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### Author Topic: Is the energy needed to create pairs of identical particles always the same?  (Read 13580 times)

#### JP

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##### Is the energy needed to create pairs of identical particles always the same?
« Reply #25 on: 16/07/2011 21:29:02 »
We use a given amount of energy to create pair particles at a high gravitational potential at the top of a tower and let them fall.  In falling they gain kinetic energy which we use to operate an electric generator at the bottom of the tower.  The electricity generated is used to produce more photons at the top of the tower which are then used to create pair particles which fall gaining kinetic energy and so on."

Ah, but to create more photons at the top of the tower you have to spend at least as much energy as you just gained by having their decay products fall to the bottom.  Still not perpetual motion.  Still Newtonian gravity.

#### MikeS

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##### Is the energy needed to create pairs of identical particles always the same?
« Reply #26 on: 17/07/2011 04:49:15 »
We use a given amount of energy to create pair particles at a high gravitational potential at the top of a tower and let them fall.  In falling they gain kinetic energy which we use to operate an electric generator at the bottom of the tower.  The electricity generated is used to produce more photons at the top of the tower which are then used to create pair particles which fall gaining kinetic energy and so on."

Ah, but to create more photons at the top of the tower you have to spend at least as much energy as you just gained by having their decay products fall to the bottom.  Still not perpetual motion.  Still Newtonian gravity.

The energy of creation equals the mass energy equivalence for the production of the particles plus the extra energy required to create them in a higher gravitational potential.  The KE gained in the fall being equal to the original PE.  I agree with all of that.

What I was trying to get at before in the experiment I outlined is how does Newtonian gravity explain what is happening in that particular instance.  In the downward leg of the loop particles gain KE in every loop.  On the upward leg of the loop, the electrical circuit, there would appear to be neither gain nor loss as gravity acts equally on both sides of the electrical circuit.  There seems to be an excess of energy left over on every loop.
In a simple two wire electrical circuit the current flow in any part of the circuit is the same.  As we have assumed the circuit to be physically lossless, both the current flow and the voltage at the top of the tower is the same as at the bottom of the tower.  It would appear there is no loss.

As I mentioned before, there is a loss but it's due to time dilation which Newtonian gravity does not take into account.

I don't doubt that Newtonian gravity covers the situation, in that energy has to be conserved.  I just don't see how it can explain it.

#### MikeS

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##### Is the energy needed to create pairs of identical particles always the same?
« Reply #27 on: 17/07/2011 05:15:28 »
I guess, it depends from what reference frame you mean gain energy.  I maintain that from the reference frame of the photons, they do not gain energy.
Then you mean absolutely nothing, since there is no coherent way to define the reference frame of a photon.
Not so, They contain the energy of their creation and even if annihilated the energy still remains.
Wrong.  Energy can neither be created or destroyed, it can only be transformed into another kind of energy.
Here you appear to be contradicting yourself. Either the pair has no net energy or it has the energy of its creation, not both.

If we consider that the pair contains the energy of its creation, then we must take this energy into account gravitationally. The pair does not simply fall towards the surface of a planet, the planet also falls towards the pair. Given the equivalence of mass-energy, the same would have happened if that energy has not generated a pair of particles.

If we consider the energy of the pair to be zero, then there is no creation of energy with regards to the pair and there is not destruction of energy when the pair is destroyed.

This does not mean 'nothing' simply we do not know how to define the reference frame of a photon.  Not being able to define a reference frame for a photon does not mean it hasn't got one.  Actually, I can't see why we can not define a reference frame lets say the reference frame is travelling along with the photon.

Obviously the pair have the energy of their creation.  I don't see any contradiction.

I agree.
I don't understand what this is saying.

Why would we consider the energy of the pair to be zero?

#### PhysBang

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##### Is the energy needed to create pairs of identical particles always the same?
« Reply #28 on: 17/07/2011 14:50:06 »
This does not mean 'nothing' simply we do not know how to define the reference frame of a photon.  Not being able to define a reference frame for a photon does not mean it hasn't got one.  Actually, I can't see why we can not define a reference frame lets say the reference frame is travelling along with the photon.
Well, can you use your photon reference frame to do any calculations of the energy of a photon?
Quote
I don't understand what this is saying.
In general relativity, the action of gravity depends on the presence of mass and energy. The energy that goes into creating the pair also interacts with gravity. It doesn't matter whether that energy is in the form of energy or in the form of particles. You are introducing a difference where there is none.
Quote
Why would we consider the energy of the pair to be zero?
If we are talking about virtual particles, then the net energy of those particles is zero. The come into being and are then annihilated.

