Does an electric charge "curve spacetime"?

[Gonzolo]

If theorists (starting with A. E.) can make a theory about spacetime curvature caused by mass (GR), couldn't there be a similar theory where some spacetime curvature is caused by electric charges? Both are F = k/r^2 in elementary physics.

A postulate could be that an electron in an elevator (made of electons, or a negatively charged inside surface) cannot tell the difference whether :

1. the elevator is stopped and that there is a large + charge underneath or :

2. the elevator is accelerating upwards

Where does this lead? What happens if you take magnetism into account?

[Garth]

Mass and energy cause curvature. Einstein spent much of the rest of his life after developing GR to try and include E-M theory as well, a unified field theory, but was unsuccessful; probably because he didn't know about the strong and the weak nuclear forces at the time, which have to be included as well.

However this question may be the opportunity to consider the following, "According to the EEP a stationary electron on a laboratory bench is accelerating w.r.t. the local Lorentzian freely falling inertial frame of reference. According to Maxwell’s theory of electromagnetism an accelerating electric charge, such as an electron, radiates. So why doesn’t it? Or, if it is thought that such an electron actually does radiate, what is the source of such radiated energy?"
Garth

[Gonzolo]

I don't know what EEP and w.r.t. stand for.

That's interesting but it's not unification that I want to talk about. I am wondering whether a theory similar to GR can be developped from the two postulates of my first post (or similar ones), instead of the familiar ones with gravity and mass.

Based on the premise that charge and mass are equally important, why does mass curve spacetime, but not charges? "What happens if we replace m by q in GR's equations" and consider attractions and repulsions?" I believe that if mass can curve spacetime, then so should + and - charges. So a GR-like theory should be able to be developped talking about charges instead of mass (whether its useful or not).

Perhaps this thread belongs in the theory developped section. I let GR specialists out there be the judges.

[jcsd]

It's a fair question. The reason is that gravity affects all particles in the same way; charge doesn't. Accelration due to gravity is independentof the particular properties of the particle whereas acceleration due to the electromagnetic force depends on it's mass to charge ratio.

There was an attempt to include the electromagnetic force as the curvature of spacetime - Kaluza-Klein theory, which needed the additon of an extra spacetime dimension. Later some of the ideas of Kaluza-Klein thoery were used in string theory.

[Garth]

I don't know what EEP and w.r.t. stand for.


EEP = Einstein Equivalence Principle :w.r.t = with respect to.

That's interesting but it's not unification that I want to talk about. I am wondering whether a theory similar to GR can be developped from the two postulates of my first post (or similar ones), instead of the familiar ones with gravity and mass.

Einstein had a go and couldn't develop one, but perhaps you will succeed!

Garth

[humanino]

Based on the premise that charge and mass are equally important, why does mass curve spacetime, but not charges? "What happens if we replace m by q in GR's equations" and consider attractions and repulsions?" I believe that if mass can curve spacetime, then so should + and - charges. So a GR-like theory should be able to be developped talking about charges instead of mass (whether its useful or not).
The problem is that a spin 2 field, like the supposedly graviton, always causes attraction. It cannot yield to repulsion.

This requires negative energy. You will get time travel and loose causality.

[Gonzolo]

I believe that spins and gravitons were un-thought of when A. E. developped GR and am not aware that they are considered in GR today. The theory I have in mind should be able to be developped from nothing else than classical physics, as GR was.

I'm living in 1916 for this thread and telling Einstein that there is a (perhaps parallel) spacetime associated with charges (mass not considered). I'm asking him to prove to me that a very strong electric field cannot curve a light beam as would a massive star. And am suggesting that electric forces can be explained with differential geometry.

[Dburghoff]

Now, I'm not a physicist (just a first-year engineering student), but I'd like to venture a guess why. I've actually thought of something like this, Gonzolo, but I came up with an explanation which satisfies me, and I think will satisfy you as well.

The whole premise behind GR is that it is impossible to distinguish between gravity and acceleration. That is, if you stood in an elevator in space which was accelerating at 9.8 m/s² without access to the outside world, there is no experiment that you could do to determine that you are not on Earth (assuming that on Earth, slight fluctuations in g near the surface are impossible to measure). Essentially, gravity and acceleration are the same thing. Now, everything else in GR is based on this premise, including the curved space around massive objects.

However, the same cannot be said of a electric fields. Let's go back to our closed elevator scenario. It's incredibly easy to determine the difference between electromagnetism and "regular" acceleration. All you have to do is place an electron on one side of you and a proton on the other. If both fall to the floor of the elevator, you know that you're both "regularly" accelerating. If one falls to the floor and the other rises towards the ceiling, you know that you're under the influence of a charged object. Therefore, the basic premise of GR is false when considering EM, and nothing else can be derived from it.

Hope I helped!

[pervect]

If theorists (starting with A. E.) can make a theory about spacetime curvature caused by mass (GR), couldn't there be a similar theory where some spacetime curvature is caused by electric charges? Both are F = k/r^2 in elementary physics.


So far, this sounds like Kaluza-Klein theory, which is an attempt to get electromagnetism from a geometrical theory. It didn't work out on its own, but it helped inspire string theory

http://en.wikipedia.org/wiki/Kaluza-Klein_theory

However, elevators and equivalence principles are not involved in Kaluza-Klein theory. Instead, one contemplates a 5-d spacetime, and eventually one concludes that one of the dimensions may be small and "curled up". One gets gravity and electromagnetism and a scalar field (which hasn't been observed) out of such a theory. You'll see some of the ideas in Kaluza-Klein theory which are later used by string theory (the extra spatial dimensions and the way they are handled).

BTW, magnetism and electrostatic forces are unified by special relativity - a magnetic field is basically just the consequence of an electric field as seen by a moving observer.

[Gonzolo]

The whole premise behind GR is that it is impossible to distinguish between gravity and acceleration. That is, if you stood in an elevator in space which was accelerating at 9.8 m/s² without access to the outside world, there is no experiment that you could do to determine that you are not on Earth (assuming that on Earth, slight fluctuations in g near the surface are impossible to measure). Essentially, gravity and acceleration are the same thing. Now, everything else in GR is based on this premise, including the curved space around massive objects.


Yup, I pretty much agree with that.


However, the same cannot be said of a electric fields. Let's go back to our closed elevator scenario. It's incredibly easy to determine the difference between electromagnetism and "regular" acceleration. All you have to do is place an electron on one side of you and a proton on the other. If both fall to the floor of the elevator, you know that you're both "regularly" accelerating. If one falls to the floor and the other rises towards the ceiling, you know that you're under the influence of a charged object. Therefore, the basic premise of GR is false when considering EM, and nothing else can be derived from it.

That is true. But if you only have an electron, how can you tell? Couldn't one recreate GR with this case? And then another GR for the proton case? Perhaps that by then mixing the 2 "new" GR theories, we would arrive to the same conclusions than by having both charge in the elevator. I do not know.

pervect, Kaluza-Klein theory may be what I'm trying to talk about, I am not sure. I would try to avoid mass and gravity to begin with, to see where it goes. Perhaps KK did that, and eventually added mass to complete the theory. Honestly, I would need rigourous GR and KK introductions. I am wondering where the premise of my first post would lead if I gave it to A. E. or knew how to demonstrate his equations.

[hellfire]

However this question may be the opportunity to consider the following, "According to the EEP a stationary electron on a laboratory bench is accelerating w.r.t. the local Lorentzian freely falling inertial frame of reference. According to Maxwell’s theory of electromagnetism an accelerating electric charge, such as an electron, radiates. So why doesn’t it? Or, if it is thought that such an electron actually does radiate, what is the source of such radiated energy?"

Radiation is due to a reconfiguration of the electric field lines due to a change in the motion of the charge (the new field lines emerging from the charge do not match with the old ones). Inside a gravitational field the field lines are static and thus there is no radiation. May be this explanation is too simple and I am missing something...?

[da_willem]

That is true. But if you only have an electron, how can you tell? Couldn't one recreate GR with this case? And then another GR for the proton case? Perhaps that by then mixing the 2 "new" GR theories, we would arrive to the same conclusions than by having both charge in the elevator. I do not know.

all particles fall at the same rate in a gravitational field because both inertia and the gravitational attraction are proportional to the mass of the falling object: F=ma=GMm/r^2. So the acceleration a=Gm/r^2 is independent of the mass of the object.

For an electric charge this is different: F=ma=CqQ/r^2 so a=CqQ/mr^2. So it depends both on the mass and the charge of the object. So there is no equivalence principle like that of GR.

[pervect]

all particles fall at the same rate in a gravitational field because both inertia and the gravitational attraction are proportional to the mass of the falling object: F=ma=GMm/r^2. So the acceleration a=Gm/r^2 is independent of the mass of the object.

For an electric charge this is different: F=ma=CqQ/r^2 so a=CqQ/mr^2. So it depends both on the mass and the charge of the object. So there is no equivalence principle like that of GR.

Yes, this is what makes the geometrical interpretation very "natural" for gravity. Since all particles behave the same way because of the equivalence principle, it's easy to describe the motion of a particle geometrically by making the natural motion of a particle a geodesic.

However, the geometrical POV can be used to handle forces as well, as Kaluza-Klein theory shows. The mechanism for doing so is a little "tricky" (extra spatial dimensions, often rolled up into a small curve).

[Garth]

Inside a gravitational field the field lines are static and thus there is no radiation.
Define "static" - static in the laboratory non-inertial frame, or static in the freely falling, inertial Lorentzian frame?

Which should the electron not radiate in?

-Garth

[Gonzolo]

all particles fall at the same rate in a gravitational field because both inertia and the gravitational attraction are proportional to the mass of the falling object: F=ma=GMm/r^2. So the acceleration a=Gm/r^2 is independent of the mass of the object.

For an electric charge this is different: F=ma=CqQ/r^2 so a=CqQ/mr^2. So it depends both on the mass and the charge of the object. So there is no equivalence principle like that of GR.

That makes sense (with a = GM/r^2 instead). I understand better. The ratio q/m suggests why Kaluza-Klein unifies gravity and EM. KK is probably the simplest complete geodesic EM theory.

Now this may be an insult to Newton (and to myself), but mathematically, we could use F = qE instead of F = ma. This gives rise to E playing the role of a. But E = d?/dt = F/q = (m/q)*(dx^2/dt^2) = (m/q)*a... the same ratio. The ratio q/m might suggests why Kaluza-Klein unifies gravity and EM.

For the sake of exploring what happens if I consciously look away from inertia, mass and gravity, I'll keep going :

What about if I shove m into the C (or use m = 1 for simplicity)? The electron in our EM elevator is alone and has constant mass anyway. We then got a = CqQ/r^2, just like gravity. C's units have changed though, so this may start controversy. I may be departing physics.

[da_willem]

Now this may be an insult to Newton (and to myself), but mathematically, we could use F = qE instead of F = ma. This gives rise to E playing the role of a

Physically (not only mathematically) they are both true but have a very different meaning! F=ma is an incomplete law that describes how an object responds to a force. F=ma is incomplete in the sense that it needs the input of a force to tell you something about the movement of he object. F=qE is an example of such a force, but is not an equation of motion like Newtons second law, and thus has (with all the other formulas describing all other forces) an entirely different status...

