Is there a difference between normal and anti-gravitating geodesics?

[Careful]

Hi Vanesch,

I thought the latter was your version (and I have some comments which might be useful):

**
1) there is a 4-dim manifold, called spacetime, and I have a way of describing a finite patch around me, with 4 coordinates. I can of course consider different ways to describe these coordinates and then there is of course a smooth mapping between these two sets of coordinates (4 functions of 4 variables).

2) concerning gravity, all I need to know is given by the metric tensor expressed in my 4 coordinates, over this finite patch. All what is "in gravity" is encoded in this metric. As this is a tensor, its transformation under a smooth coordinate transformation is given by the normal rules.
**

The equivalence principle rather consists out of many more parts :
(a) inertial mass = passive gravitational mass, or : the motion of a (electrically) neutral test particle is independent upon the internal structure of the particle (although this version is open for obvious criticism)
(b) covariance principle : the laws of nature have to be written in tensorial form (criticism : no spinors )

(a) + (b) = weak equivalence principle

(c) the result of a local, non gravitational experiment is independent of the speed of the free falling reference system in which it is done (local Lorentz invariance)

(d) the result of a local and non - gravitational experiment is independent of the place and time in which it is performed (local position invariance).

All four together form the strong equivalence principle (or Einstein equivalence principle) which is broken in many ``gravitational´´ theories. For abbreviation, let's speak about WEP and SEP and let me show how violation of SEP - but not of WEP - has some bearing upon your point (2).

For example (d) is very strong and practically restricts severely the coupling of matter to gravitation (no coupling to the curvature tensor). Take the simple case of a point particle, with the standard lagrangian:

m int( dt sqrt( g( dx/dt , dx/dt ) )

one could change this to

int (dt m sqrt( 1 + a R( x(t) ) ) sqrt( g( dx/dt , dx/dt ) )

with the dimension of a = length^2. Actually, in this way you can get the one particle schroedinger equation out of Weyl geometry. The latter lagrangian violates (d) but not WEP. SEP is also violated by the inflaton for example.

Another way to violate (d) would be the addition of a torsion field to the connection (this again was one of the old geometric attempts to incoorporate QM / there were of course other attempts by adding an anti symmetric part to the metric tensor). Both examples violate your second principle : the former since it is not sufficient to know just the metric to determine the coupling to the gravitational field while no other physical field is present, the latter since we add an extra physical field.

Now, torsion does not contribute to the geodesic equation, hence I thought in the beginning she might have added some curvature form to the action of the anti gravitating particle. But of course, she does none of these : basically what happens in the paper is that a rather unconventional isomorphism (that shuffles through contractions and so on) is set up between the two bundles (hence nothing changes).

Furthermore, it needs to be said that such changes (involving the curvature) in the equation of the free particle will involve violations of causality (cfr the de Broglie mass problem). Anyway, I hope I added something here to your understanding of the different versions of the equivalence principle.

 

[vanesch]

one could change this to

int (dt m sqrt( 1 + a R( x(t) ) ) sqrt( g( dx/dt , dx/dt ) )

with the dimension of a = 1/length^2. Actually, in this way you can get the one particle schroedinger equation out of Weyl geometry. The latter lagrangian violates (d) but not WEP. SEP is also violated by the inflaton for example.

I fail to see how this violates my "2)" that the local equation of motion is not derivable from the local metric over a finite patch. After all, under the hypothesis of no torsion, I can DERIVE the Riemann tensor from the metric tensor (over a finite patch of course, not in 1 point).
So it seems that this R(x(t)) can be calculated, if the metric is given over a finite patch, no ? From g, I calculate the Riemann tensor, and hence the Ricci scalar (that's what R(x(t)) is, no ?). So R(x(t)) is just a complicated expression of the metric and its derivatives, or am I wrong here ?

 

[Careful]

I fail to see how this violates my "2)" that the local equation of motion is not derivable from the local metric over a finite patch. After all, under the hypothesis of no torsion, I can DERIVE the Riemann tensor from the metric tensor (over a finite patch of course, not in 1 point).
So it seems that this R(x(t)) can be calculated, if the metric is given over a finite patch, no ? From g, I calculate the Riemann tensor, and hence the Ricci scalar (that's what R(x(t)) is, no ?). So R(x(t)) is just a complicated expression of the metric and its derivatives, or am I wrong here ?

It violates your (2) in the sense that even if the metric is the only dynamical variable involved in gravitation, then still the Einstein equations (``that what is in gravity´´) are not fixed. If your statement was kinematical, then there is not contradiction. But if you meant the latter, then the concerns you expressed towards the a priori impossibility for a distinction between gravitating and anti gravitating matter are false. The fact that they are the same in THIS case requires more that simply your version of the (weak + (c)) equivalence principle.

 

[vanesch]

It violates your (2) in the sense that even if the metric is the only dynamical variable involved in gravitation, then still the Einstein equations (``that what is in gravity´´) are not fixed. If your statement was kinematical, then there is not contradiction.

I'm only involved here with the kinematics, indeed, with small test particles.
I didn't mean to imply that the Einstein equations are fixed!

EDIT: but even then, I don't understand your remark. After all, the lagrangian IS now expressed purely in terms of the metric and its derivatives, so how does this not fix the dynamics ?

 

[Careful]

I'm only involved here with the kinematics, indeed, with small test particles.
I didn't mean to imply that the Einstein equations are fixed!

Ok, then there is no misunderstanding I only wanted to say why I was careful in the beginning with my comments towards the existence of possible differences due to tidal effects or not.



Careful

 

[Careful]

Reply to your EDIT: how to build the right hand side of the Einstein equations ? You start out with the energy momentum tensor for the dynamical object at hand in Minkowski and apply the minimal substitution principle (this garantuees that (d) is satisfied). So, the SEP + the knowledge of the dynamics in Minkowski fixes the dynamical coupling to gravity. However, if you abandon (d), this is not the case anymore. Concretely, the free particle has a unique coupling to gravitation given SEP, but given WEP one has more exotic possibilities such as the one I gave. Both alternatives are ``free particles´´ but they respond differently to a gravitational field.

 

[Rade]

...I'm in fact not so much discussing your paper, I'm discussing the simple fact that you claim that there can be particles falling UP at the surface of the earth, that this is not in contradiction with the equivalence principle, and that this is a gravitational effect...

