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I've read a number of the helpful Q&As on photons that mention the mass/mass-less issue. Do I understand correctly that the idea of mass-less (a rest mass of 0) may be just a convention to make the equations work?

From a layperson's view, it's difficult to understand how a particle of light (photon) can be mass-less. A physical object (everyday world-large or quantum-small) must have a mass. Yet, if my understanding is correct, the mass of a moving object/particle increases in proportion to its speed/velocity...so that at the speed of light, its mass would be infinite. A photon travels at the speed of light, but it obviously doesn't have infinite mass, right? Can someone formulate a practical explanation that can be understood by middle-school to high school kids? Much thanks for the help.


Wow--your answers to my original Q below clear up much of my confusion. I now have the daunting task of going over these nuggets and working up an equation-less (hopefully) explanation of the mass-less photon for non-physicist types.

Yes, from a layperson's view, it does seem remarkable that an existing piece of matter-- which has to be made of physical substance--could have zero mass at rest (though a photon is never at rest). It would be almost understandable if a piece of matter made of nothing had zero mass, but that seems to be an oxymoron, and "nothing" would equate to nonexistent, right?

In case you might find it interesting: I'm working on a writing project that posits we inhabit a universe that consists of matter (physical stuff) only, and that the NON-physical (aka supernatural) does not (and cannot) exist. For instance, if a purported supernatural phenomenon is found to actually exist, then by definition, its existence is proof that it is mundane/natural. All it would take to disprove this premise is reliable proof that ONE supernatural event has occurred. Despite thousands of such claims, that's never yet happened.

Who else better than physicists to confirm my premise? However, I do wish the TV physicists would explain the terms they throw about, some of which mislead/confuse their lay viewers. Case in point: "The universe is made up of matter and energy" (without properly defining the term "energy" as a property of matter).

The result is that laypersons are left with the impression that energy must therefore be something apart from or independent of matter (ie, nonphysical). Their use of the term "pure energy" without specifying exactly what that means adds to the confusion. (Thanks to your replies on this forum, I now understand that "pure energy" refers to photon particles.) However, "psychics" and other charlatans take advantage of such confusion by hijacking terms like energy (as in "psychic energy"), frequencies, vibrations, etc to give perceived scientific legitimacy to their claims that a supernatural spirit world, etc., exists. As you may realize, the majority of people in the US (per 2009 Harris Poll) and around the world believe in the existence of nonphysical/supernatural stuff such as ghosts and spirits.

My purpose is to give laypersons the information they need to distinguish what's real from what's not.

Thanks so much for help...And, PLEASE, add any further comments you think might be helpful/insightful to better inform laypersons.

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I don't get this question. Photons are massless, full stop. –  Noldorin Jan 22 '11 at 2:33
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Also, why are you supposing that mass is such a fundamental property that all particles must possess? The Standard Model suggests that mass arises from interaction with the Higgs field. –  Noldorin Jan 22 '11 at 2:34
    
@Noldorin, +1. The question should be "How can a photon have no mass and still travel at the speed of light?" –  voix Jan 22 '11 at 8:38
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It seems that your essay really is more about what you mean by 'matter' 'energy', 'natural' and 'supernatural'. Since a photon has the property of having no rest mass, there are some people who refer to photons as being particles with "energy" but no mass. Similarly, your definition of nautral/supernatural is tautological. More important than the atoms/void distinction that we inherit from Democritus, I think, is asking whether phenomena follow predictable rules. –  Jerry Schirmer Feb 26 '11 at 15:47
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It isn't wise to postulate that « exists » = « matter » = « particle ». Quantum Mechanics (and Quantum Field Theory) calls into question our normal intuitions about particle and matter. Virtual particles, for example. Even more fundamentally, the idea that « field » is more fundamental than particle. I can understand why you don't want to postulate that « supernatural » = violates a scientific law, because one has Hume's arguments on causality and law... but what you are trying isn't any more immune to problems. I advise you to give it up and study pollution and climate change instead... –  joseph f. johnson Jan 16 '12 at 5:43
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There is absolutely nothing conventional about the mass of different particle species. For any particle moving in the vacuum, you may measure the total energy $E$ (including the latent energy) and the momentum $p$. It turns out experimentally - and Einstein's special theory of relativity guarantees - that the combination $$E^2 - p^2 c^2$$ doesn't depend on the velocity but only on the type of the particle. It is a quantity describing the particle type and we call it $$E^2 - p^2 c^2 = m_0^2 c^4$$ This determines the rest mass $m_0$ of the particle. The formula above works for any particle in the vacuum, any speed, and is always non-singular. Photons have $E=pc$ which implies that $m_0=0$. The rest mass of a photon is equal to zero.

