Do Photons Have Mass? - Debate & Questions

[rbj]

I think the answers are:
5) 20N

i don't think that's correct. i believe that, whether you look at the photons as having relativistic mass (a deprecated term) which is E/c^2=\hbar \omega/c^2 or as the energy density, i think that GR says they affect space-time curvature or are affected by it just as if was the equivalent mass.

 

[DrGreg]

I think the answers are:
5) 20N

i don't think that's correct. i believe that, whether you look at the photons as having relativistic mass (a deprecated term) which is E/c^2=\hbar \omega/c^2 or as the energy density, i think that GR says they affect space-time curvature or are affected by it just as if was the equivalent mass.

Everything you've said supports the answer being correct. I don't understand why you think your reasons make it incorrect.

 

[Dale]

But will it show on the scales? Is it the kind of mass that is attracted by Earth's gravitation?

Yes. Remember that photons blueshift as they go down and redshift as they go up, and also remember that a photon's momentum is proportional to its frequency. So the photons that hit the top of the box will have less momentum than those hitting the bottom, resulting in a net downwards force on the box.

 

[Goodison_Lad]

i don't think that's correct. i believe that, whether you look at the photons as having relativistic mass (a deprecated term) which is E/c^2=\hbar \omega/c^2 or as the energy density, i think that GR says they affect space-time curvature or are affected by it just as if was the equivalent mass.

Your line of reasoning is actually pretty similar to the one I used in order to arrive at the answers in the first place.

Your reasoning suggests that the answers to both (6) and (7) will be 'No', in which case the answer to (5) has to be that the weight of the box-photon system is unchanged i.e. 20N.

 

[rbj]

Everything you've said supports the answer being correct. I don't understand why you think your reasons make it incorrect.

the reasons are that sometimes i don't read through the problem sufficiently. i was thinking the 20 N was the tare weight of the box and not of the contents inside.

ooops.

 

[haruspex]

this isn't entirely true as gravity does exert a force. in the example you used youre correct, but if that applied everywhere then two masses at rest relative to each other would feel no gravitational force as there is no inital motion, which I am pretty sure isn't true (correct me if I am wrong anyone)

.

Not so. The curved spacetime explanation does not require any initial relative motion in space. All objects are moving through time. Curving spacetime can mean that the time direction in spacetime may not be the same for the observed object as for the observer.

 

[Fastman99]

When a photon falls into a black hole, the black hole increases in mass by m = E/c^2, where E was the energy of the photon. This is required by the conservation of energy and momentum. Energy can be thought of as the zeroth component of the momentum 4-vector, and each component of the 4-vector is conserved locally.

In general relativity, the curvature of spacetime is not caused by rest mass, but by the energy-momentum tensor.
http://en.wikipedia.org/wiki/Stress–energy_tensor

To answer the question of the box containing matter and antimatter:
5) The energy density of the box would not change, therefore the scale would still read as 20N
6) My question is, what is meant by "inertia"? This is a subjective question, because it doesn't refer to a measurable quantity, but to a phenomenon that is actually kinda complicated. This is usually called the conservation of energy and momentum. If you took the sum of energy and momentum of all the particles in the box before and after, then you would find it didn't change. In order to induce a change in the total energy/momentum of the box, there would have to be an opposite change in the energy/momentum of something else.

A better question that might be asked, suppose you had two boxes, one box full of normal matter, and one box made of half matter and half light. Say they are both placed on scales, and give the same number, 20N in this case. If both of them have the same total momentum not equal to zero in a particular direction (say the x-axis), then do they both travel at the same apparent velocity? Or does the half matter/light box have a higher velocity because it is "lighter?" Lol puns.

My intuitive answer is that they both move at the same velocity, but I haven't worked it out yet.

7) Mass is a loaded word, and not universally defined. If you're asking about the rest mass, then yes, the rest mass has changed. Photon have no rest mass, therefore the total rest mass of the box has decreased. However the photons do have energy and momentum. The energy and momentum of the photon could be said to have "gravitation mass" and that is conserved. The "gravitation mass" has not changed.

