How much electric charge do electromagnetic waves carry?

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Since electromagnetic waves have both electric and magnetic field components, which oscillate in phase perpendicular to each other and perpendicular to the direction of energy propagation. How much is that electric charge? Can you possibly get a shock in some way? electromagnetic-radiation charge share|cite|improve this question edited Jun 29 '14 at 12:54 Qmechanic♦ 50.3k765167 asked Jan 17 '12 at 14:42 Ryan 123149

Related: physics.stackexchange.com/q/41680/2451 – Qmechanic♦ Sep 12 '14 at 16:04 You could get a shock if the EM waves kicked electrons off of something and you touched it. But that almost never happens (don't touch leads to large solar panels) – Jim May 8 at 16:57 add a comment |

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Electromagnetic waves are generated from charges and currents, but the former aren't associated with the latter. There is no electric charge "in" electromagnetic waves. share|cite|improve this answer answered Jan 17 '12 at 14:47 Siyuan Ren 2,6031020

It just makes me wonder how does radio translates them into a message, isn't that the electrons flow through in antenna at receiver end? – Ryan Jan 17 '12 at 14:58 @Ryan Yes, but these are the electrons that are already inside the piece of metal that makes the antenna. The electromagnetic wave provides the electric field which makes the electrons move (more precisely: which biases their movements in the direction of the field since the electrons already move across the metal in random directions). – Adam Zalcman Jan 17 '12 at 16:51 Can I assume that the magnetic field is converted to electric field which causes the electrons to move? Or is the electric field in EMF itself? I think I'm assuming the electric field to be the source of electric charge which is what causing the confusion. – Ryan Jan 17 '12 at 17:31 Indeed, this assumption isn't correct: electric charges generate electric field, not the other way round. They're not the only source of electric field, though. Varying magnetic field generates varying electric field (and vice versa). – Adam Zalcman Jan 17 '12 at 18:50 @Ryan: Electric field is not the source of charges; rather it is the other way around. And contrary to what you think, both electric and magnetic field exert forces on electric charges (minor caveat though, magnetic field does not exert force on stationary charges, nor charges moving parallel to the field). – Siyuan Ren Jan 19 '12 at 9:16 add a comment |

up vote2down vote	Exactly zero. Electromagnetic waves may be generated by moving charges, but they carry no charge themselves. share|cite|improve this answer answered Jan 17 '12 at 14:51 Adam Zalcman 3,042818
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To answer your question, Electromagnetic (EM) waves carries no electric charge at all. EM waves originate from electromagnetic disturbances, and according to Maxwell's equation, changing electric fields gives rise to magnetic fields, and changing magnetic fields gives rise to electric fields, and so on. That's how EM waves propagate. You can get a shock if EM waves are strong enough (i.e. the oscillation amplitude is large, which is unlikely), and you have a loop of wire which radius and the EM wavelength are of same order of magnitude. That's how antennas and hypothetical EMP weapons work. share|cite|improve this answer answered Jan 17 '12 at 15:29 Sawi 33619

Burning up is also a way of electric fields from waves interacting with matter, our hands for example, never put them in a microwave oven, that would be shock enough. – anna v Jun 29 '14 at 13:07

(rest) mass of a system is not conserved in relativity.an electron and a positron can interact to produce two photons:

Is (rest) mass conserved in special relativity?

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I don't understand why it is said that the (rest) mass of a system is not conserved in relativity. I mean, the momentum of a system is conserved (i.e.: it remains constant in a frame of reference without any external influence). Also the energy of a system is conserved in relativity: it doesn't change without any external agency and the (rest) mass of a system is just m 2 =E 2 −p 2   E  is constant, p  is constant, how can m  change? For example, entropy is not conserved in a system, that means that the entropy of the system will increase spontaneously with time, which is really the case. But, is this true for mass? here is the problem from Griffiths and exact solution as given in the Griffiths, Two lumps of clay, each of rest mass m  , collide head-on at 3c/5  and they stick together. Question: what is the rest mass M  of the composite lump? Solution: In this case conservation of momentum is trivial it is zero before and zero after. The energy of each lump prior to the collision is mc 2 /1−v 2 /c 2  − − − − − − − −  √ =5mc 2 /4.  The energy of the composite lump after the collision is Mc 2   (since its at rest). so the conservation of energy says: 5mc 2 /4+5mc 2 /4=Mc 2   and hence M=5m/2  . Notice that this is greater than the sum of the initial masses! Mass was not conserved in this collision; kinetic energy was converted into rest energy, so the rest mass increased. special-relativity mass conservation-laws share|cite|improve this question edited Apr 11 at 14:48 Qmechanic♦ 50.4k765167 asked Apr 11 at 6:41 Paul 1,9441622