Indeed, that's also the reason why one can't really have a photon at rest, $v=0$. If a speed of something is $c$ in one reference frame, it will stay $c$ in any (non-singular) reference frame - that's another postulate of the special theory of relativity. So one can't ever make the speed of the photon zero by switching to another (non-singular) reference frame.

But if you want to see some values for all quantities, you may imagine that a photon at rest could exist and its total mass would be zero. At speed $v$, the mass is increased to $$m_{total} = \frac{m_{0}}{\sqrt{1-v^2/c^2}}$$ For $m_0=0$ and $v=c$, the expression above is clearly a $0/0$ indeterminate form and its proper result may be anything. In particular, the correct value is any finite number. At the right speed, $v=c$, the massless photons can have any finite energy.

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There is a problem with this derivation, namely we can't use $E=pc$ (without making a circular argument) since AFAIK there is no way to derive $E=pc$ other than by setting $m=0$ in the $E^2=(m_0c^2)^2+(pc)^2$. Even with De Broglie's equation. –  حكيم الفيلسوف الضائع Feb 18 at 12:46
    
Dear @clicky-hacky-clicky-hacky, I wasn't trying to claim that one may derive the identity out of "nothing". Some formulae or principles always have to be assumed - and defended by a successful comparison of the results with the experiments. That's why criticisms of "circularity" in physics completely miss the epistemical status of natural sciences in general. –  Luboš Motl Feb 18 at 13:39
    
@LubošMotl You claimed that photons have $E=pc$, (the thing that can be derived using the energy-momentum formula only by setting $m=0$), you then plugged that result into that same equation and showed that $m=0$. Isn't this circular logic and thus isn't it wrong? en.wikipedia.org/wiki/Circular_reasoning And thank you by the way for the awesome 'c-h-c-h' call, you're very kind. –  حكيم الفيلسوف الضائع Feb 18 at 20:58
    
Dear @clicky-hacky-clicky-hacky, sorry for using a transliterated version of your name via Czech (orig: klikyháky, klikykáky). :-) I don't have the right keyboard to write your name and copy-and-paste would be mindless and disrespectful... If individual claims in a set of claims in physics imply each other so that you may find a loop that you call "circular reasoning", it surely doesn't mean that any of the claims is wrong. Instead, everything I write and everything I have written in my life, more or less, is true. In physics, some claims have to be assumed/extracted from observations. –  Luboš Motl Feb 19 at 6:21
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Do I understand correctly that the idea of mass-less (a rest mass of 0) may be just a convention to make the equations work?

No, a photon really does have zero mass. You can think of it as a particle of "pure energy" if it helps you make sense of it, but the only sense in which that is valid is that a photon has energy but no mass.

Yet, if my understanding is correct, the mass of a moving object/particle increases in proportion to its speed/velocity...

No, actually mass is a relativistic invariant. Each object has its own particular intrinsic mass that is always the same no matter what speed the object moves at. What increases is the energy. For massive objects, they are related by the formula Luboš cited,

$$E = \frac{mc^2}{\sqrt{1 - v^2/c^2}}$$

(note that my $m$ corresponds to his $m_0$).

Here's a little historical motivation. In the early days of relativity, physicists quickly realized that as an object speeds up, it becomes harder and harder to increase its speed by a given amount. This is one way to express the reason why no object can exceed the speed of light. Now, we say that an object that is harder to move has more inertia, and the prevailing convention was that inertia should be related to mass, so it made sense that an object moving faster should have more mass, in some sense. Physicists coined the term "relativistic mass" to refer to this quantity.

However, they also recognized that something was special about the relativistic mass of an object at rest, since (for example) that was the minimum possible relativistic mass the object could ever have; it's the amount of mass that is intrinsic to the object. Appropriately enough, they labeled this the "rest mass."

In later years, prompted in part by the discovery of general relativity, physicists realized that it makes more sense to say that inertia is related to energy, not just mass. After all, GR tells you that there are other things besides mass that contribute to gravity, so it makes sense that there are other things besides mass that contribute to inertia. So the term "relativistic mass" fell out of favor (well, sort of; there are still a fair number of people who do use it) and physicists just started talking about energy instead. The two are simply related by the equation

$$E = m_\text{rel}c^2$$

This is the famous equation that people associate with Einstein, although they often forget the subscript "rel" and that leads to a lot of confusion ;-)

Now that "mass" isn't being used to describe the total thing-that-is-related-to-inertia of a moving object, we can just use that word to describe the "intrinsic mass" or "rest mass" without any confusion. (Again, sort of; there are still a fair number of people who do talk about "rest mass" and it's possible to get confused when different groups with different conventions talk among each other.) So these days "mass" refers to what was previously known as "rest mass." As I said, each object has its own particular mass, and for a photon, that happens to be zero. As Luboš said, only objects with zero mass can travel at the speed of light, otherwise the energy would be undefined.