 

[dustinthewind]

The way I came to understand how photons can have mass and yet have no mass is that they have apparent mass because of their energy. You can calculate their energy as stated above E=\hbar\omega. So then I always took E=mc^{2}. And sovle for the mass of the electron. You can then consider it has momentum mc. So that when it hits say a solar sail and is reflected you receive twice the momentum of caching that photon and throwing it back. However it is stated that photons actually have no mass other than their energy and so immediately reach the speed of light. That is the energy applied to a massless object would immediately reach infinite speed for E=1/2 mv^{2} As you would divide a finite energy by zero mass. However we would percieve it as having the speed c instead due to space time distortion. I think even in the relativisitc formula this would be the case v=\sqrt{\left(\frac{k_i}{c m_l}\right){}^2+\frac{k_i 2 }{m_l}}

 

[Goodison_Lad]

7) Mass is a loaded word, and not universally defined. If you're asking about the rest mass, then yes, the rest mass has changed. Photon have no rest mass, therefore the total rest mass of the box has decreased. However the photons do have energy and momentum. The energy and momentum of the photon could be said to have "gravitation mass" and that is conserved. The "gravitation mass" has not changed.

Regarding inertia, I suppose I should have been more specific and referred to ‘inertial mass’. Since the Principle of Equivalence is based on the equality of inertial mass and gravitational mass, the answer to (6) would be that the inertial mass is unchanged. We could measure the inertial mass of the box system by attaching it to a spring that could oscillate in the horizontal direction, with the period of oscillation being used to determine the mass. This mass would be unchanged after the annihilation process.

As for the rest mass, I would say that the rest mass of the system is therefore also unchanged, since the gravitational/inertial mass is unchanged and rest mass is simply the mass of the system determined when it is at rest. So even though the box is now full of photons, each with zero rest mass, the box system is a closed one and so its rest mass cannot have altered.

 

[yuiop]

...
7) Mass is a loaded word, and not universally defined. If you're asking about the rest mass, then yes, the rest mass has changed. Photon have no rest mass, therefore the total rest mass of the box has decreased. However the photons do have energy and momentum. The energy and momentum of the photon could be said to have "gravitation mass" and that is conserved. The "gravitation mass" has not changed.

The total rest mass of the box has not changed. Here is a simple proof that a system of photons can have rest mass:

Consider a single photon and the momentum energy expression:

E = \sqrt{ (m_0 c^2)^2 + (pc)^2 }

where E is the total energy of the photon and m_0 is the rest mass and p is the momentum of the photon.

Now it is generally accepted that the total energy of a photon is E = pc, where p = hf and h is the Planck constant and f is the frequency of the photon. Since E = pc it is immediately obvious from the above equation that the rest mass of a single photon is zero.

Now consider the case of a pair of photons with equal energy but going in opposite directions.

Since they momentum of equal magnitude and opposite directions, the total momentum of the photon pair is zero so we can now say:

E = \sqrt{ (m_0 c^2)^2 + (pc)^2 }

E = \sqrt{ (m_0 c^2)^2 + 0 }

E = m_0 c^2

m_0 = E/c^2

Now the rest mass of the photon pair as a system is non zero and proportional to the total energy of the photon pair.

Q.E.D.

=======================================

Now I would argue (although I have not seen it generally accepted) that rest mass is what gives a system its active gravitational properties. A single photon has no rest mass and is not a source of gravity on its own but it does have passive gravitational properties in that it responds to a gravitational field. A single photon also has the property of momentum and while this is normally associated with mass a photon clearly demonstrates that a particle can momentum without having rest mass. It would seem that the unique property of rest mass that distinguishes it from other forms of mass (other than being invariant under transformation) is that it has active gravitational mass. This is further supported by the accepted observation that photons going in the same direction are not attracted gravitationally towards each other while photons going in opposite directions are.

