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?

Quote from: Courier of darkness

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.

Quote from: Courier of darkness

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.

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?

Quote from: Courier of darkness

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.

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.

Quote from: Courier of darkness

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

Quote from: Courier of darkness

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

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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.

Quote from: Courier of darkness

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.

Quote from: Courier of darkness

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

Not to someone who understands Feynman diagrams.

Quote from: Courier of darkness

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.

Quote from: Courier of darkness

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.

Quote from: Courier of darkness

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.

Quote from: Courier of darkness

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.

Quote from: Courier of darkness

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.

Quote from: Courier of darkness

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

Yes. Correct.

Quote from: Courier of darkness

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.

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 me according to Prof Einstein.

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.

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

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.

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.

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=0/0You 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]

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

Quote from: Courier of darkness

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

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 photonbetween 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 photonbetween them. So if I asked the same question now, it would be: What are the two particles interactingwith 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 formationof a photon with rest-mass m0 = 0

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.

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.

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

But that is incorrect, because that photonhas 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).

...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!

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.

Quote from: evan_au

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.

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.

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Quote from: evan_au

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.

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?

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.

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.

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.

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)

This is true.

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because of this scenario, the equation E=mc² comes into play.

Wrong. The particle has non-zero velocity so the equation you have to use is:

E2 = (mc2)2 + (cp)2.

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The energy E of the particle remains the same

And who said it? It could diminish, it could increase or stay the same. All depends on how it loses mass: if, e.g., it shoots mass away as in a rocket, in the same direction of its velocity, then its energy will diminish; if it shoot it away in the opposite direction, the velocity will increase; from side, velocity will stay the same. The rest of what you say is just your idea (not the existence of tachions but the fact a negative mass object should accelerate further for the reason you say), not supported by physical reasons.

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

That was a joke and wasn't meant to be taken seriously.

Quote from: Courier of darkness

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

That's incorrect. Anti-matter is just the same as matter. It's impossible to look at a particle and say "this is antimatter." What the term antimatter refers to is the fact that to every particle there is another particle with the same mass but opposite charge and spin. It's a matter of history which one got chosen to be called matter and the other antimatter.

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

Was it a joke, or is it just something that has been distorted with time and repetition?

I think what Feynman actually said was: “… the math describing anti-particles moving forward in time is the same as the math describing particles moving backward in time.” This is a somewhat different statement from that which is commonly attributed to him.

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