The Naked Scientists

The Naked Scientists Forum

Author Topic: Is the energy needed to create pairs of identical particles always the same?  (Read 13591 times)

Offline MikeS

  • Neilep Level Member
  • ******
  • Posts: 1044
  • The Devils Advocate
    • View Profile
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.


 

Offline MikeS

  • Neilep Level Member
  • ******
  • Posts: 1044
  • The Devils Advocate
    • View Profile
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.
 

Offline JP

  • Neilep Level Member
  • ******
  • Posts: 3366
  • Thanked: 2 times
    • View Profile
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.  :)
 

Online yor_on

  • Naked Science Forum GOD!
  • *******
  • Posts: 12001
  • Thanked: 4 times
  • (Ah, yes:) *a table is always good to hide under*
    • View Profile
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 »
 

Offline MikeS

  • Neilep Level Member
  • ******
  • Posts: 1044
  • The Devils Advocate
    • View Profile
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?
 

Online yor_on

  • Naked Science Forum GOD!
  • *******
  • Posts: 12001
  • Thanked: 4 times
  • (Ah, yes:) *a table is always good to hide under*
    • View Profile
"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.
 

Offline JP

  • Neilep Level Member
  • ******
  • Posts: 3366
  • Thanked: 2 times
    • View Profile
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.
 

Offline MikeS

  • Neilep Level Member
  • ******
  • Posts: 1044
  • The Devils Advocate
    • View Profile
"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.
 

Offline MikeS

  • Neilep Level Member
  • ******
  • Posts: 1044
  • The Devils Advocate
    • View Profile
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.
 

Offline MikeS

  • Neilep Level Member
  • ******
  • Posts: 1044
  • The Devils Advocate
    • View Profile
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 »
 

Offline JP

  • Neilep Level Member
  • ******
  • Posts: 3366
  • Thanked: 2 times
    • View Profile
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?
 

Offline MikeS

  • Neilep Level Member
  • ******
  • Posts: 1044
  • The Devils Advocate
    • View Profile
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?   ;)
 

Offline JP

  • Neilep Level Member
  • ******
  • Posts: 3366
  • Thanked: 2 times
    • View Profile
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?
 

Offline MikeS

  • Neilep Level Member
  • ******
  • Posts: 1044
  • The Devils Advocate
    • View Profile
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 »
 

Offline JP

  • Neilep Level Member
  • ******
  • Posts: 3366
  • Thanked: 2 times
    • View Profile
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.
 

Offline MikeS

  • Neilep Level Member
  • ******
  • Posts: 1044
  • The Devils Advocate
    • View Profile
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.
 

Offline JP

  • Neilep Level Member
  • ******
  • Posts: 3366
  • Thanked: 2 times
    • View Profile
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 »
 

Offline MikeS

  • Neilep Level Member
  • ******
  • Posts: 1044
  • The Devils Advocate
    • View Profile
JP
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.
 

Offline PhysBang

  • Hero Member
  • *****
  • Posts: 588
  • Thanked: 14 times
    • View Profile
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".
 

Offline PhysBang

  • Hero Member
  • *****
  • Posts: 588
  • Thanked: 14 times
    • View Profile
“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.
 

Offline JP

  • Neilep Level Member
  • ******
  • Posts: 3366
  • Thanked: 2 times
    • View Profile
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.
 

Offline MikeS

  • Neilep Level Member
  • ******
  • Posts: 1044
  • The Devils Advocate
    • View Profile
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.
 

Offline MikeS

  • Neilep Level Member
  • ******
  • Posts: 1044
  • The Devils Advocate
    • View Profile
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".

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 »
 

Offline MikeS

  • Neilep Level Member
  • ******
  • Posts: 1044
  • The Devils Advocate
    • View Profile
“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.
 

Offline PhysBang

  • Hero Member
  • *****
  • Posts: 588
  • Thanked: 14 times
    • View Profile
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.
 

The Naked Scientists Forum


 

SMF 2.0.10 | SMF © 2015, Simple Machines
SMFAds for Free Forums