How do you arrive at this equation? Can you tell me what does each variable stands for? E.g s stands for second etc thanks. – user6760 Apr 11 at 7:30 add a comment |

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Mass, or more correctly, rest mass is not conserved in special relativity. Particles are able to be created and annihilated in special relativity, for instance, an electron and a positron can interact to produce two photons: e + +e − →2γ  Here mass is clearly not conserved, because both the electron and positron are massive but photons are massless. In more detail, take the situation where e +   and e −   have opposite momentum, p 0   . The total energy of the system is then: E 2 T =2m e +2p 2 0   where m e   is the electron/positron mass. To conserve energy-momentum in the collision, we require that the photons afterwards have equal and opposite momenta p 1   , and also that E 2 T =2p 2 1   which means that p 1 =m e +p 2 0  − − − − − − −  √ .  On the other hand, the electron and positron both have mass m e   , but the two photons are massless, and hence mass is not conserved. The error in your reasoning in the question is that the formula m 2 =E 2 −p 2   only holds for the energy, mass, and momentum of a single particle, and so does not work when you are talking about the energy and momentum of a system as a whole. share|cite|improve this answer edited Apr 11 at 7:48 answered Apr 11 at 7:32 djbinder 1666

The expression for E T   is not correct. it should be 2m 2 e +p 2 0  − − − − − − −  √   . – M. Zeng Apr 11 at 7:47 7   I think it's worth pointing out that the reason mass is not conserved is that it is not an "additive" quantity in relativity, i.e. the mass of the system is not equal to the sum of masses of constituents. Mass of each particle (elementary or composite) is conserved as long as its internal state remains unchanged. – xaxa Apr 11 at 8:12 2   @xaxa That's excellently stated, and it would be great if you could work that into an answer: it certainly complements all these answers well, since all the ones so far are correct yet seem contradictory. The lack of additivity summarizes John Rennie's answer well. Einstein's famous paper "Ist die Trägheit eines Körpers ...." (the one after the famous relativity one) looks at the "inertia" of the system of two photons, and in this way would conclude that the "inertia" of the electon and photon system in djbinder's answer would be the same: at the system level rest mass has been conserved. – WetSavannaAnimal aka Rod Vance Apr 11 at 8:35 1   The energy-momentum relation E^2 = m^2+p^2 can be seen to define the "mass" of a system. For single particles, the rest mass is simply the energy of the particle at rest. For composite systems, we can define the "mass" of the system as E^2-p^2 where E is the total energy of the system, and p is the total momentum, however, this "mass" is not simply the addition of the individual particle masses in the system. Hence E^2-p^2 is conserved, as you point out, but the sum of the masses in the system isn't this quantity. – djbinder Apr 11 at 11:29 1   When you say m 2 =E 2 −p 2   only holds for the energy, mass, and momentum of a single particle, I find that very misleading. The energy-momentum for any system or subcollection has an invariant length. And it makes the everyday experience of measuring the mass of a proton somehow undefined, since it is the quarks that are the elementary particles with mass, and what we think of as the mass of a proton is just the invariant length of the total energy-momentum vector for the proton as a system of quarks and gluons. – Timaeus Apr 12 at 3:01  |  show 4 more comments

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The rest mass of the system is conserved, it's just that the rest mass of the system isn't the sum of the masses of the parts. The rest mass of a system is just the length of the total energy-momentum vector. And that vector is conserved, so the length is conserved. The sum of the rest masses of the parts is not conserved. But that simply isn't the rest mass of the system. The reason that mass appears to be the sum of the masses of the parts in non relativistic physics is because in non relativistic physics the energy-momentum vectors of the parts point in almost the same direction. That is because the energy-momentum vectors point in the same direction in spacetime as the motion of the particle in spacetime. For vectors pointing in almost the same direction the length of the sum is approximately the sum of the lengths. share|cite|improve this answer answered Apr 11 at 16:03 Timaeus 12.3k1931

