If light has no weight, how can it push objects?

In summary, light carries momentum despite having no mass. The momentum carried by a photon is given by p=E/c, where E is the energy of the photon and c is the speed of light. This means that light can exert a force and push objects, just like an electric field can push charges. The formula p=m*v is only valid for massive objects at low speeds, whereas the formula E2=(mc2)2+(cp)2 is always valid in special relativity. Therefore, even massless objects can have momentum if they have energy.
  • #36
I like Serena said:
Mass is not Lorentz invariant.
The greater the speed of an object is, relative to your frame of reference, the greater its mass is.
Rest mass, however, is Lorentz invariant and as such an intrinsic characteristic of an elementary particle.

So, what's the 'real mass' of a photon?
 
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  • #37
Dickfore said:
So, what's the 'real mass' of a photon?

I'm going with DaleSpam here.
What we have is 'rest mass' which is 0.
And 'relativistic mass' which is E/c2.
 
  • #38
I like Serena said:
I'm going with DaleSpam here.
What we have is 'rest mass' which is 0.

Exactly, and when we have 0 of something, we say we have none. Hence, photons are massless.
 
  • #39
Can somebody explain to an ignorant lay man: where is gone mass of electron and positron in their proces of so called " annihilation "? plus what about magnetic property of electron, from where it come? I will be very gratefull.
mquirce
 
  • #40
I like Serena said:
I'm going with DaleSpam here.
What we have is 'rest mass' which is 0.
And 'relativistic mass' which is E/c2.
If you are going with DaleSpam, why didn't you take note of that key phrase "which is rarely"?

Relativistic mass is just a synonym for energy. So what value does the concept add?
 
  • #41
mquirce said:
Can somebody explain to an ignorant lay man: where is gone mass of electron and positron in their proces of so called " annihilation "?
It hasn't gone anywhere. The invariant mass of the electron-positron system is equal to the invariant mass of the photon-photon system.

https://www.physicsforums.com/showpost.php?p=2745393&postcount=56

Note that the mass of a system is not equal to the sum of the masses of the consitutent particles.
 
  • #42
D H said:
If you are going with DaleSpam, why didn't you take note of that key phrase "which is rarely"?

Relativistic mass is just a synonym for energy. So what value does the concept add?

I get the feeling you feel affronted by me.
 
  • #43
Sorry for my stupidity, but i don,t understand sentence ' invariant mass of photon photon, and the meaning of summa of masses. If i bother you live with no more.
mquirce.
 
  • #44
mquirce said:
Sorry for my stupidity, but i don,t understand sentence ' invariant mass of photon photon, and the meaning of summa of masses. If i bother you live with no more.
mquirce.

Basically there are certain measurable quantities that have a defined global meaning over a manifold. There are other quantities like energy which depend on the metric being used to describe the geometry of the space - time and one cannot say that the same metric field covers the entire manifold. There can be different metrics being used to describe different parts of the manifold. Quantities that don't depend on the coordinate system being used are invariant.
 
  • #45
D H said:
If you are going with DaleSpam, why didn't you take note of that key phrase "which is rarely"?

Relativistic mass is just a synonym for energy. So what value does the concept add?

I don't really believe in what people "usually" mean, or what "all accounts" say, or "which is rarely".
If it matters what you mean, it needs to be specified if it's not clear from the context.
In non-relativistic contexts it doesn't matter what you mean when you mention "mass".
Otherwise it needs to be specified.

I guess I've been neglecting to be clear myself, using the word "mass" for "relativistic mass". My apologies for that - I'll be more careful.

What does the concept of "relativistic mass" add?
Well, for instance, you can predict the effect of a gravitational lens on photons.
Or you can predict how much photons add to dark matter mass.
Or you can say something about conservation of momentum in relativistic situations since you can't really ignore "mass-less" photons in quantum physical reactions.

If I've insulted you somehow, please enlighten me.
Otherwise accept my apologies for it was not my intention to insult anyone.
 
