[Courier of darkness (OP)]

Consider the electron-positron annihilation


If the positron has a negative mass -m, and the electron has a positive mass +m, their sum is
-m+m=0

Can we explain why the photon mass is zero (its rest-mass is 0)?
Is it 0 because the photon is the particle into which the pair positron-electron is transformed in the annihilation?

[PmbPhy]

If the positron has a negative mass -m, and the electron has a positive mass +m, their sum is
-m+m=0

First off, welcome to the forum!

Next: That's a really big if. What gave you the impression that such a thing is possible?

Can we explain why the photon mass is zero (its rest-mass is 0)?

I don't follow. The diagram you showed is a Feynman diagram but it is not the diagram of an electron annihilating a positron because you have an electron and positron coming out of the reaction. In between where you have the single photon that's a virtual photon and virtual photons have a mass which is off-the-shell of a real photon's energy-momentum curve. That means that the proper mass (i.e. rest mass) of the photon can be non-zero. Virtual photons can't be observed.

Is it 0 because the photon is the particle into which the positron-electron pair is transformed in the annihilation?

I don't see that from the actual diagram and its correct interpretation.

[Courier of darkness (OP)]

First off, welcome to the forum!

Thanks.

Next: That's a really big if. What gave you the impression that such a thing is possible?

Are you asking why I think that whether mass can be negative?
I told that negative mass can explain why the photon mass is 0, or rather why its rest-mass is 0.

Can we explain why the photon mass is zero (its rest-mass is 0)?

I don't follow. The diagram you showed is a Feynman diagram but it is not the diagram of an electron annihilating a positron because you have an electron and positron coming out of the reaction. In between where you have the single photon that's a virtual photon and virtual photons have a mass which is off-the-shell of a real photon's energy-momentum curve. That means that the proper mass (i.e. rest mass) of the photon can be non-zero. Virtual photons can't be observed.

Yes it is true that my picture is a Feynman diagram.
It is true that Feyman diagrams always repesent an action at distance, or an interaction between two particles.

Feynman diagrams are not supposed to represent a head-on collision of two particles.

The problem is, my picture represents a head-on collision of a positron and an electron.

In other words, a Feynman diagram can represent a collision of two particles, although it is not generally admitted or understood. My picture should not be understood as representing an interaction between two particles transmitted with a virtual photon. Because if that were true, what are the particles interacting with each others?

Is it 0 because the photon is the particle into which the positron-electron pair is transformed in the annihilation?

I don't see that from the actual diagram and its correct interpretation.

A pair creation happens when an electron and a positron are created out of the photon.
If there was during the "annihilation" a transformation of the pair positron-electron into a photon,
then the correct interpretation of the diagram is that there did not happen any actual annihilation.

Instead there was a transformation of the pair positron-electron into a photon.

Therefore the photon can transform back into the pair positron-electron during the pair creation.

[UltimateTheory]

First of all,
total energy of particles prior annihilation is equal to total energy of particles after annihilation,
so,
2*me*c^2 = total energy of two-three rarely more photons

where
me=0.510999 MeV/c^2
y=0.510999 MeV

so
2*0.510999 MeV (electron-positron) = 2*0.510999 MeV (2 photons)
or
2*0.510999 MeV (electron-positron) = 3*0.340 MeV (3 photons)

If electron-positron have kinetic energy (particles not at rest before annihilation), photons after annihilation will have also more energy accordingly.



If positron would have me=-0.510999 MeV
Then equation would looks like:
-0.510999 MeV +0.510999 MeV = 0
so photons would have to have 0 total energy (!),
and all would be screwed.

[PmbPhy]

Are you asking why I think that whether mass can be negative?

No. It's clear to me that you think that it can be negative. I'm asking you exactly what I posted, i.e. What gave you the impression that such a thing is possible? I.e. we already know that you believe that it's true. What we don't know is why you believe its true.

I told that negative mass can explain why the photon mass is 0, or rather why its rest-mass is 0.

But, as I explained above, your diagram and your argument don't support your assertion.

Can we explain why the photon mass is zero (its rest-mass is 0)?

If it wasn't then photons wouldn't be able to travel at the speed of light. That's the explanation.

It is true that Feyman diagrams always repesent an action at distance, or an interaction between two particles.

There's no such thing as action at a distance. Charged particles create electric fields. When another charged particle is placed in that field it exchanges virtual photons with the original charge. However it takes time for the virtual photons to move. But this can't be seen as action at a distance. See: http://math.ucr.edu/home/baez/physics/Quantum/virtual_particles.html

Forces don't happen because of any sort of action at a distance, they happen because of virtual particles that spew out of things and hit other things, knocking them around.  However, this is misleading.  Virtual particles are really not just like classical bullets.

Feynman diagrams are not supposed to represent a head-on collision of two particles.