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For the rest mass we have m 2 =p 2 =p μ p μ   where p μ =(E,p ⃗ )  . It is Lorentz invariant, which means the rest mass of the particle is always the same no matter in which frame the observer is in. While for relativistic mass, it's simply equivalent to the total relativistic energy, which is always conserved. Note the difference between conservation and invariance. share|cite|improve this answer answered Apr 11 at 7:55 M. Zeng 560111

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Yes. The so called rest mass m 0   is the magnitude of energy-momentum four vector and is always conserved. It is not just conserved but is also invariant since it is the magnitude of a four vector. Perhaps you are confused between rest mass and relativistic mass. The relativistic mass is the total Energy divided by c 2   . m≡γm 0   This relativistic mass is also conserved in all frames but is not invariant like rest mass. EDIT : After referring to the example you added, I would like to add that conservation of rest mass is a matter of how you define rest mass. If one defines rest mass of the system as mass in the rest frame of the system then the mass will not be conserved in the kind of examples you mention. But if one defines rest mass as the Energy divided by c 2   in the frame in which the total momentum is zero then the rest mass will be conserved. You can check that for the example you mention in question. share|cite|improve this answer edited Apr 11 at 7:58 answered Apr 11 at 6:50 user40330 31119

Yeh,but Griffiths says mass is not conserved. – Paul Apr 11 at 6:55 1   I would be helpful if you could post that passage from Griffiths in the question. – user40330 Apr 11 at 6:56 I know about the concept of relativistic mass and rest mass,I also know that relativistic mass is very misleading so I dont use it. but I am confused in the problem 12.7 Griffiths fourth edition. – Paul Apr 11 at 6:59 There are two types of masses. Rest mass and the relativistic mass. Both of them have to be conserved. The rest mass being invariant is conserved. The relativistic mass being related to Energy by a constant forces it to be conserved since energy is also conserved. If the relativistic mass is not conserved, the conservation of total energy would break down. – user40330 Apr 11 at 6:59 2   @Paul let me also say that it would be very helpful if you quote the specific passage from Griffiths that you're referring to. Any time there is a specific passage that confuses you, it really helps if you include it in the question when you ask about it. – David Z♦ Apr 11 at 7:10  |  show 4 more comments

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Here is my comment in more details For any system or single elementary particle mass M  is defined as M=E 2 −P 2  − − − − − − −  √   where E  is total energy and P  is total momentum. For an elementary particle (like electron) mass is always conserved. For a system M  is conserved as long as EdE−PdP=0  , in particular mass of a closed system is conserved. However for a composite system mass of the system is not generally equal to sum of masses of individual constituents (mass is not an "additive" quantity): M≠m 1 +m 2 +...  . E.g. consider a system of two non-interacting particles: M 2 =(E 1 +E 2 ) 2 −(P 1 +P 2 ) 2 =E 2 1 −P 2 1 +E 2 2 −P 2 2 +2(E 1 E 2 −P 1 P 2 )=m 2 1 +m 2 2 +2(E 1 E 2 −P 1 P 2 )  If m 1 ≠0  you can choose rest-frame of particle 1, where particle 2 moves with some relative speed then E 1 E 2 −P 1 P 2 =m 1 E 2   , and so M 2 =m 2 1 +m 2 2 +2m 1 E 2 =(m 1 +m 2 ) 2 +2m 1 (E 2 −m 2 )  So you see that in this case M>m 1 +m 2   because of energy of relative motion E 2 −m 2 >0  . You can "extract" energy M−(m 1 +m 2 )  from this system. The most striking example is probably a system of two photons of the same energy ω  moving at angle θ  . In this case M=2ωsinθ2   In case of a bound system of two particles M<m 1 +m 2   and you need to input at least energy m 1 +m 2 −M  (bound energy) to "break" the system apart. This can be easily generalised for arbitrary number of particles. Now if particles were to interact their relative motion and/or composition can change. So P 1 P 2 ≠P ′ 1 P ′ 2   and/or m 1 ≠m ′ 1   (e.g. H  -atom emitted a photon and moved to a lower energy state thus decreasing its mass). This is why if you simply add up masses before and after the interaction you will have m 1 +m 2 +...≠m ′ 1 +m ′ 2 +...  . However if the system as a whole remains closed during the interaction M=M ′   . P.S. I don't consider "relativistic mass" because I find this concept useless and misleading. share|cite|improve this answer edited Apr 11 at 15:29 answered Apr 11 at 15:22 xaxa 42728