  • #46
I like Serena said:
What does the concept of "relativistic mass" add?
I like Serena said:
Well, for instance, you can predict the effect of a gravitational lens on photons.
Photons travel along null geodesics in cuverd spacetime (since they are massless), which we interpret as bending of their trajectory. No need for relativstic mass. In fact, if you try to describe it with relativistic mass, you will encounter trouble because Newton's Law of Universal Gravitation is not relativistically covariant.

I like Serena said:
Or you can predict how much photons add to dark matter mass.
'
Dark matter is defined as that part of the matter content responsible for the extra gravitational field of galaxies, but which does not interact via the electromagnetic interaction, hence it's dark. Photons do not satisfy this criterion.

I like Serena said:
Or you can say something about conservation of momentum in relativistic situations since you can't really ignore "mass-less" photons in quantum physical reactions.

What does momentum have to do with mass? Those are distinct physical quantities. In fact, the OP made a non-sequitur. Photons DO push objects because they carry MOMENTUM (and transfer it), not because of their mass.
 
  • #47
Dickfore said:
What does momentum have to do with mass? Those are distinct physical quantities. In fact, the OP made a non-sequitur. Photons DO push objects because they carry MOMENTUM (and transfer it), not because of their mass.
What he said.

This discussion of the intrinsic mass of a collection of photons has diverted this thread from the original question. The intrinsic mass of a single photon is identically zero. The momentum of a single photon is not zero. It is instead given by p=hf/c=h/λ where h is Planck's constant, f is the photon's frequency, and λ is its wavelength.

I like Serena, you have used E=mc2 multiple times in this thread. A much better version of this equation is

[tex]E^2 = (m_0c)^2 + (pc)^2[/tex]

where m0 is the rest mass and p is momentum. For photons, which have a rest mass of zero, this reduces to E=pc. No mention of mass whatsoever.
 
  • #48
Dickfore said:
Photons travel along null geodesics in cuverd spacetime (since they are massless), which we interpret as bending of their trajectory. No need for relativstic mass. In fact, if you try to describe it with relativistic mass, you will encounter trouble because Newton's Law of Universal Gravitation is not relativistically covariant.

Interesting.
Can you say how much Newton is "off"?

Dickfore said:
Dark matter is defined as that part of the matter content responsible for the extra gravitational field of galaxies, but which does not interact via the electromagnetic interaction, hence it's dark. Photons do not satisfy this criterion.

True enough.
So how about the contribution of photons to the known mass in the universe?

Dickfore said:
What does momentum have to do with mass? Those are distinct physical quantities. In fact, the OP made a non-sequitur. Photons DO push objects because they carry MOMENTUM (and transfer it), not because of their mass.

Doesn't the equation p = mrelativistic v = γ m0 v still hold?
 
  • #49
D H said:
A much better version of this equation is

[tex]E^2 = (m_0c)^2 + (pc)^2[/tex]

where m0 is the rest mass and p is momentum. For photons, which have a rest mass of zero, this reduces to E=pc.

I agree that this is a better version.
My point is mainly that it is a matter of perspective.
Afaik both forms are true, or in other words, it's "relative", which is what relativity theory is all about.
 
  • #50
In plain english a photon has momentum because of the energy it carries. If it loses ALL of that energy then there is no photon, therefore it is massless. Some may argue that a virtual photon remains, but it is still massless.
 
  • #51
"Light, photons, is an electromagnetic field which exerts force on charges --> Push objects."

OK, I understand this. The light is an electromagnetic field, and the electromagnetic field interacts with the electrons, and the protons of the material, pushing them back and forth. But why would that push the solar sail away from the light source? The electrons (and protons) are going to vibrate perpendicularly to the direction of the ray of light that shines on the sail, right?, because the magnetic and electric fields of the light ray are perpendicular to the direction the light ray is traveling. The 'pressure' on the electron would be back and forth perpendicularly to the direction the light ray is traveling. I see that causing heat, warming up the sail. But, how does that cause the sail to move away from the light source?
 
  • #52
Jarfi said:
And also when I shine a light on paper and it heats the paper(gives it momentum?) does it loose energy and change wavelength?




Does heating up the paper cause the photon to change wavelength, or to change amplitude?
 
  • #53
dlr said:
"Light, photons, is an electromagnetic field which exerts force on charges --> Push objects."

