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I just want to point out that I'm still in highschool and I don't really have any advanced education within physics, however, this is something that has been on my mind. I may just be completely wrong with my knowledge regarding waves as a whole though.


From my understanding of waves and objects, waves have a few defining features: being massless and having energy. Objects have mass and energy. So according to Einstein's equation $$E=mc^2$$ mass and energy go hand in hand (sort of). So, why do waves, which have energy, not have mass?

I am aware of wave-particle duality, but (now I may be wrong with this as well), to my understanding, that is regarding particles giving off weak waves and not the other way around.

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    $\begingroup$ What are weak waves? Can you provide more context? Mechanical waves carry energy but they do not cause a net displacement of matter. $\endgroup$ – Yashas Mar 24 '17 at 9:44
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    $\begingroup$ They kind of do have mass: if I have a box with mirrored internal walls, which may or may not be full of light, I can tell that by weighing the box (if I have a very sensitive scale, or if the light inside it is extremely bright). $\endgroup$ – tfb Mar 24 '17 at 10:46
  • $\begingroup$ If any of these answers are too detailed for your present level of understanding, think of it like this — gross simplification: waves are a way to understand the form of energy which passes from one particle with mass to another. That energy is conceived as a coherent thing of itself, but it can also be understood as particles trading energy to their neighbors. In other words: it is the energy which travels across particles that themselves do not move to follow the wave. Wave–particle duality is something else, and it only occurs on the quantum scale. $\endgroup$ – can-ned_food Mar 24 '17 at 12:10
  • $\begingroup$ @can-ned_food This feels oversimplified to the point of being misleading. Is a wave in the ocean 'energy that passes from one particle to another'? What are the two particles in that case? What about the fact that the wave itself is made out of the collective motions of particles? $\endgroup$ – knzhou Mar 24 '17 at 22:09
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    $\begingroup$ @Qmechanic I think your edit of the title goes to changing the purpose of the question — and, as such, diverges from the body of the question rather much. Of course, if you were to squeeze rest mass of particles and photons into the question, you might as well be asking a different one. I think you guys didn't really understand the nebulous nature of the question regarding the asker's conception of ‘waves’. $\endgroup$ – can-ned_food Mar 29 '17 at 3:33
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Equation $E=mc^2$ is incomplete. The proper form is (in units with $c=1$) $E=\sqrt{m^2+p^2}$. When an object is at rest then $E=m$ is recovered. But for massless objects $E=p$. So this means that even objects which have no mass can have energy because they have momentum, and waves carry momentum. Massless objects can never be at rest.

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    $\begingroup$ You can't have the energy to be quadratic in p, instead of the square root dependence. This is the energy momentum relation for the linear theory. Maybe a typo? And I don't think we need to use negative or complex energies to address the question. $\endgroup$ – Salvatore Baldino Mar 24 '17 at 16:31
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    $\begingroup$ Although I understand the benefits of considering $c = 1$, I think that for this question it will only be an extra source of possible confusion. $\endgroup$ – Pedro A Mar 24 '17 at 19:14
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    $\begingroup$ Actually, the equation $E=mc^2$ is complete, just that the mass $m$ is not the rest mass. With that in mind, the full equation is $E^2 = m_0^2c^4 + p^4c^4$, where $m_0$ is the rest mass, that is the mass of the particle at rest. $\endgroup$ – jure Mar 24 '17 at 21:15
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    $\begingroup$ @jure Just to mention it: While it is true that $E=mc^2$ is complete when you take $m=\gamma m_0$, we don't really do the "mass increases with speeds close to $c$" thing anymore. In most modern physics texts, all $m$s refer to the rest mass. Also, see answers in physics.stackexchange.com/q/1686/19979 $\endgroup$ – Wojciech Morawiec Mar 24 '17 at 21:49
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    $\begingroup$ @knzhou which are which? I am not from "here" and I find that statement confusing. But by your statement im wrong, so whatever. $\endgroup$ – The Great Duck Mar 25 '17 at 3:45
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Well, think about photons. The electromagnetic field can cange the energy of charged matter, so it has to store energy. EM waves interact with matter and can accelerate or decelerate it, as evident from everyday life.