[hellfire]

Define "static" - static in the laboratory non-inertial frame, or static in the freely falling, inertial Lorentzian frame?

Which should the electron not radiate in?

-Garth
I would say the field lines are static in the laboratory frame in which the electron is at rest. Does this mean the electron should radiate in a free falling frame? I have to admit that my first post was only a guess, but I really dont know. May be you could elaborate a little bit.

[Garth]

I would say the field lines are static in the laboratory frame in which the electron is at rest. Does this mean the electron should radiate in a free falling frame? I have to admit that my first post was only a guess, but I really dont know. May be you could elaborate a little bit.

According to the equivalence principle of GR the natural inertial Lorentzian frame is the freely falling one. The table-top is being accelerated upwards wrt this frame by the force pushing on it by the floor. The electron itself on the lab table is therefore accelerating upwards wrt this inertial frame, being 'pushed upwards' by the table and according to Maxwell ought to be radiating.

I have had the opinion expressed in a university physics community that in fact such electrons do radiate, but at such a low power that it has not been detected. In which case the second part of my question comes into play, if so, where does this energy of radiation come from? The electron is just sitting there minding its own business!

In my view this question is tied up with the problem that energy is not locally conserved in GR, and there has been some discussion about that on these forums.
- Garth

[selfAdjoint]

I understood it perfectly, but a newby might have thought you were supporting absolute motion, which we both know isn't so.

[Garth]

I understood it perfectly, but a newby might have thought you were supporting absolute motion, which we both know isn't so.
Understood. However, and here may be the beginning of a new thread, I quote pensively "you were supporting absolute motion, which we both know isn't so" ??

My questioning ?? about a preferred (absolute is too strong a term) frame is the question of Mach's Principle.

SR is rightly configured for empty space, we have space-time and a set of test particles that define that space and time by their interactions, yet do not perturb it. In such an empty space the principle of relativity, i.e. no preferred frames, holds its own. This is codified in the conservation of energy-momentum, or 'rest energy', or mass defined by the equation of 4-momentum.

We now introduce matter and their associated gravitational fields, which are interpreted in GR as a curvature of that space-time and carry forward the SR principle of no-preferred frames, the conservation of energy-momentum.

However if we now introduce Mach's Principle, which suggests the phenomenon of inertia ought to arise from accelerations with respect to the general mass distribution of the entire universe, then we might indeed choose a particular or preferred frame when masses are introduced, that which is the Centre of Mass frame of the system under investigation.

Cosmologically this will be that in which the CMB is globally isotropic, co-moving with the surface of last emission, and in the laboratory this will be the Centre of Mass of the Earth.

The interesting observation here is that the electron sitting on a lab table might be accelerating wrt to the freely falling frame preferred by GR as the Lorentzian inertial frame, but it is at rest wrt to the Earth preferred by Mach's Principle as the Centre of Mass frame.

Hence if both Maxwell and Mach are correct the electron should not be radiating and there is no problem over where the energy of any such radiation might come from!

Just food for thought.

- Garth

[selfAdjoint]

Well, I don't believe in Mach's principle, because GR has had such success that it seems to rule it out. To reintroduce it now at this late date is to say that GR is insufficient among classical theories, and what would lead you to think that?

[juju]

Hi all,

It might be possible to form a curvature theory of electromagnetics if you start with the idea that each pair of charges is the source of the curvature rather than single particles.

juju

[pervect]

I have had the opinion expressed in a university physics community that in fact such electrons do radiate, but at such a low power that it has not been detected. In which case the second part of my question comes into play, if so, where does this energy of radiation come from? The electron is just sitting there minding its own business!


I'm surprised this issue hasn't been resolved. It seems to me that the static nature of the charge relative to the observer at infinity means that it should not radiate from the viewpoint of the observer at infinity.

[Entropy]

Mass and energy cause curvature. Einstein spent much of the rest of his life after developing GR to try and include E-M theory as well, a unified field theory, but was unsuccessful; probably because he didn't know about the strong and the weak nuclear forces at the time, which have to be included as well.

Did you quote that from Brian Green's book: The Fabric of The Cosmos? Heh, I was just reading that today and I notice that theres a passage where he says the exact same thing you said. Good book.

[Garth]

Did you quote that from Brian Green's book: The Fabric of The Cosmos? Heh, I was just reading that today and I notice that theres a passage where he says the exact same thing you said. Good book.
No I didn't - a case of "great minds think alike, fools seldom differ" I guess!
Garth

[Garth]

I'm surprised this issue hasn't been resolved. It seems to me that the static nature of the charge relative to the observer at infinity means that it should not radiate from the viewpoint of the observer at infinity.
So your observer at inifinity is co-moving with the Earth is she? A case for Aristotelian relativity?

Garth

[Garth]

Well, I don't believe in Mach's principle, because GR has had such success that it seems to rule it out. To reintroduce it now at this late date is to say that GR is insufficient among classical theories, and what would lead you to think that?

So your electron on a lab table is accelerating, and radiating I presume (if you still believe in Maxwell), so where do you think its energy of radiation is coming from?

Garth

[selfAdjoint]

So your electron on a lab table is accelerating, and radiating I presume (if you still believe in Maxwell), so where do you think its energy of radiation is coming from?

Garth

The downward force of gravity on the electron is balanced by the upward force of the table, so the net acceleration of the electron is zero, and it doesn't radiate.

In general the Standard model accounts for electromagnetic radiation and conserves energy. Wherever the electrical energy "comes from", it can't be Mach's principle because that, in denying GR, contradicts experiment.

[Garth]

The downward force of gravity on the electron is balanced by the upward force of the table, so the net acceleration of the electron is zero, and it doesn't radiate.
hmmm.. Should not the frame of reference in which the electron is not accelerating be the locally Lorentzian freely falling one? The one in which physics is simple when analyzed locally? [MTW pg 4]


In general the Standard model accounts for electromagnetic radiation and conserves energy. Wherever the electrical energy "comes from", it can't be Mach's principle because that, in denying GR, contradicts experiment.

Unless that is there is another gravitational theory that fully includes Mach's Principle, which does not contradict experiment. May I commend Self Creation Cosmology which claims to be such a theory? [See posts/threads about it on these Forums]

[Garth]

It might be possible to form a curvature theory of electromagnetics if you start with the idea that each pair of charges is the source of the curvature rather than single particles.
Hi juju;
The Earth is full of charges and we would be either completely squashed by them or propelled into space, the electromagnetic force is so much stronger (~10^40) than the gravitational force. However we are not squashed/launched into space because the like and unlike charges cancel each other out. So I guess that a pair of charges probably wouldn't be a source of curvature because the nett 'force' or curvature would be zero.
Garth

[pervect]

So your observer at inifinity is co-moving with the Earth is she? A case for Aristotelian relativity?

Garth

I'm not sure if you are saying that there will be some radiation due to the earth's acceleration due to it's orbit around the sun (probably true, IMO, I wasn't being sufficiently nit picky when I responded) or whether you are talking about the relative velocity between the observer at infinity and the earth. The relative velocity shouldn't matter as far as the radiation issue goes - an electron moving at a constant velocity shouldn't radiate.

[pervect]

hmmm.. Should not the frame of reference in which the electron is not accelerating be the locally Lorentzian freely falling one? The one in which physics is simple when analyzed locally? [MTW pg 4]


Whether or not radiation exists depends on the coordinates, it's not a physical invariant. Photon number is conserved by the Lorentz boost, but not by arbitrary coordinate changes. It's possible for the observer accelerating and co-moving with the electron not to see any radiation, while another observer accelerating with respect to it will see radiation.

[Garth]

I'm not sure if you are saying that there will be some radiation due to the earth's acceleration due to it's orbit around the sun (probably true, IMO, I wasn't being sufficiently nit picky when I responded) or whether you are talking about the relative velocity between the observer at infinity and the earth. The relative velocity shouldn't matter as far as the radiation issue goes - an electron moving at a constant velocity shouldn't radiate.
Thank you - however it is the relative acceleration that is important.

I don't think you can solve the issue by taking the r -> infinity limit, unless that is that boundary is co-moving with the Earth and thus make the Earth the centre of the universe in some sense!

Whether or not radiation exists depends on the coordinates, it's not a physical invariant. Photon number is conserved by the Lorentz boost, but not by arbitrary coordinate changes. It's possible for the observer accelerating and co-moving with the electron not to see any radiation, while another observer accelerating with respect to it will see radiation.
Two questions: 1. "Has the latter been demonstrated experimentally?"

I am thinking here of a situation in which the mutual acceleration is not gravitational (that is an added complication) but caused by the observer physically being boosted by a force.

and 2. "What is the source of the energy of such radiation?"

When you accelerate an electric charge you are doing work on it and some of that re-appears as radiation, however when the acceleration is "passive" either because it is the observer who accelerates, or because it is sitting passively and stationary in a gravitational field on a laboratory bench, then I question the source of such energy.

Garth

[pervect]

I don't think you can solve the issue by taking the r -> infinity limit, unless that is that boundary is co-moving with the Earth and thus make the Earth the centre of the universe in some sense!


We are on different wavelengths here.

I'm a bit surprised you don't see why I've been talking about the observer at infinity.

Let me jog your memory a bit

In which case the second part of my question comes into play, if so, where does this energy of radiation come from?


The reason I brought up the observer at infinity was to answer the second part of your question. Do I really need to go through the whole spiel on asymptotic flatness and energy in GR again? I will if it serves some useful purpose - If I recall correctly you have your own theory with it's own view on energy conservation, but I'd hope you'd be interested in understanding the mainstream view. I believe I'm presenting the mainstream view reasonably fairly, but I'm not, alas, infallible. Anyway, if you want me to clarify this or talk about it more I will, but I'm hoping that pointing out my previous remarks on this topic will be enough.

Oh, yes, I guess I haven't mentioned what I see as "the solution". The main solution is that the detection or non-detection of radiation is observer dependent, it's not a physical invariant. It's also not strictly speaking a local pheomenon at all. Google finds (amusingly enough) pmb's webpage with a wide variety of quotes from the literature pointing out the observer dependent nature of the existence of radiation

http://www.geocities.com/physics_world/falling_charge.htm


Two questions: 1. "Has the latter been demonstrated experimentally?"


Not as far as I know. I believe there were some experiments proposed to measure Unruh radiation, but I don't think they have been carried out, they will be very difficult. Unruh radiation is also a digression from the topic, the mechanism is different, but it illustrates the main point that the existence or non-existence of radiation is observer dependent.


I am thinking here of a situation in which the mutual acceleration is not gravitational (that is an added complication) but caused by the observer physically being boosted by a force.


I *think* that such an observer should see fields that look like radiation, but since the problem is notoriously tricky, and since I haven't actually carried out any calculations, take this with a grain of salt.


and 2. "What is the source of the energy of such radiation?"

