What is a Charge? Exploring Positive & Negative Ions

[Libohove90]

hi,
what exactly is a charge? how do you define it? why is a positively charged ion at a higher potential difference than a negatively charged ion? why is work needed to be done on a positively charged ion to move it out of an electric field of another positively charged ion?

i'm very confused with electricity, and i hope i can understand it better if these doubts get cleared :)

thanks in advance

Think of it this way.

Why do we fall? Because of gravity. What is gravity? Gravity is a force. What causes this force? A gravitational field that permeates through space. What causes the presence of this gravitational field? The Earth itself (its mass).

Now apply it to electricity.

Why do charges particles accelerate? Because of electricity. What is electricity? Electricity is a force. What causes this force? An electric field that permeates through space. What causes the presence of this electric field? Charge itself.

Just like the mass of the Earth is the source of a gravitational field, the charge of a particle is the source of an electric field. Simple as that.

Charge is a fundamental property of matter than cannot be described in simpler terms, and that's what makes it tough to know what it is.

In fact, what is mass? Mass is a measure of inertia. But if you ask anyone what causes inertia, no one really knows. There is no real answer to that. So in a sense, mass is just as mysterious as charge. The difference is that we think we know what mass is because we experience its effects on our daily lives. But deep down, both mass and charge are quite interesting properties of matter. They are responsible for the presence of their respective force fields (gravity and electromagnetism). No mass no gravity, no charge no electricity (not entirely true, but for now, just think of it this way).

 

[Zentrails]

Think of it this way.

Why do we fall? Because of gravity. What is gravity? Gravity is a force. What causes this force? A gravitational field that permeates through space. What causes the presence of this gravitational field? The Earth itself (its mass).

I'm not trying to start an argument, but Einstein described gravity not as a force, but as an illusion of an attractive force that was in reality just caused by distortions in space-time that we humans can't see.

His famous "spooky action at a distance" is sometimes attributed to GR (other sources say he was referring to something else), but IIRC even Newton was mystified by the apparent attraction that he saw in the astronomical data of his time and wondered how that force could reach so far out through apparently empty space.

 

[Libohove90]

I'm not trying to start an argument, but Einstein described gravity not as a force, but as an illusion of an attractive force that was in reality just caused by distortions in space-time that we humans can't see.

His famous "spooky action at a distance" is sometimes attributed to GR (other sources say he was referring to something else), but IIRC even Newton was mystified by the apparent attraction that he saw in the astronomical data of his time and wondered how that force could reach so far out through apparently empty space.

Do I really need to use General Relativity if my goal is to convey the parallelisms between electricity and gravity? I am trying to give him a better sense of understanding what electricity and charge is by showing an analogous example using gravity and mass.

If I started talking about General Relativity and curved spacetime, that would further complicate things, wouldn't it?

 

[starfish99]

Comparison of charge to mass can be confusing. There is a positive and negative charge, but there is no positive and negative mass. In fact antimatter is composed of subatomic particles with opposite charges(negative protons and positive electrons) compared to conventional matter.
The electron and positron have identical mass and spin but opposite charge. We can't say that a positron has "something" that an electron lacks because the names of the charges are arbitrary. We do know that we need a rest mass to have charge. Photons and gravitons never have charge. We just don't know. Maybe we will know by the end of the century.

 

[Dickfore]

We do know that we need a rest mass to have charge. Photons and gravitons never have charge.

I don't mean to derail the thread or anything, but I was struck by your statement quoted above. I have wondered about it myself:

Is it possible to have a massless charged particle, and, if not, is there a general theoretical argument that precludes this?

 

[Andrew Mason]

I don't mean to derail the thread or anything, but I was struck by your statement quoted above. I have wondered about it myself:

Is it possible to have a massless charged particle, and, if not, is there a general theoretical argument that precludes this?

Not yet, as far as I can tell. But there would be an inherent detection problem. Consider a charged massless particle passing through an electric field. What would happen? It would not experience a detectable force, F = ma, since m = 0 and, since it always travels at the same speed, the speed of light, a = dv/dt = 0.

AM

 

[mquirce]

Dickfore hi. I wonder that even photon has two charges with different sign, which evolve toward each other in the same time that this structure move with C velocity.Electromagnetic wave is, in this concept, the field created by both electric charges in a helicoidal field, that can be decomposed in two sinusoidal fields vertical and horisontal, named by physicants electric and magnetic fields.
As for the fact that charge is linked with mass, i think must be wrong. For my opinion mass is created by particle of matter which is different from particle of mass.
Two particle of mater my create a particle of mass, or a particle massles.And the electric "charge" and "gravity" charge are intrisic ability of particle of mater to atract or repeal the same partner according their signs.

 

[Zentrails]

Do I really need to use General Relativity if my goal is to convey the parallelisms between electricity and gravity? I am trying to give him a better sense of understanding what electricity and charge is by showing an analogous example using gravity and mass.