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What does conservation of mass mean in classical mechanics? Weight and mass are the same , we know the mass by weighing it, and if we add 1 kilo of sugar to another kilo of sugar, we will have two kilos of sugar. That is what is meant classically that the mass is conserved. Dissolving a kilo of sugar to a kilo of water will give you two kilos of sirop. The same is not true at the elementary particle level and where the energies are high enough for special relativity to hold. In special relativity as you state , m 2 =E 2 −p 2   As one of the other answers state, this m^2 is the "length" of the four vector in the four vector space of Lorenz transformations. In a similar way that the length of a ruler is invariant in three dimensions the mass of an elementary particle does not change. But in three dimensions also adding vectors can have a variable result, depending on the angles of the vectors. Similarly in the four dimensions of special relativity the addition of other particles with their energy and momentum will give an m', but this m' will not be the sum of the constituent particle masses, but the "length" of the new four vector. For this specific system of particles this length will be conserved , if no new four vectors enter the problem. The masses though of the elementary particles will add up to less than the invariant mass of the system they compose, the sum will be the lowest limit(, if all are at rest in the center of mass of the system.) Thus there is no conservation of mass a la classical physics.

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I don't understand why it is said that the (rest) mass of a system is not conserved in relativity. I mean, the momentum of a system is conserved (i.e.: it remains constant in a frame of reference without any external influence). Also the energy of a system is conserved in relativity: it doesn't change without any external agency and the (rest) mass of a system is just m 2 =E 2 −p 2   E  is constant, p  is constant, how can m  change? For example, entropy is not conserved in a system, that means that the entropy of the system will increase spontaneously with time, which is really the case. But, is this true for mass? here is the problem from Griffiths and exact solution as given in the Griffiths, Two lumps of clay, each of rest mass m  , collide head-on at 3c/5  and they stick together. Question: what is the rest mass M  of the composite lump? Solution: In this case conservation of momentum is trivial it is zero before and zero after. The energy of each lump prior to the collision is mc 2 /1−v 2 /c 2  − − − − − − − −  √ =5mc 2 /4.  The energy of the composite lump after the collision is Mc 2   (since its at rest). so the conservation of energy says: 5mc 2 /4+5mc 2 /4=Mc 2   and hence M=5m/2  . Notice that this is greater than the sum of the initial masses! Mass was not conserved in this collision; kinetic energy was converted into rest energy, so the rest mass increased. special-relativity mass conservation-laws share|cite|improve this question edited Apr 11 at 14:48 Qmechanic♦ 50.4k765167 asked Apr 11 at 6:41 Paul 1,9441622

How do you arrive at this equation? Can you tell me what does each variable stands for? E.g s stands for second etc thanks. – user6760 Apr 11 at 7:30 add a comment |

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Mass, or more correctly, rest mass is not conserved in special relativity. Particles are able to be created and annihilated in special relativity, for instance, an electron and a positron can interact to produce two photons: e + +e − →2γ  Here mass is clearly not conserved, because both the electron and positron are massive but photons are massless. In more detail, take the situation where e +   and e −   have opposite momentum, p 0   . The total energy of the system is then: E 2 T =2m e +2p 2 0   where m e   is the electron/positron mass. To conserve energy-momentum in the collision, we require that the photons afterwards have equal and opposite momenta p 1   , and also that E 2 T =2p 2 1   which means that p 1 =m e +p 2 0  − − − − − − −  √ .  On the other hand, the electron and positron both have mass m e   , but the two photons are massless, and hence mass is not conserved. The error in your reasoning in the question is that the formula m 2 =E 2 −p 2   only holds for the energy, mass, and momentum of a single particle, and so does not work when you are talking about the energy and momentum of a system as a whole. share|cite|improve this answer edited Apr 11 at 7:48 answered Apr 11 at 7:32 djbinder 1666