When you accelerate an electric charge you are doing work on it and some of that re-appears as radiation, however when the acceleration is "passive" either because it is the observer who accelerates, or because it is sitting passively and stationary in a gravitational field on a laboratory bench, then I question the source of such energy.

Garth

Well, let me jog your memory again here. Just to be sure we're communicating, do you recall what I think is a necessary and sufficient condition for energy to be conserved in GR? (You don't have to wade through all my posts, just this one, to answer this question).

[juju]

Hi Garth,

I was thinking of a situation where each pair of charges only curved the space between the charges. Thus, in the aggregate these would cancel, but still exist locally.

juju

[Garth]

We are on different wavelengths here.

I'm a bit surprised you don't see why I've been talking about the observer at infinity.
Earlier I'm surprised this issue hasn't been resolved. It seems to me that the static nature of the charge relative to the observer at infinity means that it should not radiate from the viewpoint of the observer at infinity.
But the observer in question is the one in the local gravitational field - in the laboratory. Indeed with the Earth itself in the Sun's/galactic/intergalactic gravitational fields is not the concept of an observer co-moving at inifinity rather hypothetical?

The reason I brought up the observer at infinity was to answer the second part of your question. Do I really need to go through the whole spiel on asymptotic flatness and energy in GR again? I will if it serves some useful purpose - If I recall correctly you have your own theory with it's own view on energy conservation, but I'd hope you'd be interested in understanding the mainstream view. I believe I'm presenting the mainstream view reasonably fairly, but I'm not, alas, infallible. Anyway, if you want me to clarify this or talk about it more I will, but I'm hoping that pointing out my previous remarks on this topic will be enough.
It is the physical reality that I am trying to understand, whether that is modelled by the mainstream view, or mine, or anybody else's individual theories.
GR is an example of Noether's improper energy theorems and conserves energy-momentum and not in general energy. If a time-like killing vector exists it is possible to define a concept that behaves like energy, the covariant time-component of the 4momentum vector but in many cases it is the contravariant time component that is defined as energy, especially when considering the total energy of a static gravitating body and field as measured by an observer ‘at infinity’. The situation is confusing because energy is not conserved in GR and our natural inclination to want it be so forces an unnatural definition on the theory.
Oh, yes, I guess I haven't mentioned what I see as "the solution". The main solution is that the detection or non-detection of radiation is observer dependent, it's not a physical invariant. It's also not strictly speaking a local pheomenon at all. Google finds (amusingly enough) pmb's webpage with a wide variety of quotes from the literature pointing out the observer dependent nature of the existence of radiation

http://www.geocities.com/physics_world/falling_charge.htm

Thank you for Pete’s link. If I can quote from the sources he quotes;
1. ‘Classical Radiation from a Uniformly Accelerated Charge, Thomas Fulton, Fritz Rohrlich, Annals of Physics: 9, 499-517 (1960)’
"An electron which falls freely in a uniform gravitational field embedded in an inertial frame will radiate, and one which sits at rest on a table in the same field will not radiate; and these two statements do not contradict each other."
2. ‘Radiation from an Accelerated Charge and the Principle of Equivalence, A. Kovetz and G.E. Tauber, Am. J. Phys., Vol. 37(4), April 1969’
“A nonvanishing energy flux is found only if the charge is freely falling and the observer supported, or vice versa”
So 2 is saying that an inertial observer i.e. freely falling – with no forces acting on her – will observe radiation from the desk bound (supported) charge. Whereas 1. says not, the observer in case 1 is presumeably supported.
But again whence the energy? The fact the GR does not conserve energy does not in itself explain where the energy received by the inertial observer comes from.

The situation is confused, which is why Kirk T. McDonald in “Hawking Unruh radiation and radiation of a uniformly accelerated” (Pete’s link) concludes, “We
now see that the quantum view is richer than anticipated, and that Hawking- Unruh radiation provides at least a partial understanding of particle emission in uniform acceleration or gravitation.”

A partial understanding is better than none – but the case is not closed!

Garth

[pervect]

I started to write a rather long and technical response, but it seems to me that I simply wind up repeating myself.

It's really very simple. If one is talking about system energy in GR, one is(or should be) talking about an asymptotically flat space-time to apply the usual formulas.

Let's take a very contrived example. Suppose we have a universe that consists of a single star (with optional planets), and a rocketship.

If the rocket accelerates, the star will move faster and faster. It will have more energy. The rocket's exhaust will have to exist, and it will have some energy, but by taking the limit where the star is a lot bigger than the rocket, it is obvious that the total energy of the rocket exhaust can be ignored, as can the rocket itself. So, in the limit, the energy of the universe from the viewpoint of the rocket is going up.

This is not a GR issue. The same thing happens in Newtonian mechanics. The only issue is, that we can no longer simply say "the rocket is not in an inertial frame" the way we used to in Newtonian mechanics. So we have to do something more complicated.

There's a perfectly good defintion for energy in GR with asymptotically flat space-times. You can also define energy in GR when you have only a time-like killing vector without asymptotic flatness, to some extent, but you have a calibration problem witht the formulas I've seen. (It's easy enough to come up with a conserved quantity when you have a time-like killing vector, but it's hard to scale this quantity correctly without an asymptotically flat space-time to serve as a reference. A conserved quantity remains conserved when multiplied by an arbitrary constant - setting the value of the constant in the usual manner requires asymptotic flatness).

In short, to define system energy, it is sufficient to have an asymptotically flat space-time (and it's close to necessary, as well as being sufficient).

If you really do want to understand reality, you should take a look at how asymptotic flatness defines energy in GR. You come up with oddball defintions of energy, then complain that they don't act the way you want. At least use one of the accepted notions of energy (such as Bondi mass, or ADM mass) - then complain that *they* doesn't work the way you want :rofl: .

As far as practical issues go, they aren't as bad as you think. Compare and contrast

acceleration of earth's sun due to galactic rotation (250 million year rotation period, 26,000 light years from center): 10^-10 m/s^2

moon's orbital accelration around earth: .002 m/s^2
earth-moon's acceleration around sun: .006 m/s^2

You're welcome to work out the galaxies acceleration due to the local group, if you can figure out how - I think you'll find that it's smaller than the sun's acceleration due to galactic rotation, so there's a general downward trend.

The trend is generally downward for the same reason that the night sky is black. (Of course, the reason the night sky is black isn't particularly obvious).

[Garth]

If the rocket ship in your example turns off its engine and free falls towards the star then in its frame of reference the star/universe's energy continues to increase.

In GR energy is not conserved, in general, even for inertial observers.
That is what I have been saying all along! It is one of the starting points of SCC.

However the issue with the supported charge and free falling observer is whether the radiation actually exists and can be measured. Therefore, can energy be taken out of the system? Is any work being done, either by the support of the charge or the deviation of the observer from its geodesic - and if not whence the energy? In other words, is this a free lunch?

I am glad to see you seem to have the situation sorted, others do not!

Garth

[pervect]

I'm going to take another shot at the more technical approach. But I've said it all before. I can still hope that it will "make sense" this time.


If a time-like killing vector exists it is possible to define a concept that behaves like energy, the covariant time-component of the 4momentum vector but in many cases it is the contravariant time component that is defined as energy


It's true that E0 is a constant of motion for particles following a geodesic. But this is a red herring. It does not lead to good defintion of the energy of a system. This should be no surprise, it's not even coordinate independent.

I'm assuming that is what we are interested in, yes? The energy of a system, not a constant of motion of a particle following a geodesic?

Several good and equivalent defintions for the energy M of a system with a timelike killing vector ka are:

a surface intergal

M = -\frac{1}{8 \pi} \int_S \epsilon_{abcd} \nabla^c k^d

here \epsilon_{abcd} is the Levi-civita tensor normalized to be a volume element of the space-time, this is the surface intergal of a two form.

a volume intergal

M = \frac{1}{4 \pi} \int_{\Sigma} R_{ab} n^a k^b dV


here Rab is the Ricci, and na is a unit future perpendicular to the volume element dV.


M = 2 \int_{\Sigma} (T_{ab} - \frac{1}{2} T g_{ab}) n^a k^b dV


here Tab is the stress energy tensor, gab is the metric tensor, and na remains the unit future perpendicular to the volume element dV.

Note that this last expression illustrates why you can't just intergrate Tab overe a volume of space and expect to come up with a system energy . (Except as an approximation that's only valid in the weak field case).

These can all be found in Wald, pg 288-291

Now that we've talked about the easy case of a system with a timelike killing vector, let's talk about the harder case - what if you don't have a timelike killing vector.

Well, the idea is simple. If you don't have a killing vector that's timelike everywhere, maybe you have a killing vector that's asymptotically timlike, at infinity. We can then apply the first definition directly. If turns out that if you have an asymptotically flat space-time that's a vacuum at infinity, you do have killing vectors which are asymptotically timelike. To formulate this rigorously requires a defintion of conformal infinity. In this case, we also have to specify "which infinity" (it's null infinity, in the jargon of conformal infinity). It turns out


Fortunately ... the asymptotic symmetry group of null infinity has a preferred 4 parameter subgroup of translations, so the notion of "an asymptotic time translation" is well defined.


So, if I may conclude with a few words

asymptotic flatness, asymptotic flatness, and asymptotic flatness, are the three keys to energy conservation in standard GR.

[Garth]

Agreed - which goes to show how slippery a concept energy is in GR.

However for ordinary energy 'book keeping', in experiments in the laboratory etc. we do not have the luxury of being able to integrate over the whole gravitational system out to an asymptotically flat null infinity.
We do experiments within a limited volume in which we keep an energy account and quite naturally expect energy to be conserved in such a closed system. It is in these cases that it is possible to question whether the approach of GR is adequate or not, and the detailed study of a supported electric charge in a gravitational field may be helpful in answering that question.

Garth

[Garth]

Hi Garth,

I was thinking of a situation where each pair of charges only curved the space between the charges. Thus, in the aggregate these would cancel, but still exist locally.

juju
Hi juju - your question and ideas have become a little lost in the discussion here!

You are taking the same approach as Einstein did when he tried to formulate a unified field theory. He wanted to explain the electromagnetic force in the same way as he explained the gravitational force by space-time curvature only he couldn't get it to work.
There is of course repulsion as well as attraction to explain together with the electric and magnetic fields. As pervect said earlier there is another theory, the Kaluza-Klein theory, which uses an extra dimension, and this helped towards developing string theory.
So yours are good fertile ideas.
Garth

[selfAdjoint]

You are taking the same approach as Einstein did when he tried to formulate a unified field theory. He wanted to expalin the electromagnetic force in the same way as he expalined the gravitational force by space-time curvature only he couldn't get it to work.


See this guy's website (http://www.einstein-schrodinger.com/) . He got it to work. He makes some striking assumptions (a very large cosmological constant, etc.) but it's all legitimate EUFT.

[Garth]

See this guy's website (http://www.einstein-schrodinger.com/) . He got it to work. He makes some striking assumptions (a very large cosmological constant, etc.) but it's all legitimate EUFT.
Thank you for that very interesting link, I shall download E & S's original papers and study them. Do you have any idea why, "This was supposedly disproven way back in 1953, but there are a few stubborn souls such as myself who still think it is correct, and who work to prove it", it was supposedly disproven?