If I started talking about General Relativity and curved spacetime, that would further complicate things, wouldn't it?

Yes, that's the rub. It is indeed more complicated.

 

[Dickfore]

Not yet, as far as I can tell. But there would be an inherent detection problem. Consider a charged massless particle passing through an electric field. What would happen? It would not experience a detectable force, F = ma, since m = 0 and, since it always travels at the same speed, the speed of light, a = dv/dt = 0.

AM

Having constant speed $v = |\mathbf{v}|$ does not imply zero acceleration $\mathbf{a} = d \mathbf{v}/d t$. All you can say with certainty is that the dot product:

$$v^{2} = \mathbf{v} \cdot \mathbf{v}$$
$$2 \, v \dot{v} = 2 (\mathbf{v} \cdot \dot{\mathbf{v}})$$
$$(\mathbf{v} \cdot \mathbf{a}) = 0$$
is zero.

Furthermore, 2nd Newton's Law has to be modified in relativistic limit:
$$\mathbf{F} = d \mathbf{p}/ d t, \; \mathbf{p} = m \, \gamma \, \mathbf{v}, \gamma = (1 - \beta^{2})^{-\frac{1}{2}}, \beta = v / c$$
Using the chain rule:
$$\dot{\mathbf{p}} = m \, \left(\gamma \, \dot{\mathbf{v}} + \dot{\gamma} \, \mathbf{v}\right)$$
$$\dot{\gamma} = \frac{d \gamma}{d v} \, \dot{v} = \left(-\frac{1}{2}\right) \, \gamma^{3} \, \left(-\frac{2 \, v \, \dot{v}}{c^{2}}\right) = \frac{\beta \, \gamma^{3} \, \dot{v}}{c}$$
Unfortunately, when $m = 0$, we have an indeterminate expression $0 \cdot \infty$. That is why we eliminate $\mathbf{v}$ in terms of momentum $\mathbf{p}$:
$$p = m \, c \, \gamma \, \beta \Rightarrow \gamma^{-2} = \left(\frac{p}{m \, c \, \beta}\right)^{-2}$$
$$1 - \beta^{2} = \left(\frac{m \, c}{p}\right)^{2} \, \beta^{2} \Rightarrow \beta^{2} = \frac{p^{2}}{p^{2} + (m \, c)^{2}}, \; 1 - \beta^{2} = \frac{(m \, c)^{2}}{p^{2} + (m \, c)^{2}} \Rightarrow \gamma = \frac{\sqrt{p^{2} + (m \, c)^{2}}}{m \, c}$$
Substituting everything and simplifying, we get:
$$\mathbf{F} = m \, \frac{\sqrt{p^{2} + (m \, c)^{2}}}{m \, c} \, \left( \mathbf{a} + \dot{v} \, \frac{p \, \mathbf{p}}{m \, c} \right)$$
Taking $m \rightarrow 0$, $\dot{v} \rightarrow 0$ and we still have an indeterminate form $0/0$. Let us evaluate $\dot{v}$ in terms of $\dot{\mathbf{p}}$:
$$\dot{v} = \frac{d v}{d p} \, \dot{p}$$
$$\frac{d v}{d p} = c \, \frac{d \beta}{d p} = c \, \frac{1 \cdot \sqrt{p^{2} + (m \, c)^{2}} - p \frac{p}{\sqrt{p^{2} + (m \, c)^{2}}}}{p^{2} + (m \, c)^{2}} = \frac{m^{2} \, c^{3}}{\left(p^{2} + (m \, c)^{2}\right)^{3/2}}$$

So,
$$\frac{\dot{v}}{m} = \frac{m \, c^{3} \, \dot{p}}{\left(p^{2} + (m \, c)^{2}\right)^{3/2}} \rightarrow 0, \; m \rightarrow 0$$
the second term in the parentheses in the expression for the relativistic force becomes zero in the massless limit and we have:
$$\mathbf{F} = \frac{p \, \mathbf{a}}{c}, \; m = 0$$
Thus, a finite force can give rise to a finite acceleration of a massless particle in the same direction as the force. But, a non-zero acceleration must be perpendicular to the velocity. Thus, the force must be perpendicular to the velocity of the particle. Lorentz force always fulfills this condition, but an electrostatic field might not.

 

[Drakkith]

Dickfore hi. I wonder that even photon has two charges with different sign, which evolve toward each other in the same time that this structure move with C velocity.Electromagnetic wave is, in this concept, the field created by both electric charges in a helicoidal field, that can be decomposed in two sinusoidal fields vertical and horisontal, named by physicants electric and magnetic fields.
As for the fact that charge is linked with mass, i think must be wrong. For my opinion mass is created by particle of matter which is different from particle of mass.
Two particle of mater my create a particle of mass, or a particle massles.And the electric "charge" and "gravity" charge are intrisic ability of particle of mater to atract or repeal the same partner according their signs.