OK, I understand this. The light is an electromagnetic field, and the electromagnetic field interacts with the electrons, and the protons of the material, pushing them back and forth. But why would that push the solar sail away from the light source? The electrons (and protons) are going to vibrate perpendicularly to the direction of the ray of light that shines on the sail, right?, because the magnetic and electric fields of the light ray are perpendicular to the direction the light ray is traveling. The 'pressure' on the electron would be back and forth perpendicularly to the direction the light ray is traveling. I see that causing heat, warming up the sail. But, how does that cause the sail to move away from the light source?

You are ignoring that while photons are massless particles, they have non-zero momentum.

When a photon interacts with a surface, one of three things will happen.
  • Specular reflection.
    The photon (or another one just like the incoming photon) will bounce off the surface as if the surface were a mirror. "Angle of incidence = angle of reflection". The imparted momentum will be twice that of the photon if the photon hits the surface squarely, less than that for the incidence angle less than 90 degrees.

  • Diffuse reflection.
    If the surface is microscopically rough, the reflection can be diffuse. Modeling diffuse reflection is a bit ad hoc. One widely used scheme is that the incoming photons impart all their momentum to the object, and the outgoing photons are distributed uniformly over the hemisphere pointing away from the object.

  • Absorption / thermal emission.
    Some photons are absorbed rather than reflected. The absorbed photons impart all of their momentum to the object. This heats the object. The heated object will emit thermal photons based on the local temperature. If the object is at a uniform temperature, the momentum transfer due to this thermal emission averages out to zero. If the object is not at a uniform temperature there will be a non-zero force on the object pointing roughly in the direction of the coolest point on the object. The hotter side emits more photons, and more energetic photons, than does the cooler side.
 
  • #54
D H said:
You are ignoring that while photons are massless particles, they have non-zero momentum.

When a photon interacts with a surface, one of three things will happen.
  • Specular reflection.
    The photon (or another one just like the incoming photon) will bounce off the surface as if the surface were a mirror. "Angle of incidence = angle of reflection". The imparted momentum will be twice that of the photon if the photon hits the surface squarely, less than that for the incidence angle less than 90 degrees.

  • Diffuse reflection.
    If the surface is microscopically rough, the reflection can be diffuse. Modeling diffuse reflection is a bit ad hoc. One widely used scheme is that the incoming photons impart all their momentum to the object, and the outgoing photons are distributed uniformly over the hemisphere pointing away from the object.

  • Absorption / thermal emission.
    Some photons are absorbed rather than reflected. The absorbed photons impart all of their momentum to the object. This heats the object. The heated object will emit thermal photons based on the local temperature. If the object is at a uniform temperature, the momentum transfer due to this thermal emission averages out to zero. If the object is not at a uniform temperature there will be a non-zero force on the object pointing roughly in the direction of the coolest point on the object. The hotter side emits more photons, and more energetic photons, than does the cooler side.


But what is actually happening, physically, at the molecular level?

1) Specular Reflection When the photon "bounces off the surface" that means the photon was absorbed by an electron, and then emitted again, in some random direction, right? Same wavelength, so same energy, so how could any momentum (energy) have been imparted to the sail?

2) Thermal What I don't see is how heating translates to linear motion at the molecular level, with individual electrons, protons and photons. The energetic photon comes in and it's electromagnetic field causes all of the electrons (and protons?) it passes to begin vibrating -- but back and forth, not in any continuous direction. The electrons are accelerating/decelerating as they bounce back and forth, so they are giving off photons. But how does that process of giving off photons cause the molecule move? And why consistently in one direction, instead of randomly? I mean, what is going on physically, at the level of the individual electron or molecule?
 
  • #55
dlr said:
But what is actually happening, physically, at the molecular level?
Don't worry about it for now. Worrying about it at this point is hindering your understanding.

1) Specular Reflection When the photon "bounces off the surface" that means the photon was absorbed by an electron, and then emitted again, in some random direction, right? Same wavelength, so same energy, so how could any momentum (energy) have been imparted to the sail?
No! Specular reflection is "mirror-like". You are describe diffuse reflection.