A large cosmological constant might tie in with present questions about DE for example.
Garth

[juju]

Hi Garth,

Here's are some link to information about other theories that unify gravity and electromagnetics within the general framework of GR.

http://www.americanantigravity.com/davidmaker.shtml

http://www.compukol.com/mendel/

http://www.aias.us/

Also, I understand that Roger Penrose also believes that the path to a unified field theory lies through GR, rather than QM.

Thanx

juju

[selfAdjoint]

Thank you for that very interesting link, I shall download E & S's original papers and study them. Do you have any idea why, "This was supposedly disproven way back in 1953, but there are a few stubborn souls such as myself who still think it is correct, and who work to prove it", it was supposedly disproven?

A large cosmological constant might tie in with present questions about DE for example.
Garth

Neither Einstein not Schroedinger, I believe, were able to derive Maxwell's equations as a limiting case of their theories. But also consider Hlavaty's Geometry of Einstein's Unified Theory circa 1958, which is cited in the papers I linked to. Hlavaty claimed to have derived the "Maxwell" (i.e. Faraday) tensor, which turned out to be not at all obvious.

[pmb_phy]

If theorists (starting with A. E.) can make a theory about spacetime curvature caused by mass (GR), couldn't there be a similar theory where some spacetime curvature is caused by electric charges? Both are F = k/r^2 in elementary physics.
If there is a configuration of charges which has an associated energy distribution then the mass associated with that energy will curve spacetime. I am not conviced that a single charge can do that since the energy of a charge distribution is defined as the potential energy of the distribution and that energy is the energy required to assemble the distribution. One does not assemble a point charge. As a matter of fact if one follows the derivation of the ennergy in the electric field then one sees that it starts by assembling point charges by bringing them from infiity to a finite distance to each other. One starts with a single point charge and a starting energy of zero. One then ends up with a relationship for the energy in terms of a sumation. One then takes the limit and goes to an integral and one can then show that there is a one to one relationship between the total potetial energy (i.e. energy required to assemble the distribution) and the E field. But one assumes that if there are finite discrete charges then the initial energy of the configuration is zero. So there is no energy in the field of a point charge - At least in my opinion ... today ... as of 7:41am. However I had some percocet not to long ago so it might be the drugs. :biggrin:

I can justify it though. If one takes a dumbell consisting of two identical point charges held together by a rod then the momentum of the rod at low speed will be p = Mv where

M = 2me + mrod + mem

The first term represents the bare mass of the charges, the second the bare mass of the rod and the last is the mass of the EM field as measured in the rest frame. If the mass of the EM field includes the energy of the point charges then the mometum is zero. If the mass of the EM field includes only the mutual potential energy then the mass is finite. This makes sense for many reasons. If this actually true then EM field of a point charge carries no momentum either. For details see Griffiths and Owen's paper Mass renormalization in classical electrodynamics, Am. J. Phys.


A postulate could be that an electron in an elevator (made of electons, or a negatively charged inside surface) cannot tell the difference whether :

1. the elevator is stopped and that there is a large + charge underneath or :

2. the elevator is accelerating upwards

Where does this lead? What happens if you take magnetism into account?
I have no idea what your saying here but I will say this. The weight of a charged particle will depend on the spacetime curvature. If there is no spacetime curvature then the weight of the charge will not tell you if you are in a curved spacetime. But if there is spacetime curvature then the weight of the charge will be different given the same local acceleration. This is due to the fact that a charged particle does not follow a geodesic in spacetime. The field of the charge is not localized and thus the field can "feel out" the surrounding spacetime and is thus not a locall phenomena. Clifford Will wrote a paper on the weight of a charged particle in a Scharzchild spacetime.
It turns out that the weight is a function of charge (I think I recall that the charge weighed less but am not sure). Thus if one is in a box in a Schwarzchild spacetime then you can tell if you're not in an accelerating box in flat spacetime by using a charged particle and weighing it. This is not cheating the equivalence principle since the fields are not local and the equivalece principle, when applied to a curved spacetime, is a local phenomena and the field of a charge is not a local phenomena. Think of this as using the field of a charge to probe spacetime for curvature.


Pete

[Gonzolo]

I have no idea what your saying here but I will say this...Pete

Just trying to make an electric analogy with a mass in an elevator, which cannot tell whether it is accelerating or near a planet. I have pretty much given up on this (idea and thread) for the time being, due to some of the points that were mentioned + the fact I believe that there is a strict asymetry between masses and charges : m -> \gamma m, while q -> q (invariant?). So the analogy has many limits.

[pmb_phy]

Just trying to make an electric analogy with a mass in an elevator, which cannot tell whether it is accelerating or near a planet. I have pretty much given up on this (idea and thread) for the time being, due to some of the points that were mentioned + the fact I believe that there is a strict asymetry between masses and charges : m -> \gamma m, while q -> q (invariant?). So the analogy has many limits.

As I've explained, you can distinguish whether you're accelerating or near a planet (i.e. Schwarzchild spacetime, curved spacetime etc.). Just weigh the charge. The weight is spacetime curvature dependant.

Pete

[pervect]

Agreed - which goes to show how slippery a concept energy is in GR.

However for ordinary energy 'book keeping', in experiments in the laboratory etc. we do not have the luxury of being able to integrate over the whole gravitational system out to an asymptotically flat null infinity.
We do experiments within a limited volume in which we keep an energy account and quite naturally expect energy to be conserved in such a closed system. It is in these cases that it is possible to question whether the approach of GR is adequate or not, and the detailed study of a supported electric charge in a gravitational field may be helpful in answering that question.

Garth

From a practical or experimental point of view, I think the problem is more or less the opposite - finding strong enough fields / high enough curvatures that these effects manifest.

For instance, the weak-field approximation of integrating Tab to get the total energy is "good enough" to handle anything in the solar system, via the PPN approximation even though it's not the actual correct expression for the total energy of a system. Really strong fields (like those near a black hole), would be needed to find any departure even in theory.

Furthermore, there are experimental problems in finding, for instance, Tab in the Sun experimentally to carry out the intergration to verify the above statement. It's much more practical to compute the mass of the sun by observing satellite orbits and studying it's far gravitational field than it is to actually measure its stress-energy tensor, or even its conventional density, everywhere in its interior and integrate.

[nrqed]

Whether or not radiation exists depends on the coordinates, it's not a physical invariant. Photon number is conserved by the Lorentz boost, but not by arbitrary coordinate changes. It's possible for the observer accelerating and co-moving with the electron not to see any radiation, while another observer accelerating with respect to it will see radiation.

Let's say the charge is surrounded by detectors (let's say photomultipliers). Either the detectors will register a count can't be observer dependent. Either they click or they don't. But you are saying that some obersever could see the detector click while no radiation is emitted by the charge.

So I guess that from the point of view of that observer, the detector would be picking up vacuum fluctuations? Is that the usual explanation?

If this is correct, then it seems to me an amazing fact that such a simple consideration (thinking about a charge in the context of GR) leads to a quantum physics concept (quantum fluctuations)! It's as if trying to marry GR and E&M points to the need for quantum physics, and I have never seen things presented this way.

Pat

[nrqed]

The weight of a charged particle will depend on the spacetime curvature. If there is no spacetime curvature then the weight of the charge will not tell you if you are in a curved spacetime. But if there is spacetime curvature then the weight of the charge will be different given the same local acceleration. This is due to the fact that a charged particle does not follow a geodesic in spacetime. The field of the charge is not localized and thus the field can "feel out" the surrounding spacetime and is thus not a locall phenomena. Clifford Will wrote a paper on the weight of a charged particle in a Scharzchild spacetime.
It turns out that the weight is a function of charge (I think I recall that the charge weighed less but am not sure). Thus if one is in a box in a Schwarzchild spacetime then you can tell if you're not in an accelerating box in flat spacetime by using a charged particle and weighing it. This is not cheating the equivalence principle since the fields are not local and the equivalece principle, when applied to a curved spacetime, is a local phenomena and the field of a charge is not a local phenomena. Think of this as using the field of a charge to probe spacetime for curvature.


Pete

Very interesting point.

But then I got thinking and it seems to me that the wwight of *anything* will also depend on the curvature because, as far as I can tell, there is no way (even as a thought experiment) to measure the weight of anything in a purely local way (again, afaik). Any weight measurement involves a displacement (of a spring for example) so is not purely local. Therefore, in principle, any weight measurement will distinguish a curved spacetime from a noncurved one.

Unless I am missing something.

Pat

[pervect]

Let's say the charge is surrounded by detectors (let's say photomultipliers). Either the detectors will register a count can't be observer dependent. Either they click or they don't. But you are saying that some obersever could see the detector click while no radiation is emitted by the charge.

So I guess that from the point of view of that observer, the detector would be picking up vacuum fluctuations? Is that the usual explanation?

If this is correct, then it seems to me an amazing fact that such a simple consideration (thinking about a charge in the context of GR) leads to a quantum physics concept (quantum fluctuations)! It's as if trying to marry GR and E&M points to the need for quantum physics, and I have never seen things presented this way.

Pat

The phenomenon of radiating charges can be understood entirely classically. In fact, it's probably a bit easier to understand that way. One doesn't think of a wave as having a definite location, like a particle does. When you think of the wave as having particle properties, of arriving in a "packet" that is either there or not there, you are introducing the quantum element into an otherwise classical problem.

In quantum terms, I would describe Unruh radiation as particles coming from the vacuum, but I would not describe the radiation of accelerating charges in that way. There are some similarities in the mechanism, though. What's happening (in my interpretation, anyway) is that photons that are virtual from one viewpoint are no longer virtual from another, so that the virtual photons that produce the columb force in one coordinate system are seen as "real" (non-virtual) photons in another coordinate system. So the photons arise not from the charge and not the vacuum for the problem of radiating charges - but the mechanism is similar to Unruh radiation in that photons that are virtual in one coordiante system are real in another.

[Garth]

. What's happening (in my interpretation, anyway) is that photons that are virtual from one viewpoint are no longer virtual from another, so that the virtual photons that produce the columb force in one coordinate system are seen as "real" (non-virtual) photons in another coordinate system. So the photons arise not from the charge and not the vacuum for the problem of radiating charges - but the mechanism is similar to Unruh radiation in that photons that are virtual in one coordiante system are real in another.
Thank you for that pervect.

So the inertial observer freely falling past a supported electric charge receives real photons, whereas for the supported observer they are only virtual?
Is the inertial observer then drawing (real) energy from the false vacuum?

If so, would it be that for the supported observer, co-moving with the Centre of Mass (the 'Machian' frame) of the gravitating body, energy is conserved; whereas for the inertial freely falling observer it is not?

That would be my understanding of the situation, concordant with the principles of SCC.

