter Dourmashkin
Prof. Bruce Knuteson
Prof. Gunther Roland
Prof. Bolek Wyslouch
Dr. Brian Wecht
Prof. Eric Katsavounidis
Prof. Robert Simcoe
Prof. Joseph Formaggio

Course Co-Administrators:
Dr. Peter Dourmashkin
Prof. Robert Redwine

Technical Instructors:
Andy Neely
Matthew Strafuss

Course Material:
Dr. Peter Dourmashkin
Prof. Eric Hudson
Dr. Sen-Ben Liao


The TEAL project is supported by The Alex and Brit d'Arbeloff Fund for Excellence in MIT Education, MIT iCampus, the Davis Educational Foundation, the National Science Foundation, the Class of 1960 Endowment for Innovation in Education, the Class of 1951 Fund for Excellence in Education, the Class of 1955 Fund for Excellence in Teaching, and the Helena Foundation. Many people have contributed to the development of the course materials. (PDF)

2. Para Dosen Pendidikan Fisika, FPMIPA, Universitas Pendidikan Indonesia.

Terima Kasih Semoga Bermanfaat dan mohon Maaf apabila ada kesalahan.

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