Yes. I'm well aware of that, thank you. Most people here or in any physics forum are in fact. They represent an interaction between two charges regardless of their motion.

The problem is, my picture represents a head-on collision of a positron and an electron.

Not to someone who understands Feynman diagrams.

In other words, a Feynman diagram can represent a collision of two particles, although it is not generally admitted or understood.

I wouldn't make such an assumption if I were you. People who know what Feynman diagrams are probably know what they mean and what the representation means too.

My picture should not be understood as representing an interaction between two particles transmitted with a virtual photon.

That's incorrect. That's exactly what your diagram means.

Because if that were true, what are the particles interacting with each others?

I don't understand that question. It was poorly stated. I.e. what does "..., what are the particles interacting with each others?"
 mean? I.e. "interacting with each others" I have no clue what that's supposed to mean and I doubt anybody else does. I think you must have made a mistake and not have seen it.

A pair creation happens when an electron and a positron are created out of the photon.

While I know what you meant, it may not be clear to everyone else. You should have made it clear that it's a virtual photon because pair creation cannot happen with a single real photon. It has to happen with a virtual photon. That's because there's no frame of reference in which a real photon's momentum can be zero while there always is with an electron/positron pair.

If there was during the "annihilation" a transformation of the pair positron-electron into a photon,..

No. The transformation was into a virtual photon because a virtual photon has non-zero rest mass and it can't be observed and only last an incredibly short period of time.

Why are you bothering with this anyway? Is it because you thought that when the electron and positron annihilated each other than the end result was zero rest mass? If that's what you thought then you were mistaken because the result wasn't a photon, which has zero rest mass, it was a virtual photon, which doesn't have zero rest mass.

then the correct interpretation of the diagram is that there did not happen any actual annihilation.

Yes. Correct.

Instead there was a transformation of the pair positron-electron into a photon.

No. There was a momentary transformation of the two particles into a virtual photon.

[alancalverd]

The misunderstanding starts at the beginning. The e-p annihilation produces two photons, not one. Symmetry and conservation of momentum results in both having an energy of 511 keV, so neither can initiate further pair production.

The threshold for pair production is (fairly) easily measured as 1.022 MeV (you can use it to calibrate a VandeGraaf accelerator) which neatly translates as 2 x m_e according to Prof Einstein.

No negative mass is required.

[PmbPhy]

The misunderstanding starts at the beginning. The e-p annihilation produces two photons, not one.

His mistake was that he thought he was talking about electron/positron annihilation, probably because that's what the Feynman diagram looked like to him. The Feynman diagram he showed was merely an electron interacting with a positron but not annihilating. Such a thing can happen in positronium.

[ScientificSorcerer]

It is unclear to me what you mean by negative mass.  Do you mean mass which has a negative charge or anti mass I think your mixing the 2 together which is why your question dose not make sense.

[PmbPhy]

It is unclear to me what you mean by negative mass.  Do you mean mass which has a negative charge or anti mass I think your mixing the 2 together which is why your question dose not make sense.

Nope. He means exactly what he said, i.e. negative mass. There are several types. The one he's talking about is negative inertial mass which is defined as p = mv where m < 0 and v is the 3-velocity and p is the 3-momentum. Herman Bondi addressed these concepts in an article in the journal Reviews of Modern Physics. The article is

Negative Mass in General Relativity, Herman Bondi, Rev. Mod. Phys., 29(3), July 1967
I uploaded it onto my companies website for you to read. See
http://www.newenglandphysics.org/Science_Literature/Journal_Articles/Bondi.pdf

[JohnDuffield]

Courier of darkness: the mass of a body is a measure of its energy-content. That's what Einstein said, see his E=mc² paper. He also said a radiating body loses mass. That's what happens in electron-positron annihilation. Two radiating bodies lose mass, all of it, and then they're not there any more. They can't lose more mass or energy than what they've got. There are no bodies with negative mass. This light is heavy article is worth a read. It's about light in a mirror-box. Open the box, and it's a radiating body that loses mass. The electron is a bit like light in a box, minus the box. NB: the 't Hooft isn't the Nobel 't Hooft, it's some other guy.

[Courier of darkness (OP)]

Are you asking why I think that whether mass can be negative?

No. It's clear to me that you think that it can be negative. I'm asking you exactly what I posted, i.e. What gave you the impression that such a thing is possible? I.e. we already know that you believe that it's true. What we don't know is why you believe its true.

I already said why: negative mass explains why the photon has zero rest-mass.

But, as I explained above, your diagram and your argument don't support your assertion.

If what I say is properly understood, then my diagram supports my assertion.

Can we explain why the photon mass is zero (its rest-mass is 0)?

If it wasn't then photons wouldn't be able to travel at the speed of light. That's the explanation.

