Is the energy needed to create pairs of identical particles always the same?

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Offline MikeS

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All matter within the universe is within the gravitational field of the universe.
When a matter/antimatter particle pair are created they therefore are created with gravitational potential energy.

My question is this:-
Does this extra energy have to be accounted for in the pair particle creation?
It seems to me that in order for energy to be conserved it has to.

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Offline MikeS

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Was it something I said?

I know that some of you, perhaps all of you do not agree with me but if I am wrong then I would like to know in what way and I need a better explanation that just to say that I am wrong.

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Offline JP

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My gut reaction is yes, that it has to be, but my secondary reaction is twofold:

1) Energy conservation in general relativity isn't always a nice clean thing: http://www.desy.de/user/projects/Physics/Relativity/GR/energy_gr.html

2) Gravity is such a weak force that we don't really know how it interacts with things on a quantum mechanical level. 

So my final answer is that I don't know.  :)

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Offline yor_on

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Are you asking if particles exist in gravity?

SpaceTime is defined by 'space' in where we find matter and radiation.
Space as such is 'curved' depending on 'gravity'. Where 'gravity' is stronger particles might 'bend their path' and blue, or red, shift, as defined by some 'inertial observer'.

You might want to define stretches between galaxies as 'flat', just as you when looking very closely at a sphere might find it to be 'flat' too. Neither of those is any proof of 'space' existing without gravity though. Although there is a state wherein you can define 'space' as being 'flat', called a geodesic, it is related to a very special type of relative motion we call 'uniform motion'. The best way to see that motion is using radiation, a photon, as we then don't have to discuss any mass, as it is defined from main stream physics as being mass less.

It depends also on what you think give us a 3D space. You might define it as gravity. Then the next question becomes in what way gravity comes to be? If you define it as 'radiating' from primary mass and accelerations, and to that add that it takes a 'time' 'c' to 'propagate, you might assume that the space between galaxies could be seen as flat. But that would then also mean that we no longer can talk about space as 3D except in those cases where gravity has come to reach. That would make a  galactic space trip rather tricky, like traveling a one dimensional string.

On the other hand you can postulate that space is 'gravity', meaning that a 'one dimensional' space can not be. That one better fits the universe we see, and in that one any particle will be defined by the point it is created in, in some 4D position relative you. That 'potential' you discuss would then be observer dependent.
==

The observer dependence is in which way you observe it, being at rest with it, or as defined by not being 'at rest', like standing on a planet observing it/them getting created in space. In a geodesic there is no gravity to be observed, excepting the gravity you and your ships invariant mass create locally.

And that is the wonder of a 'uniform motion'. That, and the fact that all 'uniform motions' are the same in a 'black room scenario' making it impossible for you to differ them through experiments. Which to me is a proof that our definitions of 'motion' are slightly lacking, to me it seems as we haven't really described it as it should be described.

That one is discussable though, as what I do here is to define 'gravity' as equivalent to 'energy'. If we assume that 'space warps' around a particle, then that 'warping' should be there no matter your motion, uniform or accelerating. When it comes to 'energy', as in 'potential', my statement is correct though, as you being at rest with that particle will find it of a less energy than when accelerating. Maybe this one is possible to use for a argument that 'gravity' isn't observer dependent, it's a tricky one as you still will find it to be observer dependent locally, but not between frames of reference. But energy will always be observer dependent, differing between frames of reference, as to compared locally. So 'energy' and 'gravity' can not be the exact equivalent, as I see it now.

Which, if my reasoning is correct? Should mean that there is no such thing as a 'potential gravity'. You can only find a 'same gravity' as compared between 'frames of reference' and locally. That as I then would expect that particle to 'warp space', just as good with you in that geodesic, as when watching it from that planet. Which leaves us the question of how 'gravity' can disappear in that geodesic. If you define it so then a 'free fall' is the best description leaving you at rest with the overall 'gravity' surrounding you, even though your, and the particles, invariant 'rest mass' still will influence the space near you.
==

That's also why it is simplest to discuss this from 'photons propagating', as we then can leave out 'rest mass'. So in my universe space now is defined by gravity(s metric), and gravity can't be 'potential'. Although the 'energy' expressed is dynamically changing, with invariant 'rest masses relative motion' and accelerations.