I had mentioned before the possibility of "up" and "down" particle interactions between gravity "down interaction" and antigravity "up interaction" when you consider the particles that interact to be clusters of nucleons (protons and neutrons). Thus a matter cluster [PNP] (helium-3) may bind with an antimatter cluster [NP] (anti deuteron) via a gravity + antigravity interaction--but of course there is neither mathematics nor experiments ever attempted to falsify such a claim--thus remains an open question for future scientific study. Who has time to develop the mathematics for such interaction of rotating nucleon clusters using gravity and antigravity ?

 

[Chronos]

Pardon me for being dense, but does not an anti-gravitating particle wreak havoc on the energy conservancy principle? It looks very unphysical to me.

 

[vanesch]

What I understand of Sabine's work is that she arrives at a potentially different equation of motion for a "free" anti-gravity particle (further on called a sabinon ), than the usual geodesic which is postulated for a normal particle in GR.
I do not mind this! The problem I have is not so much that there could be a different equation of motion than the geodesic. The problem I have with her proposal is that this equation of motion seems not to be derivable from the metric in a local patch of spacetime. Now, or she admits this - and then I have no problem with it - but it is hard to see how she can claim then not to shatter to pieces ENTIRELY the equivalence principle. For instance, you could introduce also instantaneous action in GR, using an extra field that tells you how to make this unambiguous (essentially an ether theory then).
Or she claims that this is not the case, that you CAN derive this equation of motion from the local metric over a finite patch ; and then I don't believe her and want to see this derivation. The reason why I don't believe her is that there is, to a good approximation, no difference between the metric at the surface of the earth, and the metric of a uniformly accelerated observer in flat space. So both should find the SAME equation of motion (up to eventually some SMALL correction). The small difference between both metrics resides in tidal effects which are present at the Earth surface, and which are not present for a uniformly accelerated observer, but these effects are 1) very small compared to g and 2) more importantly, can be made as small as I wish, by going to a bigger planet, keeping the 1 g and having less and less tidal effects.
It is difficult to imagine that for observers with two metrics who are nearly the same (surface of earth/big planet versus accelerated observer in flat space), one will find a particle that falls UP and the other will find a particle that falls DOWN.
So that's why I want to see the explicit derivation.
I suspect that the whole "trick" is the "fixation" of this famous "tau". Because in the paper is discussed how tau *transforms*, but not how you calibrate it (in what frame, and why ?). If there is some arbitariness in this choice, then this tau is a NEW FIELD that is not derivable from the metric, locally. This can very well be: that she fixes this tau totally differently in a Schwarzschild metric than in a Rindler metric.
But this then, comes down to what I claimed: that the equation of motion is NOT derivable from a local finite patch of metric, in which case not much remains from the equivalence principle and in which case we have in fact a kind of ether theory.

Another possibility exists, in which the actual equation of motion of a sabinon does NOT indicate that it falls up at the surface of the earth. But then I don't know why this is called "antigravity" and it goes against Sabine's claim. I didn't check it starting from the elements of her paper. The path between the metric and the equation of motion is rather long (and involves this tau).

And maybe I'm wrong, in which case I'm in for a big surprise and I'll learn something, which is also positive, and Sabine will have had an exercise in explaining her idea.

But in the mean time I remain sceptical, for the reason I set out, that it will be possible, from a given metric over a finite patch of space, to derive an upward falling particle equation of motion.

 

[hossi]

Rindler and Einstein's elevator

So why should you bother talking with an idiot ?
The reason is that if you want to have any chance for your idea to be valued, you need to be able to explain it to idiots ! If you can't, your idea is worthless, in a way.

Hi vanesh,

I take it very seriously to make my work accessible to as many people as possible. I genuinely try the best I can, this is part of the reason why I am here in this forum (another reason is that I like arguing).

Let me start again with the equivalence principle. A particle locally experiences gravitation the same way as acceleration in flat space. Place your particle in Einstein's elevator. In a free falling frame in a gravitational field that means, when you accelerate the elevator with the appropriate acceleration, the particle will not not notice anything. For an a-particle the elevator has to be accelerated in the oppiside direction. Both free-falling frames are not identical, but for both seperately, the eq. principle holds. That's why I speak of a relaxation of the equivalence principle, not of a generalization. You are of course completely right that in a gravitational field, there is no elevator that could make BOTH particles at once not notice it. That is exactly the point. Both particles will be indistinguishable if and only if spacetime is globally flat.

If you want to apply that to your accelerated observer in (globally) flat
space you have to compare the one particle to acceleration a, the other one to acceleration -a. That is, you can not map both to the same Rindler-observer, since the ratio of intertial to gravitational mass is +1 for the one, and -1 for the other.

The answer to your question is of course yes, the motion of the a-particle can be derived from the local metric (over a patch of spacetime). Just that this is not exactly the easiest way to do it. The reason is that in a global coordinate system, the geodesics are related by g being the inverse of g, in a local frame, the relation between both is not that apparent (as careful has found, more about this later).

The "metric" of the accelerated observer in flat space you look at

ds^2 = -(1+aZ)^2 dT^2 + dX^2 + dY^2 + dZ^2,

Is not in space-time coordinates (see e.g. MTW 6.18), therefore you can't just take the inverse and get the geodesics. But I guess I don't get the point here, if you have an accelerated observer in flat space, he apparently does not move on a geodesic.

Does that help?

B.

 

[hossi]

Pardon me for being dense, but does not an anti-gravitating particle wreak havoc on the energy conservancy principle? It looks very unphysical to me.

Indeed the energy conservation is an important point. You have to look at the covariant version. The conservation law for the kinetic energy is modified - essentially because the covariant derivative you use has to be appropriate for the a-particle. A usual particle that falls down in a graviational field gains kinetic energy. An a-particle gains kinetic energy in the same gravitational field, when it falls up. (This actually is identical to the equations of motions).

B.

 

[hossi]

Hi Careful,

your objection was the toughest one. It took me some while to figure out the reason for my confusion. The point is, your comment is right, but you have asked the wrong question.

The essence of the approach I have proposed is that there are two ways a field can behave under coordinate trafos. The one is mapped to the other by \tau. Whether you like that or not, it is handy. I personally don't like it, but I couldn't come up with anything better. Note that the elements of both spaces don't transform alike. How could they possibly have the same properties under parrallel transport?! They can't (unless spacetime is globally flat.)

