Dear Gavaskar Muppavaram

I first i would like to explain how a wave travels in a medium. we know that a wave is a disturbance and it has certain energy associated, so when this wave enters a medius it basically sets the particles of the medium to vibrate this vibration is passed from one particle to other thus the wave is transferred in a medium.

Now in care of a an Electromagnetic wave in a free-space/vacuum, as we know from the Maxwell's equations that a changing magnetic field produces an electric field and vice versa. EM wave requires no medium as electric fields and magnetic fields constantly generate each other as the wave propagates.

With regards 

Dhanu Chettri

Dear Gavaskar Muppavaram,

the answer given by Mr. Dhanu Chettri is quite clear and convincing. In pursuing your studies, you will come across more sophisticated explanations, as those involving potentials, or photons. As you are studying engineering, I just wish to point out to you that theoretical explanations are purely logical accounts (models). You should be aware that the vacuum you are speaking of in the context of electromagnetism is not a physical thing, i.e. not something you can obtain once you have completely removed the air from your room.

Electromagnetive waves do not need a material medium to propagate. Generally, electromagnetic and  gravitationnal  interactions do need material medium too. In electrostatics, for example, Coulomb's law establishes the possibility of action at a distance without contacting material. The universal law of attraction of Newton also.  In this case, this explains the movement of planets in the universe that is essentially made of vacuum.  Further, at the atomic level, the matter is lacunar; it is made mostly of vacuum. This does not prevent the electromagnetic interaction to maintain the electrons in orbit around the nucleus. 

In contrary, the sound needs a material medium (air, metals,...) to propagate. The acoustic waves are very different of electromagnetic waves.

An electromagnetic wave is the oscillation of magnetic and electric fields in space and time. However, a magnetic or electric field is only a CONCEPT used to describe the force acting on a test magnet or test charge. As long as there is no test particle, there will be no force, no way to detect the wave. Actually, physicists do not know how the force can propagate through a vacuum. It is an open question. Physicists can only speak in metaphors. The question could be restated: Why does an electric field even exist in a vacuum? Noone knows.

Hi Gavaskar, your question reminded me of two sentences written by Heaviside: "There are only two Things going, Matter and Energy. Nothing else is a thing at all; all the rest are Moonshine, considered as Things." While this might be an oversimplifikation, I find it helpful most of the time. Applied to your question:

A classical vacuum is space without anything else in it. While this space is traversed by an electromagnetic wave it is space filled with energy. So it's no longer a vacuum.

Of course, as Sascha pointed out, you can only measure radiation by its effect on matter. The idea behind the concept of classical wave propagation is that energy does not "jump" but has to traverse space in order to come from point A to point B.

Sorry, after adding my first answer I realized that "A vacuum with an electromagnetic wave in it is a paradox" might not answer your original question. So, another point of view:

The Maxwell equations state that

the curl of E is accompanied by a changing B (and/or a current of magnetic charge)

the curl of H is accompanied by a changing D and/or a current of electric charge.

the divergence of D is accompanied by electric charge

B has no divergence

So, without electric charge the "lines" of the field D have no ends (in the absence of magnetic charge, the lines of B have never ends). If there were no curl either, the only solution would be homogeneous fields filling the whole universe, i. e. no waves, like unlimited static fields. But since the curls of E and H can exist without currents we can have fields with a structure (= waves) provided the structure is moving (in order to result in changing B and D). Hope this helps.

Hi, Gavaskar Muppavaram,

The contradiction in your reflections  can be easily eliminated, if we think about the displacement current postulated by Maxwell. AC, unlike constant ones, passes through the circuit with the capacitor. Around a conductor with a capacitor continuous magnetic field (including the space around condenser!) is formed. Consequently, an alternating electric field in the capacitor produces magnetic field just as electric current consisting of  moving charges. That is why genius Maxwell called an alternating electric field a special current. So an alternating electric field is a reason for alternating magnetic field, and vice versa. Thus the needs in electric charges for propagation of alternating electromagnetic field disappears as soon as it was created , and nothing prevents its propagation in a vacuum. 

Sincerely,

Alexander G. 

The view that electric and magnetic fields generate each other is misleading. The view that an electromagnetic field oscillates is at risk of being misleading.  An electromagnetic wave glides through space outward from the (oscillating) source without any other motion.  But an observer of the field as it glides by will perceive an oscillation in the local measurement.  See "The Electromagnetic field in space-time"  at          users.rcn.com/eslowry   .

I'd like to support Edward; this is the reason I wrote "is accompanied" instead of "is caused". As both Jefimenko, "Causality, electromagnetic induction, and

gravitation" and Rosser, "Interpretation of classical electromagetism" pointed out, Maxwell's equations concerned with the curl of the fields in empty space do not

describe cause-effect relations. Sadly, most textbooks have it wrong.