If this now makes sense :)
=

Da*n, I will need to think about this one some more. It's easier when you think of it in form of energy, gravity is tricky. Still, a lot of the confusion comes from comparing between 'frames of reference'. Defined locally all particles will have one 'energy' and one 'gravity', defined between frames it seems to me that 'gravity' still will hold, but 'energy' will change depending on motion.
==

Assume that you accelerate one kg. That acceleration will according to Einsteins equivalence principle now 'frame drag' the space surrounding it, and warp it. If you were on it and felt one gravity act on you constantly, would the 'warping of that space' be the same at all times? That can't be true, showing us that even though the gravity is equivalent to earth, the warping of space is not. Then assume that you accelerate this way for a week (one G) closing your engines each day to 'coast' uniformly. What happens with that warping surrounding your rocket as you do that? Will the relative 'speed' you build up between each time you coast make a difference?

As far as I can see it won't. The only 'gravity' you will find coasting, no matter your 'speed' is the one created from your own, and the ships, mass.

And that one will mean that although the invariant 'rest mass' might warp 'space' the same, no matter if you observe it locally, or not (frames of reference), the 'mass', and 'gravity', created by your acceleration can not be the same as a invariant 'rest mass'. Also it tells us that a uniform motion is the place where your 'rest mass' is easiest to define, as it won't differ, no matter your 'relative speed'.
« Last Edit: 12/07/2011 07:08:42 by yor_on »
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Offline MikeS

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JP
Thanks for your honest answer.

yor_on
Let me re-phrase the question.
Would it take more energy to create pair particles high above the Earth than at ground level?

From a previous post
“Consider the following experiment.
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.

In your little experiment in last two posts you can even remove the need to generate paricles - just have a laser light in high orbit connected with a wire to a photovoltaic cell in low orbit.  We can assume very low inefficiency and superconductivity etc - and a huge drop in potential.  I do not know the mechanism - but clearly the transfer of electrical energy thru a wire from low to high grav pot cannot be lossless as this would entail a net energy gain.
[/quote imatfaal]

Both of the above examples, if they worked would be perpetual motion.  Clearly this cannot be.  Assuming super conductors etc.  Its a two wire circuit, electrons go up one wire presumably slowed by gravity accelerated down the other.  The gravitational effect being cancelled.  So where is the energy going? 

The electricity is being produced in a low gravitational potential where time is dilated relative the high GP where electric is being used.  Due to time dilation at source and contraction at sink the current generated at source is more than at sink.  There is a loss of energy due to gravitational time dilation.”

Clearly, to me at least, it requires more energy to create pair particles in a high gravitational potential.  The mass energy equivalence principle holds but extra energy is required if particles are to be created in a higher gravitational potential.

It costs energy, lots of energy to raise an object to a higher gravitational potential, take the space shuttle for instance.  I don't see why pair particle production would be any different?

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Offline yor_on

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"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."

Assume it was a man jumping down on some device instead, generating some electricity as it compressed. How would he get back up to do it again? Would it cost him energy to get up to jump?

when it comes to "it requires more energy to create pair particles in a high gravitational potential." I'm not sure. If you assume that the particles are in a 'free fall' then the gravity as such will be the same everywhere, and the 'gravitational field' will be negligible. And to assume that the particles comes into creation fighting gravity doesn't seem probable, at least not if we're discussing spontaneous pair production. If you assume something expending a lot of energy creating them, having a vector in 3D +1 space/time, they should still find themselves in a uniform motion at creation. To accelerate them would be creating a LHC in space.

What I think you're thinking of is if energy and gravity is the same. That's a question we both share if so, and the honest answer is as always, wish I knew? :)

But if we look at the first post we can see that energy and gravity must differ. If you think of that kg accelerating it will deliver one G locally, but it will not 'warp Space' around it as Earth can. As it accelerate uniformly it should increase that warp though, and it must have a relation to the energy it would express if colliding with something uniformly moving. A big problem there is uniform motion, at least to me. All uniform motions are  the same, being 'free falls' and being 'at rest' with the gravitational potential surrounding them. But they all differ motion wise, relative some arbitrarily chosen 'inertial frame'.

To me it becomes a sort of 'magic', you can by any uniform motion nullify all gravity existing, except the one you are expected to express yourself in your amount of invariant rest mass. It's about motion, 'energy' and gravity.

You might want to assume that not following a geodesic should take energy though, and that I think is correct.
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Offline JP

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MikeS, I'm not clear where the perpetual motion paradox comes in.