You see that most easily by looking at the covariant derivative. For a field A in TM (or higher products), you have two ways to introduce a covariant derivative. The one is the usual \nabla A. The other one is \tau \nabla \tau^{-1} A. (You have to distribute indices according to what A is.) The latter means that you charge-conjugate your field, transport it, then push it back to the manifold. Since \nabla \tau does not vanish, the result of both is not identical.

You can therefore derive two curves following from them, the usual geodesic and Eq (38).

Now what you have done is you have made a coordinate transformation on the manifold from \partial to \tau \partial, and derived the geodesic motion in that system. I am not even sure what that system means. It's neither a coordinate system nor a LOB. But that is not essential. The point is that of course this transformation does not change the geodesic equation. It just makes it look more ugly (as you point out, the commutators don't vanish, neither is the metric locally constant). What you have to consider instead is taking the derivatives \tau \nabla \tau^{-1} (which have the same commutation behaviour as the usual ones).

Thanks for the brain stretch.

B.

 

[hossi]

I had mentioned before the possibility of "up" and "down" particle interactions between gravity "down interaction" and antigravity "up interaction" when you consider the particles that interact to be clusters of nucleons (protons and neutrons). Thus a matter cluster [PNP] (helium-3) may bind with an antimatter cluster [NP] (anti deuteron) via a gravity + antigravity interaction--but of course there is neither mathematics nor experiments ever attempted to falsify such a claim--thus remains an open question for future scientific study. Who has time to develop the mathematics for such interaction of rotating nucleon clusters using gravity and antigravity ?

I completely fail to see how anti-gravity would bind the system. Wouldn't it in the contrary push it apart? Also, in my model the both types of particles don't interact except for the gravitational interaction. Indeed, it turned out to be impossible - all other coupling terms (electro, strong, etc) either violate Lorentz- or gauge-invariance (or both). That has seriously depressed me for quite some while (I mean, how could we ever detect the a-particles?!). Thus, I don't see why such a study would be interesting, or what model it should be based on. I personally doubt you can cook up a reasonable setup for the investigation.



B.

 

[Rade]

I completely fail to see how anti-gravity would bind the system. Wouldn't it in the contrary push it apart?

Thank you for the time of your response. The model I study predicts that there is a neutral coexistence between gravity and anti-gravity, with gravity related to "matter" clusters, and anti-gravity related to "anti-matter" clusters. Because the two clusters must have different mass, they bind. This differs from the common understanding that matter and antimatter when they meet annihilate. From your answer I take it then that you are not aware of any research that has attempted to bind matter helium-3 with antimatter deuteron. I guess I do not understand why such an experiment would not be of interest mathematically and/or theoretically. If the model I study is correct, when these two isotopes meet, the reaction should not yield complete annihilation energy. The reason being that anti-gravitation is present within the system, coexisting with gravity at the sub-atomic level.

 

[hossi]

Thank you for the time of your response. The model I study predicts that there is a neutral coexistence between gravity and anti-gravity, with gravity related to "matter" clusters, and anti-gravity related to "anti-matter" clusters. Because the two clusters must have different mass, they bind.

Hi Rade,

I understand the matter clusters, and the anti-grav. matter clusters. But what force do they bind with? Strong? Electric?

This differs from the common understanding that matter and antimatter when they meet annihilate. From your answer I take it then that you are not aware of any research that has attempted to bind matter helium-3 with antimatter deuteron. I guess I do not understand why such an experiment would not be of interest mathematically and/or theoretically. .

Wait, wait, are we talking anti-matter (conjugation of electic charge), or are we talking anti-gravitating-matter (negative gravitational mass), or both? I know that there is effort going into making clusters out of anti-matter (like nuclei etc), which is certainly interesting to see whether they have - or don't have - the same properties as the usual nuclei. But where does the anti-g come in? And, how do you avoid it annihilates into nothing with the usual matter (that better shouldn't happen, or you could also make them both out of nothing).

Besides this, what is the model you study, and could you give me some references?



B.

 

[twin]

Hi Sabine,

I think there cannot be any difference between geodesics for "ordinary" and "anti-gravitating" particles for the following reasons:

In (3) you define the (ordinary) tangent bundle as a bundle of "column vectors" with diffeomorphisms operationg from the left, and the (ordinary) cotangent bundle as bundle of "row vectors" with diffeomorphisms operating from the right. Whether you define your bundles as row or column vectors is of course only a matter of convention.
The underlined bundles defined in (4) are nothing else but the cotengent bundle and the tangent bundle, but with the opposite conventions: The underlined TM is the cotangent bundle, considered as bundle of column vectors, and the underlined TM* is the tangent bundle, considered as bundle of row vectors. Therefore, your anti-gravitating particles essentially live in the cotangent bundle.
Your map \tau, which identifies the ordinary and underlined bundles, is nothing else but the canonical identification of tangent and cotangent bundle provided by the metric (and therefore it comes as no surprise that the composite map defined in (7), which identifies the two tangent bundles, does not depend on the metric and is merely given by a transposition).
You go on and define the covariant derivative for the underlined bundles. However, since you use no extra data apart from the metric, this must result in the ordinary covariant derivative in the cotangent bundle (modulo transpositions perhabs).
The geodesics for this covariant derivative are nothing else than the ordinary geodesics.


Twin

 

[Careful]

**
your objection was the toughest one. It took me some while to figure out the reason for my confusion. The point is, your comment is right, but you have asked the wrong question.

The essence of the approach I have proposed is that there are two ways a field can behave under coordinate trafos. The one is mapped to the other by \tau. Whether you like that or not, it is handy. I personally don't like it, but I couldn't come up with anything better. Note that the elements of both spaces don't transform alike. **

I was already way past that issue.

** How could they possibly have the same properties under parrallel transport?! They can't (unless spacetime is globally flat.) **

Their properties under parallel transport map to each other under \tau.

** You see that most easily by looking at the covariant derivative. For a field A in TM (or higher products), you have two ways to introduce a covariant derivative. The one is the usual \nabla A. The other one is \tau \nabla \tau^{-1} A. (You have to distribute indices according to what A is.) The latter means that you charge-conjugate your field, transport it, then push it back to the manifold. Since \nabla \tau does not vanish, the result of both is not identical. **

Well, two comments. First, you say that A is in TM, so I cannot apply \tau^{-1} on it. Second, I presume you just mean that I should consider \nabla_{\beta} g_{\gamma \kappa} A^{\kappa} instead of \nabla{\beta} A^{\kappa}. But that does not change zip, since the metric goes through the covariant derivative.