#### MikeS

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##### Is the energy needed to create pairs of identical particles always the same?
« Reply #29 on: 17/07/2011 16:02:54 »
This does not mean 'nothing' simply we do not know how to define the reference frame of a photon.  Not being able to define a reference frame for a photon does not mean it hasn't got one.  Actually, I can't see why we can not define a reference frame lets say the reference frame is travelling along with the photon.
Well, can you use your photon reference frame to do any calculations of the energy of a photon?
Quote
I don't understand what this is saying.
In general relativity, the action of gravity depends on the presence of mass and energy. The energy that goes into creating the pair also interacts with gravity. It doesn't matter whether that energy is in the form of energy or in the form of particles. You are introducing a difference where there is none.
Quote
Why would we consider the energy of the pair to be zero?
If we are talking about virtual particles, then the net energy of those particles is zero. The come into being and are then annihilated.

No but I don't think the photon cares.  I use reference frame of the photon to differentiate it from all other reference frames.

Are you saying that both energy (photons) and mass interact with gravity in the same way?

We are not taking about virtual particles, I thought that was clear.

#### PhysBang

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##### Is the energy needed to create pairs of identical particles always the same?
« Reply #30 on: 17/07/2011 17:00:44 »
No but I don't think the photon cares.  I use reference frame of the photon to differentiate it from all other reference frames.
Only in a poetic sense. You are not doing anything that we can make sense of physically.
Quote
Are you saying that both energy (photons) and mass interact with gravity in the same way?
Photons are not energy, but, yes, energy and mass interact with gravity in the same way.

#### MikeS

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##### Is the energy needed to create pairs of identical particles always the same?
« Reply #31 on: 17/07/2011 19:12:14 »
No but I don't think the photon cares.  I use reference frame of the photon to differentiate it from all other reference frames.
Only in a poetic sense. You are not doing anything that we can make sense of physically.
Quote
Are you saying that both energy (photons) and mass interact with gravity in the same way?
Photons are not energy, but, yes, energy and mass interact with gravity in the same way.

If not energy then what are they, they're not mass?

#### PhysBang

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##### Is the energy needed to create pairs of identical particles always the same?
« Reply #32 on: 17/07/2011 19:24:00 »
A photon is a type of particle in the quantum mechanical sense. Some particles have mass; all particles have energy. Particles can transfer energy between each other.

#### MikeS

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##### Is the energy needed to create pairs of identical particles always the same?
« Reply #33 on: 17/07/2011 19:44:14 »
A photon is a type of particle in the quantum mechanical sense. Some particles have mass; all particles have energy. Particles can transfer energy between each other.

I believe all known particles have mass with the exception of the photon.
All particles have gravitational potential energy as by definition they are all within a gravitational field.
I believe that a photon is pure energy as in E=mc2.

Going back to your previous post, if you are saying that photons and mass interact with gravity in the same way then I would have to disagree.

#### PhysBang

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##### Is the energy needed to create pairs of identical particles always the same?
« Reply #34 on: 17/07/2011 23:58:46 »
I believe all known particles have mass with the exception of the photon.
You are free to believe that. Those of us that believe in gluons will have to disagree.
Quote
All particles have gravitational potential energy as by definition they are all within a gravitational field.
I believe that a photon is pure energy as in E=mc2.
Contemporary demonstrations of this theorem begin with proving something similar with regards to the momentum of a photon.

e.g., http://terrytao.wordpress.com/2007/12/28/einsteins-derivation-of-emc2/
Quote
Going back to your previous post, if you are saying that photons and mass interact with gravity in the same way then I would have to disagree.
Well, you are free to disagree, but those of us that believe in the general theory of relativity will have to disagree.

#### MikeS

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##### Is the energy needed to create pairs of identical particles always the same?
« Reply #35 on: 18/07/2011 07:14:04 »
I believe all known particles have mass with the exception of the photon.
You are free to believe that. Those of us that believe in gluons will have to disagree.[/color]
Quote
All particles have gravitational potential energy as by definition they are all within a gravitational field.
I believe that a photon is pure energy as in E=mc2.
[/color]
Contemporary demonstrations of this theorem begin with proving something similar with regards to the momentum of a photon.

e.g., http://terrytao.wordpress.com/2007/12/28/einsteins-derivation-of-emc2/
Quote
Going back to your previous post, if you are saying that photons and mass interact with gravity in the same way then I would have to disagree.
Well, you are free to disagree, but those of us that believe in the general theory of relativity will have to disagree.

As I understand it a gluon is still a hypothetical particle?

You have not made clear what you don't agree with here.