(I'm not sure I understood what you said correctly, so I apologize if this doesn't fit your post)
There is no "particle of mass". ALL matter has mass and is composed of particles. In addition, we have a particle with zero Rest Mass, the photon. Various types of mass can be used to describe the energy of matter and light, however using anything but Rest Mass causes ALOT of confusion, as I have been told AND seen here on PF. As such a Photon has 0 mass, but it does have momentum and energy.

Like mass, charge is a property of matter that we have defined as such.

Yes, that's the rub. It is indeed more complicated.

The basic effects of gravity and electromagnetism are indeed similar enough to compare in his example. It's just an example and isn't trying to say both are the same thing or exactly alike. There is no need to bring in the complexities of GR into it.

 

[Dickfore]

Correction, the formula for the force is:
$$\mathbf{F} = \frac{\sqrt{p^{2} + (m \, c)^{2}}}{c} \, \left( \mathbf{a} + \dot{v} \, \frac{p \, \mathbf{p}}{(m \, c)^{2}} \right)$$
and
$$\dot{v} = \frac{m^{2} \, c^{3} \, \dot{p}}{\left(p^{2} + (m \, c)^{2}\right)^{3/2}}$$
$$\dot{p} = \frac{\mathbf{p} \, \cdot \dot{\mathbf{p}}}{p}$$

We get an expression:

$$\mathbf{F} = \frac{\sqrt{p^{2} + (m \, c)^{2}}}{c} \, \mathbf{a} + \frac{\mathbf{p} \, (\mathbf{p} \cdot \mathbf{F})}{p^{2} + (m \, c)^{2}}$$

which, in the massless limit reduces to:

$$\mathbf{F}_{\bot} = \mathbf{F} - \hat{\mathbf{p}} \, (\hat{\mathbf{p}} \cdot \mathbf{F}) = \frac{p}{c} \, \mathbf{a}$$

just as before, but without any contradiction, because the parallel force cancels out exactly!

 

[Drakkith]

Dickforce, i don't think your posts are being converted correctly.

 

[Dickfore]

For me they are.

 

[Drakkith]

For me they are.

Hrmm. Maybe it's my browser. I'm using Firefox, the newest version. Oh well.

 

[Drakkith]

Ahah! Found the problem! My addon for firefox, NoScript was blocking it for some reason. Fixed now!

 

[Zentrails]

(I'm not sure I understood what you said correctly, so I apologize if this doesn't fit your post)
There is no "particle of mass". ALL matter has mass and is composed of particles. In addition, we have a particle with zero Rest Mass, the photon. Various types of mass can be used to describe the energy of matter and light, however using anything but Rest Mass causes ALOT of confusion, as I have been told AND seen here on PF. As such a Photon has 0 mass, but it does have momentum and energy.

Like mass, charge is a property of matter that we have defined as such.



The basic effects of gravity and electromagnetism are indeed similar enough to compare in his example. It's just an example and isn't trying to say both are the same thing or exactly alike. There is no need to bring in the complexities of GR into it.

Then why talk about photons as particles?

Once you do that then you must realize that both the apparent magnetism and electrostatic repulsions/attractions are nothing more than transfer of momentum between two objects via photon exchange between the two objects.

I think that is actually an easier and simpler way to understand "charge" and it neatly eliminates the "spooky action at a distance" complaint.

I would dispute your claim that "all matter is composed of particles." Matter, just like photons, has wave-like AND particle-like properties, which is dependent on the experiment you use to observe them, but are actually something different than either.

It's just a convenient, dangerously intuitive way of describing something that goes against common sense.

 

[Dickfore]