The expression for E T   is not correct. it should be 2m 2 e +p 2 0  − − − − − − −  √   . – M. Zeng Apr 11 at 7:47 7   I think it's worth pointing out that the reason mass is not conserved is that it is not an "additive" quantity in relativity, i.e. the mass of the system is not equal to the sum of masses of constituents. Mass of each particle (elementary or composite) is conserved as long as its internal state remains unchanged. – xaxa Apr 11 at 8:12 2   @xaxa That's excellently stated, and it would be great if you could work that into an answer: it certainly complements all these answers well, since all the ones so far are correct yet seem contradictory. The lack of additivity summarizes John Rennie's answer well. Einstein's famous paper "Ist die Trägheit eines Körpers ...." (the one after the famous relativity one) looks at the "inertia" of the system of two photons, and in this way would conclude that the "inertia" of the electon and photon system in djbinder's answer would be the same: at the system level rest mass has been conserved. – WetSavannaAnimal aka Rod Vance Apr 11 at 8:35 1   The energy-momentum relation E^2 = m^2+p^2 can be seen to define the "mass" of a system. For single particles, the rest mass is simply the energy of the particle at rest. For composite systems, we can define the "mass" of the system as E^2-p^2 where E is the total energy of the system, and p is the total momentum, however, this "mass" is not simply the addition of the individual particle masses in the system. Hence E^2-p^2 is conserved, as you point out, but the sum of the masses in the system isn't this quantity. – djbinder Apr 11 at 11:29 1   When you say m 2 =E 2 −p 2   only holds for the energy, mass, and momentum of a single particle, I find that very misleading. The energy-momentum for any system or subcollection has an invariant length. And it makes the everyday experience of measuring the mass of a proton somehow undefined, since it is the quarks that are the elementary particles with mass, and what we think of as the mass of a proton is just the invariant length of the total energy-momentum vector for the proton as a system of quarks and gluons. – Timaeus Apr 12 at 3:01  |  show 4 more comments

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The rest mass of the system is conserved, it's just that the rest mass of the system isn't the sum of the masses of the parts. The rest mass of a system is just the length of the total energy-momentum vector. And that vector is conserved, so the length is conserved. The sum of the rest masses of the parts is not conserved. But that simply isn't the rest mass of the system. The reason that mass appears to be the sum of the masses of the parts in non relativistic physics is because in non relativistic physics the energy-momentum vectors of the parts point in almost the same direction. That is because the energy-momentum vectors point in the same direction in spacetime as the motion of the particle in spacetime. For vectors pointing in almost the same direction the length of the sum is approximately the sum of the lengths. share|cite|improve this answer answered Apr 11 at 16:03 Timaeus 12.3k1931

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For the rest mass we have m 2 =p 2 =p μ p μ   where p μ =(E,p ⃗ )  . It is Lorentz invariant, which means the rest mass of the particle is always the same no matter in which frame the observer is in. While for relativistic mass, it's simply equivalent to the total relativistic energy, which is always conserved. Note the difference between conservation and invariance. share|cite|improve this answer answered Apr 11 at 7:55 M. Zeng 560111

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Yes. The so called rest mass m 0   is the magnitude of energy-momentum four vector and is always conserved. It is not just conserved but is also invariant since it is the magnitude of a four vector. Perhaps you are confused between rest mass and relativistic mass. The relativistic mass is the total Energy divided by c 2   . m≡γm 0   This relativistic mass is also conserved in all frames but is not invariant like rest mass. EDIT : After referring to the example you added, I would like to add that conservation of rest mass is a matter of how you define rest mass. If one defines rest mass of the system as mass in the rest frame of the system then the mass will not be conserved in the kind of examples you mention. But if one defines rest mass as the Energy divided by c 2   in the frame in which the total momentum is zero then the rest mass will be conserved. You can check that for the example you mention in question. share|cite|improve this answer edited Apr 11 at 7:58 answered Apr 11 at 6:50 user40330 31119