Specular reflection:

Imagine you are the catcher in a baseball warmup. If the pitcher throws you a ball you throw it right back to the pitcher. If the first baseman throws you a ball you throw it to third. If the shortstop throws the ball you throw it back to the second baseman (who is between 1st and 2nd base). Wherever the ball comes from, you throw it at just that angle, but reversed. Just like a mirror.

The act of catching the ball imparts a certain amount of momentum to you. So does the act of throwing the ball. Regardless of where the ball came from, the net momentum imparted to you is always directly away from the pitcher.

Suppose that instead of balls coming from a random direction, the balls always come from third base. This is a solar sail: The vast, vast majority of the photons hitting the solar sail come from the Sun. An ideal solar sail is 100% specular. The net momentum transferred to the sail will be against the outward normal to the sail.


Diffuse reflection:

Again with the catcher analogy: Now when the third baseman throws the ball, you might throw the ball to first, or to third, or maybe even the manager who is lounging in the dugout. The direction is random, but it is always somewhere in front of you. Catching the ball imparts a certain momentum to you, as does throwing it. Since your throw is in a random direction in front of you, over the long haul the momentum transfer from your throws will be aray from the pitcher. If the balls are coming from one particular direction (e.g., the Sun), you will also get a momentum transfer from the incoming balls. This is a non-ideal solar sail. Most of the incoming photons are reflected specularly, some are reflected diffusely, and some are absorbed.

2) Thermal What I don't see is how heating translates to linear motion at the molecular level, with individual electrons, protons and photons. The energetic photon comes in and it's electromagnetic field causes all of the electrons (and protons?) it passes to begin vibrating -- but back and forth, not in any continuous direction. The electrons are accelerating/decelerating as they bounce back and forth, so they are giving off photons. But how does that process of giving off photons cause the molecule move? And why consistently in one direction, instead of randomly? I mean, what is going on physically, at the level of the individual electron or molecule?
Suppose the object in question is a perfect absorber of photons. It is so very hungry for photons that it doesn't even emit any! That isn't possible of course; the intent here is to separate absorption and emission into two distinct processes. This perfect absorber will feel a thrust directly against the Sun because these gobbled-up photons transfer momentum to the object.

Now let's look at emission only. Suppose the object is in deep space, far from any star, and has some heat source inside it so it doesn't cool off as it is emitting photons. Each emitted photon will impart a thrust on the object. If the object has a uniform temperature, this imparted thrust will average out to nothing because it is radiating uniformly. That is no longer true if the heat source is very close to one surface. Now one side will be much hotter than the opposite side. The radiation will no longer be uniform. The object will experience a thrust directed away the hot side of the object.

Now let's put absorption and emission together. Suppose the object is keeping one side always facing the Sun. This sunlit side will be warmer than the perpetually shaded side, the there will be thrust away from the Sun due to this non-uniform temperature. The object is absorbing photons, so this also generates a thrust away from the Sun.
 
  • #56
D H said:
Don't worry about it for now. Worrying about it at this point is hindering your understanding.

No, I understand what is happening at the macroscopic level. What I don't understand is what is physically happening at the molecular level.

Specular reflection: If Specular Reflection doesn't involve an electron absorbing and emitting a photon, how does it actually work at the molecular level? ie, how, physically, , is the momentum transferred from the photon to the electron?

Is the electromagnetic field of the photon being repelled by a charged particle in the material (ie, an electron)? ie, is the bounce the same bounce you get by shoving a charged particle into an electromagnetic field?

Physically that would mean the photon's magnetic field would repel and be repelled by the first charged particle it meets (which would undoubtably be an electron). The electron moves a little bit when it is repelled by the photon. The movement of the electron generates a little magnetic pulse of it's own, which would repel nearby electrons(?) and attract nearby protons(?), which would all move a little bit and thus generate little emf fields of their own, and do the same, in decreasing amplitudes all through the material...

Is that the actual underlying mechanism?









By the way, thanks for answering.
 
  • #57
IMO specular reflection simply does not happen at the molecular level. In order to get specular reflection you need structures which are significantly larger than a wavelength and are smooth over that scale. That simply doesn't happen at the molecular level, so a specular reflection implies an interaction of a photon with a continuum of molecules. If you understand how specular reflection happens at the macroscopic level then you understand how it happens.