Let me "rederive" the equation that determines the acceleration $\mathbf{a} = \dot{\mathbf{v}}$ in terms of the force $\mathbf{F}$, but eliminating the velocity $\mathbf{v}$ from all intermediate calculations. Assuming that the relativistic version of 2nd Newton's Law is:
$$\dot{\mathbf{p}} = \mathbf{F}$$
and the relation between momentum and velocity is:
$$\mathbf{p} = m \, \gamma \, \mathbf{v}$$
we can eliminate the velocity in favor of momentum:
$$p = m \, c \, \gamma \, \beta$$
$$\gamma^{-2} = 1 - \beta^{2} = \left(\frac{p}{m \, c \, \beta}\right)^{-2} = \left(\frac{m \, c}{p}\right)^{2} \, \beta^{2}$$
$$\beta^{2} = \frac{p^{2}}{p^{2} + (m \, c)^{2}}, \; 1 - \beta^{2} = \frac{(m \, c)^{2}}{p^{2} + (m \, c)^{2}}, \gamma = \frac{\sqrt{p^{2} + (m \, c)^{2}}}{m \, c}$$
$$\mathbf{v} = \frac{c \, \mathbf{p}}{\sqrt{p^{2} + (m \, c)^{2}}}$$
Taking the derivative:
$$\dot{\mathbf{v}} = c \, \left((p^{2} + (m \, c)^{2})^{-\frac{1}{2}} \, \dot{\mathbf{p}} + (-\frac{1}{2}) (p^{2} + (m \, c)^{2})^{-\frac{3}{2}} 2 \, p \, \dot{p} \right)$$
Also:
$$p^{2} = (\mathbf{p} \cdot \mathbf{p}) \Rightarrow 2 p \, \dot{p} = 2 \, (\mathbf{p} \cdot \dot{\mathbf{p}})$$
so, we can write:
$$\dot{\mathbf{v}} = \frac{c}{\sqrt{p^{2} + (m \, c)^{2}}} \, \left( \dot{\mathbf{p}} + \frac{\mathbf{p} \, (\mathbf{p} \cdot \dot{\mathbf{p}})}{p^{2} + (m \, c)^{2}}\right)$$
Writing $\mathbf{p} = p \, \hat{\mathbf{v}}$ (we have used the fact that momentum and velocity are collinear vectors so that $\hat{\mathbf{p}} = \hat{\mathbf{v}}$) and the energy-momentum relation:
$$\frac{\mathcal{E}}{c} = \sqrt{p^{2} + (m \, c)^{2}} \Rightarrow \frac{p^{2}}{p^{2} + (m \, c)^{2}} = \frac{\left(\frac{\mathcal{E}}{c}\right)^{2} - (m \, c)^{2}}{\left(\frac{\mathcal{E}}{c}\right)^{2}} = 1 - \left(\frac{m \, c^{2}}{\mathcal{E}}\right)^{2}$$
we have:
$$\mathbf{a} = \frac{c^{2}}{\mathcal{E}} \, \left[\mathbf{F} + \left(1 - \left(\frac{m \, c^{2}}{\mathcal{E}}\right)^{2}\right) \, \hat{\mathbf{v}} (\hat{\mathbf{v}} \cdot \mathbf{F})\right]$$
We denote:
$$\mathbf{F}_{||} = \hat{v} \, (\hat{v} \cdot \mathbf{F})$$
as the component of the force parallel to the velocity vector and $\mathbf{F}_{\bot}$ as the perpendicular component ($\mathbf{F} = \mathbf{F}_{||} + \mathbf{F}_{\bot}$). Then, the relation between acceleration and force for a relativistic particle can be written most transparently as:
$$\boxed{ \mathbf{a} = \frac{c^{2}}{\mathcal{E}} \, \left[ \left(\frac{m \, c^{2}}{\mathcal{E}}\right)^{2} \, \mathbf{F}_{||} + \mathbf{F}_{\bot}\right] }$$

Finally, for a massless particle, the parallel term drops out and we simply have:
$$\boxed{ \mathbf{a}_{||} = 0, \ \mathbf{a}_{\bot} = \frac{c^{2}}{\mathcal{E}} \, \mathbf{F}_{\bot} }$$
The parallel component of the acceleration is zero, as it should be since the speed of the massless particle must be $c$. The perpendicular force changes the direction of the velocity vector. Thus, we can visualize the motion of a massless particle with a constant speed $c$ (and energy $\mathcal{E}$) and the curvature $\kappa$ of the trajectory of the particle is:
$$\boxed{ a_{\bot} = \kappa \, c^{2} = \frac{c^{2}}{\mathcal{E}} \, F_{\bot} \Rightarrow \kappa = \frac{F_{\bot}}{\mathcal{E}} }$$

It is interesting to use the above formula to determine the trajectory of a massless particle moving in a central force field. But, I will do that in some other post.

 

[James A. Putnam]

Electric charge is not an exchange of momentum. Its effect results in changes of momentum. However, if one wishes to get to the point about: What is electric charge? It is a given. It exists as a cause for no reason that comes afterwards. It is a part of the beginning before effects. It is not the intermediaries of effects nor the effects themselves. Electric charge has units of coulombs because it is an indefinable property. That means it cannot be defined in terms of pre-existing properties. Its units are not definable in terms of pre-existing units.

James

 

[DragonPetter]

I think everyone is getting at that charge is an inherent property to fundamental particles. Charge describes one property of a particle when it is described in its quantum state mathematically, just like spin or mass. As particles have differences, a proton to an electron and so on, the mathematical descriptions of these particles are different and this is where you start to see the interactions of particles like you noted about positive and negative ions having different potentials.

Charge has rules and interactions with the universe attached to it, but its still just an intrinsic property of many particles. There is not much more we can do to describe this property than to define its mathematical and observed characteristics. If you try to go any deeper, I'm afraid you're hitting a physics road block and venturing into philosophical territory, with questions like "why does the universe exist?" that may turn out to be invalid questions to begin with.

 

[James A. Putnam]

I think everyone is getting at that charge is an inherent property to fundamental particles. Charge describes one property of a particle when it is described in its quantum state mathematically, just like spin or mass.

Charge has rules and interactions with the universe attached to it, but its still just an intrinsic property of many particles. There is not much more we can do to describe this property than to define its mathematical and observed characteristics. If you try to go any deeper, I'm afraid you're hitting a physics road block and venturing into philosophical territory, with questions like "why does the universe exist?" that may turn out to be invalid questions to begin with.