Any gain in power for a motor on the ground as compared to a motor in orbit is due to the energy gained by the electrons when they fall from orbit to the ground.  This energy has to come from somewhere, and it comes from transporting the electrons (or their parent particles) up into orbit to begin with, which takes energy.

Back to your original question, yes, it does take more energy to create particles in orbit than it does on the earth's surface because you have to get those particles into orbit and then collide them, and moving them up into orbit takes energy. 

Also, general relativity is a red herring here.  You don't need it to work this out--Newtonian gravity works just fine since this effect doesn't require the precision of GR.

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Offline MikeS

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"when it comes to "it requires more energy to create pair particles in a high gravitational potential." I'm not sure. If you assume that the particles are in a 'free fall' then the gravity as such will be the same everywhere, and the 'gravitational field' will be negligible.

At the moment of creation, the particles can be considered to be stationary at a certain height within the gravity well.  Therefore, they immediately have a finite gravitational potential.  They are not under free fall so gravity does have to be considered.  Even if they were born in free fall, it would take time for them to convert some of their potential energy into kinetic energy as they start to accelerate.  Initially they are still at rest.

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Offline MikeS

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MikeS, I'm not clear where the perpetual motion paradox comes in.

Any gain in power for a motor on the ground as compared to a motor in orbit is due to the energy gained by the electrons when they fall from orbit to the ground.  This energy has to come from somewhere, and it comes from transporting the electrons (or their parent particles) up into orbit to begin with, which takes energy.

Back to your original question, yes, it does take more energy to create particles in orbit than it does on the earth's surface because you have to get those particles into orbit and then collide them, and moving them up into orbit takes energy. 

Also, general relativity is a red herring here.  You don't need it to work this out--Newtonian gravity works just fine since this effect doesn't require the precision of GR.

JP
Some people think that it costs no more energy to create particles in a higher gravitational potential than a lower one.  I didn't agree, which was why I mentioned what would have been a paradox.

Although it seems obvious that it costs energy to raise an object to a greater gravitational potential, I don't see how Newtonian gravity can adequately explain what is happening.
Quote from my previous post

"Both of the above examples, if they worked would be perpetual motion.  Clearly this cannot be.  Assuming super conductors etc.  Its a two wire circuit, electrons go up one wire presumably slowed by gravity accelerated down the other.  The gravitational effect being cancelled.  So where is the energy going?"

I don't see how Newtonian gravity can explain that.  There is a gain of energy from PE being converted into KE but why is there a loss on the upward leg?  Gravity is pulling on the rising electrons (trying to slow them) and pulling on the descending electrons (trying to accelerate them).  The gravitational effects cancel.  This still leaves the question where has the energy gone?

I think it requires GR to explain the situation.
My interpretation was this from a previous quote.
"The electricity is being produced in a low gravitational potential where time is dilated relative the high GP where electric is being used.  Due to time dilation at source and contraction at sink the current generated at source is more than at sink.  There is a loss of energy due to gravitational time dilation.”

In more detail.
At the source deep within the gravity well (generator or photovoltaic cell) electricity is being created and its current flow is at the speed of light.  X number of electrons flow past a point in the cable per second.
At the current sink at high altitude time is contracted.  Less electrons flow past a given point in the cable per second.  The current is lower at a high gravitational potential than at a lower gravitational potential.

I would be interested to know in what way you think Newtonian gravity can explain the situation and secondly do you agree with my GR explanation and if not, why not. 
Thanks very much.

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Offline MikeS

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It could be argued that photon blue-shift which is usually interpreted as a gain of energy is nothing more than an effect of observing the photons from a dilated time frame.  Photons do not gain energy falling within a gravitational field, they only appear to.  If a photon can not accelerate how can it gain energy?  There is no gain of energy.

Matter on the other hand does gain KE free falling (accelerating) in a gravitational field by converting its PE into KE.  Overall the energy remains the same.
« Last Edit: 14/07/2011 08:50:06 by MikeS »

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Offline JP

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Mike, you seem to have changed the question a bit from particles to currents.  Here's why Newtonian gravity explains all these cases, though.  Newtonian gravity and energy conservation says that if it takes Y energy units to raise something to a given height, that thing will gain Y units of energy falling from that height.  If I want to collide particles in orbit and that collision requires X units energy be put into the particles, doing it in a lab will take X energy, but taking those particles from the lab into orbit and then colliding them will take X+Y energy (Y to raise them up and X to collide them.) 