During my first encounter with electromagnetic theory, the notion that a radiation field itself contains an infinite and still growing number of sources kept me from digging deeper into the meaning of the formulas. Later on, I was relieved to learn that things are in fact simpler.

Addendum: The energy is moving, the (phase of the) wave is moving, but the fields themselves are not moving.

Perhaps it would be good to answer this question in several steps for being clear:

1. Maxwell's equations in vacuum without sources are equivalent to a wave equation for the fields, whose velocity corresponds to the light in vacuum.

2. Do these waves propagate energy and momentum? Yes the Poynting vector for radiation fields, i.e. which depend of the inverse of the distance (not of its square), for a closed surface, gives a conserved linear momentum for such a waves. The same happens for its energy.

3. The waves have orthogonal electric and magnetic fields always,i.e. they are transversal waves without loses in vacuum. Thus the electric energy transforms in magnetic and vice verse keeping the velocity constant.

4. There is a particular case where they interact with the vacuum, it is when its photon associated have doble energy (very high frequency), or more, than the mass of the electron. In such a case the pairs electron-positron have to be consiered.

  I hope that this can helps.

Dear colleagues

Hi all 

Really it is a good philosophical discussion.

As mentioned before the propagation of electromagnetic waves either in a certain medium or in vacuum is due the the mutual changes between the electric and magnetic field. Referring to the four maxwell's equation and from the wave equations of the electric and magnetic fields, any one can conclude that the spatial variation of of the electric field is always accompanied with a time varying of the magnetic field and hat the spatial variation of of the magnetic field is always accompanied with a time varying of the electric field. These variations in electric and magnetic fields fields lead to the transfer of the energy which is carried by the EM wave. Therefore, EM waves will constantly propagate with non stopping until meet an obstacle or barrier.

https://en.wikipedia.org/wiki/Maxwell%27s_equations

http://www.physicsclassroom.com/mmedia/waves/em.cfm

Christian raises a very good point.

If energy propagated as waves in a medium, "we should expect dissipation and then we would not be able to see distant galaxies".

There is a solution that explains how electromagnetic energy can propagate any distance in vacuum without dissipation. It was first hypothesized by Louis de Broglie in the 1930's. Here is a recently published paper describing this solution.

http://www.omicsonline.com/open-access/on-de-broglies-doubleparticle-photon-hypothesis-2090-0902-1000153.php?aid=70373

We must realized that EM waves are different from sound waves. I would like to give two simple answers which also recapitulates all the above answers.

1) EM wave is also a particle (from the wave particle duality property). therefore, it can travel in vacuum as if a ball is thrown into open space.

2) EM wave is also a traveling electromagnetic field. Like any other basic field, e.g gravitational field, an electromagnetic field doesn't need a medium to establish itself. Therefore, EM fields as well as EM waves can establish themselves in vacuum.

  Dear Pragnan,

  I understand your answers, but they are not true or at least not the whole true. Let me try to respond to every of your answers.

  1. If you through balls into empty space, it is almost impossible that they can go from one galaxy to another. They are going to trapped by the massive bodies in between. Thus, fortunatly, the photons have rest mass zero and they are independent of the gravitation interaction.

  2. The electromagnetic waves, even they belong to a "basic field" they have losses when they are in contact with matter having permability and permittivity as the free space has. Then it seems logic to explain the lack of losses for such values and that is not possible, in all that I know. Notice that even we have the concept of vacuum impedance for one electromagnetic field in vacuum, used mainly by engineers.

  3. The answer, besides the one of the transversality of the electromagnetic waves, which make them very different of the sound waves, is that the equation of motion

               Da Tab=0

i.e. the conservation of the energy-momentum tensor in vacuum (currents are zero), tell us that the interaction of the waves must be zero with the such medium. How does it?. That is more complex to explain and it was the contribution of many body in this post. 

Dear Prof. Baldomir,

I appreciate your interest in my answers. The answers were intended for a basic question asked by an undergraduate student that why EM waves can travel in vacuum whereas sound waves can't? Quite contrary to the discussions carried out here, it is not so complex to answer this question. The ability of EM waves to travel in vacuum is neither a mystery nor a myth. I can't agree with your comment that my answers are not true or are half-true. Both my answers are based on established physical phenomenon and are absolutely true. You can't disagree with the fact that EM waves exhibit duality whereas sound waves don't. Similarly, sound waves are not driven by fundamental fields/forces unlike EM waves (driven by electromagnetic fields/forces).

Unfortunately, most of the answers given in this post digressed from the original question and got entangled in answering another; i.e. whether the electromagnetic waves suffer losses during their travel through vacuum? I believe Maxwell's equations in complex medium has answer to this question too. Believe me, its not so complex.