I don't think that asking what electric charge is or attempting to answer it is a philosophical question. it is a physics question. In other words, either we say what it is or we say it is an unknown cause. There is nothing wrong with saying that it is an unknown cause. The fact of the matter is that no one knows what cause is. Empirical physics is the practice of gathering empirical evidence. Empirical evidence consists of patterns in changes of velocity. We learn, from that evidence that there are various patterns that appear to be unconnected. That is where theoretical physics becomes relevent. theoretical physics attaches educated guesses about what cause may be, and, if that entails multiple fundamental causes, then theoretical physics proposes that such multiple causes exist. These possible causes are given names. We do not yet know what causes are real and which ones are not. Finally, there are no invalid questions about the operation of the universe when speaking about causes and effects.

James

 

[Dickfore]

Charge is intimately connected to the $\mathrm{U}(1) \times \mathrm{SU}(2)$ gauge symmetry of the electroweak interactions. It has something to do with hypercharge and isospin, but I do not the exact details of the relation.

Of course, one may ask what is gauge symmetry. For now, it is a fundamental symmetry of the world, just as Lorentz invariance.

 

[James A. Putnam]

First there is cause. Secondly there are effects. Afterwards there is theory. No cause is the result of theory. No theory explains cause. Theory invents types of cause and uses its concepts to explain effects. Higher level theory can be useful, but, not for explaining cause.

James

 

[DragonPetter]

I don't think that asking what electric charge is or attempting to answer it is a philosophical question. it is a physics question. In other words, either we say what it is or we say it is an unknown cause. There is nothing wrong with saying that it is an unknown cause. The fact of the matter is that no one knows what cause is. Empirical physics is the practice of gathering empirical evidence. Empirical evidence consists of patterns in changes of velocity. We learn, from that evidence that there are various patterns that appear to be unconnected. That is where theoretical physics becomes relevent. theoretical physics attaches educated guesses about what cause may be, and, if that entails multiple fundamental causes, then theoretical physics proposes that such multiple causes exist. These possible causes are given names. We do not yet know what causes are real and which ones are not. Finally, there are no invalid questions about the operation of the universe when speaking about causes and effects.

James

When we say something is fundamental, and then try to describe its cause, are we not admitting doubt that it is truly a fundamental property? When something is fundamental or indivisible, it should have no cause, or else it is no longer fundamental. I'm not stating this as fact, but I think you can't attach a cause to something that is fundamental, or else you're getting into these infinity chick/egg arguments that I was referring to with the philosophical remark.

Of course you can say charge's cause is not known, but I don't think we can assume there is a cause with what we know, and our mathematics thus far shows that it is intrinsic to fundamental particles like electrons and muons as far as I know, but I'm not a theoretical physicist.

 

[DragonPetter]

First there is cause. Secondly there are effects. Afterwards there is theory. No cause is the result of theory. No theory explains cause. Theory invents types of cause and uses its concepts to explain effects. Higher level theory can be useful, but, not for explaining cause.

James

In our theory the postulate is that quarks, electrons, etc. have a charge associated with them. The postulate comes from how our math works out to show this so, and it is confirmed experimentally. Does a postulate have a cause?

edit: Does a fundamental postulate have a cause? If so, we must modify the theory and no longer say it is fundamental.

 

[James A. Putnam]

In our theory the postulate is that quarks, electrons, etc. have a charge associated with them. The postulate comes from how our math works out to show this so, and it is confirmed experimentally. Does a postulate have a cause?

edit: Does a fundamental postulate have a cause? If so, we must modify the theory and no longer say it is fundamental.

I have to go out on a limb to answer this question. The forum prefers book answers. However, I will risk saying something that I think is important for theorists to know. All guesses have a high potential for leading to misleading theory. Most importantly, the guesses start all the way back at f=ma. The very first one is the choice to make mass an indefinable property. I am not saying that force should have been the indefinable property. I am saying that neither should have been made arbitrarily into an indefinable property. The basis for this objection of mine is that empirical evidence consists of measures involving changes of distance and time. All other properties are inferred from this evidence. Therefore, all properties of theoretical physics should be ultimately expressible in the terms of the evidence from which they were inferred. What this means is that all properties should be expressible in terms of combinations of meters and seconds. That level is the level that I would call fundamental. Since this kind of message is probably unwanted here, I will let it stand by itself without further comment. It was simply meant to be helpful.

James

 

[DragonPetter]

I have to go out on a limb to answer this question. The forum prefers book answers. However, I will risk saying something that I think is important for theorists to know. All guesses have a high potential for leading to misleading theory. Most importantly, the guesses start all the way back at f=ma. The very first one is the choice to make mass an indefinable property. I am not saying that force should have been the indefinable property. I am saying that neither should have been made arbitrarily into an indefinable property. The basis for this objection of mine is that empirical evidence consists of measures involving changes of distance and time. All other properties are inferred from this evidence. Therefore, all properties of theoretical physics should be ultimately expressible in the terms of the evidence from which they were inferred. What this means is that all properties should be expressible in terms of combinations of meters and seconds. That level is the level that I would call fundamental. Since this kind of message is probably unwanted here, I will let it stand by itself without further comment. It was simply meant to be helpful.