**
What you have to consider instead is taking the derivatives \tau \nabla \tau^{-1} (which have the same commutation behaviour as the usual ones). **

But that does not make sense : \tau is supposed to be a map from TM to \underline{TM} so, it should map a basis in TM to a basis in \underline{TM}. The \underline{\partial} you wrote down here is the \partial in the operational sense ON \underline{TM}. But that is meaningless since the latter is not an intrinsic operation.

Look, it is really very simple. Given the metric g, there is only one prescription to go to the connection which does not require extra fields INDEPENDENT of it and that is the Christoffel prescription. For example: the introduction of spinors requires *extra structure*, that is a spinor bundle attached to a preferred tetrad.

Cheers,

Careful

 

[Careful]

Hi Sabine,

I think there cannot be any difference between geodesics for "ordinary" and "anti-gravitating" particles for the following reasons:

In (3) you define the (ordinary) tangent bundle as a bundle of "column vectors" with diffeomorphisms operationg from the left, and the (ordinary) cotangent bundle as bundle of "row vectors" with diffeomorphisms operating from the right. Whether you define your bundles as row or column vectors is of course only a matter of convention.
The underlined bundles defined in (4) are nothing else but the cotengent bundle and the tangent bundle, but with the opposite conventions: The underlined TM is the cotangent bundle, considered as bundle of column vectors, and the underlined TM* is the tangent bundle, considered as bundle of row vectors. Therefore, your anti-gravitating particles essentially live in the cotangent bundle.
Your map \tau, which identifies the ordinary and underlined bundles, is nothing else but the canonical identification of tangent and cotangent bundle provided by the metric (and therefore it comes as no surprise that the composite map defined in (7), which identifies the two tangent bundles, does not depend on the metric and is merely given by a transposition).
You go on and define the covariant derivative for the underlined bundles. However, since you use no extra data apart from the metric, this must result in the ordinary covariant derivative in the cotangent bundle (modulo transpositions perhabs).
The geodesics for this covariant derivative are nothing else than the ordinary geodesics.


Twin

Exactly, that is the ``non-conventional isomorphism´´ I was talking about.

Cheers,

Careful

 

[vanesch]

Both free-falling frames are not identical, but for both seperately, the eq. principle holds. That's why I speak of a relaxation of the equivalence principle, not of a generalization. You are of course completely right that in a gravitational field, there is no elevator that could make BOTH particles at once not notice it. That is exactly the point. Both particles will be indistinguishable if and only if spacetime is globally flat.

It depends on what you mean by "globally". If you mean by that, the *entire* manifold, ok, but then there's nothing that remains of the equivalence principle (we now have something like an ether theory).
But if you mean by "globally" in a finite patch of spacetime, instead of a point, then I don't follow you.

If you want to apply that to your accelerated observer in (globally) flat
space you have to compare the one particle to acceleration a, the other one to acceleration -a. That is, you can not map both to the same Rindler-observer, since the ratio of intertial to gravitational mass is +1 for the one, and -1 for the other.

?? So for a GIVEN Rindler metric, does the a-particle fall as a normal one, or not ?

If your answer is yes (which means that in the local Lorentzian metric over the patch - which is possible, it is a flat space - both describe a uniform motion. But then I have the problem that we have THE SAME METRIC for an observer at the surface of the earth, so same metric -> same equations -> same geodesic, and hence your a-particle falls just as well down as a normal particle.

Now, if your answer is no, it means that in the local Lorentzian metric over the patch, we DO NOT have uniform motion of the a-particle, and this in flat space, which is against what you postulated ? So I take it that this is impossible.

The answer to your question is of course yes, the motion of the a-particle can be derived from the local metric (over a patch of spacetime). Just that this is not exactly the easiest way to do it.

Difficult or not, if this is *in principle* possible, I don't see how this can lead to an "up falling" equation of motion at the surface of the earth.

In the case your answer was "yes" to the first discussion, I'd think that the SAME metric must give rise to the SAME equation of motion, no ? If this is the only thing from which it is derived. And there, it was "downfalling" in a Rindler set of coordinates.
Otherwise we have the funny situation that, in flat space, the particle, in a Rindler metric, falls down, and at the surface of the earth, in just as flat a space, in the Rindler metric, the particle falls up.

Because at the surface of the eath, there is (almost) no difference between this situation, and the free, flat space situation.

How can you obtain a totally different equation of motion, starting from exactly the same inputs (the metric over a local patch) ?

The "metric" of the accelerated observer in flat space you look at

ds^2 = -(1+aZ)^2 dT^2 + dX^2 + dY^2 + dZ^2,

Is not in space-time coordinates (see e.g. MTW 6.18), therefore you can't just take the inverse and get the geodesics.

That's a coordinate system as any other, no ?
It is in fact the coordinate system you would normally obtain "in a rocket".

For small a.z, you can expand this, and this comes down to the Newtonian transformation:

T = t
X = x - a/2 t^2
Y = y
Z = z

and this coordinate system, and its metric, is just as "real" as the Lorentzian coordinate system in flat spacetime, right ?

 

[hossi]

In (3) you define the (ordinary) tangent bundle as a bundle of "column vectors" with diffeomorphisms operationg from the left, and the (ordinary) cotangent bundle as bundle of "row vectors" with diffeomorphisms operating from the right. Whether you define your bundles as row or column vectors is of course only a matter of convention.
The underlined bundles defined in (4) are nothing else but the cotengent bundle and the tangent bundle, but with the opposite conventions: The underlined TM is the cotangent bundle, considered as bundle of column vectors, and the underlined TM* is the tangent bundle, considered as bundle of row vectors. Therefore, your anti-gravitating particles essentially live in the cotangent bundle.

In globally flat space it is a matter of convention (when I say globally, I mean on the whole manifold). In a curved-space you usually don't go from the tangential space to the co-tangential space by changing "row vectors" into "column vectors". Instead, you use the metric. Therefore, the space TM^* is not identical to TM. You can eighter use the metric to go from TM to TM^*, or you use \tau to go from TM to TM^* .

You see from the first some equations, that the elements of both spaces do not transform identically. You get an element of TM to look like that of TM^*, by applying the transposition. You call it operating from the left or from the right: but this is not how, in GR, you usally identify tangential and co-tangential space.