But only the rest mass m₀ of photons is zero. What can explain how photons can travel at the speed of light if their mass m is non-zero?
The relativistic mass formula is m= m₀/
Insert m₀ = 0 and v=c
It leads to m=0/0
You can't use the relativistic mass formula to explain the mass of photons. Anyway photons should have mass because they have energy.

There's no such thing as action at a distance. Charged particles create electric fields. When another charged particle is placed in that field it exchanges virtual photons with the original charge. However it takes time for the virtual photons to move. But this can't be seen as action at a distance. See: http://math.ucr.edu/home/baez/physics/Quantum/virtual_particles.html [nofollow]

There is such a thing as action at distance. Look at
https://www.hep.ucl.ac.uk/undergrad-projects/3rdyear/photons-at-HERA/guage.htm [nofollow]

The fact that gauge bosons are invisible to us, and that we can only observe the effects they produce rather then the particles themselves, has led to the phrase "action at a distance" to describe the way in which they work

My picture should not be understood as representing an interaction between two particles transmitted with a virtual photon.

That's incorrect. That's exactly what your diagram means.

Because if that were true, what are the particles interacting with each others?

I don't understand that question. It was poorly stated. I.e. what does "..., what are the particles interacting with each others?"
 mean? I.e. "interacting with each others" I have no clue what that's supposed to mean and I doubt anybody else does. I think you must have made a mistake and not have seen it.

I wonder why did you not understand what I said. I try to explain again:

If my first picture represents two particles which interact with each others by exchanging a virtual photon
between them, then what are these two particles?

Let me show you another picture:


The above picture represents two electrons which interact with each others by exchanging a virtual photon
between them. So if I asked the same question now, it would be: What are the two particles interacting
with each others by exchanging a virtual photon between them? And the answer is in this case: two electrons.

Why are you bothering with this anyway? Is it because you thought that when the electron and positron annihilated each other than the end result was zero rest mass? If that's what you thought then you were mistaken because the result wasn't a photon, which has zero rest mass, it was a virtual photon, which doesn't have zero rest mass.

As I said, there happens no annihilation. What happens is a transformation leading into a formation
of a photon with rest-mass m₀ = 0

[JohnDuffield]

Anyway photons should have mass because they have energy.

They do have a non-zero inertial mass and a non-zero active gravitational mass. But nowadays when people say mass they usually mean rest mass. A photon doesn't have any rest mass because it isn't at rest, it's always travelling at c. However when it's bouncing back and forth in the mirror-box, it's effectively at rest, and it adds to the mass of the system. If you've heard of things like electron diffraction and magnet moment and the Einstein-de Haas effect and electron spin and atomic orbitals where electrons "exist as standing waves", you can hopefully think of photon momentum as resistance to change-in-motion for a wave going along at c, and electron mass as resistance to change-in-motion for a wave going round and round at c. Ditto for positron mass. It doesn't have a negative mass, it has a positive mass.

[Ethos_]

 Ditto for positron mass. It doesn't have a negative mass, it has a positive mass.

Agreed John.....................the positive and negative signs here only relate to charge and not to the mass of the particle. Should be very logical and simple to understand that charge and mass are separate and singular attributes of these particles.

[lightarrow]

I already said why: negative mass explains why the photon has zero rest-mass.

But that is incorrect, because that photon has not zero mass. PmbPhy has already written it to you: that photon is a virtual photon WITH mass.
(a real photon have zero mass; but unfortunately that diagram does not represent any real photon).

--
lightarrow

[evan_au]

If the positron has a negative mass -m...

...then you would expect antimatter to fly upwards if you released it (and I mean release it in a vacuum - ordinary Hydrogen manages to fly upwards in air!).

Measuring the acceleration of antimatter in the Earth's gravitational field is an experiment that has been attempted several times before, but with very large error bars. At least one team is hoping to get a fairly accurate measurement with anti-Hydrogen sometime this year.

They expect that if you dropped an atom of Hydrogen and an atom of anti-Hydrogen simultaneously, they will hit the bottom at the same time.
But the beauty of  the experiment is that you don't really know until you try it!

[JohnDuffield]

Agreed John.....................the positive and negative signs here only relate to charge and not to the mass of the particle. Should be very logical and simple to understand that charge and mass are separate and singular attributes of these particles.

And moreover, positive and negative charge is merely a convention. If you google on positron chirality you appreciate that the positron has the opposite chirality or "handedness" to the electron, and that this is related to its opposite charge. Now look at your hands. They have the opposite handedness. But your left hand isn't actually some negative version of your right hand.

They expect that if you dropped an atom of Hydrogen and an atom of anti-Hydrogen simultaneously, they will hit the bottom at the same time.

They know it. General relativity is one of the best-tested theories we've got, and its energy that interacts gravitationally, not just matter. It doesn't matter what form it takes. Light curves down, we make electrons and positrons out of light in pair production, they will both fall down. I'm afraid does antimatter fall up? is just a soundbite to attract the attention of the popscience media.