But let's think about it in a relativistic field theory approach. It is a change in perspective that you have to make, when you want to study a relativistic field theory. And, at speeds that approach (or reach, in the case of the photon) the speed of light, relativistic field theory is the way to go, so it's good to develop some intuition.

Let me start with a massive field, of mass $m$. Think about it just as a field, like the EM field: a law associating to each point in space a value, a vector or any kind of item. I will clarify the concept of the field "having mass" shortly. Now, in the simplest case (free field) we can say that the field has zero energy when it vanishes everywhere. Any excited configuration (nonvanishing field) has a positive energy. Consider a plane wave propagating: from the Lagrangian (I won't go into details), you have that the energy of this plane wave in relation to its impulse $p$ is $$ E(p)=\sqrt{p^2c^2+m^2c^4}. $$ Now, the group velocity of those waves (roughly speaking, the velocity at which an envelope moves) is $$ v_g=\frac{\partial E}{\partial p}=\frac{p}{E}c. $$ As it's obvious that $E>p$, this number is always smaller than $c$. The group velocity of this field cannot be greater than the speed of light, if you have a non zero mass. You can define the mass through the $E(p)$ relation.

This is for massive waves. Massless waves are different: for those, the dispersion relation is $$ E(p)=|p|c. $$ Calculating the group velocity as a derivative, you have $v_g=c$, massless waves move with group velocity $c$, the speed of light. But they still have energy, due to the fact that a wave configuration is a non zero configuration of the field. You can see this as a limit $m\to0$, even if this is not the most correct way to think about it. Massless waves always travel at the speed of light, and their dispersion relation is totally different from the massive dispersion relation.

Still, massless waves have energy, because they can interact with matter exchanging energy. So, in relativistic field theory, it's not strange at all to have massless energetic fields, as energy is a way to quantify your "energetic distance" from the vacuum configuration, when the field is 0 everywhere (and does not interact with other objects).

EDIT: to give an example about how the EM field accelerates objects in everyday life, we can go to nonrelativistic theory of interaction of light with matter. In this case one should really use QM, but we will stick with the classical model to give an intuitive example. You can describe a solid quite roughly as a set of electrons, of an effective mass $m^*$. You can mimic the interactions between electrons using a relaxation parameter $\tau$. Now, let an EM wave $\vec E$ hit the solid: the motion equations for the electron are $$ m^*\ddot{\vec x}+\frac{m^*}\tau\dot{\vec x}=-e\vec E $$ Now, with an oscillating wave you can try an oscillating solution, and find out that the electrons will oscillate around their initial position, with a damping. The electrons are accelerated, as they absorb enegy from the EM field.

This is the basic of the Drude model, that is described in every good book of statistical physics or solid state physics. This model explains macroscopic properties of a material by describing it microscopically and applying statistical tools. Now, the Drude model quite fails at low temperatures as it is based on classical mechanics, but its lesson is still valid: EM field interacts with an object, exciting (or accelerating, in a classical viewpoint) the charges inside the material and causing the conduction of electrical current inside the material and emission of photons outside of it, photons that will allow you to see the body when they reach your eye. You see accelerated electrons everyday when you $\textit{see anything}$: simply, you perceive them through the photons that they emit. But the emission of those photons is due to excitation (or acceleration) of the components of the material from external sources, like the massless EM waves that come from the Sun or a simple lamp.

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  • $\begingroup$ Obviously, I accept that EM waves interact with matter, since I'd be unable to see if they didn't. But I can't think of an everyday example of EM waves accelerating an object. Am I missing something obvious? $\endgroup$ – David Richerby Mar 24 '17 at 13:41
  • $\begingroup$ I will edit the answer to give you another kind of example and way to see it. $\endgroup$ – Salvatore Baldino Mar 24 '17 at 15:01
  • $\begingroup$ @DavidRicherby If you accept that EM waves can exert forces on matter, then it's literally just $F = ma$. It's just that the $a$ typically oscillates very fast. $\endgroup$ – knzhou Mar 24 '17 at 22:01
  • $\begingroup$ @David At the macroscopic scale the Japanese deployed a test-scale light sail on a recent Venus probe. It saved them more than a km/s in delta V. At an intermediate scale the interaction between electromagnetic waves and the free electron in an antenna is how radios work. Then at the atomic scale Salvatore's example of the Drude model was what I was reaching for as well. $\endgroup$ – dmckee Mar 26 '17 at 1:39
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, from my understanding of waves and object, waves have a few defining features: being massless and having energy.