You go on and define the covariant derivative for the underlined bundles. However, since you use no extra data apart from the metric, this must result in the ordinary covariant derivative in the cotangent bundle (modulo transpositions perhabs).
The geodesics for this covariant derivative are nothing else than the ordinary geodesics.

They are identical if and only if the covariant derivative on \tau vanishes. In general, this will not be the case. E.g. in terms of the tetrads (Eq 10), \tau is (EE^T){-1}, which simplifies if E is diagonal to E^-2, the covariant derivative on which usually doesn't vanish. The covariant derivative in the cotangent bundle is the same as always. You can of course formulate the geodesic equation for tangential or co-tangential vectors. It looks different, but is the same curve, the reason for which is essentially that \nabla g_munu vanishes.



B.

 

[hossi]

Their properties under parallel transport map to each other under \tau.

Well, two comments. First, you say that A is in TM, so I cannot apply \tau^{-1} on it. Second, I presume you just mean that I should consider \nabla_{\beta} g_{\gamma \kappa} A^{\kappa} instead of \nabla{\beta} A^{\kappa}. But that does not change zip, since the metric goes through the covariant derivative.

I am sorry, I might have mixed up \tau with \tau^-1 . Let us define TM: A -> \tau A \in TM. Okay, then I agree, it should have been \tau^-1 \nabla \tau acting on A in TM. The metric goes through the covariant derivative but \tau doesnt. That's the reason why you can of course 'relate' the properties of elements in TM to those in TM under parallel transport, but the resulting curves are not identical (see Eq (39).

But that does not make sense : \tau is supposed to be a map from TM to \underline{TM} so, it should map a basis in TM to a basis in \underline{TM}. The \underline{\partial} you wrote down here is the \partial in the operational sense ON \underline{TM}. But that is meaningless since the latter is not an intrinsic operation.

It is meaningless, or is it \partial in the operational sense? I didn't use the basis in \underline TM as an operator, that's why I explicitly wrote it's a notation.

Look, it is really very simple. Given the metric g, there is only one prescription to go to the connection which does not require extra fields INDEPENDENT of it and that is the Christoffel prescription. For example: the introduction of spinors requires *extra structure*, that is a spinor bundle attached to a preferred tetrad.

Look, is it really very simple? The prescription to go the Christoffelsymbols is unique. But it matters what field you transport whether the Christoffelsymbols are the symbols to use or not. The extra structure is the properties of elements in TM to react to general coordinate transformations. If I remember that correctly, you can express the connexion for spinors in terms of the usual Christoffels by using the generators of the approriate spinor representation. In which sense is this more independent of the usual Christoffels than in my scenario?



B.

 

[hossi]

It depends on what you mean by "globally". If you mean by that, the *entire* manifold, ok, but then there's nothing that remains of the equivalence principle (we now have something like an ether theory).
But if you mean by "globally" in a finite patch of spacetime, instead of a point, then I don't follow you.

By globally I always mean on the entire manifold. I guess we have some disagreement on the equivalence principle. As I understand it it says: the local effects of gravity in curved space are identical to those that an uniformly accelerated observer in globally flat space experiences. The uniformly accelerated observer in globally flat space is your Rindler observer. Now you have two of them, with positive or negative acceleration. The point is to relate the local properties in curved space-time (General Relativity) to those of Special Relativity (which does not require dealing with curved spaces, otherwise you haven't won anything).

Since space is globally flat for the accelerated observer, you can of course then say, well, then it's the same everywhere and I only look at a local piece again. That I think, is what you are doing (?). But that doesn't change the fact that the space for the Rindler observer is globally flat, and, in particular, if this observer is uniformly accelerated, he doesn't move on a geodesic.

?? So for a GIVEN Rindler metric, does the a-particle fall as a normal one, or not ?

If your answer is yes[...]

Now, if your answer is no[...]

So I take it that this is impossible.

The Rindler space is flat and therefore the particle, no matter which, does not 'fall' at all.

Otherwise we have the funny situation that, in flat space, the particle, in a Rindler metric, falls down, and at the surface of the earth, in just as flat a space, in the Rindler metric, the particle falls up.

Because at the surface of the eath, there is (almost) no difference between this situation, and the free, flat space situation.

How can you obtain a totally different equation of motion, starting from exactly the same inputs (the metric over a local patch) ?

You are repeating your mistake. I have the same metric over a local patch. In the one case, I attribute it (using the equivalence principle) to an Rindler observer (I am trying to use your reasoning) with positive acceleration, in the other case with negative acceleration. The prescription you need to 'simulate' gravitational effects in flat space is different for both particles.

That's a coordinate system as any other, no ?
It is in fact the coordinate system you would normally obtain "in a rocket".

It's not a global coordinate system, therefore you can't just take the coefficients in ds^2 and say, it's a metric of the manifold. Look, if spacetime is flat, you can accelerate your observer as you like, it better stays flat.

You can of course use local coordinates like the ones you use, but you have to be careful what conclusions you draw from the coefficients, that was all I wanted to say.

B.

 

[Careful]

** I am sorry, I might have mixed up \tau with \tau^-1 . Let us define TM: A -> \tau A \in TM. Okay, then I agree, it should have been \tau^-1 \nabla \tau acting on A in TM. The metric goes through the covariant derivative but \tau doesnt. **

But, \tau is constructed trough transposition of indices and/or application of g^{\alpha \beta} so \tau goes through the covariant derivative! If you contest this, give us a detailed calculation as well as your definitions.
Another reason why both geodesics are equal is that there exists no nonvanishing (1,2) tensor which is constructed from the metric and first derivatives alone. In more detail: you claim that we have to look at

A \tau^(-1) \nabla \tau A = A \nabla A + A \tau^{-1} ( \nabla \tau ) A

so \tau^{-1} ( \nabla \tau ) must be a (1,2) tensor depending upon g and \partial g only. Hence it must be zero.



**
Look, is it really very simple? The prescription to go the Christoffelsymbols is unique. But it matters what field you transport whether the Christoffelsymbols are the symbols to use or not. The extra structure is the properties of elements in TM to react to general coordinate transformations. **

But the elements in TM react identically to those in \underline{TM}^* and vice versa ! In a spinor bundle you DO add something, for example you ENLARGE the gauge group (= Lorentz group) by going over to the universal cover. Here, there is no enlarged gauge group whatsoever.




**If I remember that correctly, you can express the connexion for spinors in terms of the usual Christoffels by using the generators of the approriate spinor representation. In which sense is this more independent of the usual Christoffels than in my scenario? **

This gauge connection uses more than simply a local property of the metric !