The same goes for electron flows.  Let's say I have an electron gun that I charge with X units of potential energy and then it fires electrons with X units of kinetic.  If I fire it in my lab, I get electrons with X units of energy when I put in X units of energy.  If I fire these electrons from orbit down to my lab, I charge the gun with X units of energy and get electrons with X+Y units of energy.  Yippee!  Free energy!  But when you think about it, you had to transport those electrons up to the gun to begin with, which takes Y units of energy per electron.  The total cost of firing an electron from an orbital gun is going to be X+Y units of energy and you get an electron with ... X+Y units of energy.  No gain.  And entirely explained by Newtonian gravity.

Of course GR can explain this too, but GR doesn't treat gravity as a force and you're stuck doing coordinate transformations from frame to frame rather than using conservation of energy (which isn't even a law in GR as it is in Newtonian gravity).  I'm very wary of attempts to explain GR effects with hand waving and the invocation of terms like "time dilation" and "length contraction."  Sure, these are things, but they can often be misleading and confusing, especially when in the hands of someone (like myself) who doesn't fully understand the physical implications of the coordinate transformations involved.

By the way, because the earth is a very weakly gravitating source, any GR results had better match very well with Newtonian results.  In that case, why bother using the more complex theory?

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Offline MikeS

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JP
I agree Newtonian gravity does explain the example you gave but I don't think it explains the example that I gave.  In a two wire circuit from low to high gravitational potential where does the energy loss arise?  Saying that it requires energy to put something into a higher gravitational potential is correct but does not explain what is happening in my example.

Time dilation, I believe, explains exactly what is happening in very simple terms as different parts of the circuit are time dilated at different rates.

Although you agreed that GR can explain this you did not answer my question of whether you agree with my interpretation or not and if not why not? 

I am looking at the problem purely from the gravitational time dilation aspect and as far as I can see my conclusions must be correct.  It's so simple and obvious, not in the slightest way complicated, it must be right.  Unless you know differently of course?   [;)]

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Offline JP

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You send current up the wire.  It takes more energy to do so because it's rising against gravity.  Where's the problem with that?

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Offline MikeS

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You send current up the wire.  It takes more energy to do so because it's rising against gravity.  Where's the problem with that?

The other wire of the circuit has electricity flowing down it assisted by gravity.  They cancel each other out.  There is the problem.
« Last Edit: 15/07/2011 14:13:10 by MikeS »

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Offline JP

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Why is that a problem?  The energy can, and should, exactly cancel out.  If you throw a ball up in the air, the energy needed to toss it up to a given height is exactly the same as the energy it gains falling from that height, so it reaches the ground with the same speed and kinetic energy it left with (minus losses due to air resistance.) 

The exact same thing happens in a wire.  The electrons returning to the ground do so with exactly the energy they left with (minus losses due to resistivity in the wire), so there's no gain of energy and no perpetual motion.

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Offline MikeS

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Why is that a problem?  The energy can, and should, exactly cancel out.  If you throw a ball up in the air, the energy needed to toss it up to a given height is exactly the same as the energy it gains falling from that height, so it reaches the ground with the same speed and kinetic energy it left with (minus losses due to air resistance.) 

The exact same thing happens in a wire.  The electrons returning to the ground do so with exactly the energy they left with (minus losses due to resistivity in the wire), so there's no gain of energy and no perpetual motion.

Previously you said this, they can't both be right.
You send current up the wire.  It takes more energy to do so because it's rising against gravity.  Where's the problem with that?

JP
Thanks for your honest answer.

yor_on
Let me re-phrase the question.
Would it take more energy to create pair particles high above the Earth than at ground level?

From a previous post
“Consider the following experiment.
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.

In your little experiment in last two posts you can even remove the need to generate paricles - just have a laser light in high orbit connected with a wire to a photovoltaic cell in low orbit.  We can assume very low inefficiency and superconductivity etc - and a huge drop in potential.  I do not know the mechanism - but clearly the transfer of electrical energy thru a wire from low to high grav pot cannot be lossless as this would entail a net energy gain.
[/quote imatfaal]


Both of the above examples, if they worked would be perpetual motion.  Clearly this cannot be.  Assuming super conductors etc.  Its a two wire circuit, electrons go up one wire presumably slowed by gravity accelerated down the other.  The gravitational effect being cancelled.  So where is the energy going? 