In order to make sense of the photon, remember Einstein's full equation

$$E^2 = m^2 c^4 + p^2 c^2$$

which says that energy comes from both mass and momentum. So even if a particle has no mass, it can still get its energy from momentum, and photons do indeed have both energy and momentum.

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David -- I don't want to be nitpicking, and I know mathematically there is no difference, but from a physics (and didactically) point of view there is a huge difference: we should not write the mass-energy-momentum relation in Pythagorean form. This is an equation for the norm of of the energy-momentum vector for which the proper format is the invariant format: mass on one side and energy-momentum on the other. This makes the Minkowski signature explicit and prevents lay readers to get wrong footed. –  Johannes Jan 21 '11 at 23:37
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@Johannes: objection noted, but in this instance I am expressing the energy as the combination of contributions from mass and momentum. So the form I used is the one that best matches my purpose here. For other purposes, other arrangements of the terms will be most appropriate. I don't think you can reasonably argue that there is one proper form for all applications, though. –  David Z Jan 22 '11 at 0:19
    
"a photon really does have zero mass". Then explain $E = mc^2$. If there is a mass ($m$) of 0, then there is no energy ($E = 0c^2$ -> $E = 0$). –  Cole Johnson Sep 6 '13 at 7:24
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@Cole see the last paragraph of my answer. –  David Z Sep 6 '13 at 16:41
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Another point of view. The smaller rest mass of the particle, the easier it is to be accelerated to the same speed. Light electron is easy to accelerate to the speed $0.9999c$ than heavy nucleus. Photon with rest mass $< 10^{-18}eV$ may be considered as the lightest particle.

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What's special about $10^{-18}eV$? –  Trevor Alexander Dec 28 '13 at 10:21
    
Look at the article and the main link –  voix Dec 29 '13 at 14:24
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You ask for a practical explanation that can be understood by interested middle/high-school kids. I think this comes close: What's Wrong With $E=mc^2$?

Mass is the norm (=length) of the energy-momentum vector, and therefore invariant and conserved. The square norm of the energy-momentum vector for an object with energy $E$ and momentum $p$ is $E^2-p^2$. (see note below) The square root of this expression is mass ($m$). A photon has $E = p$, and hence zero mass. A massive particle at rest has $p = 0$ and hence $E = m$, Einstein's most famous equation (see footnote).

Forget about 'relativistic mass'. This is a confusing term that does not add any understanding. Whenever you see the term 'relativistic mass', replace it with the term 'energy', as that is what it really is. A photon has energy (in proportion to it's frequency) but no mass.

Do you really think it is remarkable that there can be energy without mass?

Note: In the above insert factors $c$ when not working in natural units (i.e. replace $m$ with $m c^2$, and $p$ with $p c$)

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I think you are confused. Photons do not have mass, they have momentum.

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In response to discussion, "Yes, from a layperson's view..." It is important to notice the supernatural was given a definition, i.e. non-physical, then it was said that it must not be natural – natural possibly meaning physical. One of the axioms of the view was that, 'Only what is physical can be found,' and, 'What exists is physical.' Although the author makes it sound like a conclusion, it is an axiom, therefore there is circular reasoning. Instead of 'by definition', it should be said, 'By definition and by the assumption that something's existence means it is is physical, I conclude that the supernatural does not exist.'

It is also important to notice that any definitions given can create axioms, and that any axioms given can create undefined terms (as in an axiomatic approach to geometry). Also what is the author's definition of 'found'? Is it that it is found by experimental physics, or is it that is found by theoretical physics?

About the energy, all I have to say is that it is generally accepted that particles would need energy to be created, namely a photon. The author argues that the photon is matter itself. The axioms include: 'Everything that we can think of as a particle is matter.' Actually, this is either an axiom that leads (possibly together with other axioms) to the conclusion that everything is physical, or a conclusion that is a consequence of the axiom that everything is physical. The author does not prove these ideas independently. Again, circular reasoning.