I would also like to comment on some of your responses to my answers, point wise:

1). Photons do get trapped by massive bodies. Bending of light towards celestial bodies like black-holes is well known.

2) I never disagreed or talked about losses in my preceding answers. Vacuum impedance (intrinsic impedance of around 377ohms) is not a metric of losses in any medium. In fact,unlike the intrinsic impedance of vacuum, a medium needs to have a complex impedance to make itself lossy.

3) It is not so complicated to explain the conservation of energy-momentum either. You just need to see it from the perspective of electromagnetic engineers.

I sincerely hope that we being part of the common scientific community will try to make our answers as simple as their questions. Let us also hope that we shall put our egos in trash cans before we  authoritatively put forth our answers. I humbly look forward to your experienced association in dealing with the mysteries of our world.

Hope this addresses your concerns,

With Best Regards,

Pragnan Chakravorty

    Dear Pragnan,

    The main problem of these discussions is the difficulty to focus in only one answer. Let me to start for your last comments:

   1). Photons do get trapped by massive bodies. Bending of light towards celestial bodies like black-holes is well known.

   - This hasn't anything to do with if they are balls or if they are waves of electromagnetic fields. This is only due to have special geodesics of the space-time in such region.

2) I never disagreed or talked about losses in my preceding answers. Vacuum impedance (intrinsic impedance of around 377ohms) is not a metric of losses in any medium. In fact,unlike the intrinsic impedance of vacuum, a medium needs to have a complex impedance to make itself lossy.

We agree in this point. If I wrote:

 "Notice that even we have the concept of vacuum impedance for one electromagnetic field in vacuum, used mainly by engineers".

 Was only for saying that the vacuum is not the same that "nothing". In fact the vacuum is provided with one physical structure. One photon with energy higher than twice of the electron mass, produces two charged particles electron-positron.

3) It is not so complicated to explain the conservation of energy-momentum either. You just need to see it from the perspective of electromagnetic engineers.

  The perspective of electromagnetic engineers is very interesting, but this question needs more things that a "perspective" to be understood. Notice that for a long time the main justification to have aether is that light of the sun came here,i.e, the waves needed a medium where they oscillate.

Finally, let me to tell you that most of the people try to answer as well as they can, and it is quite complex for me to know their intentions which always I presume of scientific interest.

  If you want, just for testing your electromagnetic engineering perspective, how do you understand that two electric charges of the same sign repels while the ones of opposite attract, just taking into account the exchanging of photons ("balls") between them?

Dear Prof. Baldomir,

I agree about the lack of focus. I'll try to ask some focused questions on these issues which may require your expertise for an answer. Please look for those.

Sincerely,

Pragnan

When mentioning "photons", there is need to clearly distinguish between "virtual photons" and real electromagnetic photons.

"Exchange" of "photons" between repelling or attracting electrons does not refer to physically existing electromagnetic photons, but to "virtual photons" as defined by Feynman as a mathematical artefact to allow calculating force/energy interaction levels between charged particles by means of the Lagrangian. He specifically cautioned in his 1949 seminal paper about the need not to confuse what  he termed "real quanta" and "virtual quanta".

"Virtual photons" bundle together the Coulomb force and the amount of kinetic energy induced at the particular intensity of the force applicable to each distance considered into individual "virtual exchange quanta", which were meant to represent the electric interaction between elementary particles.

Real electromagnetic photons do not bundle together the Coulomb force and the kinetic energy induced by the force, but are made of only kinetic energy in motion at the speed of light.

http://authors.library.caltech.edu/3523/1/FEYpr49c.pdf

http://dx.doi.org/10.4172/2090-0902.1000153

 Dear Andrè,

   You are right that it is necessary to distinguish real photons from virtual ones, if we want to use this quantum excitation (photon) of the electromagnetic field, as responsable of the interaction besides the propagation of the electromagnetic fields by themselves. The virtual photons can be thought as mathematical artefacts useful to calculate perturbative diagrams of Feynman within a Dyson serie for giving us the amplitud of probability of a certain scattering.

  The question that I have put to distinguish attraction of repulsion needs this separation of concepts and if we have photons on or off the mass shell. But even if you have this concepts clearly separated, the photons works very differently than balls between the charges exchanging momentum and energy. Notice that if you have exchanging balls with momentum p, you always are going to get repulsion, no attraction. In this point the Heisenberg indeterminacy plays one important role if we take into account that a monochromatic photon is not at all localized and so on. But this concept of attraction and repulsion is not clearly solved and only statistically can be justified.

  Summarizing, I only tried to show with this question that the concept of photon is quite far of the classical balls.

Daniel,

From the QED perspective, you are absolutely right.

Looks like I tend to react on automatics when the word photon is being used without it being clearly identified as virtual or electromagnetic. I noticed that many students are not aware of the distinction. This is why I gave the Feynman paper in reference.