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


by Jarfi
Tags: light, mass, momentum, wtf
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#37
Mar31-11, 12:07 PM
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Quote Quote by Dickfore View Post
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.
Dickfore
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#38
Mar31-11, 12:09 PM
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Quote Quote by I like Serena View Post
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.
mquirce
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Mar31-11, 12:30 PM
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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
D H
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Mar31-11, 12:43 PM
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Quote Quote by I like Serena View Post
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?
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Mar31-11, 12:46 PM
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Quote Quote by mquirce View Post
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.

http://www.physicsforums.com/showpos...3&postcount=56

Note that the mass of a system is not equal to the sum of the masses of the consitutent particles.
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Mar31-11, 01:50 PM
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Quote Quote by D H View Post
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.
mquirce
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Mar31-11, 01:58 PM
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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.
WannabeNewton
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Mar31-11, 05:07 PM
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Quote Quote by mquirce View Post
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.
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Apr1-11, 04:38 PM
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Quote Quote by D H View Post
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.
Dickfore
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Apr1-11, 05:46 PM
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Quote Quote by I like Serena View Post
What does the concept of "relativistic mass" add?
Quote Quote by I like Serena View Post
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.

Quote Quote by I like Serena View Post
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.

Quote Quote by I like Serena View Post
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.
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Apr1-11, 06:12 PM
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Quote Quote by Dickfore View Post
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.
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Apr1-11, 06:39 PM
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Quote Quote by Dickfore View Post
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"?

Quote Quote by Dickfore View Post
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?

Quote Quote by Dickfore View Post
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?
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Apr1-11, 06:45 PM
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Quote Quote by D H View Post
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.
pawprint
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#50
Jun22-11, 04:41 AM
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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.
dlr
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#51
Sep21-11, 02:24 AM
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"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?
dlr
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#52
Sep21-11, 02:27 AM
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Quote Quote by Jarfi View Post

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?
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#53
Sep21-11, 03:38 AM
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Quote Quote by dlr View Post
"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.
dlr
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Sep21-11, 06:17 PM
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Quote Quote by D H View Post
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?


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