If you disagree then try to think of any possible way to determine which specific molecule in a mirror a given photon actually reflected off.
 
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  • #58
dlr said:
No, I understand what is happening at the macroscopic level. What I don't understand is what is physically happening at the molecular level.

Specular reflection: If Specular Reflection doesn't involve an electron absorbing and emitting a photon, how does it actually work at the molecular level? ie, how, physically, , is the momentum transferred from the photon to the electron?

Is the electromagnetic field of the photon being repelled by a charged particle in the material (ie, an electron)? ie, is the bounce the same bounce you get by shoving a charged particle into an electromagnetic field?

Physically that would mean the photon's magnetic field would repel and be repelled by the first charged particle it meets (which would undoubtably be an electron). The electron moves a little bit when it is repelled by the photon. The movement of the electron generates a little magnetic pulse of it's own, which would repel nearby electrons(?) and attract nearby protons(?), which would all move a little bit and thus generate little emf fields of their own, and do the same, in decreasing amplitudes all through the material...

Is that the actual underlying mechanism?

By the way, thanks for answering.

Optical conductivity/reflectivity is not as trivial as most of us think.

There is no one single mechanism that can explain everything in the case of reflection. You can get Bragg-type of reflection, which has more to do with the spacing of lattice points in a solid. You can also have lattice vibration (phonons) being involved in optical reflectivity. In this case, it is why optical reflectivity is used to probe phonon modes in experiments such as FTIR, etc. Even as you change the wavelength, the mechanism that causes the same thing (reflection) could change. And certainly, the reflection of visible light off a mirror (i.e. metallic surface) is different than the reflection of light off a wooden desk.

Note that this is no longer a topic in "Relativity".

Zz.
 
  • #59
Can anyone link to a write-up of an experiment that proves absorbtion of photons results in a momentum exchange to the object absorbing the photons?

Like dlr, post #54, I would like to get to the bottom of what mechanisms are actually happening. For example, is the following a reasonable or unreasonable explanation, complete or incomplete?; photons mediate electromagnetic forces, so if we imagine an emitter of an electromagnetic force causing electromagnetic effects on a distant object that pushes it away (viz. magnetic forces on the electrons push the material away, so the macroscopic effect of momentum transfer is actually a distant electromagnetic force) then the photon can be seen as the mediation between the equal and opposite reactions of an emitter and receiver of EM radiation.
 
  • #60
cmb said:
Can anyone link to a write-up of an experiment that proves absorbtion of photons results in a momentum exchange to the object absorbing the photons?
Compton scattering experiments, for one. Arthur Compton won the Nobel Prize for his discovery of the Compton effect. Satellites in geosynchronous orbit, for another. If you want a high precision orbit determination you had better model the solar radiation pressure on the satellite. 6489 Golevka, for yet another. This asteroid drifted more than ten km from its predicted position over the span of a decade or so. Chesley et al., "Direct Detection of the Yarkovsky Effect via Radar Ranging to Asteroid 6489 Golevka", Science 302 (5651): 1739–1742. Finally, the Japanese space agency (JAXA)'s solar sail experiment IKAROS. IKAROS is the first successful solar sail experiment.

That photons have momentum is right there in the relativistic energy equation,

[tex]E^2 = (pc)^2 + (m_0c^2)^2[/tex]

Like dlr, post #54, I would like to get to the bottom of what mechanisms are actually happening.
You are asking us to write a book. Unless you are a junior in an undergrad physics program taking the introductory condensed matter course, any quantum explanation of reflection is going to be hand-waving at best. I gave the hand-waving, semiclassical description above.
 
  • #61
cmb said:
Can anyone link to a write-up of an experiment that proves absorbtion of photons results in a momentum exchange to the object absorbing the photons?

D.S. Weiss et al., "Precision measurement of the photon recoil of an atom using atomic interferometry" Phys. Rev. Lett. 70, 2706 (1993).

G.K. Campbell et al., "Photon Recoil Momentum in Dispersive Media" Phys. Rev. Lett. 94, 170403 (2005).

Zz.
 