James

Well that would be cool if everything could be described in 2 units, and its possible, although that's purely wishful thinking or unproven at this point and does not have much to do with what OP asked and I think it has gotten too much into semantics at this point.

OP asked about what charge is, and the current position is that it is an intrinsic property of elementary particles as I said in my first post of this thread. As far as I know, there has not been any accepted experimental evidence that charge is not elementary and we are assuming there is no cause of charge in our definition of it; it is elementary at this point and there is nothing more we can say on its "fundamentalness" until proven/theorized otherwise. Anything else is a different discussion and should be in its own thread I think.

 

[Drakkith]

I have to go out on a limb to answer this question. The forum prefers book answers. However, I will risk saying something that I think is important for theorists to know. All guesses have a high potential for leading to misleading theory. Most importantly, the guesses start all the way back at f=ma. The very first one is the choice to make mass an indefinable property. I am not saying that force should have been the indefinable property. I am saying that neither should have been made arbitrarily into an indefinable property. The basis for this objection of mine is that empirical evidence consists of measures involving changes of distance and time. All other properties are inferred from this evidence. Therefore, all properties of theoretical physics should be ultimately expressible in the terms of the evidence from which they were inferred. What this means is that all properties should be expressible in terms of combinations of meters and seconds. That level is the level that I would call fundamental. Since this kind of message is probably unwanted here, I will let it stand by itself without further comment. It was simply meant to be helpful.

James

It isn't helpful because that isn't how it is done. Time and distance are just as fundamental as mass and charge, not more so. How do we measure time? By the change in something. From hourglasses to pendulum clocks to the atomic clock, all these define the duration of our unit of time. So our measurement of time itself is based on other concepts like distance and mass and charge. (The cesium atom wouldn't vibrate like it does if it had a different mass!)

How about distance? Based on time and change in position, which depends on things like mass.

The reason that all of these qualities are fundamental is because they cannot be broken down into more basic ideas that explain what they are or what they do.
And no, all guesses do not have a high potential to lead to misleading theory. Only the guesses that don't take into account evidence do.

 

[vanesch]

Is it possible to have a massless charged particle, and, if not, is there a general theoretical argument that precludes this?

The problem with massless charged particles (the way coulomb charge behaves) is as follows:

As the field of a charge is infinitely extended, it corresponds to a finite amount of energy. So a charge represents an energy by itself, so that can't be massless. You can of course say that the mass belongs to the field, and not to the charge that is it's source, but the field "goes with" the particle that is charged, so in any case it would behave as an *effective* mass. And it might very well be that ALL mass is just "effective" interaction-energy.

I don't know how "strong" this argument is in fact.

There DO exist massless "charged" particles, namely gluons. They are color-charged, so they are the QCD equivalent of a charged, massless particle in electrodynamics. But the color interaction is such that it is always perfectly "shielded": no naked color charge seems to exist. That is, charged gluons always are "compensated" by nearby other color charges so that they are neutral at any macroscopic distance (and these conglomerates ARE massive, most probably for an important part because of the energy included in the color field - this is the "effective" mass I was talking about).

 

[Dickfore]

The problem with massless charged particles (the way coulomb charge behaves) is as follows:

As the field of a charge is infinitely extended, it corresponds to a finite amount of energy. So a charge represents an energy by itself, so that can't be massless. You can of course say that the mass belongs to the field, and not to the charge that is it's source, but the field "goes with" the particle that is charged, so in any case it would behave as an *effective* mass. And it might very well be that ALL mass is just "effective" interaction-energy.

I don't know how "strong" this argument is in fact.

I think what you are calculating is the zero-point energy of the electromagnetic field, which is not interpreted as a mass of anything. In fact, it is always subtracted from the Hamiltonian of the system.

In the language of QFT, i think mass of charged particles would be spontaneously generated in every order of perturbation theory, since the loop correction to the propagator, corresponding to the following Feynman diagram:

​

would give a contribution to the self-energy and generate mass.

I don't know if this is the right argument, or, if it has some non-perturbative generalization (like the Ward-Takahashi identity).

There DO exist massless "charged" particles, namely gluons. They are color-charged, so they are the QCD equivalent of a charged, massless particle in electrodynamics.

I was referring specifically to electric charge, but your argument is still valid. However, if you want to make the analogy, gluons would actually correspond to the photon, which is massless in QED. Furthermore, both photons and gluons are gauge bosons, i.e. force carrier particles and cannot describe the matter content of the Universe. In fact, the "colored" fermion elementary particles - quarks are quite massive even at the bare level.