You didn't answer the previous question but I believe you are saying that photons and mass interact with gravity in the same way.  As I understand it they don't.  The only way that a photon is affected by gravity is that gravity causes time dilation and that appears to change the frequency and hence energy level of the photon.  The photon has not been affected in any way it just looks like it has.  Gravity does not affect the geodesic of a photon which continues on a straight line in curved space time.
What part of that does not agree with general relativity?

#### PhysBang

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##### Is the energy needed to create pairs of identical particles always the same?
« Reply #36 on: 18/07/2011 11:33:02 »
Gravity does not affect the geodesic of a photon which continues on a straight line in curved space time.
What part of that does not agree with general relativity?[/color]
Gravity is what determines the geodesic of a photon!

#### MikeS

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##### Is the energy needed to create pairs of identical particles always the same?
« Reply #37 on: 18/07/2011 12:35:16 »
Gravity does not affect the geodesic of a photon which continues on a straight line in curved space time.
What part of that does not agree with general relativity?[/color]
Gravity is what determines the geodesic of a photon!

Gravity is what determines the warp of space time.  A photon travels a straight line in curved space-time.  The photon is is not directly affected by gravity, space time is.

#### PhysBang

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##### Is the energy needed to create pairs of identical particles always the same?
« Reply #38 on: 18/07/2011 12:41:15 »
Gravity is what determines the warp of space time.  A photon travels a straight line in curved space-time.  The photon is is not directly affected by gravity, space time is.
Well, you are free to have your own theory of gravity. Good luck.

#### MikeS

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##### Is the energy needed to create pairs of identical particles always the same?
« Reply #39 on: 19/07/2011 06:51:26 »
Gravity is what determines the warp of space time.  A photon travels a straight line in curved space-time.  The photon is is not directly affected by gravity, space time is.
Well, you are free to have your own theory of gravity. Good luck.

Thank you and yes I do have my own theory of gravity available here
http://www.thenakedscientists.com/forum/index.php?topic=39216.0

Where you have quoted me above, is not from my own theory but is from general relativity.
"A photon travels a straight line in curved space-time."

"Space tells mass how to move" while "mass tells space how to curve" -- This well known phrase was coined by J.A. Wheeler.  It describes Einstein's space time.  It follows that a photon travels a straight line in curved space-time.

#### MikeS

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##### Is the energy needed to create pairs of identical particles always the same?
« Reply #40 on: 19/07/2011 07:16:57 »
JP.

Our debate had just become very interesting when you stopped posting.

This is where you left it
We use a given amount of energy to create pair particles at a high gravitational potential at the top of a tower and let them fall.  In falling they gain kinetic energy which we use to operate an electric generator at the bottom of the tower.  The electricity generated is used to produce more photons at the top of the tower which are then used to create pair particles which fall gaining kinetic energy and so on."

Ah, but to create more photons at the top of the tower you have to spend at least as much energy as you just gained by having their decay products fall to the bottom.  Still not perpetual motion.  Still Newtonian gravity.

The energy of creation equals the mass energy equivalence for the production of the particles plus the extra energy required to create them in a higher gravitational potential.  The KE gained in the fall being equal to the original PE.  I agree with all of that.

What I was trying to get at before in the experiment I outlined is how does Newtonian gravity explain what is happening in that particular instance.  In the downward leg of the loop particles gain KE in every loop.  On the upward leg of the loop, the electrical circuit, there would appear to be neither gain nor loss as gravity acts equally on both sides of the electrical circuit.  There seems to be an excess of energy left over on every loop.
In a simple two wire electrical circuit the current flow in any part of the circuit is the same.  As we have assumed the circuit to be physically lossless, both the current flow and the voltage at the top of the tower is the same as at the bottom of the tower.  It would appear there is no loss.

As I mentioned before, there is a loss but it's due to time dilation which Newtonian gravity does not take into account.

I don't doubt that Newtonian gravity covers the situation, in that energy has to be conserved.  I just don't see how it can explain it.

The downward leg of the loop gains energy through gravity.  The upward leg neither gains nor looses energy.  Therefore there is an excess of energy left over.
Can you explain how Newtonian gravity can account for the above apparent excess of energy in this experiment?
Thanks

#### JP

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##### Is the energy needed to create pairs of identical particles always the same?
« Reply #41 on: 19/07/2011 16:53:27 »
Mike, I stopped responding because it's not a debate and you continue to ignore or misinterpret most of what I say.  I've said repeatedly that the current going up the wire loses energy due to pushing upward against gravity.  It loses the same amount of energy, in fact, that it gained coming down the wire.

If you don't understand this, it's beyond the scope of a forum for me to teach you a course in introductory mechanics and gravity.  However, any textbook for "Physics 101," i.e. introductory mechanics, will cover this in detail.