Now when a massive particle or pair of massive particles decay into photons, momentum is conserved and decay photons are normally produced in pairs going in opposite directions. For a box of particles that is stationary in a given coordinate system, the average momentum of the box of particles is zero and when the box of particles decays into photons conservation of momentum dictates that the total momentum of all the photons in the box is zero and therefore the system of photons has rest mass and active gravitational properties.

 

[Goodison_Lad]

Nice proof, yuiop.

I wasn't aware, though, about photons going in the same direction not attracting each other, and this set me thinking...

This is further supported by the accepted observation that photons going in the same direction are not attracted gravitationally towards each other while photons going in opposite directions are.

What would happen if the box experiment in post 47 were repeated, but this time the two annihilation photons are diverted vertically by mirrors? After the first reflection, the mirrors are retracted (as in the attached image), and the photons continue to reflect up and down in step, somewhat like the proverbial light clock. Presumably the scales would continue to register the combined mass of the box and the photons - there is still an energy density present in the box.

So if we had an identical box-photon system side-by-by side with the first one (again, resting on a set of scales), with its photons reflecting up and down in step with the photons in the first box, is it the case that the two boxes would not attract each other gravitationally (ignoring the masses of the boxes themselves for the sake of argument), even though each set of scales would register the masses of the photons? Or would they?

 

[yuiop]

... So if we had an identical box-photon system side-by-by side with the first one (again, resting on a set of scales), with its photons reflecting up and down in step with the photons in the first box, is it the case that the two boxes would not attract each other gravitationally (ignoring the masses of the boxes themselves for the sake of argument), even though each set of scales would register the masses of the photons? Or would they?

Good question! (Which is another way of saying I am not really sure of the answer ) I imagine that with the photons reflecting off the top and bottom of the box, that they exert a pressure on the container and that in turn causes tension and stresses in the walls. These stresses/pressures/tensions add to the total gravitational mass of the boxes and so the two boxes would attract each other gravitationally with the force you would expect from observing their weights on the scales. However I know nothing about GR tensors, so I cannot give a definitive answer.

 

[wmsmith]

Hello all, I am a chemistry graduate student at UCI. This question (do photons have mass) came up in a discussion section for general chemistry. The student gave the following scenario and I'm a bit stumped.

Suppose 2 photons were emitted from two points on a plane separated by a distance, d and traveling with identical velocity, v , parallel to the normal of the plane.

The question is, would the paths of the two photons ever intersect?
I.e. would their relativistic masses yield a gravitational interaction, causing their paths of motion to be distorted, or would they continue on in a straight path, forever parallel?

Or, a more physical interpretation, if a measurement were taken of their diffraction pattern at a distance r from the plane, would there ever be a large enough distance such that there would no longer be two distinct loci of maximal intensity in the diffraction pattern, but instead a single locus of maximal intensity (suppose that d was such that each photon's emission location was centered upon a node for the other photon)

One overlapping intensity maximum

Two distinct intensity maxima


My gut reaction was that they would stay parallel (two distinct intensity maxima), but I'm not 100% certain.

If either one were a particle with rest mass, the answer would be that the path of the photon would be altered...

Of course a particle with rest mass would never achieve the speed of light, so the question of their paths crossing would be mute unless the particle were so massive and dense that it acted like a black hole...

 

[pervect]

GR is a classical theory, but it's fairly well known that "photons" (classically electromagnetic waves and usually abstracted as a "null dust" in General Relativity) don't attract each other when going in the same direction.

They will attract each other if they have anti-parallel paths though.

See for instance http://prola.aps.org/abstract/PR/v37/i5/p602_1
Phys. Rev. 37, 602–615 (1931) "On the Gravitational Field Produced by Light", Richard C. Tolman, Paul Ehrenfest, and Boris Podolsky.

Tolman's result might be easier to look up in http://books.google.com/books/about/Relativity_Thermodynamics_and_Cosmology.html?id=1ZOgD9qlWtsC

Wiki also has a reference, which I haven't studied closely, but looks OK at first glance that gives the metric:

http://en.wikipedia.org/w/index.php?title=Bonnor_beam&oldid=479934048

About the only thing valid about the "relativistic mass" argument is that the "relativistic mass" of photons ( which is rather more likely to be called "energy", see any of the way-too-long threads on this point) is part of the stress-energy tensor. It's actually the stress energy tensor that causes gravity in GR, not mass, or energy.