Waves have been observed in water, to start with, and then it was found that sound can be described with waves, i.e. as solutions of differential wave equations, as also strings can display wave behavior.

A water wave can be described by the energy it carries and the height/amplitude of the wave. But it is riding on a medium (which has mass), a huge number of molecules of water whose mass is not relevant to the classical water wave description (it is implicit in the constants that describe the behavior of the fluid).

Waves

Waves may be graphed as a function of time or distance. A single frequency wave will appear as a sine wave in either case. From the distance graph the wavelength may be determined. From the time graph, the period and frequency can be obtained. From both together, the wave speed can be determined.

Light waves are classical, and they transfer energy, but it has been experimentally found that they are not riding on a medium.

This brings us to the quantum mechanical frame where $E = mc^2$ is relevant and defined.

Actually, for particles at the quantum mechanical framework the energy is $E^2 = p^2 + m_0^2$, where special relativity has to be used and there are four vectors, $(E, p_x,p_y,p_z)$ and the $m_0$ is the "length" of this four-vector. (Here we assume the c, the velocity of light, is 1).

Object have mass and energy.

Classical objects.

So according to Einstein's equation $E=mc^2$, mass and energy go hand in hand (sort of)

See above and links given for the quantum mechanical objects.

so why do waves, which have energy not have mass?

Classical waves ride on atoms and molecules, which have mass.

Electromagnetic waves classically have varying electric and magnetic fields in space and time, and quantum mechanically they are riding on photons which have mass 0 by the special relativity equations; they just have energy equal to momentum.

I am aware of wave-particle duality but (now I may be wrong with this as well) to my understanding that is regarding particles giving off weak waves and not the other way around.

You are confusing the two frameworks, the classical and the quantum mechanical. The wave particle duality is a quantum mechanical phenomenon, but the particles are not spread out like a classical water wave. The quantum mechanical probability density of finding by measurement a particle at a particular (x,y,z) obeys wave equations and can show interference patterns like classical waves. This experiment of sending single electrons at a double slit may help to understand the difference between probability density wave behavior and particle itself.

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  • $\begingroup$ Well, a lot of people tend to forget that scientific method never demonstrates anything, except insofar as it demonstrates the absence of something else. Those experiments demonstrated that the expected medium was absent, not that there was no other medium which had different characteristics that were not detected or detectable. Otherwise, the ultimate paragraph is concise and excellent. $\endgroup$ – can-ned_food Mar 27 '17 at 8:17
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(I am repeating what others have said, but hopefully in a simpler and more accessible language)

The problem is that both energy and mass can mean more than one thing.
Energy can mean rest energy or total energy. (or a number of other things)
Mass can mean rest mass or relativistic mass. $$E_{rest}=m_{rest}c^2$$ $$E_{total}=m_{rel}c^2$$

For a normal particle, like a uranium atom, the first equation says that it has energy even when it is sitting still. For light, which has zero rest mass, it says that light sitting still has zero energy. Or, in other words, light cannot sit still.

The second equation says that all forms of energy has relativistic mass, including kinetic energy. For normal particles this means that they get heavier as they get faster. For light, it means that it has both non-zero energy and non-zero relativistic mass.

Now, both these equations are simple multiplications with a constant. This suggests that energy and mass is the same thing, only measured in different units. When other answers have said to set $c=1$, they are basically saying that things get easier if you measure $E$ and $m$ in the same units.

Physicists have taken this to heart and no longer talk about both rest mass and rest energy as that is redundant. The term used is "rest mass" or just "mass".

Likewise, they rarely talk about "relativistic mass" but prefer to say "total energy" or just "energy".

When using this convention, you can no longer say $E=mc^2$ since total energy and rest mass does not fit together in this way.