Garth

[pervect]

Google turns up

http://prola.aps.org/abstract/PRD/v29/i6/p1047_1


The nature of the interaction between a quantum field and an accelerating particle detector is analyzed from the point of view of an inertial observer. It is shown in detail for the simple case of a two-level detector how absorption of a Rindler particle corresponds to emission of a Minkowski particle. Several apparently paradoxical aspects of this process related to causality and energy conservation are discussed and resolved.


unfortunately I don't have access to the full article :-(. But from what I can recall reading in Wald about the general topic of Unruh radiation (aka acceleration radiation), there is some mapping between positive and negative frequencies when one goes from the Rindler coordinate system of an accelerating observer to the Minkowski coordinate system of an inertial observer. So the terse result quoted in the abstract above is more or less what I'd expect, one observer sees a negative frequency particle as being absorbed, another observer sees a positive frequency particle as being emitted.

[Chronos]

Reflecting upon the original question, a simple experiment. Project a tightly focused laser on a sufficiently distant, sensitive target. Place a strong electromagnet near the source and in the path of the beam. Power up the electromagnet and rapidly alternate polarity and see if focal spot moves.

[Antonio Lao]

If theorists (starting with A. E.) can make a theory about spacetime curvature caused by mass (GR), couldn't there be a similar theory where some spacetime curvature is caused by electric charges?
If we view mass as the sum of equal amount of positive and negative electric charges then we can say that the curvature is caused by electric charges (equal amount of pluses and minuses).

It is a fact that black holes cause spacetime curvature. But it is also a fact that there are black holes with electric charges therefore we can also say that electric charges indirectly affect curvature. The question really is how much charges must be needed to be localized in a small region of spacetime in order to be comparable in strength with relativistic mass located at the same region.

[Gonzolo]

Reflecting upon the original question, a simple experiment. Project a tightly focused laser on a sufficiently distant, sensitive target. Place a strong electromagnet near the source and in the path of the beam. Power up the electromagnet and rapidly alternate polarity and see if focal spot moves.

Ok, but I would have to know what to expect as a result, before actually investing time in doing this. What strength of electromagnet to use? And how much of a distance shift should I expect if there is curvature (assuming there is a theory out there that predicts it). I would expect someone to have tried similar things already, so there might relevant litterature on the subject.

I have yet to find the time to formally introduce myself to general relativity, diff. goem. and these curvature theories, so if you can tell me and calculate an expected shift as a function of field strength, I'll see if its possible for me to do this.

Just to make it clear to all (I don't follow everything that has been said in the thread), I'm looking for an electric curvature that is independant of mass (and I would rather not bother with magnetic effects if possible). If an electric field curves a massless particle, that's pretty much up the alley I'm loitering.

[Antonio Lao]

Assuming magnetic field is zero, then the localized charge needed for unit electric field along one component of the energy-momentum tensor is given by

q= \frac{8\pi T_{11}}{r}

as r approaches zero-point region, the charges needed approaches infinity. This type of disparity in the charge polarization cannot be found anywhere in the universe and also is technologically impossible to construct. Not even the infinite energy of spontaneous vacuum polarization can be harnessed in an effective time period. Because of the infinitesimal time interval and the Heisenberg's Uncertainty Principle, the infinite energy from the vacuum polarization of electric charges cannot be used to effect the curvature of spacetime. Therefore, spacetime is flat if devoid of mass but not devoid of polarized electric charges.

This seems to indicate that at a deeper level, charge and mass are really the composites of a positive and a negative spacetime curvatures where the curvature is given as the inverse of Planck length. These two structures of curved spacetime cannot be separated in order to create an electric potential comparable to the gravitational potential which causes spacetime curvature in the large domain of GR. GR is really a macroscopic theory while the infinite curvature of spacetime is a theory of quantum gravity.

[Gonzolo]

Thanks Antonio. "Therefore, spacetime is flat if devoid of mass but not devoid of polarized electric charges." If this is according to regular GR, and everyone agrees, I'm pretty happy.

[pervect]

Thanks Antonio. "Therefore, spacetime is flat if devoid of mass but not devoid of polarized electric charges." If this is according to regular GR, and everyone agrees, I'm pretty happy.

I'm not quite sure what this is supposed to mean. Certainly charge affects the space-time geometry, as per the Reissner-Nordstrom metric

http://encyclopedia.thefreedictionary.com/Reissner-Nordstr%F8m

Let me try and explain this in English.

The charge on a black hole, Q, affects the metric, or if you prefer, the curvature, of space-time. This means that _uncharged_ particles, following a geodesic, will not follow the same path near a charged black hole as they will near an uncharged one with the same mass.

I should also add that it is usually believed that Q < M (in geometric units) for the above metric.

Note that the stress-energy tensor (and hence the Einstein and Ricci tensors) are _not_ zero as they are with the Schwarzschild solution, because the electric field E acts as a source term in Einstein's field equations. I.e. electric fields generate curvature

G^a{}_{ b} = 8 \pi T^a{}_{ b} = {\begin{array}{cccc}
- {\displaystyle \frac {Q^{2}}{r^{4}}} & 0 & 0 & 0 \\ [2ex]
0 & {\displaystyle \frac {Q^{2}}{r^{4}}} & 0 & 0 \\ [2ex]
0 & 0 & {\displaystyle \frac {Q^{2}}{r^{4}}} & 0 \\ [2ex]
0 & 0 & 0 & - {\displaystyle \frac {Q^{2}}{r^{4}}}
\end{array}}


Thus the various components of Ta b are proportional to the square of the electric field E, as E=Q/r^2, which is what one would expect.

I'm afraid I can't make any sense of Antonio's second post, I'm not sure what he's trying to say.

[Antonio Lao]

I'm afraid I can't make any sense of Antonio's second post, I'm not sure what he's trying to say.
what i mean, without going into a lot of mathematical jargons, is that infinite mass of GR leads to a singularity but infinite spacetime curvatures of positive and negative structures from polarization of localized gauge fields (hence structures for electric charge, weak charge, and color charge) lead to a viable theory of quantum gravity.

Note:

GR deals with only one singularity but a workable quantum theory of gravity will have to deal with two singularities for both positive and negative curvature.

[selfAdjoint]

AFAIK ther's no infinite mass in GR. The singularity inside a black hole is a finite mass confined in zero volume; infinite density.

[Antonio Lao]

selfAdjoint,

Thanks for your clarification.

[Gonzolo]

"This means that _uncharged_ particles, following a geodesic, will not follow the same path near a charged black hole as they will near an uncharged one with the same mass."

This is interesting. Are these charges necessarily paired up as dipoles? Or does it include the case of a net charge (+ or -)? This would apply no only to black holes, but also to a body with finite and measurable volume (i.e. a balloon or static-charged balloon), wouldn't it?

[pervect]

"This means that _uncharged_ particles, following a geodesic, will not follow the same path near a charged black hole as they will near an uncharged one with the same mass."

This is interesting. Are these charges necessarily paired up as dipoles? Or does it include the case of a net charge (+ or -)? This would apply no only to black holes, but also to a body with finite and measurable volume (i.e. a balloon or static-charged balloon), wouldn't it?

The metric I gave the URL for (Reisner-Nordstrom, abbreviated as R-N from here on) describes a charged black hole, a black hole with a net charge. Physically, if the BH has a positive charge, the negative charges would have to be located somewhere. Such charges are _not_ included in the metric I gave, however.

You would use the R-N metric to describe the external gravitational field of a highly charged spherically symmetric body, just as you would use the Schwarzschild metric to describe the gravitational field of an uncharged spherically symmetric body.

You'd model the electric field of such a black hole as Q/r^2, (just like the columb field) in the R-N coordinates.

This is a purely classical solution - it doesn't include quantum effects, such as particle pair production from intense electric fields (the Schwinger critical field limit).

[Gonzolo]

Thanks, this is just about my level of talk when it comes to GR.

(although, I would tend to believe it is in great part because I haven't chosen to completely dive into it yet...the 1916 paper is the only formal litterature I have, and I can confirm that it is not the best for an introduction to the subject...)

[pervect]

what i mean, without going into a lot of mathematical jargons, is that infinite mass of GR leads to a singularity but infinite spacetime curvatures of positive and negative structures from polarization of localized gauge fields (hence structures for electric charge, weak charge, and color charge) lead to a viable theory of quantum gravity.


This looks like a case where the mathematical jargon might be more comprehensible than the English :-).

From what I can make out you are not doing an analysis based on General Relativity. Exactly what it's based on isn't too clear at this point - here's an opportunity for a simple description, such as - string theory, M theory, loop quantum gravity, or none of the above.

[Antonio Lao]

pervect,

Can You help me visualize the various intersections of a surface of positive curvature and a surface of negative curvature?

[pervect]

I'll assume that the answer is "other theory" until I hear different.

The usual visual aid/3d representation for/of a negative curvature is a saddle surface, for a positive curvature a sphere. So I guess you'd have to visualze a sphere intersecting a saddle surface, though I"m not sure why or what you are visualizing.

[Antonio Lao]

pervect,

Since both the saddle and sphere are both described by some sort of parametric equations of differential geometry (a subject of which I have little knowledge of), their intersections can become simultaneous solutions of their interactions when the parameters are quantitified as time, mass, force, or energy functions. Can I take this for granted? I am looking for some minimum geodesics. Perhaps, a geodesic that is equivalent to the Planck length?

[raodhananjay]

I am a statistician, I wanted to know whether light is a wave or a particle ? And what is light in string theory ?

[selfAdjoint]

I am a statistician, I wanted to know whether light is a wave or a particle ? And what is light in string theory ?

Short answers. Yes. and The Same.

Longer answer, since early in the twentieth century light has been viewed as having both particle and wave properties. Which property you see depends on what experiment you do; if your experiment concerns, say, frequency or interference you will detect waves, and if you do an experiment based on momentum, you will detect particles. This idea actually came out of experiments, not theory, and it was only later includied in the growing theory of quantum mechanics.

String theory builds particles at the wave level out of string vibrations, so it basically supports this wave/particle duality.

[juju]

This means that _uncharged_ particles, following a geodesic, will not follow the same path near a charged black hole as they will near an uncharged one with the same mass.


This should, in theory, be testable using neutral particles in a large electric field. If it is true for black holes it should be true everywhere.

Here's another thought.

The curvatures (positive and negative) do not need to be thought of as being in the same direction as the gravitational curvature. They can be looked at as being orthogonal to it.

They can also be looked as a twisting of space/time, rather than a curvature.

juju

[gptejms]

I have a question:-from the little bit that I know,any physical field except the gravitational is a part of T_{mu,nu} and causes curvature.Has any experiment
ever been done which shows that the electromagnetic field causes space-time
curvature?Are static fields from charges also included in T_{mu,nu}?

[raodhananjay]

Thanks for the reply. Sorry I would love fire in some more so please ...
In the duality priniciple , how does it explain an energy dissipation ?

Another question is as how does Light travel continously for an infinitely long distance (say sun to earth )?

Atmosphere starts at some good distance above earths surface then in absence of vacuum , shouldn't the light get decayed ?
If you answer that its too powerful so it reaches us then shouldn't be too bright and hot above earth atmosphere.

[Nereid]

Welcome to Physics Forums raodhananjay!