The fact that the photons don't attract when moving parallel, and do attract when moving in anti-parallel might suggest some sort of gravitational effect that mimics magnetism. This idea turns out to have some merit - in the weak field, one can draw a useful analogy between gravity and Maxwell's equations. See for example http://en.wikipedia.org/w/index.php?title=Gravitoelectromagnetism&oldid=509366342

 

[georgir]

tl;dr
But the idea that something needs mass to be affected by gravity is obviously false - all things that have mass are affected absolutely identically by gravity, they receive exactly the same acceleration, regardless of their mass. So even if they had zero mass, it would be normal to assume they will still be affected in the same manner and get the same acceleration.

 

[wmsmith]

Thanks for the quick reply.

I get the idea of conceptualizing gravity as a warping of time-space curvature. In that context it makes perfect sense that a photon (or any other object) path would be distorted by the presence of a gravitational field. I just wasn't sure if this effect would be produced by objects with relativistic mass as well.

The fact that photons traveling in parallel will not attract but photons traveling anti-parallel will is quite surprising though. I will have to look into the stress energy tensor formulation in more detail. Thanks for the links.

 

[robinpike]

If that is such a glaring evidence for light having a mass, then would something as obvious as that be missed by the whole physics community. I mean, let's get real here. How dumb do you think physicists are to miss such a thing?

This issue has been discussed to death in several threads in both the Quantum physics forum, and the SR/GR forum. Please do your search there and figure out how gravity is a spacetime curvature, and why light follows the "geodesic" of that spacetime curvature. It has NOTHING to do with light having a mass.

Zz.

Sorry I don't understand - If gravity is the consequence of following a curved spacetime line, doesn't that explanation require another, unmentioned, force to curve the spacetime line?

 

[jtbell]

The curvature of spacetime is caused by the local energy and momentum densities (specified by the stress-energy tensor), via the Einstein field equations of general relativity.

 

[robinpike]

The curvature of spacetime is caused by the local energy and momentum densities (specified by the stress-energy tensor), via the Einstein field equations of general relativity.

Sorry I don't understand that - is there a picture showing what that means, that you can upload?

 

[andrien]

http://en.wikipedia.org/wiki/General_relativity

 

[robinpike]

http://en.wikipedia.org/wiki/General_relativity

Thanks for the picture.

If, in that picture, gravity is a consequence of following the curved spacelines, then what is the force is pulling the spacelines down in the picture?

 

[Nugatory]

If, in that picture, gravity is a consequence of following the curved spacelines, then what is the force is pulling the spacelines down in the picture?

There isn't one. That picture can be used to give you a mental model of what curved space does, but doesn't help with why the space curves.

 

[HallsofIvy]

Space around a mass is curved as a result of the General Theory of Relativity, there is no force necessary.

 

[harrylin]

Thanks for the picture.

If, in that picture, gravity is a consequence of following the curved spacelines, then what is the force is pulling the spacelines down in the picture?

It is the opposite: the properties of space have an influence on the motion of everything that results in a contact force if you counter it. Newton called such a motion changing influence a force, but not everyone uses the exact same definition.

 

[robinpike]

There isn't one. That picture can be used to give you a mental model of what curved space does, but doesn't help with why the space curves.

So curved spacetime lines are a means to perform the calculation of how things behave near objects with mass, but the spacetime line concept in itself is not the explanation for gravity?

 

[harrylin]

So curved spacetime lines are a means to perform the calculation of how things behave near objects with mass, but the spacetime line concept in itself is not the explanation for gravity?

A spacetime line is a graphical sketch of the mathematics. The underlying concept is that space has properties that are influenced by nearby matter - in other words, GR is a field theory.

 

[georgir]

posts are getting deleted by database errors?

 

[pervect]

Good news, bad news...