But the color interaction is such that it is always perfectly "shielded": no naked color charge seems to exist. That is, charged gluons always are "compensated" by nearby other color charges so that they are neutral at any macroscopic distance (and these conglomerates ARE massive, most probably for an important part because of the energy included in the color field - this is the "effective" mass I was talking about).

You are right. In fact, what you are describing are glueballs that are predicted to have mass. Unfortunately, they had not been experimentally identified as of now. However, this is an exclusive property of QCD which is built up on an anomalous non-Abelian gauge theory (SU(3)). For theories with anomaly = 0, this cannot happen since the force carriers do not interact with each other.

 

[vanesch]

I think what you are calculating is the zero-point energy of the electromagnetic field, which is not interpreted as a mass of anything. In fact, it is always subtracted from the Hamiltonian of the system.

I was going to say "no that's not what I mean", but now I'm not so sure actually. I was of course thinking of the *classical* argument of electron mass by the EM field energy (argument which is quantitatively erroneous, but of which I thought the principle was still valid).

Aren't you mixing up the sum over bubble diagrams on one hand - the vacuum energy or the zero point energy of a sourceless field - on one hand, and the "mass correction terms" terms on the other hand ?

But I don't even know whether this includes the field energy of a field with source.

(you know, the equivalent of the classical integral over |E|^2)

In the language of QFT, i think mass of charged particles would be spontaneously generated in every order of perturbation theory, since the loop correction to the propagator, corresponding to the following Feynman diagram:

​

would give a contribution to the self-energy and generate mass.

I don't know if this is the right argument, or, if it has some non-perturbative generalization (like the Ward-Takahashi identity).

What surprises me here is that one would need higher-order corrections to describe a classical concept (integral of |E|^2).

I was referring specifically to electric charge, but your argument is still valid. However, if you want to make the analogy, gluons would actually correspond to the photon, which is massless in QED. Furthermore, both photons and gluons are gauge bosons, i.e. force carrier particles and cannot describe the matter content of the Universe. In fact, the "colored" fermion elementary particles - quarks are quite massive even at the bare level.

My point was to find an example of a massless particle (fermion or boson) which carried "charge". And I could only think of the gluon, as indeed the quarks are massive (and also electrically charged apart from being color charged).

 

[Dickfore]

I was going to say "no that's not what I mean", but now I'm not so sure actually. I was of course thinking of the *classical* argument of electron mass by the EM field energy (argument which is quantitatively erroneous, but of which I thought the principle was still valid).

Aren't you mixing up the sum over bubble diagrams on one hand - the vacuum energy or the zero point energy of a sourceless field - on one hand, and the "mass correction terms" terms on the other hand ?

But I don't even know whether this includes the field energy of a field with source.

The term 'source' in QFT has a special meaning. For example, the EM field is described by the 4-potential $A^{\mu}(x)$. Then, we introduce an artifical current density $J^{\mu}(x)$ and include an extra interaction term in the Lagrangian:

$$\mathcal{L}(A, J) = \mathcal{L}_{0}(A) - J_{\mu} \, A^{\mu}$$

Using this Lagrangian, one calcualates the partition function:

$$Z[J] = \exp{\left[i \, \int{d^{d}x \, \mathcal{L}(A, J)} \right]} = Z_{J = 0} \, \exp{\left[\frac{i}{2} \, \int{d^{d}y \, d^{d}z \, J_{\mu}(y) \, D^{\mu \nu}(y - z) \, J_{\nu}(z)}\right]}$$

In fact, it is by taking (functional) derivatives w.r.t. to these external currents that we get the propagators:

$$\frac{1}{i} \, D^{\mu \, \nu}(x - y) = \langle T A^{\mu}(x) \, A^{\nu}(y) \rangle = \frac{1}{i} \, \frac{\delta}{\delta J_{\mu}(x)} \, \frac{1}{i} \, \frac{\delta}{\delta J_{\nu}(y)} \, \left. Z[J] \right|_{J = 0}$$

At the end of the day we always set the sources equal to zero!

There is another famous calculation of the total energy of the field for a given external source distribution. This energy in fact diverges for a point source. However, in QFT, this divergence is meaningless, since it is an artifice of the 'weird apparatus' the experimenter was using to probe the field.

Another thing is when the current that couples to the field is a Noether current of another field that is conserved due to the intrinsic symmetry of the other field (for QED it is U(1)). Then, one understands the EM field as a gauge field that is necessary to be introduced in the 'gauge covariant derivative' in order to preserve the local gauge invariance.

However, now the field that creates the (microscopic) current is itself a quantum object and one cannot treat it as a parameter over which we can differentiate. It has its own dynamics described by its own Lagrangian and we arrive at an interacting field theory. In this way, loop corrections to the propagators play role. And, in fact, the integrals involved in these corrections actually diverge even at first order. The procedure to make these integrals convergent in a controlled manner is called regularization and the procedure of canceling out divergences order by order in perturbation theory through the redefinition of the bare parameters (mass, charge) is called renormalization.