A suggestion if I may: your questions are good ones, but are not related to relativity, or to whether (and to what extent) electric charges curve spacetime - perhaps you could start a thread with these questions, in Quantum Mechanics?In the duality priniciple , how does it explain an energy dissipation ? It doesn't, directly; dissipation involves absorption (and, maybe, re-emission).Another question is as how does Light travel continously for an infinitely long distance (say sun to earth )? As far as we can tell, there is no limit to how far light can travel. The furthest we've seen - and indeed, the further we can ever see - is the surface of last scattering (http://nedwww.ipac.caltech.edu/level5/Glossary/Essay_lss.html), approx 13 billion light-years.Atmosphere starts at some good distance above earths surface then in absence of vacuum , shouldn't the light get decayed ?Some of the photons incident on the top of the Earth's atmosphere are indeed absorbed or scattered; astronomers measuring the apparent magnitude of a star include a correction - called 'air mass' - in their data reductions.

[pervect]

I have a question:-from the little bit that I know,any physical field except the gravitational is a part of T_{mu,nu} and causes curvature.Has any experiment
ever been done which shows that the electromagnetic field causes space-time
curvature?Are static fields from charges also included in T_{mu,nu}?

T_{mu,nu} = T_{\mu \nu} is the stress energy tensor, and as was previously mentioned, yes, the static electric field generates terms in the stress-energy tensor. As was also mentioned, the contribution is a trace-free one. This means that Baez's ball of perfectly electrical neutral coffee grounds

http://math.ucr.edu/home/baez/gr/outline2.html

don't change in volume (gravitate together) because of the electric field, as the trace of the stress-energy tensor determines R00. However, the presence of the electric field does curve space-time.

[gptejms]

pervect,thanks for your answer.I repeat the second part of my question:-has any experiment ever been done which shows that the electromagnetic field causes space-time curvature?

[Nereid]

pervect,thanks for your answer.I repeat the second part of my question:-has any experiment ever been done which shows that the electromagnetic field causes space-time curvature?If you don't mind, I'd like to ask a slightly different question ... with our current experimental capabilities, *could* we do an experiment which would show that an 'electromagnetic field causes space-time curvature'? In principle, what would an experiment to test the idea look like?

[gptejms]

Your question 'suggests' to me that with our present capabilities,we can't produce such high intensity electromagnetic fields---enough to cause any significant spacetime curvature.Can you give an order of magnitude calculation to give us an idea?An experiment to test the idea could look like this---if light bends in region of such a high intensity electromag. field(perhaps we could concentrate on just the magnetic field and try producing extraordinarily strong magnets)then we would know that the curvature has been produced.But I am sure there must be better ways around.

[pervect]

There are some relevant experimental tests, but they may not be very satisfying. For instance, Eotvos type tests comparing aluminum and gold.

Columb energy, (i.e. the energy in the electrostatic field), which is proportional to [nuclear charge]^2, amounts in a gold nucleus to .4 percent of the mass, but only .1 percent of the mass in an aluminum nucleus. Very sensitive experiments have been done to detect any differences in gravitational forces on aluminium and gold, (testing the principle of equivalence), but none have been found.

[Nereid]

Your question 'suggests' to me that with our present capabilities,we can't produce such high intensity electromagnetic fields---enough to cause any significant spacetime curvature.Can you give an order of magnitude calculation to give us an idea?An experiment to test the idea could look like this---if light bends in region of such a high intensity electromag. field(perhaps we could concentrate on just the magnetic field and try producing extraordinarily strong magnets)then we would know that the curvature has been produced.But I am sure there must be better ways around.Let's think about this from the PoV of what the most intense EM fields and greatest spacetime curvature is, in the present universe, and later we'll examine whether any of these are amenable to tests - even in principle - of well-formed hypotheses. OK?

First, the most intense magnetic fields, in the present universe, are likely to be magnetars (http://solomon.as.utexas.edu/~duncan/magnetar.html), whose fields are likely to be as many OOM stronger than the strongest we can generate here on Earth is greater than the Earth's own field.

Next, the most energetic gammas may well have a comparable 'curvature effect' to objects whose gravity we can measure. For example, we have 'seen' (http://icrhp9.icrr.u-tokyo.ac.jp/results.html) TeV gammas from the Crab pulsar (or nebula), and expect that GRBs emit PeV or higher gammas (such energetic gammas probably cannot propogate clear across the universe, but as we've 'seen' a couple of nearby AGN in TeV, perhaps a nearby GRB might be 'visible' in PeV gammas). Homework question: if a TeV gamma were converted (magically) into a lump of baryons (yes, it would be magic!), what would the mass of that lump be?

Next (2), it may be that a 'long duration' GRB results in the formation of black hole whose mass is several times that of the Sun. If the progenator star had a magnetic field, it may be that, in the last few microseconds before the BH formed, that field reached an intensity which makes a magnetar's field look like a fridge magnet. Too, the spacetime curvature would be far more extreme than that around the Sun (which has provided the most sensitive tests of some aspects of GR to date).

Finally, as the two neutron stars in a binary merge/collide (they lose energy as gravitational waves; without some external event, a collision is certain), many kinds of extreme environments will likely be created.

Closer to home, you might like to read the reports of the Fundamental Physics Working Group (http://sci.esa.int/science-e/www/object/index.cfm?fobjectid=35858) (it's in the middle of the page), from the recent ESA Cosmic Visions workshop. Do you feel that any of the proposed (local) experiments would test a hypothesis related to spacetime curvature resulting from EM?

[Gonzolo]

You lost me with GRB and AGN.

But surely, one could calculate the deviation of a beam of light due to an earth-built electric or magnetic field, even if the deviation is in order of femtometers or less.

I finally received my first book on tensors. GR is next. Hopefully, I'll see what the difficulty resides.

[Nereid]

You lost me with GRB and AGN.Gamma ray burst (http://antwrp.gsfc.nasa.gov/apod/ap030414.html), active galaxy nuclei (http://antwrp.gsfc.nasa.gov/apod/ap020202.html) (another site (http://www.ulo.ucl.ac.uk/~diploma/year_one/heasarc.gsfc.nasa.gov/docs/objects/agn/agntext.html)).But surely, one could calculate the deviation of a beam of light due to an earth-built electric or magnetic field, even if the deviation is in order of femtometers or less.You mean, for example, moving a strong permanent magnet close to and away from a LIGO (http://www.ligo.caltech.edu/LIGO_web/PR/scripts/facts.html) arm?

[raodhananjay]

Thanks NEREID . Sorry , you are right , I guess should be asking related questions. Lets start now, I had been inspired by one of the posts , hence I started my queries here. Otherwise although the header " electric .. " of this thread made me curious , I am kind of lost here. Ok Is Somebody ready to do small revision or introduction of things you do here. PLEASE somebody do that for me.

[Gonzolo]

Gamma ray burst (http://antwrp.gsfc.nasa.gov/apod/ap030414.html), active galaxy nuclei (http://antwrp.gsfc.nasa.gov/apod/ap020202.html) (another site (http://www.ulo.ucl.ac.uk/~diploma/year_one/heasarc.gsfc.nasa.gov/docs/objects/agn/agntext.html)).You mean, for example, moving a strong permanent magnet close to and away from a LIGO (http://www.ligo.caltech.edu/LIGO_web/PR/scripts/facts.html) arm?


Exactly, but perhaps an electrically-generated magnet instead (MRI systems etc.). I would think that superconducting AC would generate the strongest human-made B-fields possible.

[Nereid]

Re-reading some of the posts near the start of this thread gave me the idea that perhaps we should be a little more precise about what any experiment is trying to test.

Gonzolo started this thread with a question about whether electric charge could 'curve spacetime', and in the last few posts we've been discussing whether a magnetic field can 'deflect' a beam of light. Would someone please be kind enough to explain why a non-null result in the latter would lead one to conclude 'yes, it can' for the former?

[Gonzolo]

I assume that in free space, all that can deflect a beam of light is space-time curvature. So whether the electric charge is moving (B-field) or not (E-field) can be seen as a detail (although certainly not for calculations). The question is for EM in general.

[juju]

Hi,

If EM fields curve space/time then time dilation effects should be seen. These would manifest in the changing of particle decay rates.

juju

[Calculex]

Re-reading some of the posts near the start of this thread gave me the idea that perhaps we should be a little more precise about what any experiment is trying to test.

Gonzolo started this thread with a question about whether electric charge could 'curve spacetime', and in the last few posts we've been discussing whether a magnetic field can 'deflect' a beam of light. Would someone please be kind enough to explain why a non-null result in the latter would lead one to conclude 'yes, it can' for the former?

I have a similar problem with the question. Curvature of space-time necessarily means that the frames of reference of all inertial observers would we affected. Since only charged masses would be affected by the presence of another electric charge, electrically neutral masses would measure time and space without being affected. How can this be a curvature of space-time?

Besides, Maxwell's equations state that the speed of propagation of an electromagnetic wave depends only on the values for \epsilon_0 and \mu_0 (the permittivity and permeability constants for space). These are experimentally derived and and neither of these constants depend on the magnitude of the respective electric or magnetic fields.

Calculex

[raodhananjay]

As light is an EMW I think (feel) the magnetic fields (extremely strong of course) may be able to deflect the light.

[pervect]

I have a similar problem with the question. Curvature of space-time necessarily means that the frames of reference of all inertial observers would we affected. Since only charged masses would be affected by the presence of another electric charge, electrically neutral masses would measure time and space without being affected. How can this be a curvature of space-time?


Uncharged masses *are*affected by the presence of electric charge (according to GR). This is why the Reissner-Norsdstrom metric for a charged black hole is different from the Schwarzschild metric for an uncharged black hole. Note the additional Q^2/r^2 terms in the Reissner-Nordstrom metric.

http://reissner-nordstrom-black-hole.wikiverse.org/

http://en.wikipedia.org/wiki/Schwarzschild_metric

This happens because electromagnetic fields are a form of energy, and thus contribute to the gravitatioanl field.

In GR, electromagnetic fields contribute to the stress energy tensor (the RHS of Einstein's field equations), which implies they contribute to curvature (the LHS of Einstein's field equations)

http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html


Now, the Einstein field equations are

Gmu,nu = 8pi Tmu,nu

Here Gmu,nu is the Einstein curvature tensor, which encodes information about the curvature of spacetime, and Tmu,nu is the so-called stress-energy tensor, which we will meet again below. Tmu,nu represents the energy due to matter and electromagnetic fields, but includes NO contribution from "gravitational energy".

[juju]

Hi all,

There is a problem when dealing with the EM energy of the field created by a charged particle,

If it is only potential energy activated by another charge, then there would be no effect on neutral particles and probably no curvature in the GR sense,

However, if it results from a scaler energy field, then even if the force vectors cancel from the fields of two or more charges, the scaler energy fields from all the particles still exist and would add, producing more curvature than would be calculated just from a resultant EM vector field.

juju

[Andrew Mason]

Uncharged masses *are*affected by the presence of electric charge (according to GR). ....