Yeh,but Griffiths says mass is not conserved. – Paul Apr 11 at 6:55 1   I would be helpful if you could post that passage from Griffiths in the question. – user40330 Apr 11 at 6:56 I know about the concept of relativistic mass and rest mass,I also know that relativistic mass is very misleading so I dont use it. but I am confused in the problem 12.7 Griffiths fourth edition. – Paul Apr 11 at 6:59 There are two types of masses. Rest mass and the relativistic mass. Both of them have to be conserved. The rest mass being invariant is conserved. The relativistic mass being related to Energy by a constant forces it to be conserved since energy is also conserved. If the relativistic mass is not conserved, the conservation of total energy would break down. – user40330 Apr 11 at 6:59 2   @Paul let me also say that it would be very helpful if you quote the specific passage from Griffiths that you're referring to. Any time there is a specific passage that confuses you, it really helps if you include it in the question when you ask about it. – David Z♦ Apr 11 at 7:10  |  show 4 more comments

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Here is my comment in more details For any system or single elementary particle mass M  is defined as M=E 2 −P 2  − − − − − − −  √   where E  is total energy and P  is total momentum. For an elementary particle (like electron) mass is always conserved. For a system M  is conserved as long as EdE−PdP=0  , in particular mass of a closed system is conserved. However for a composite system mass of the system is not generally equal to sum of masses of individual constituents (mass is not an "additive" quantity): M≠m 1 +m 2 +...  . E.g. consider a system of two non-interacting particles: M 2 =(E 1 +E 2 ) 2 −(P 1 +P 2 ) 2 =E 2 1 −P 2 1 +E 2 2 −P 2 2 +2(E 1 E 2 −P 1 P 2 )=m 2 1 +m 2 2 +2(E 1 E 2 −P 1 P 2 )  If m 1 ≠0  you can choose rest-frame of particle 1, where particle 2 moves with some relative speed then E 1 E 2 −P 1 P 2 =m 1 E 2   , and so M 2 =m 2 1 +m 2 2 +2m 1 E 2 =(m 1 +m 2 ) 2 +2m 1 (E 2 −m 2 )  So you see that in this case M>m 1 +m 2   because of energy of relative motion E 2 −m 2 >0  . You can "extract" energy M−(m 1 +m 2 )  from this system. The most striking example is probably a system of two photons of the same energy ω  moving at angle θ  . In this case M=2ωsinθ2   In case of a bound system of two particles M<m 1 +m 2   and you need to input at least energy m 1 +m 2 −M  (bound energy) to "break" the system apart. This can be easily generalised for arbitrary number of particles. Now if particles were to interact their relative motion and/or composition can change. So P 1 P 2 ≠P ′ 1 P ′ 2   and/or m 1 ≠m ′ 1   (e.g. H  -atom emitted a photon and moved to a lower energy state thus decreasing its mass). This is why if you simply add up masses before and after the interaction you will have m 1 +m 2 +...≠m ′ 1 +m ′ 2 +...  . However if the system as a whole remains closed during the interaction M=M ′   . P.S. I don't consider "relativistic mass" because I find this concept useless and misleading. share|cite|improve this answer edited Apr 11 at 15:29 answered Apr 11 at 15:22 xaxa 42728

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What does conservation of mass mean in classical mechanics? Weight and mass are the same , we know the mass by weighing it, and if we add 1 kilo of sugar to another kilo of sugar, we will have two kilos of sugar. That is what is meant classically that the mass is conserved. Dissolving a kilo of sugar to a kilo of water will give you two kilos of sirop. The same is not true at the elementary particle level and where the energies are high enough for special relativity to hold. In special relativity as you state , m 2 =E 2 −p 2   As one of the other answers state, this m^2 is the "length" of the four vector in the four vector space of Lorenz transformations. In a similar way that the length of a ruler is invariant in three dimensions the mass of an elementary particle does not change. But in three dimensions also adding vectors can have a variable result, depending on the angles of the vectors. Similarly in the four dimensions of special relativity the addition of other particles with their energy and momentum will give an m', but this m' will not be the sum of the constituent particle masses, but the "length" of the new four vector. For this specific system of particles this length will be conserved , if no new four vectors enter the problem. The masses though of the elementary particles will add up to less than the invariant mass of the system they compose, the sum will be the lowest limit(, if all are at rest in the center of mass of the system.) Thus there is no conservation of mass a la classical physics.