 

[Andrew Mason]

Finally, for a massless particle, the parallel term drops out and we simply have:
$$\boxed{ \mathbf{a}_{||} = 0, \ \mathbf{a}_{\bot} = \frac{c^{2}}{\mathcal{E}} \, \mathbf{F}_{\bot} }$$
The parallel component of the acceleration is zero, as it should be since the speed of the massless particle must be $c$. The perpendicular force changes the direction of the velocity vector. Thus, we can visualize the motion of a massless particle with a constant speed $c$ (and energy $\mathcal{E}$) and the curvature $\kappa$ of the trajectory of the particle is:
$$\boxed{ a_{\bot} = \kappa \, c^{2} = \frac{c^{2}}{\mathcal{E}} \, F_{\bot} \Rightarrow \kappa = \frac{F_{\bot}}{\mathcal{E}} }$$

It is interesting to use the above formula to determine the trajectory of a massless particle moving in a central force field. But, I will do that in some other post.

Thanks for your analysis, which appears to be correct. What this shows though is that for such a theoretical massless charged particle with charge q passing through a perpendicular electric field, $\vec{E}$, the acceleration it would experience is:

(1) $$a_{\bot} = \frac{c\lambda}{h}q\vec{E}$$

Such a particle would be deflected laterally a distance (lateral speed << c):

(2)$$\Delta s = \frac{1}{2}at^2$$

where t is the length of time spent by the particle in passing through the perpendicular electric field $\vec{E}$. This time t would be equal to L/c where L is the length of the path through the perpendicular field. Using the expression for a in (1) and substituting L/c for t in 2, 2 becomes:

$$\Delta s = \frac{q\vec{E}L^2\lambda}{2ch}$$

For a massless particle with the wavelength of .1 nm ($10^{-10}$ m) and charge e passing through a million volt/metre field ($10^6$ N/C) for a distance of 1 metre the deflection would be:

$$\Delta s = \frac{1.6*10^{-19}*10^6*1*10^{-10}}{6.2*10^{18}*2*3*10^8*6.6*10^{-34}} = \frac{1.6*10^{-23}}{245*10^{-8} } = 6.5*10^{-18}$$ m.

In this example, one would have to observe a deflection of about 1/1000th the diameter of a proton. So detection of a charged massless particle would be challenging.

AM

 

[Dickfore]

Actually, I was able to solve for the exact trajectory of an ultra-relativistic particle incident perpendicular to a uniform electric field. The unit of length is:

$$L = \frac{\mathcal{E}}{|q \, \vec{E}|}$$

Then, if the x-axis is in the initial direction and the electrostatic force acting on the particle is in the positive y-direction, we get:

$$\frac{y}{L} = \ln{\left|\sec\frac{x}{L}\right|}$$

This trajectory has a vertical asymptote at:

$$\frac{x_{\mathrm{max}}}{L} = \frac{\pi}{2} \Rightarrow x_{\mathrm{max}} = \frac{\pi \, L}{2}$$

so the particle cannot go further than this and it wil get effectively deflected by 90⁰. In your numerical example:

$$L = \frac{12.4 \times 10^{3} \, \mathrm{V}}{10^{6} \, mahtrm{V} \cdot \mathrm{m^{-1}}} = 12.4 \, \mathrm{mm}$$

So, the particle will get effectivly deflected by 90^o in a length $1.57 \times 12.4 \, \mathrm{mm} = 19.5 \, \mathrm{mm}$. The electric field acts like some 45^o mirror.

 

[Andrew Mason]

Where does the bolded number come from?

Oops. i was intending to use force/unit charge e. But in that case the numerator should be 1 not 1.6 x 10-19 C. ...Well, I was only out by a factor of 10^19!

So a massless charged particle should be rather easily detectable after all...

Also, a De Broglie wavelength $\lambda = 0.1 \, \mathrm{nm}$ corresponds to an energy of $12.4 \, \mathrm{keV}$ of an ultrarelativistic particle.

?? It has 0 rest mass so it travels at c. I was thinking that if a massless particle could have a charge, it would have to have atomic or subatomic dimensions.

AM

 

[Zentrails]

Electric charge is not an exchange of momentum. Its effect results in changes of momentum. However, if one wishes to get to the point about: What is electric charge? It is a given. It exists as a cause for no reason that comes afterwards. It is a part of the beginning before effects. It is not the intermediaries of effects nor the effects themselves. Electric charge has units of coulombs because it is an indefinable property. That means it cannot be defined in terms of pre-existing properties. Its units are not definable in terms of pre-existing units.

James

Who says electric charge is an exchange of momentum?

It's the apparent attraction or repulsion that is a result of change in momentum, IF you consider photons to be particles.

An electric charge is not a given unless it is measured. It is measured by experiments that detect attraction or repulsion.
That is not the same thing as saying electric charge is due to photon exchange momentum transfers.