The electricity is being produced in a low gravitational potential where time is dilated relative the high GP where electric is being used.  Due to time dilation at source and contraction at sink the current generated at source is more than at sink.  There is a loss of energy due to gravitational time dilation.”

Clearly, to me at least, it requires more energy to create pair particles in a high gravitational potential.  The mass energy equivalence principle holds but extra energy is required if particles are to be created in a higher gravitational potential.

It costs energy, lots of energy to raise an object to a higher gravitational potential, take the space shuttle for instance.  I don't see why pair particle production would be any different?

Your answer would seem to take us back full circle to the above experiment which implies that perpetual motion can exist.  Clearly, it can't.  So far, Newtonian gravity in the above experiment has failed to show me why not.  Time dilation does.

We agree that gravitational effects in the circuit are cancelled in the above experiment?
Therefore there is no gain or loss of energy in the electrical circuit transporting energy from a low GP to a high GP.  Clearly, this is wrong but Newtonian gravity fails to explain it.
We agree that the falling pair particles will convert PE to KE?
We agree that the KE can be used to do work ie produce electricity?
Therefore there would seem to be a gain of energy?

I still don't see how Newtonian gravity can explain this apparent discrepancy.

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Offline JP

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One of the predictions of Newtonian gravity is that gravity is a conservative force.  That means that if you take any closed path in a gravitational field so that you end up back where you started, your total energy gain due to gravity is exactly zero (you will have losses due to friction, but not to gravity.)  This holds for the electrons in your wire, so when they complete a loop, they come out with zero net energy gain.

Another way of looking at it is to compute the energy gained when they come down the wire and the extra energy it takes to push them up the wire.  Let's say they gain Y units of kinetic energy coming down the wire.  To go up the wire will take them exactly Y extra units of kinetic energy due to gravity resisting the upward push.  The total energy gain from a complete loop is +Y energy from coming down -Y energy from going back up or +Y-Y=0 energy gain, as I said above.

These are both results from dealing with Newtonian gravity as a conservative force without using any results from general relativity.  Where in these explanations is there a problem?
« Last Edit: 15/07/2011 18:36:11 by JP »

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Offline MikeS

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From a previous post
“Consider the following experiment.
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.

This would seem to be perpetual motion.  There has to be an energy loss to balance the energy gain of the particle pairs in falling down the tower.  We agree that any gravitational effects on the upward and downward paths of the electricity are cancelled (there is no loss or gain of energy).  There is an energy gain on the downward leg (falling particles) and no gain or loss on the upward leg (electrical circuit).  This still leaves the original question of where has the excess energy gone?

The energy has to be lost on the upward leg, the electrical circuit.  This is what you said originally that energy was required to lift the electrons against gravity.  When I pointed out that the same gravity is also pulling electrons in the other wire of the circuit down, you agreed that energy is conserved in the electrical circuit. 

JP quote
"One of the predictions of Newtonian gravity is that gravity is a conservative force.  That means that if you take any closed path in a gravitational field so that you end up back where you started, your total energy gain due to gravity is exactly zero (you will have losses due to friction, but not to gravity.)  This holds for the electrons in your wire, so when they complete a loop, they come out with zero net energy gain."

The above has to remain true whilst still showing a loss of energy at a higher gravitational potential.  Time contraction allows that to happen.  A second at high altitude is shorter so less electrons flow past a certain point in the circuit per second.  The current is less at a high gravitational potential than at a lower one.  Energy has been lost.

I still fail to see how Newtonian gravity can explain what is happening in this experiment, whereas time dilation does.

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Offline PhysBang

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It could be argued that photon blue-shift which is usually interpreted as a gain of energy is nothing more than an effect of observing the photons from a dilated time frame.
These are the same thing. They have the same physical meaning, all that changes is the particular means of description that one is using.
Quote
Photons do not gain energy falling within a gravitational field, they only appear to.
If by "appear to" you mean, "behave in absolutely all interactions as if they have a higher energy", then yes. But that's a funny "appear to".

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Offline PhysBang

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“Consider the following experiment.
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.
Why would we think that it is possible to harvest energy from these particles? These particles either self-annihilate or one of them annihilates with another particle. (This is essentially how black holes lose mass and energy.) In any case, the net energy that they introduce into the system is zero. If they fall and gain energy, then they annihilate that energy, too. No detailed thinking about gravity or time dilation required.