 

[hossi]

But, \tau is constructed trough transposition of indices and/or application of g^{\alpha \beta} so \tau goes through the covariant derivative! If you contest this, give us a detailed calculation as well as your definitions.

Definitions are in the paper. Detailed calculations are to follow - I am still waiting for feedback from some friends/colleagues. For now, take the above example for a diagonal metric, then \tau is essentially E^-2 and \tau^-1 \nabla \tau is -2 E^-1 \nabla E.

Another reason why both geodesics are equal is that there exists no nonvanishing (1,2) tensor which is constructed from the metric and first derivatives alone. In more detail: you claim that we have to look at

A \tau^(-1) \nabla \tau A = A \nabla A + A \tau^{-1} ( \nabla \tau ) A

so \tau^{-1} ( \nabla \tau ) must be a (1,2) tensor depending upon g and \partial g only. Hence it must be zero.

Why do you assume \tau is a function of g? In certain cases you can express elements of \tau as functions of elements of g - as you have also done (how I conclude from your post above), but that does not mean \tau is a function of g in general. You can write it as a function of E and E^T, but the ^T is something you don't usually do in GR, so I am not sure how your argument goes with it.

But the elements in TM react identically to those in \underline{TM}^* and vice versa ! In a spinor bundle you DO add something, for example you ENLARGE the gauge group (= Lorentz group) by going over to the universal cover. Here, there is no enlarged gauge group whatsoever.

No, it is an additional discrete symmetry. If you want to get that formally right, you probably had some direct product of twice the gauge groups. The elements in TM do not react identically to those in \underline TM^* to general diffeomorphism.

This gauge connection uses more than simply a local property of the metric !

Right! It also uses the transformation behaviour of the fields, which differs whether it's a vector or spinor - even which type of (Weyl) spinor for that matter.



B.

 

[Rade]

...are we talking anti-matter (conjugation of electic charge), or are we talking anti-gravitating-matter (negative gravitational mass), or both? ... what is the model you study, and could you give me some references?

It is my understanding that we talk about both, anti-matter clusters with conjugation of electric charge, and also having negative gravitational mass. For details of what is known please see this link:http://d2800693.u47.phoenixrising-web.net/downloads/antimttr.pdf
Additional information is available at this web page:http://www.brightsenmodel.phoenixrising-web.net/Download.html

 

[vanesch]

By globally I always mean on the entire manifold. I guess we have some disagreement on the equivalence principle. As I understand it it says: the local effects of gravity in curved space are identical to those that an uniformly accelerated observer in globally flat space experiences.

Yes, which comes down to saying that the effects of gravity (the kinematical effects: we're only concerned with test particles here, not with experiments where the dynamics of gravity enters into account (where the test masses MODIFY the metric or something else) are solely dependent on the metric on a local, finite patch around where we try to establish them.

The uniformly accelerated observer in globally flat space is your Rindler observer. Now you have two of them, with positive or negative acceleration.

Eh, no. I only have one of them: it is my right to consider one, or the other and that choice is set up by the experimental situation. I mean: in a rocket, there are not TWO but only ONE coordinate system which is "fixed to the rocket", and in this system of coordinates, the acceleration is well-defined: it is defined by the behavior of NORMAL particles (of which the experimental existence doesn't need any explanation).
With these normal particles, in this rocket frame (X,Y,Z,T) we can establish the local metric, and it is the Rindler metric, and the sign of the acceleration in this metric is unambiguously established. In fact, it is established by "letting the particles fall" in this rocket frame. They will "fall down" (in the rocket frame = Rindler metric) in the sense that their equation of motion will be something of the kind: Z(T) = Z0 + V0 T - a/2 T^2 (Newtonian approximation). These are the geodesics of the Rindler metric. These same geodesics (world lines) are of course straight lines because when we transform this to the Lorentz frame, we have free particles in flat space. Now, this transformation between the Lorentz frame and the rocket frame is a transformation of coordinates on the manifold, and one doesn't have any choice there in the sign of the acceleration.
And if a-particles 1) describe a world line and 2) follows uniform motion in the Lorentz frame, then its world lines, are geodesics and hence will remain geodesics of the Rindler metric in the X,Y,Z,T coordinate system.

If you insist that in the coordinate frame XYZ,T (fixed to the rocket, so experimentally entirely unambiguous), the equation of motion of an a-particle is not a geodesic Z(T) = Z0 + V0 T - a/2 T^2, but rather
Z(T) = Z0 + V0 T PLUS a/2 T^2, then the only possibilities are:
a) that this equation of motion does not describe a world line but is observer-dependent OR
b) that the equation of motion in the Lorentz frame is NOT uniform motion.

The reason is of course that a world line cannot be a geodesic of a metric in one coordinate system and NOT be a geodesic in another coordinate system (a geodesic is a concept that is coordinate-system independent).

Now, both claims are denied by you, so this cannot be true.

Hence, the only possible motion for an a-particle, in the rocket frame, is given by Z(T) = Z0 + V0 T - a/2 T^2 (= a geodesic of the metric), which can be called "falling down in the rocket".

Since space is globally flat for the accelerated observer, you can of course then say, well, then it's the same everywhere and I only look at a local piece again. That I think, is what you are doing (?). But that doesn't change the fact that the space for the Rindler observer is globally flat, and, in particular, if this observer is uniformly accelerated, he doesn't move on a geodesic.

Yes, but an *observer* doesn't have to move on a geodesic! An "observer" is a coordinate system, meaning: mapping 4 numbers on a patch of spacetime. That's the whole idea of coordinate systems on a manifold.

The Rindler space is flat and therefore the particle, no matter which, does not 'fall' at all.

Well, it "falls" of course from the POV of the coordinate system of the rocket (with the Rindler metric), where the equation of motion is:
Z(T) = Z0 + V0 T - a/2 T^2 (in the Newtonian limit)
which is the equation of motion of a "falling" particle.

You are repeating your mistake. I have the same metric over a local patch. In the one case, I attribute it (using the equivalence principle) to an Rindler observer (I am trying to use your reasoning) with positive acceleration, in the other case with negative acceleration.