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When you start to think about what mass is at a deeper level it becomes easier to separate it from the idea of energy. The two aren't necessarily mutually exclusive. One might think of mass as something that has a resistance to changes in velocity. One might also think of energy as what's required to change the velocity of an entity, that entity doesn't necessarily have to have an intrinsic resistance to changes in velocity.

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Your assumption that waves are massless is incorrect.

A photon has zero rest mass, not zero mass. In effect, this means that a photon can only exist when moving at the speed of light (which varies depending on the material the photon is passing through).

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    $\begingroup$ According to Einstein's equation (E=mc^2) , if a photon has zero mass it must also have zero energy (as its energy would be its mass, i.e. zero, multiplied by its speed, and zero multiplied by anything equals zero), so the implication in the question that a wave is massless must - in some sense - be incorrect, given that the question implies that the wave has energy. If its speed is zero, i.e. it is at rest, then the equation is satisfied: zero energy equals zero mass multiplied by zero speed. This would be a purely theoretical state: a state of nil energy. That, I think, is what is meant. $\endgroup$ – Ed999 Apr 3 '17 at 12:28
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In a real sense, an electromagnetic wave is energy. Where a particle emits energy, it principally does so by emiting it in the form of electromagnetic waves. The more energy it emits, per second, the greater the frequency of the wave (the more peaks in the waveform per second) and the shorter its wavelength (the distance between those peaks).

As only particles couple/bind to the fabric of spacetime (as Einstein terms it), only particles have mass. Energy does not couple to spacetime. It is the coupling-bond which gives a particle mass (and thus inertia). As energy does not have a coupling bond, it has no inertia, which is what permits it to move at the speed of light.

In technical terms, a particle interacts with the Higgs field, thereby acquiring inertia. A wave does not interact with that field, so has no inertia. It is the Higgs field which gives an object inertia and mass. In a sense, mass is only a measurement of the strength of the coupling charge by which the particle adheres to the spacetime fabric (to what, in modern terms, I suppose we should simply call the Higgs field).

If an object (i.e. a particle) couples to spacetime, a certain amount of energy must be applied to break the coupling bond, and we term that quantum of energy "mass", a particle's mass being equivalent to the amount of energy required to break the bond which is holding it in place. But all we are really saying (from Newton) is that a particle will only move if acted upon by an outside force. Energy does not obey Newton's laws of motion, in the sense that electromagnetic energy (waves) move although no outside force is acting on them.

Hence an object which has mass, and thus obeys Newton's laws of motion, we define as a particle. And an object which does not, we define as energy. Energy has no mass by definition, since mass implies inertia, but a waveform has no inertia (when did you last stand in the sunlight and experience being knocked to the ground by it?)

It is, perhaps, wrong to say that waves have energy. It is probably less inacurate to say that waves are energy. Mass has energy, too: Einstein explained that energy equals mass multiplied by the square of the speed of light. Mass is energy is a confined state. But free energy is free in the sense of being unconstrained, and that degree of freedom comes from not having mass, which is really only another term for inertia.

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    $\begingroup$ Paragraph 1 is wrong. It's very dangerous to say something "is" energy. In standard physics language, energy is a property of objects, not an object itself. What you're doing is like saying an expensive painting is money. $\endgroup$ – knzhou Mar 24 '17 at 22:03
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    $\begingroup$ Paragraph 2 is exactly backwards. Energy affects the curvature of spacetime; in fact, it's basically the only thing that does. $\endgroup$ – knzhou Mar 24 '17 at 22:04
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    $\begingroup$ Paragraph 3 is wrong. The 'spacetime fabric' and 'Higgs field' are completely unrelated things. The Higgs mechanism doesn't involve general relativity at all. Also, there's no clear distinction between "particles" and "waves" at the quantum level. $\endgroup$ – knzhou Mar 24 '17 at 22:05
  • $\begingroup$ Paragraph 4 is wrong. Mass is not quantized, period. $\endgroup$ – knzhou Mar 24 '17 at 22:06
  • $\begingroup$ @knzhou Why not write an answer? $\endgroup$ – Stijn de Witt Mar 26 '17 at 23:03

protected by Qmechanic Mar 24 '17 at 19:15

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