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As I am a layperson, this is how I justify it to myself, as this is an opinionated Beowulf, please regard it as my "opinion" and not fact. In my view any two objects with mass that are directed toward each other will create an explosion. Taking this into account, I take two flashlights and point them at each other: oh no, is this a cheap particle accelerator? Alas, the flashlights are turned on, and well, nothing really happens. They do not collide and create some sort of particle shower. In lay terms, they just pass through the other flashlight beam; same as with two laser pointers. They pass through, not collide in a physical sense.

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The beams of can collide, it's just weak. It's no different then shooting a neutrino through the Earth. –  Ron Maimon May 15 '12 at 6:30
    
@RonMaimon: In a mathematically rigorous way I agree completely, but I the scope of this question he is asking for a more generalized idea of how a layperson can validate such a complicated ideology. –  Argus May 15 '12 at 14:22
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They don't have rest mass, or simply, mass. However, they have the so-called relatvistic mass, although the term is hardly used. The relation between the relativistic mass $M$ and the mass (Rest mass) $m$ is simply given by: $$M=\gamma m=\frac{m}{\sqrt{1-\beta^2}}$$.

Here, $\beta=\frac{v}{c_0}$ and $\gamma=\frac{1}{\sqrt{1-\beta^2}}$ is the lorentz factor. However, when $v=c_0$, and $m=0$, as for a photon of light, this equation goes to $\frac{0}{0}$, i.e., which is undefined/indeterminate. So, how do we overcome this problem? Simple! $$M=\frac{m}{\sqrt{1-\frac{v^2}{c_0^2}}}=\frac{m^2c_0^2}{c_0\sqrt{1-\frac{v^2}{c_0^2}}}=\frac{mc_0}{\sqrt{c_0^2-v^2}}$$

$$M^2=\frac{m^2c_0^2}{c_0^2-v^2}=\frac{mc_0}{\left(c_0-v\right)\left(c_0+v\right)}$$ Use partial fractions. Let $$\frac{mc_0}{\left(c_0-v\right)\left(c_0+v\right)}=\frac{U}{c_0-v}+\frac{V}{c_0+v}$$ $$\frac{mc_0}{\left(c_0-v\right)\left(c_0+v\right)}=\frac{(U+V)c_0+(U-V)v}{(c_0-v)(c_0+v)}$$ $$\left. {\begin{array}{*{20}{c}} {U + V = m} \\ {U - V = 0} \end{array}} \right\}$$ $$U=V=\frac{m}{2}$$ $$M^2=\frac{m^2c_0^2}{2(c_0-v)}+\frac{mc_0}{2(c_0+v)}$$

This isn't going anywhere. Let's try another method. $$\gamma = \frac{1}{{\sqrt {1 - {\beta ^2}} }} $$ $$\beta = \sqrt {1 - \frac{1}{{{\gamma ^2}}}} $$ $$1 - \frac{1}{{{\gamma ^2}}} = {\beta ^2} $$ $${\gamma ^2} - 1 = \frac{{{\gamma ^2}{v^2}}}{{c_0^2}} $$ $$\left( {{\gamma ^2} - 1} \right){m^2} = \frac{{{\gamma ^2}{m^2}{v^2}}}{{c_0^2}} $$ $${M^2} - {m^2} = \frac{{{p^2}}}{{c_0^2}} $$ $${M^2} = \frac{{{p^2}}}{{c_0^2}} + {m^2} $$ Now, no indeterminate / undefined stuff !.

Now, consider the case for a photon: $$M=\frac{p}{c_0}=\frac{\hbar k}{c_0}$$

So, in conclusion, a photon is massless (no rest mass) but has so called "relativistic mass" due to the reasons you mentioned.

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Since this question is about how a photon can travel at light speed and yet have no mass, I will answer by saying that photons having no mass is precisely why they can travel so fast, and without mass, it becomes intangible for anything to make it go slower.

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Compaction due to length contraction is not a valid concept for a point particle like a photon, and in any case, it is not related to increase in relativistic mass. –  Abhinav Feb 10 at 11:03
    
the compaction statement was for objects with mass, not the photon, length contraction may not be related to increase in mass but it is related to its compression which was my point. –  GammaRay Feb 10 at 11:31
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There can also exist massive point particles. The introduction of length contraction is really irrelevant here. –  Abhinav Feb 10 at 15:01
    
Ok, large bodies with atomic composition can be explained this way but you are correct. this physical process can not apply to point particles with mass, it can only be analogous to something more internal. (perhaps it was A misleading example.) –  GammaRay Feb 10 at 21:58
    
In fact I was confusing mass with relative density. –  GammaRay Feb 20 at 6:18
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