  • #62
D H said:
You are asking us to write a book. Unless you are a junior in an undergrad physics program taking the introductory condensed matter course, any quantum explanation of reflection is going to be hand-waving at best. I gave the hand-waving, semiclassical description above.

Ultimately it must be an electric-magnetic interaction. One might therefore interpret this as some sort of magnetic repulsion of two distant bodies, for which the photon is 'only' a mediation.
 
  • #63
cmb said:
Ultimately it must be an electric-magnetic interaction. One might therefore interpret this as some sort of magnetic repulsion of two distant bodies, for which the photon is 'only' a mediation.
Ultimately, it is quantum electrodynamics that explains the interaction of photons and matter, and that is way, way beyond the scope of this thread.
 
  • #64
DaleSpam said:
In order to get specular reflection you need structures which are significantly larger than a wavelength and are smooth over that scale. That simply doesn't happen at the molecular level, so a specular reflection implies an interaction of a photon with a continuum of molecules.



OK, but am I'm right in saying the underlying physical mechanism is the interaction of an electromagnetic field with charged particle(s) (ie, electrons)?

And, if I understand you correctly, you are objecting that the magnetic field of the photon is so large compared to the size of electrons and molecules that the interaction is between a photon's magnetic field and 1000's or 10's of thousands of electrons -- a sea of electrons (metals) or a lattice of electrons (a crystal) or a sea of molecules (water), instead of with just one electron. Which makes sense.

And then what happens next, after the charged particles get the magnetic "push" from the photon's magnetic field, they move of course, which gives a little push to the protons/neutrons of all the nearby molecules? This effect ultimately being caused by the effect that keeps electrons from getting too close to a proton/neutron?
 
  • #65
dlr said:
the magnetic field of the photon

What is this? I don't even
 
  • #66
It's easier to see from a classical em viewpoint. You have a positive charge sitting stationary in space. A step function em wave approaches from the left with the electric field always pointing down your computer screen and magnetic field always pointing into the screen. The charge accelerates downward due to the E field, so its velocity increases. That downward velocity interacts with the em wave's magnetic field via v X B, resulting in a force to the right, which accelerates the charge to the right. Voila, the em wave has imparted downstream velocity/momentum to the charge.

One interesting special case is when the charge is initially traveling with or against the em wave. In that case, v X B becomes -v/c * E and the resulting total initial force on the charge is (1 - v/c)*E. Looks a lot like a "Doppler Shift" of the electric field strength, eh? You'll note in the extreme case of v = c (i.e. when the charge is moving with the wave at the speed of light) there is no net force on it.

That's what classical em says ... not that I believe it (or QM). I don't have any better answer so I basically keep my mouth shut!
 
  • #67
fizzle said:
The charge accelerates downward due to the E field, so its velocity increases. That downward velocity interacts with the em wave's magnetic field via v X B, resulting in a force to the right, which accelerates the charge to the right. Voila, the em wave has imparted downstream velocity/momentum to the charge.


Yes, the force is perpendicular to the direction of travel of the light ray. So then, why does the solar sail move directly away from the light rays hitting it?
 
  • #68
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  • #69
dlr said:
Yes, the force is perpendicular to the direction of travel of the light ray. So then, why does the solar sail move directly away from the light rays hitting it?
I was showing above that a simple, constant step wave transfers downstream momentum to a charge. For light, which is a circularly polarized (CP) em wave, you don't get the down-the-screen motion because the E field rotates; so the charge rotates in a circle while being accelerated downstream by v x B. The overall path is a helix in the direction of the wave.

BTW, if you do a full relativistic/classical em analysis of a CP wave interacting with a charge you can derive the Compton Scattering equations - as Compton noted in his original paper. The problem is that the transfer a "photon" of energy to the charge requires an unrealistically high E field (10^16 V/m or more) in the CP wave, sort of a Compton catastrophe. This is the root cause of QM and its associated weirdness (e.g. a photon is a particle+wave, wave function collapse, zero-point energy with its enormous but "hidden" fields, etc.). No one was able to figure out how a realistic CP em wave could transfer so much energy to a single charge.

"speculation deleted"

Integral
 
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