[PhysBang]

Agreed John.....................the positive and negative signs here only relate to charge and not to the mass of the particle. Should be very logical and simple to understand that charge and mass are separate and singular attributes of these particles.

And moreover, positive and negative charge is merely a convention. If you google on positron chirality you appreciate that the positron has the opposite chirality or "handedness" to the electron, and that this is related to its opposite charge. Now look at your hands. They have the opposite handedness. But your left hand isn't actually some negative version of your right hand.

However, charge is some negative version, in the mathematical sense of negative.

Physics is about the actual measurements done and how well theory is able to match those measurements. Those who forgo the mathematics of physics forgo physics.

They expect that if you dropped an atom of Hydrogen and an atom of anti-Hydrogen simultaneously, they will hit the bottom at the same time.

They know it. General relativity is one of the best-tested theories we've got, and its energy that interacts gravitationally, not just matter. It doesn't matter what form it takes. Light curves down, we make electrons and positrons out of light in pair production, they will both fall down. I'm afraid does antimatter fall up? is just a soundbite to attract the attention of the popscience media.

The equivalence principle is what guarantees that particles of opposite charge fall at the same rate. It has been tested as directly as possible to a very high accuracy and it is thus a part of a great many gravitational theories, not merely GR. It is, of course, good science to ensure that it applies to antimatter when this can be tested rather than merely assuming that it applies and never doing the possible test.

[ScientificSorcerer]

Answer me this.

Lets say there is a particle with normal mass, it has some kinetic energy and is moving through space.  Like any thing with mass, this particle has a tiny gravitational field.  If this hypothetical particle's gravitational field gets weaker and weaker over time then it's mass will diminish over time.

Just because the particle is loosing it's gravitational field (Mass) does not mean that it's kinetic energy is decreasing (it would still remain the same) because of this scenario, the equation E=mc² comes into play. The energy E of the particle remains the same and M is decreasing so the velocity of the particle therefore must be increasing until the mass of the particle gets to zero (at which point the particle would be going the speed of light)

If the mass goes into a negative number then any particle of negative mass would be going faster then the speed of light (perhaps backward in time) if it had any energy in it. BUT could such a particle have Negative energy?  If so then it might as well have negative velocity meaning that the particle would be going backward in time. Right?

[Bill S]

If the mass goes into a negative number then any particle of negative mass would be going faster then the speed of light (perhaps backward in time) if it had any energy in it. BUT could such a particle have Negative energy?  If so then it might as well have negative velocity meaning that the particle would be going backward in time. Right?

SC, you seem to be introducing the idea of the tachyon, here.

 
I have still not mastered the art of inserting diagrams, so here’s a description of what this one should look like.

Start with a rectangle. Draw a horizontal line through the middle, and mark this “c”.
Now draw some arrows, as follows:
Above the rectangle; pointing left, marked “Time: Tachyon’s view”.
Below the rectangle ; pointing right, marked “Time; Photon’s view”.     
Within the rectangle;
Left side, pointing up; marked “Increasing speed of photon”.
Right side; pointing down; marked “ Increasing speed of tachyon”. 
Above centre; pointing left; marked “Tachyon”.
Below centre; pointing right; marked “Photon”.

I seem to be suggesting that a photon accelerates, but this should not be interpreted as making any claim other than that, in different media the photon may appear to travel at less than “c”.

There seems to be a general assumption, at least in Pop. Sci. that tachyons accelerate. John Gribbin says: “So if a tachyon were created in some violent event in space, it would radiate energy away furiously…..and go faster and faster, until it had zero energy ……and was travelling at infinite speed”. 

However, if the above “thought diagram” is inverted, there is a marked similarity between photons and tachyons.

Do we read too much into the idea of the tachyon?

[Courier of darkness (OP)]

Lets say there is a particle with normal mass, it has some kinetic energy and is moving through space.  Like any thing with mass, this particle has a tiny gravitational field.  If this hypothetical particle's gravitational field gets weaker and weaker over time then it's mass will diminish over time.

Just because the particle is loosing it's gravitational field (Mass) does not mean that it's kinetic energy is decreasing (it would still remain the same) because of this scenario, the equation E=mc² comes into play. The energy E of the particle remains the same and M is decreasing so the velocity of the particle therefore must be increasing until the mass of the particle gets to zero (at which point the particle would be going the speed of light)

If the mass goes into a negative number then any particle of negative mass would be going faster then the speed of light (perhaps backward in time) if it had any energy in it. BUT could such a particle have Negative energy?  If so then it might as well have negative velocity meaning that the particle would be going backward in time. Right?

You are right.

Richard Feyman said that a positron is an electron going backwards in time.

Anti-matter can have Negative energy, meaning that its mass can be negative.