Good news: One can understand GR as drawing space-time diagrams on the surface of some curved sheet of paper.

Bad news: The surface one really has to draw on doesn't look like anything the 250 px bmp attached earlier in the thread.

Good news: There's a paper by Marolf, http://arxiv.org/abs/gr-qc/9806123, that describes the surface you DO need to draw your space-time diagram on (an embedding) to model the r-t plane of a Schwarzschld black hole

Bad news: This paper, http://arxiv.org/abs/gr-qc/9806123, isn't terribly accessible to the layperson, even if one optimistically assumes that the reader is already familiar with drawing space-time diagrams and doing Lorentz transforms.

The important points about Marolf's shape is that Lorentz transforms work correctly in the flat tangent space on said curved surface, and furthermore that the natural path of matter free-falling is a geodesic path (the closest thing you can draw to a straight line) on said surface.

Good news: The "250 px bitmap" earlier mention in the thread does look a little bit like the purely spatial part of the space-time curvature around a massive object. Furthermore, the purely spatial part of the curvature explains a few interesting things about gravity.

Bad news: The interesting things spatial curvature does explain are second order effects - interesting, but not striking to the heart of the topic. Things like light deflection and Mercury's perihelion advance, which were crucial to the acceptance of the theory.

Good news: If you viewed space-curvature as an "add-on" to Newtonian theory, it might actually convey a lot of what happens - use Newtonian theory for the main predictions of gravity, and the spatial curvatrue part of "unexpected extra effects" that are due to the full GR theory and not expected by Newtonian theory or any simple "mash-up" arising from it.

Summary: There doesn't seem to me to be a really good way of describing GR without a significant prior background, which seems to include as a minimum understanding special relativity, the Lorentz transform, and enough about geodesics on curved surfaces to feel comfortable talking about them. Furthermore, diagrams which are commonly shown and do not presuppose this sort of background seem to be rather misleading.

 

[jtbell]

posts are getting deleted by database errors?

A person who was previously banned from here, reappeared under another name. We deleted his posts and the responses to them (which would have looked strange all by themselves).

 

[HmmTheCat]

If photons make up light (electromagnetic radiation) and radiation can be measured as energy, and energy can be measured as a mass, then why couldn't light have, at least, an extremely small mass?

Say you have an x-watt lightbulb

watts = Joules/second (power/sec)

Power is the conversion of energy (also radiant energy).

Energy is defined
- mechanically by ΔE=work
-- Work=Force X Distance
--- Force=mass X acceleration
- as E=mass X (speed of light)^2

 

[pervect]

If photons make up light (electromagnetic radiation) and radiation can be measured as energy, and energy can be measured as a mass, then why couldn't light have, at least, an extremely small mass?

Say you have an x-watt lightbulb

watts = Joules/second (power/sec)

Power is the conversion of energy (also radiant energy).

Energy is defined
- mechanically by ΔE=work
-- Work=Force X Distance
--- Force=mass X acceleration
- as E=mass X (speed of light)^2

We have a FAQ on this. Hopefully it will answer your question (just click on the link). If it doesn't, please give us some feedback so that the FAQ can be improved.

 

[rbj]

Hmm, there are some different definitions of mass that people have here. i, personally, like having the concept of relativistic mass and rest mass. the former is now a deprecated term (but i like it) and the latter is now usually called simply "mass" or sometimes "invariant mass".

anyway using the now deprecated terminology, photons with energy E=h\nu have a relativistic mass of m=E/c^2. but their rest mass must be zero if they move at the speed of c. any particle that moves at the speed of c must have an infinite energy if it has any non-zero mass at rest.

 

[Drakkith]

If photons make up light (electromagnetic radiation) and radiation can be measured as energy, and energy can be measured as a mass, then why couldn't light have, at least, an extremely small mass?