If you still don't believe this, you're free to have your own theory of gravity as PhysBang said.  This is the New Theories forum.  However, the claims you're making are demonstrably false.

#### MikeS

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##### Is the energy needed to create pairs of identical particles always the same?
« Reply #42 on: 19/07/2011 18:36:25 »
JP
Adopting a condescending attitude is non productive.

I have not intentionally ignored or mis-interpreted anything that you have said.

It is you that continually ignores what I say and ask.

Mike, I stopped responding because it's not a debate and you continue to ignore or misinterpret most of what I say.  I've said repeatedly that the current going up the wire loses energy due to pushing upward against gravity.  It loses the same amount of energy, in fact, that it gained coming down the wire.

Yes, I agree, I keep telling you I agree but that is only one half of the loop.  The other side of the loop is pair particles gaining energy as they fall.  Therefore, there is an excess of energy in every cycle of the loop.  You keep ignoring this.

"The downward leg of the loop gains energy through gravity.  The upward leg neither gains nor looses energy.  Therefore there is an excess of energy left over."  Can you explain how Newtonian gravity can account for the above apparent excess of energy in this experiment?

The gravitational potential energy of the pair particles is converted into kinetic energy as they fall.  They have gained kinetic energy.
Do you agree?

The other leg of the experiment is taking the generated electricity back to the top of the tower to create more pair particles.

This leg is a two wire circuit.  Gravity affects both wires equally.  It tries to slow the flow of electrons in the upward wire and accelerate the flow in the downward wire.  The gravitational effect on the electrical circuit is cancelled.  There is no gain or loss of energy.
Do you agree?

This leaves a surplus of energy created by the falling pair particles.
Do you agree?

If not why not?

You still have not accounted for where this excess energy has come from?
Repeating that
"the current going up the wire loses energy due to pushing upward against gravity.  It loses the same amount of energy, in fact, that it gained coming down the wire." Is not going to help as it is only looking at one half of the experiment.  The other half shows an energy gain.

I am not proposing anything new.  I simply know that time dilation (which is a known and tested fact) can and does simply explain the experiment in question.  You have still not explained how Newtonian gravity explains away the experiments apparent overall energy gain.

If you still don't believe this, you're free to have your own theory of gravity as PhysBang said.  This is the New Theories forum.  However, the claims you're making are demonstrably false.

I am making one claim in this experiment and that is 'time dilation' simply explains the mechanism by which energy is lost on the upward (electrical) leg of the experiment.  I have explained the mechanism in detail in previous posts.
You have made the statement that my claims are "demonstrably false".  You need to demonstrate in what way they are false.

I have also claimed that I do not understand in what way (the mechanism by which) Newtonian gravity can explain away the apparent energy gain of the experiment.
« Last Edit: 19/07/2011 18:48:01 by MikeS »

#### PhysBang

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##### Is the energy needed to create pairs of identical particles always the same?
« Reply #43 on: 19/07/2011 19:07:23 »
This leg is a two wire circuit.  Gravity affects both wires equally.  It tries to slow the flow of electrons in the upward wire and accelerate the flow in the downward wire.  The gravitational effect on the electrical circuit is cancelled.  There is no gain or loss of energy.
Do you agree?
No, since clearly some of the energy from the wire is going in to creating the photons. So there is less energy going along the current downwards and this is gaining less energy from going downwards.

#### MikeS

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##### Is the energy needed to create pairs of identical particles always the same?
« Reply #44 on: 20/07/2011 06:58:57 »
This leg is a two wire circuit.  Gravity affects both wires equally.  It tries to slow the flow of electrons in the upward wire and accelerate the flow in the downward wire.  The gravitational effect on the electrical circuit is cancelled.  There is no gain or loss of energy.
Do you agree?
No, since clearly some of the energy from the wire is going in to creating the photons. So there is less energy going along the current downwards and this is gaining less energy from going downwards.

The current in the circuit must be the same wherever measured.  That's what current does.  Therefore, the gravitational pull is the same on both the upward and downward wires.  The gravitational effect is cancelled.  The resistance of the load determines the current flow but as already mentioned this is the same in both wires.
There is not less current going downwards to be affected by gravity.
(It's pair particles being created not photons)
« Last Edit: 20/07/2011 07:02:54 by MikeS »

#### JP

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##### Is the energy needed to create pairs of identical particles always the same?
« Reply #45 on: 20/07/2011 18:30:03 »
1) X+Y is the energy supplied at the source.

2) X is the energy left after using Y units to rise to the top.

3) 0 is the energy left after giving up X units to create particles.