You can't. In a GIVEN coordinate system, (such as the one in a rocket), the acceleration is unambiguously defined. You cannot assign TWO accelerations to it. The unambiguously defined acceleration defines the sign of the acceleration in the metric expressed in this ONE AND UNIQUE coordinate system, which is nothing else but a Rindler metric. There's no way to assign TWO DIFFERENT accelerations to the same coordinate system, because the transformation between this coordinate system (of the rocket) and a given Lorentz frame is UNIQUELY DEFINED. It is a *coordinate transformation*. X,Y,Z and T are FUNCTIONS of (x,y,z,t) in a unique way.

The prescription you need to 'simulate' gravitational effects in flat space is different for both particles.

But as I outlined above, there is only ONE way to be compatible with:
a) a-particles describe world lines (coordinate-independent)
b) a-particles describe geodesics in the Lorentz frame.

If the world lines are geodesics of the metric in ONE frame, they are geodesics in ALL frames, because "geodesic" and "world line" are geometric concepts which are coordinate independent.

It's not a global coordinate system, therefore you can't just take the coefficients in ds^2 and say, it's a metric of the manifold. Look, if spacetime is flat, you can accelerate your observer as you like, it better stays flat.

Of course. The Rindler metric is a flat metric, but it is not Lorentzian. It isn't "global" because some piece of spacetime is missing, but it surely is "global enough" to define the metric over a finite local patch. (and not just in one point). By the equivalence principle, on which we both seem to agree, this should be good enough to determine uniquely the equation of motion in it.

You can of course use local coordinates like the ones you use, but you have to be careful what conclusions you draw from the coefficients, that was all I wanted to say.

But "local coordinates over a finite patch of spacetime" is all you have in GR. You have to be able to deduce everything from within local coordinates and the metric expressed in it. It's the essence of the equivalence principle. And what I'm saying is that these local coordinates are THE SAME for an observer on the surface of the earth, and in an accelerated rocket (up to tiny tidal effects).
If these are the same, then the equation of motion that you can derive of it, should be the same. And you claim it doesn't.

 

[Careful]

**Definitions are in the paper. Detailed calculations are to follow - I am still waiting for feedback from some friends/colleagues. For now, take the above example for a diagonal metric, then \tau is essentially E^-2 and \tau^-1 \nabla \tau is -2 E^-1 \nabla E. **

This is damn amazing I have worked out the definitions in your paper and came to the conclusions mentioned previously ! You first publish (miracles happen every day still), then you claim that you did not check things yet but at the same time you are telling to everyone who is kindly trying to inform you that the miracle does not happen that it somehow HAS to occur. As far as I can guess your notion of transpose : E_\mu E^T_\nu = g_\mu \nu (I remind you : I asked you to clarify this issue in the beginning), so there you have it.

**
Why do you assume \tau is a function of g? In certain cases you can express elements of \tau as functions of elements of g - as you have also done (how I conclude from your post above), but that does not mean \tau is a function of g in general. You can write it as a function of E and E^T, but the ^T is something you don't usually do in GR, so I am not sure how your argument goes with it. **

I can do the ^T in GR as much as I want to without changing anything (as twin correctly asserted). And if \tau does not depend upon g and its derivatives, then you introduce an extra field (see later) - eather, here we come.

**
No, it is an additional discrete symmetry. **

That is what you think ! But we don't see anything, and you are repeating the same issues we already know for 4 pages on this thread now.


**Right! It also uses the transformation behaviour of the fields, which differs whether it's a vector or spinor - even which type of (Weyl) spinor for that matter.
**

Correct, and hence it uses the extra (not uniquely fixed by the metric) structure (a preferred tetrad), and it has an enlarged gauge group SL(2,C) - exactly what I said and you denied.

I made in the beginning the comment that you do not need the tetrad in your construction, and you *agreed*. You still claim that everything can be done using local properties of the metric.

So, for a change I would like to have some scientific arguments, not just your beliefs about what it should be. We are not doing quantum gravity here, this topic should be simple enough.

Careful

 

[josh1]

Pair creation (and annihilation) in the antigravity sector, which by the uncertainty principle must occur everywhere, would provide a nonsensical mechanism by which black hole mass decreases but charge increases.

 

[hossi]

Dear Careful,

I sense you are getting angry at me for reasons I can't quite follow. I am sorry if this topic annoys you, I assure you I don't try to be stupid on purpose.

This is damn amazing I have worked out the definitions in your paper and came to the conclusions mentioned previously ! You first publish (miracles happen every day still), then you claim that you did not check things yet but at the same time you are telling to everyone who is kindly trying to inform you that the miracle does not happen that it somehow HAS to occur. As far as I can guess your notion of transpose : E_\mu E^T_\nu = g_\mu \nu (I remind you : I asked you to clarify this issue in the beginning), so there you have it.

I pointed out that your conclusions are right, but you have investigated the wrong question. I did not come across the question you had because I took another way, how I tried to explain above. I don't know what you mean with my (?) notion of transpose, the E's are the usual tetrads, i.e. E^i_\nu E^j_\mu \eta_ij = g_\mu\nu and there is no ^T involved here.

(The paper actually took a long time to be published and went through a quite lengthy review process. I am the first to admit that this still does not mean it's content is without doubt. However, I would say, science lives from controversy.)

I can do the ^T in GR as much as I want to without changing anything (as twin correctly asserted). And if \tau does not depend upon g and its derivatives, then you introduce an extra field (see later) - eather, here we come.

Then you are also changing the transformation from G to G^T. The second point actually is a very good point. I don't think \tau has any additional degrees of freedom, at least in my version of the model it hasn't (exept maybe some additional gauge-freedom like rotations or so that wouldn't really change anything). However, maybe it ought to be dynamical (credits for this idea don't go to me, and I am still not sure about it). I have no idea why this is an eather, it certainly has no fixed background. There are plenty of examples where additiona fields are coupled to the SM that change the dynamics without violating GR or SR.

That is what you think ! But we don't see anything, and you are repeating the same issues we already know for 4 pages on this thread now.

Well, I could say the same thing.

Correct, and hence it uses the extra (not uniquely fixed by the metric) structure (a preferred tetrad), and it has an enlarged gauge group SL(2,C) - exactly what I said and you denied.

I made in the beginning the comment that you do not need the tetrad in your construction, and you *agreed*. You still claim that everything can be done using local properties of the metric.

Hmm. Let me think. I certainly use the tetrad to make my computations. Now do I really need it? How do I find out? I admit, I am actually not sure about it I found it handy to use. Anyway, I don't deny that a spinor transforms under a different representation than a vector does.