It is important to understand that "mass" refers to "invariant mass". That is, mass that doesn't change when you switch between different frames of reference. Light has energy, and will add to the mass of the SYSTEM it is in, but it itself does not have mass as in invariant mass. That said, it is always possible light is actually massive but has an extremely small amount of mass. Current measurements have shown that if light has mass it must be below 3x10^-27 eV/c². This is a VERY small number. For example, the energy of a visible light photon is around 1.5-3.0 eV.

http://en.wikipedia.org/wiki/Photon#Experimental_checks_on_photon_mass

 

[ZapperZ]

If photons make up light (electromagnetic radiation) and radiation can be measured as energy, and energy can be measured as a mass, then why couldn't light have, at least, an extremely small mass?

Say you have an x-watt lightbulb

watts = Joules/second (power/sec)

Power is the conversion of energy (also radiant energy).

Energy is defined
- mechanically by ΔE=work
-- Work=Force X Distance
--- Force=mass X acceleration
- as E=mass X (speed of light)^2

Please read the FAQ subforum

https://www.physicsforums.com/forumdisplay.php?f=210

Zz.

 

[harve]

Do you mean that light can be accelerated by gravity? Curvating motion also means acceleration.

With regards to the black hole gravity, that would probably mean that the curvature in spacetime in that locale is so great that even the speed of light cannot escape it.

That is just my take.[/QUOTE]

 

[Nugatory]

Do you mean that light can be accelerated by gravity? Curvating motion also means acceleration.

With regards to the black hole gravity, that would probably mean that the curvature in spacetime in that locale is so great that even the speed of light cannot escape it.

There's no substitute for actually doing the math, but the above is a pretty decent summary of how gravity affects light. You might want to be careful with that word "accelerated" - are you thinking that gravity can change the speed at which light moves? It doesn't, it just changes the direction of travel; this effect has actually been observed.

 

[TomTelford]

Now if I remember what I read correctly then the photon should be increasing in energy as it falls into the gravity well, if not velocity, effictively blue-shifting it off the scale. Let me know if this is incorrect.

Tom.

 

[Nugatory]

Now if I remember what I read correctly then the photon should be increasing in energy as it falls into the gravity well, if not velocity, effictively blue-shifting it off the scale. Let me know if this is incorrect.

That is correct.

 

[harrylin]

There's no substitute for actually doing the math, but the above is a pretty decent summary of how gravity affects light. You might want to be careful with that word "accelerated" - are you thinking that gravity can change the speed at which light moves? It doesn't, it just changes the direction of travel; this effect has actually been observed.

In fact it does, as expressed in "non-local" coordinates; the change of direction was first predicted as due to the gradient in speed (Huygens construction).

 

[harrylin]

Now if I remember what I read correctly then the photon should be increasing in energy as it falls into the gravity well, if not velocity, effictively blue-shifting it off the scale. Let me know if this is incorrect.

Tom.

That is only true in "local" coordinates, which do not conserve energy. As a matter of fact, the observed blueshift is ascribed to gravitational time dilation of the clocks at lower gravitational potential. There have been several discussions with detailed clarifications on that topic in this forum.

 

[harve]

tl;dr
But the idea that something needs mass to be affected by gravity is obviously false - all things that have mass are affected absolutely identically by gravity, they receive exactly the same acceleration, regardless of their mass. So even if they had zero mass, it would be normal to assume they will still be affected in the same manner and get the same acceleration.

Do you mean that light can be accelerated overcoming its constant speed,or simply follows the space curvature? But curvating motion also means acceleration.

 

[Drakkith]

Do you mean that light can be accelerated overcoming its constant speed,or simply follows the space curvature? But curvating motion also means acceleration.

The velocity of light is always c, yet it is affected by gravity and will change its direction of propagation.

 

[pervect]

One thing should be clarified. The velocity of light is always "c" using local clocks and rulers, which means that in a coordinate independent sense, it's always "c".

The rate of change of the distance coordinate with respect to the time coordinate isn't always "c". So it's important to know how you are defining velocity before you talk about it. If you define it as being measured by local clocks and rulers, then it's always constant.

It's a separate argument about why that's the best way to define velocity - I find that it's mostly a waste of time. it may be worth mentioning - errr repeating - that the issue is one of coordinate independence.