4) Y is the energy gained on the fall down.

Total energy at the bottom: Y.

5) X:  If the particles created from the energy are massive and allowed to fall, then the total energy they have at the bottom is X.

6) X+Y: Added to the current's energy, this gives X+Y total energy which brings us back to...

1) X+Y is the energy supplied at the source.

.
.
.

#### MikeS

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##### Is the energy needed to create pairs of identical particles always the same?
« Reply #46 on: 21/07/2011 07:23:16 »
1) X+Y is the energy supplied at the source.

2) X is the energy left after using Y units to rise to the top.

3) 0 is the energy left after giving up X units to create particles.

4) Y is the energy gained on the fall down.

Total energy at the bottom: Y.

5) X:  If the particles created from the energy are massive and allowed to fall, then the total energy they have at the bottom is X.

6) X+Y: Added to the current's energy, this gives X+Y total energy which brings us back to...

1) X+Y is the energy supplied at the source.

.
.
.

You are still overlooking that in the experiment there is no apparent energy gain or loss on the upward leg of the loop, the total electrical circuit.  Therefore, Y units are not being used to rise to the top, they are surplus.

1st cycle
Energy at top is      X
Energy at bottom is    X + Y where Y is energy gained on the fall down.

2nd cycle
Energy at top is     X + Y
Energy at bottom is      X + 2Y

3rd cycle
Energy at top is     X + 2Y
Energy at bottom is      X + 3Y

#### JP

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##### Is the energy needed to create pairs of identical particles always the same?
« Reply #47 on: 21/07/2011 12:58:58 »
Well, now we've pinned down exactly where you're wrong.  It takes energy to move something uphill when gravity is pulling it down.  You can see this simply by rolling a ball up a hill.  It will arrive at the top with less energy than it had at the bottom.

#### MikeS

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##### Is the energy needed to create pairs of identical particles always the same?
« Reply #48 on: 21/07/2011 16:27:28 »
Well, now we've pinned down exactly where you're wrong.  It takes energy to move something uphill when gravity is pulling it down.  You can see this simply by rolling a ball up a hill.  It will arrive at the top with less energy than it had at the bottom.

No, we have not pined it down.  You are still not reading or answering my posts.
I agree with the above entirely but that is not what we are discussing.
What we are discussing is in the experiment we have been considering, according to Newtonian gravity there appears to be a surplus of energy left over on every loop.  Newtonian gravity seems unable to explain this surplus energy away whilst time dilation does.

If you go back and read this post you will see what I mean:-
JP

Mike, I stopped responding because it's not a debate and you continue to ignore or misinterpret most of what I say.  I've said repeatedly that the current going up the wire loses energy due to pushing upward against gravity.  It loses the same amount of energy, in fact, that it gained coming down the wire.

Yes, I agree, I keep telling you I agree but that is only one half of the loop.  The other side of the loop is pair particles gaining energy as they fall.  Therefore, there is an excess of energy in every cycle of the loop.  You keep ignoring this.

"The downward leg of the loop gains energy through gravity.  The upward leg neither gains nor looses energy.  Therefore there is an excess of energy left over."  Can you explain how Newtonian gravity can account for the above apparent excess of energy in this experiment?

The gravitational potential energy of the pair particles is converted into kinetic energy as they fall.  They have gained kinetic energy.
Do you agree?

The other leg of the experiment is taking the generated electricity back to the top of the tower to create more pair particles.

This leg is a two wire circuit.  Gravity affects both wires equally.  It tries to slow the flow of electrons in the upward wire and accelerate the flow in the downward wire.  The gravitational effect on the electrical circuit is cancelled.  There is no gain or loss of energy.
Do you agree?

This leaves a surplus of energy created by the falling pair particles.
Do you agree?

If not why not?

You still have not accounted for where this excess energy has come from?
Repeating that
"the current going up the wire loses energy due to pushing upward against gravity.  It loses the same amount of energy, in fact, that it gained coming down the wire." Is not going to help as it is only looking at one half of the experiment.  The other half shows an energy gain.

I am not proposing anything new.  I simply know that time dilation (which is a known and tested fact) can and does simply explain the experiment in question.  You have still not explained how Newtonian gravity explains away the experiments apparent overall energy gain.

#### JP

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##### Is the energy needed to create pairs of identical particles always the same?
« Reply #49 on: 21/07/2011 16:31:57 »
Mike, as I posted elsewhere, you don't understand the physics you're arguing.  I don't see the point in continuing this until you make an attempt to do so.

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##### Is the energy needed to create pairs of identical particles always the same?
« Reply #49 on: 21/07/2011 16:31:57 »