This happens because electromagnetic fields are a form of energy, and thus contribute to the gravitational field.
A charge in an electric field has energy which is the product of its charge and the electric potential of the field where the charge is located. The units of an electric field are Energy/charge (eg. volts or joules/coulomb). So is it correct to say that an electric field represents stored energy in the absence of another mass having a charge? I should think that there would be no energy represented by the field itself.

AM

[Stingray]

Yes, the field itself has a local energy density (at least in SR where such things are well-defined). Think of radiation for example.

Anyways, GR states that geometry couples to the stress-energy tensor, not "energy density," whatever that may be. Electromagnetic fields have a well-defined stress-energy tensor, so they do generate gravitational fields. They are usually very small, but were significant in the early universe. I think that the CMB experiments have verified this.

As for measuring the gravity of EM fields directly, I think that's hopeless for now. For order of magnitude purposes, it is ok to think of the mass as being due to an energy density. Then the effective mass is ~E/c^2. I don't think we can measure gravitational effects for things less than a few kilograms (I may be wrong!), which implies an energy of ~10^17 Joules. This must also be localized to within less than 10 cm or so. So the required energy density for a "tabletop" experiment is about 10^20 J/m^3. The electric field strength required is about 10^15 V/m. The requisite magnetic field strength is about 10^7 Tesla. Very large fields can be achieved for with special lasers, but they are so fleeting that I doubt any gravitational effect could be measured.

[pervect]

A charge in an electric field has energy which is the product of its charge and the electric potential of the field where the charge is located. The units of an electric field are Energy/charge (eg. volts or joules/coulomb). So is it correct to say that an electric field represents stored energy in the absence of another mass having a charge? I should think that there would be no energy represented by the field itself.


The field itself has energy, even if there is no charge in it. Consider two electric charges that repel each other. Bring them closer together. It requires work to do so. The energy to do this work is not lost, but it is stored as potential energy. Where is this energy stored? In the electric field.

The expression for the field energy of an electromagnetic field is given at


http://scienceworld.wolfram.com/physics/ElectromagneticField.html


Anyways, GR states that geometry couples to the stress-energy tensor, not "energy density," whatever that may be.


Yes, this is true.

For general information, the energy density (energy per unit volume) is one of the components of the stress-energy tensor.

If you represent a volume of space by a 4-vector orthogonal to the volume, the stress-energy tensor can be thought of as the density of momentum-energy per unit volume. Multiplying the stress energy tensor Tij by the vector Vi representing a volume element yields the energy momentum 4-vector contained in that volume.

[Nereid]

As for measuring the gravity of EM fields directly, I think that's hopeless for now. For order of magnitude purposes, it is ok to think of the mass as being due to an energy density. Then the effective mass is ~E/c^2. I don't think we can measure gravitational effects for things less than a few kilograms (I may be wrong!)AFAIK, tests (and measurements?) of gravity are now into the 100 micron, 1 mg range (e.g. this interesting summary (http://www.phys.lsu.edu/mog/mog22/node9.html))which implies an energy of ~10^17 Joules. This must also be localized to within less than 10 cm or so. So the required energy density for a "tabletop" experiment is about 10^20 J/m^3. The electric field strength required is about 10^15 V/m. The requisite magnetic field strength is about 10^7 Tesla.10^7 Tesla = 10^11 Gauss, which is far, far above what's achieved here on Earth (compression via explosives); however, it's weak for a neutron star, and mere noise near a magnetar (http://solomon.as.utexas.edu/~duncan/magnetar.html) .Very large fields can be achieved for with special lasers, but they are so fleeting that I doubt any gravitational effect could be measured.Aye, that's the real challenge - how to build an experiment to measure a gravitational effect that's both weak and fleeting? Perhaps repetition and nulling (differencing) are the answers?

[Andrew Mason]

The field itself has energy, even if there is no charge in it. Consider two electric charges that repel each other. Bring them closer together. It requires work to do so. The energy to do this work is not lost, but it is stored as potential energy. Where is this energy stored? In the electric field.

You have merely pointed out that a charged particle in an electric field has potential energy. My point was that a static electric field which surrounds a charged mass contains no energy UNLESS you bring another charge into it.

When there is a charge in an electric field, that charge has electric potential and one can think of the energy being stored in the field, with its units being that of kQq/r. When there is no charge q in the electric field created by charge Q, I don't see how the field can have any energy.

One way to ask the question is this: when a charge is created, does energy go into creating its electric field? (e.g neutron decaying to proton, electron and antineutrino?). I don't know the answer to that. I am just wondering.

AM

[Nereid]

One way to ask the question is this: when a charge is created, does energy go into creating its electric field? (e.g neutron decaying to proton, electron and antineutrino?). I don't know the answer to that. I am just wondering.It's interesting that 'charge' is created in pairs, in this example, and in all cases (?).

A somewhat similar situation is the 'destruction' of 'mass', e.g. annihilation (e- + e+ -> 2 photons, for example) ... from the GR perspective there is no loss of 'that which creates spacetime curvature', merely conversion of one form (two particles) to another (two different particles, moving at c).

So perhaps a different way of asking this is whether the 'QM' conservation laws (e.g. charge, those 'built in' to the weak and strong forces) are 'built in' to GR?

[pervect]

This gets into some interesting but advanced territory. It's fairly well known that the field energy of a charged point particle (usually, this is talked about as the self-energy of the electron) is infinite.

Google finds the archived physicsforum post

http://www.physicsforums.com/archive/t-29348_Electrostatic_Field_Energy_of_Electron.html

as a previous discussion about this point.

This issue requires quantum mechanics to resolve - unfortunately, GR isn't a quantum theory.

[juju]

This gets into some interesting but advanced territory. It's fairly well known that the field energy of a charged point particle (usually, this is talked about as the self-energy of the electron) is infinite..

This is only true if it is assumed that the field structure close to the convergence point of the field (the point particle) is the same as the field structure further away (the normal field structure.)

juju

[Garth]

This may be of interest:
"Space-time Curvature of Classical Electromagnetism"
http://arxiv.org/abs/gr-qc/0410043

Garth

[selfAdjoint]

Weyl's conformal tensor?

[pervect]

Weyl's conformal tensor?


It's the usual Weyl tensor, the traceless part of the Riemann. Someone explained to me once why it's sometimes called a "conformal tensor", but alas I don't recall the details.

http://mathworld.wolfram.com/RiemannTensor.html

reviews the breakdown of the Riemann into the Ricci and the Weyl tensors.

Unfortunately this is about as far as I've gotten with the paper so far, I don't understand the main point of it yet.

[edit]

The key term I'm missing the meaning of is
"the same support as the matter", it probably has something to do with homology which I don't know much about.
[end edit]

[Stingray]

The key term I'm missing the meaning of is
"the same support as the matter", it probably has something to do with homology which I don't know much about.


If you're asking what support means, that's the portion of a function's domain where it is nonzero. The sentence is saying that whatever it is discussing is nonzero only where there is matter.

The Weyl tensor is also called the conformal tensor because it is invariant under conformal transformations (multiplying the metric by a scalar function).

[pmb_phy]

You have merely pointed out that a charged particle in an electric field has potential energy. My point was that a static electric field which surrounds a charged mass contains no energy UNLESS you bring another charge into it.
The energy of a electric field is proportional to the integral of the square of the E-field over the region that the field doesn't vanish. Following the derivation which leads to that conclusion leads one to believe that what you say is true. However if it were true then this proves to be troublesome. It would seem to imply that the field had no momentum. But if the field has no momentum then the principle of momentum conservation would be violated.

When there is a charge in an electric field, that charge has electric potential and one can think of the energy being stored in the field, with its units being that of kQq/r. When there is no charge q in the electric field created by charge Q, I don't see how the field can have any energy.
There are various ways to express EM energy, not all of which include the integration mentioned above. As such there are different ways to look at the energy involved. Shadowitz covers this in his EM text.

To be quite literal - Look at how the association of energy<-> field came to be. One starts with a single point charge and assigns a total potential energy of this isolated system to be zero. One then brings in a charged particle from infinity to a nearby point. The work done will equal the change in the potential energy of the system. It makes no sense to say "The potential energy is at this place..etc". Bring in more charges - there will be more potential energy. Now let the charges become infinitesimal and let the number of them approach infinity. The discrete charge distribution then becomes a continuous charge distribution. Then the potential energy can then be expressed as being proportional to the integral of the square of the E-field over all space. If one then applies this relation to the field of a point charge then the integral diverges. What this means physically is that you tried to assemble a point charge. The work required to do that would be infinite. But now recall what this energy means. We *defined* the energy of a point charge to be zero. Thus it would seem reasonable to hold that the energy associated with the E-field of a point charge is zero.

However I have been unable to justify this assertion since when I attempted to do so it led to other problems (I can't recall exactly but I do remember trouble in doing so).

Pete

[4newton]

All observation indicates that there is no basic difference between forces, charge, mass, magnetic. All these forces act on the mechanisms that produce the forces in the same way. They all act over infinite distance. The only difference is the strength of the forces and their geometry.

[Nereid]

Q: "Does an electric charge 'curve spacetime'?"
A: GR predicts this, however the effect is way, way smaller than anything we can measure (or even detect) with 'lab experiments' here on Earth.

So, remaining questions? My €0.02's worth:
- what are the experimental limits on 'spacetime curvature' by electric (or magnetic) charges?
- what astronomical observations or space-based experiements, in principle, could be made/done to measure the GR predicted effect?
- how might the effect be indirectly detected (other than by experiments and observations which test GR in general)?

[pervect]

I'd add "anything we can detect directly" to your summary. Part of the mass of gold and aluminum atoms is due to their electromagnetic field, so the fact that there is no difference in the Eotovos experiments for these two elements tells us *something* about gravitation and energy.

One could also try and question whether or not gold and aluminium have the predicted mass change due to their electromagnetic energy. I'm not aware of the experimental details here, but it looks to me like there is enough precision in the measurements where the total electronic binding energy should be measuarable - though it may be tricky to separate from the nuclear binding energy.

[Nereid]

I'd add "anything we can detect directly" to your summary. Part of the mass of gold and aluminum atoms is due to their electromagnetic field, so the fact that there is no difference in the Eotovos experiments for these two elements tells us *something* about gravitation and energy.

One could also try and question whether or not gold and aluminium have the predicted mass change due to their electromagnetic energy. I'm not aware of the experimental details here, but it looks to me like there is enough precision in the measurements where the total electronic binding energy should be measuarable - though it may be tricky to separate from the nuclear binding energy.Thanks Pervect, I'd forgotten your earlier comment on this. IIRC, these experiments (there've been several, over the decades) have just the results predicted from GR (to parts per thousand? or better??), but, as you say, the interpretation is more about nuclear binding energy than electronic ... hmm, worth hunting down some of the papers?

[pervect]

I suppose the other thing we ought to mention is Mallet's idea for detecting the gravitational effect of light. The problem is that I'm not sure whether or not his idea will work. It appears (see the Mallet thread) that Mallet's analysis has some major problems, I expect the time machine results to disappear when these are corrected, but it's unclear to me whether or not the rest of his results will likewise vanish.