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Offline JP

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Mike,

1) To have a perpetual motion machine, you have to get free energy out of a system without putting any energy in.

2) I've demonstrated above that with Newtonian gravity, you don't gain energy--you break even on each loop.

3)  In reality, since there's resistivity in an actual system, you actually lose energy as the current runs.

4) So it's not perpetual motion.

QED.

If you disagree with this, please tell me which point you disagree with. 

By the way, "perpetual motion" might be a bit misleading.  In the absence of friction or other losses, a current can move forever without stopping.  This is purely theoretical, though, since there are always some losses in reality, even in the best superconductors.  But in theory, even if there is no resistance in the above wire and the current moves forever, it's not a violation of the second law of thermodynamics unless it gains energy with each loop.

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Offline MikeS

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JP
Mike,

1) To have a perpetual motion machine, you have to get free energy out of a system without putting any energy in.

2) I've demonstrated above that with Newtonian gravity, you don't gain energy--you break even on each loop.

3)  In reality, since there's resistivity in an actual system, you actually lose energy as the current runs.

4) So it's not perpetual motion.

QED.

If you disagree with this, please tell me which point you disagree with. 

By the way, "perpetual motion" might be a bit misleading.  In the absence of friction or other losses, a current can move forever without stopping.  This is purely theoretical, though, since there are always some losses in reality, even in the best superconductors.  But in theory, even if there is no resistance in the above wire and the current moves forever, it's not a violation of the second law of thermodynamics unless it gains energy with each loop.

1) You just need an excess of energy.
That's what this would appear to do.
“Consider the following experiment.
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."
2)Newtonian gravity only balances the electrical side of the loop.  There is an excess of energy left over on the other side of the loop from particles gaining KE.
3) Yes.
4) According to Newtonian gravity there is an excess of energy left over, so it is perpetual motion.
Newtonian gravity is not accounting for why there is a loss of energy on the upward (electricity) side of the loop.
The above experiment would seem to gain energy on each loop.

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Offline MikeS

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It could be argued that photon blue-shift which is usually interpreted as a gain of energy is nothing more than an effect of observing the photons from a dilated time frame.
These are the same thing. They have the same physical meaning, all that changes is the particular means of description that one is using.
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Photons do not gain energy falling within a gravitational field, they only appear to.
If by "appear to" you mean, "behave in absolutely all interactions as if they have a higher energy", then yes. But that's a funny "appear to".

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.  I don't see how they can.  When the energy they have, is measured in a dilated time frame it will appear more. If you measure it in a contracted time frame it will appear less.

If you measure the energy of a photon, it should be remembered the measurement is not instantaneous. It takes place over time.
« Last Edit: 16/07/2011 16:37:20 by MikeS »

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Offline MikeS

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“Consider the following experiment.
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.
Why would we think that it is possible to harvest energy from these particles? These particles either self-annihilate or one of them annihilates with another particle. (This is essentially how black holes lose mass and energy.) In any case, the net energy that they introduce into the system is zero. If they fall and gain energy, then they annihilate that energy, too. No detailed thinking about gravity or time dilation required.

This is a follow up on this article
http://www.desy.de/user/projects/Physics/ParticleAndNuclear/antimatter_fall.html
Its meant as a thought experiment, not to be taken literally.

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.

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Offline PhysBang

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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.

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Offline JP

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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.

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Offline MikeS

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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.

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Offline MikeS

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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?

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Offline PhysBang

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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?
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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.
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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.

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Offline MikeS

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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?
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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.
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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.

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Offline PhysBang

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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.
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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.

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Offline MikeS

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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.
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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?

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Offline PhysBang

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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.

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Offline MikeS

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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.

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Offline PhysBang

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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.
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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/
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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.

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Offline MikeS

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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]
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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/
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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?

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Offline PhysBang

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  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!

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Offline MikeS

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  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.

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Offline PhysBang

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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.

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Offline MikeS

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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.

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Offline MikeS

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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

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Offline JP

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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.

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Offline MikeS

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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.

Could you please answer the simple question in my last post.
"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 »

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Offline PhysBang

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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.

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Offline MikeS

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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 »

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Offline JP

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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.

.
.
.

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Offline MikeS

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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

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Offline JP

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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.

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Offline MikeS

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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.

Could you please answer the simple question in my last post.
"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.




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Offline JP

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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.