So, for a change I would like to have some scientific arguments, not just your beliefs about what it should be. We are not doing quantum gravity here, this topic should be simple enough.

You are aware that you are being very insulting? Try the following: take a vector field V and apply some general coordinate transformation to it. You get V' = ... Now take some field-with-one-index W and transform it under the inverse and transponed of the above matrix. Tell me where the field with this property appears in GR, what it means and how it behaves.



B.

 

[hossi]

Pair creation (and annihilation) in the antigravity sector, which by the uncertainty principle must occur everywhere, would provide a nonsensical mechanism by which black hole mass decreases but charge increases.

Pair creation (and annihilation) in the antigravity sector does not occur at the black hole horizon, because the anti-g particles don't experience a horizon at this surface. (Well, they get repelled by the thing, how could they get trapped?) Nevertheless, good question,

B.

 

[hossi]

It is my understanding that we talk about both, anti-matter clusters with conjugation of electric charge, and also having negative gravitational mass. For details of what is known please see this link:http://d2800693.u47.phoenixrising-web.net/downloads/antimttr.pdf
Additional information is available at this web page:http://www.brightsenmodel.phoenixrising-web.net/Download.html

Hi Rade,

okay, I see what you are talking about. Naturally, I came across some of these ideas. However, I have a serious problem with the idea that anti-matter is also anti-gravitating. How do you make sure the vacuum is stable and what do you do with loop-contributions in, say QED. I know that the actual MEASUREMENT of the mass has not been done with very high precision, but if anti-graviating particles exist, I think they shouldn't be able to annihilate with usual matter.

B.

 

[josh1]

Hi Hossi,

Is there an antigravity analog of a black hole?

 

[hossi]

Dear vanesh,

you are still making the same mistakes. You are not using the equivalence principle as it's use is intended.

Eh, no. I only have one of them: it is my right to consider one, or the other and that choice is set up by the experimental situation. I mean: in a rocket, there are not TWO but only ONE coordinate system which is "fixed to the rocket", and in this system of coordinates, the acceleration is well-defined: it is defined by the behavior of NORMAL particles (of which the experimental existence doesn't need any explanation).

This is completely right. If you are in your rocket and describe the acceleration you experience it's the inertial mass that plays a role and it's the same for both particles. The difference between usual and anti-g particle lies in the comparison to gravitational acceleration, which in the one case has to go in the same direction (upward acceleration equals positively gravitating mass on the bottom) in the other case in the opposide direction (upward acceleration equals positively gravitating mass on the ceiling).

Now, this transformation between the Lorentz frame and the rocket frame is a transformation of coordinates on the manifold, and one doesn't have any choice there in the sign of the acceleration.

No, of course not, the acceleration is a parameter you have to compare to the graviational acceleration. If both are identical you have GR (thats exactly what the eq. principle says). If they are identical only up to a sign, you get what I say.

You don't get the point, for the Rindler observer both particles ARE identical. There is no gravitational effect, only acceleration, they both do the same thing. The difference is how you translate the particle's behavior to it's reaction to the gravitational pull.

Of course. The Rindler metric is a flat metric, but it is not Lorentzian.

I have no idea what you are trying to say.

Yes, but an *observer* doesn't have to move on a geodesic! An "observer" is a coordinate system, meaning: mapping 4 numbers on a patch of spacetime. That's the whole idea of coordinate systems on a manifold.).

It's not and an observer is not a coordinate system. A coordinate system has to cover the whole manifold, whereas you have a collection of local maps, and locally these are always flat. You have to patch the local maps toghether to get an atlas of the manifold. Then you get what I call a coordinate system, that is an appropriate base of 1-forms dx^\nu whose coefficients are the metric tensor and actually say something about the properties of the space.

B.

 

[hossi]

Hi Hossi,

Is there an antigravity analog of a black hole?

Hi Josh ,

never thought I would be happy about a comment of yours - the old guys are really stubborn, sigh. That's quite an interesting point, I have thought about that a bit, but not really come to any conclusions. The first question is of course how the anti-grav. matter behaves, like, how does it clump and structure? And, is there enough stuff around of it to actually make such a thing? Also: would it be stable? (There was some paper about the issue recently, will try to find the reference).

Unfortunately, that does not change the fact that these anti-g black holes would be unobservable I mean, they would be even more black than black holes. Ultra-black holes so to say.



B.

 

[Careful]

**
I pointed out that your conclusions are right, but you have investigated the wrong question. I did not come across the question you had because I took another way, how I tried to explain above. I don't know what you mean with my (?) notion of transpose, the E's are the usual tetrads, i.e. E^i_\nu E^j_\mu \eta_ij = g_\mu\nu and there is no ^T involved here.
**

There are no wrong questions, there are only inaccurate answers.
So define **rigorously** your sacred T and the controversy is over (I asked this since my first post).
You are not answering my objections why the two geodesics are identical. In this case, I presume you mean by E^T_(a \mu) = \eta_ab E^b_\mu .

** I don't think \tau has any additional degrees of freedom, at least in my version of the model it hasn't (exept maybe some additional gauge-freedom like rotations or so that wouldn't really change anything). However, maybe it ought to be dynamical (credits for this idea don't go to me, and I am still not sure about it). I have no idea why this is an eather, it certainly has no fixed background. There are plenty of examples where additiona fields are coupled to the SM that change the dynamics without violating GR or SR. **

So, here you say you don't know whether you add additional degrees of freedom or not ?? Hello you were claiming since the beginning nothing was added.


**You are aware that you are being very insulting? Try the following: take a vector field V and apply some general coordinate transformation to it. You get V' = ... Now take some field-with-one-index W and transform it under the inverse and transponed of the above matrix. Tell me where the field with this property appears in GR, what it means and how it behaves.
**

Twin has already *explicitely* mentioned that you only have to take the transpose of the cotangent basis to achieve that (and I acknowledged that I did the same thing before) - it is as simple as that.

Tell us EXACTLY - using MATHEMATICS only - what you mean. In each message you are trying to tell us what it *could* be - tell us what it IS. So, you see, I am (with good reason) insulted by your messages since on the contrary to you, I am giving mathematical arguments, using real definitions, why I think it does not work and you do not respond to that at all. For example : you previously said you did not need the tetrad and now you claim you don't know ?

Furthermore, I must tell you that Vanesch' interpretation of the equivalence principle is entirely correct - as any relativist (and that includes myself) will acknowledge.

Careful