[Andrew Mason]

One could also try and question whether or not gold and aluminium have the predicted mass change due to their electromagnetic energy. I'm not aware of the experimental details here, but it looks to me like there is enough precision in the measurements where the total electronic binding energy should be measuarable - though it may be tricky to separate from the nuclear binding energy.
Is this not just a matter of proving the invariance of the speed of light? Einstein proved mathematically that the principle of relativity implies that the absorption/release of energy by a body increases/reduces its mass by E/c2

So, in a real sense, Michelson Morley confirms that electronic binding energy contributes to the mass of the atom.

But, getting back to the question, the issue here is whether a single electric charge curves space time. A body with electric charge of mass m and charge q and a non charged body of mass m will, according to GR, curve space time in exactly the same way.

Now, to the extent that the charge q contributes to the body's energy, and therefore its mass, the charge q will contribute part of that mass, m. Hence, it would contribute to the curvature of space-time. But my point is that I don't see how a single charge contributes energy. You need two charges separated by a distance in order to have electrical energy.

Andrew Mason

[pervect]

Is this not just a matter of proving the invariance of the speed of light? Einstein proved mathematically that the principle of relativity implies that the absorption/release of energy by a body increases/reduces its mass by E/c2

So, in a real sense, Michelson Morley confirms that electronic binding energy contributes to the mass of the atom.


I would be very, very, very surprised if electronic binding energy didn't obey the standard formula E=mc^2.

That said, I think one needs a stronger statement than the constancy of the speed of light to conclude that m=E/c^2. One needs at a minimum to assume that *all* physical laws, not just the speed of light, are the same for moving observers and stationary observers. I'm not positive if even that's sufficient - I think one has to assume additionally that the laws of physics are given by an action principle, and that the laws of physics do not change with time, or with location in space. The last two assumptions are more-or-less necessary for the laws of physics to be the same for differently moving observers, but we might as well be explicit about all the assumptions up front.


But, getting back to the question, the issue here is whether a single electric charge curves space time. A body with electric charge of mass m and charge q and a non charged body of mass m will, according to GR, curve space time in exactly the same way.


Look up the metric for a charged black hole. Look up the metric for an uncharged black hole. [I believe I've posted links to them elsewhere in the thread BTW.]. If what you said above were correct, then the metric of a charged black hole would have to be the same as the metric of an uncharged black hole. But they metrics are not the same.

[Chronos]

This is all consistent with the equivalence principle. A charge, or magnetic field has the same gravitational effect as the mass equivalent of the field strength, which is obviously a very minute effect and probably unmeasureable by any technology I can imagine.

[Andrew Mason]

That said, I think one needs a stronger statement than the constancy of the speed of light to conclude that m=E/c^2. One needs at a minimum to assume that *all* physical laws, not just the speed of light, are the same for moving observers and stationary observers.

Einstein made only one basic assumption: that the laws of electro-magnetism were the same to all inertial observers. I am not aware of any other assumption that is necessary to his conclusion "m=L/c2"

Look up the metric for a charged black hole. Look up the metric for an uncharged black hole. [I believe I've posted links to them elsewhere in the thread BTW.]. If what you said above were correct, then the metric of a charged black hole would have to be the same as the metric of an uncharged black hole. But they metrics are not the same.
I am not suggesting the analysis of black holes might lead to such a result. I don't pretend to understand the math involved in GR well enough to question it. I am just saying that I don't understand the physical basis for charge alone having energy. I was hoping someone might explain how it does.

AM

[pervect]

I am not suggesting the analysis of black holes might lead to such a result. I don't pretend to understand the math involved in GR well enough to question it. I am just saying that I don't understand the physical basis for charge alone having energy. I was hoping someone might explain how it does.

AM

The point isn't that charges have energy - the point is that fields have energy. Or, if you prefer, fields act just exactly as if they have energy.

The simplest discussion I could find in any of my textbooks was not particularly simple, and was based on Maxwell's equaitons.

Look for "Poynting's theorem" in an E&M textbook.

It starts with the vector calculus identity

\nabla \cdot (A \times B) = B \cdot (\nabla \times A) - A \cdot (\nabla \times B)

now we let E = A and B = H. Then we get

\nabla \cdot (E \times H) = B \cdot (\nabla \times E) - E \cdot (\nabla \times H)

and we substitute

\nabla \times E = -\frac{\partial B}{\partial t}
\nabla \times H = J + \frac{\partial D}{\partial t}

We wind up with

- \nabla \cdot (E \times H) = E \cdot J + (E \cdot \frac{\partial D}{\partial t} + H \cdot \frac{\partial B}{\partial t})

This can be expressed in intergal form

-\oint (E \times H) \cdot n \, da = \int_v E \cdot J \, dv + \int_v (E
\cdot \frac{\partial D}{\partial t} + H \cdot \frac{\partial B}{\partial t} ) \, dv

The right hand side is the rate at which work is being done on the charges. the power.

Power = Voltage * current = (E * dl) . (J * da) = (E . J) dv is the most obvious example.

We add to this two other terms - a similar voltage*current term, but with the displacement current, and finally the rate at which magnetic fields do work.

The left hand side is the Poynting vector, the surface integral of which gives the amount of energy being transfered into the volume.

The right hand side can be re-written in the usual isotropic media, where D = eo E and B = u0 H as

\frac{d}{dt} \int_v (\frac{1}{2} \epsilon E^2 + \frac{1}{2} \mu H^2) \, dv

And this term can therfore be interpreted as the total energy stored in the volume V.

Which is why the total energy is proportional to e E^2 + u H^2.

Which can also be seen to be the right answer from the bare result presented at:

http://scienceworld.wolfram.com/physics/EnergyDensity.html

[Andrew Mason]

The point isn't that charges have energy - the point is that fields have energy. Or, if you prefer, fields act just exactly as if they have energy.

The simplest discussion I could find in any of my textbooks was not particularly simple, and was based on Maxwell's equaitons.

It looks to me that you are describing time dependent electromagnetic fields. Time dependent EM fields, of course, propagate as EM radiation. So it is apparent that time dependent electro-magnetic fields represent energy. The Poynting vector represents energy flow rate through a surface.

But I thought we were talking about non-time dependent electric field of a point charge.

If so, we are talking about a situation in which current densities are 0, charge density, \rho \rightarrow \infty. So \frac{\partial B}{\partial t} = 0 and \frac{\partial E}{\partial t} = 0 and \frac{\partial D}{\partial t} = 0

In a real situation, the charge density is not infinite, so there is some charge distribution in a finite volume. This represents energy because we have charges separated by distances. The energy of such a charge distribution will, therefore, contribute mass to the charges.

AM

[pervect]

It looks to me that you are describing time dependent electromagnetic fields. Time dependent EM fields, of course, propagate as EM radiation. So it is apparent that time dependent electro-magnetic fields represent energy. The Poynting vector represents energy flow rate through a surface.

But I thought we were talking about non-time dependent electric field of a point charge.


The Poynting vector does indeed describe the rate at which energy flows, that's on the left hand side.

The right hand side has two terms. The first is just voltage*current (the volume intergal of E . J dv. The second term is

\frac{d}{dt} \int_v (\frac{1}{2} \epsilon E^2 + \frac{1}{2} \mu H^2) \, dv

When there is no energy flow, this term is zero. When there is energy flow, energy is being either stored or released from storage.

One can (and does) interpret this term as the energy stored in the electromagnetic field.

Consider a capacitor being charged, for instance, and look at the above equations to see where it is going.

We know that power is flowing into it (V * I). There is no actual current flow in the capacitor through the dielectric - J is zero - the capacitor is not dissipating any energy (like a resistor does).

But while J is zero, there is still a displacement current \frac{\partial D}{\partial t} through the dielectric. This current represents energy that is being stored in the capacitor. You can think of this energy as being stored in the electric field of the capacitor with a magnitude of .5*e0*E^2 per unit volume between the plates of the capacitor. (Plus possibly some small amount of energy stored in the fringing fields).

It's fairly clear that a capacitor does store energy. The above formula gives the right answer for the amount of energy a capacitor stores.

A charged capacitor is a "static" system, but it still has stored energy, waiting to be released - just connect a load across it, and the presence of the energy will make itself manifest.

The other term involving H^2 (often rewritten in terms of B^2 which is proportional to H^2) gives magnetic energy storage. This describes the energy stored in an inductor.

Note that as I mentioend before, there are issues with point charges. At zero radius, the energy in the electric field is infinite using classical formulas. The classical electron radius is the radius at which the energy in the electric field of the electron is equal to it's mass. It's about 2.8*10^-15 meters.

see for instance

http://scienceworld.wolfram.com/physics/ElectronRadius.html

GR doesn't deal well with point charges, just as classical electrodynamics doesn't. If one is looking at charges at distances much greater than the classical electron radius, GR and classical electrodynamics will give a good answer to the problems one formulates. One can approximate electrons as having the classical electron radius, rather than being point particles.

If one is looking at distances closer to an electron than the classical electron radius, one probably needs a different theory - quantum electrodynamics, or quantum gravity, depending on whether one is looking at the electric or magnetic field.

[Andrew Mason]

When there is no energy flow, this term is zero. When there is energy flow, energy is being either stored or released from storage.

One can (and does) interpret this term as the energy stored in the electromagnetic field.
There is no question about that. A charged capacitor represents a charge distribution - charges separated by distances. The energy is stored in the electric field between the charges. But the field surrounding a charged capacitor should be essentially 0 (there is a small electric dipole effect because the + and - charges are separated by a distance). We are talking about a large charge (+ or -) sitting in space all by itself.

It appears from all of this, that:

1. A quantity of charge Q distributed within a finite volume of space V contains energy, E.

2. This energy, E contributes to the mass within that volume of space, V by the factor mQ=E/c2

3. E is contained in the electric field within V

4. This mass contributes to the curvature of space time external to V (ie it is caused by the total mass contained in V, which includes mQ).

5. There is zero energy in, and 0 mass contributed by, and 0 contribution to the curvature of space-time by, the electric field in the space outside of V.

The classical electron radius is the radius at which the energy in the electric field of the electron is equal to it's mass. It's about 2.8*10^-15 meters.

If one is looking at distances closer to an electron than the classical electron radius, one probably needs a different theory - quantum electrodynamics, or quantum gravity, depending on whether one is looking at the electric or magnetic field.
Isn't the calculation of the energy of the electric field of the electron based on the classical model of a volume of space (the size of an electron) having point charges distributed throughout that volume?

AM

[pervect]

There's one other point I should mention. The stress-energy tensor, the right hand side of Einstein's equation, is actually a 4x4 matrix. In many situations, only one term of this tensor, the energy density per unit volume, T00, is important. This is not the case for the electromagnetic field.

For the case of an electrostatic field, additional diagonal terms T11, T22, T33 can be as large as the energy density term, T00.

You'll see negative terms (tension) along the diirection of the field lines, but positive terms (pressure) in the two directions perpendicular to the field lines. The off diagonal terms will be zero for an electrostatic field (they are given by the Poynting flux).

These pressure and tension terms also cause gravity.