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posted on Mar, 24 2015 @ 08:54 AM
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a reply to: pfishy
That's about the overall energy of the beams (600 sticks of dynamite, or so depending on the experiment). They have a beam dump which has to be able to handle that much energy. The protons carrying that energy have about the same rest mass as a grain of sand.

When two protons collide in a single collision, now that the LHC is upgraded, its about 14TeV. The mosquito in flight analogy is for only 1 TeV, so 14 TeV is like 14 mosquitoes in flight.

Despite the fact that most protons don't collide, there are still 600 million collisions per second:

lhc-machine-outreach.web.cern.ch...

we end up with around 600 million collisions per second
So that's like 14 times 600 million mosquitoes per second flying into a barrier, which is about 1300 watts, about the same rate of energy (power) used by a 1300W space heater or microwave oven, a very impressive figure when you consider how tiny a fraction of a grain of sand is colliding to produce that.

edit on 24-3-2015 by Arbitrageur because: clarification




posted on Mar, 24 2015 @ 12:56 PM
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Ok, I actually have a real question for you. Being that mass and energy are interchangeable in GR (loosely speaking), how would I convert the mass of an object to an equivalent energy quantity? Say, for instance, how does one kg convert to eV? Or is that even a plausible conversion?



posted on Mar, 24 2015 @ 01:02 PM
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originally posted by: Arbitrageur
a reply to: pfishy
That's about the overall energy of the beams (600 sticks of dynamite, or so depending on the experiment). They have a beam dump which has to be able to handle that much energy. The protons carrying that energy have about the same rest mass as a grain of sand.

When two protons collide in a single collision, now that the LHC is upgraded, its about 14TeV. The mosquito in flight analogy is for only 1 TeV, so 14 TeV is like 14 mosquitoes in flight.

Despite the fact that most protons don't collide, there are still 600 million collisions per second:

lhc-machine-outreach.web.cern.ch...

we end up with around 600 million collisions per second
So that's like 14 times 600 million mosquitoes per second flying into a barrier, which is about 1300 watts, about the same rate of energy (power) used by a 1300W space heater or microwave oven, a very impressive figure when you consider how tiny a fraction of a grain of sand is colliding to produce that.

Agreed, and thanks for the clarification. I've always been in near awe at the amounts of energy colliders are able to harness and impart on target. Even the civilian ones for radiation therapy, etc. Bringing something that minute to a level where it can unleash undiscovered fundamental particles is easily one of my favorite aspects of experimental physics.



posted on Mar, 24 2015 @ 02:16 PM
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a reply to: pfishy

The only known way thus far is annihilation using an anti-matter + matter interaction. The issue here is that you use a lot of energy in the first place to create anti-matter.

Now thats not the worse of it. The worse of it is in order to ' process' matter in this interaction, it is kind of tricky because you would really want to convert everything to protons, all i guess i am saying is that, its tricky and there is no magical switch or known process that can allow you to do that conversion on enormous scales.

in essence fusion and fission are our current ways of harnessing that power.



posted on Mar, 24 2015 @ 02:30 PM
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a reply to: PraiseTheHighestOne
The only thing I could think of and this would cause exactly those effects : being a little close to a long range radar system. The microwaves from the radar could and would produce those effects



posted on Mar, 24 2015 @ 03:22 PM
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originally posted by: pfishy
Ok, I actually have a real question for you. Being that mass and energy are interchangeable in GR (loosely speaking), how would I convert the mass of an object to an equivalent energy quantity? Say, for instance, how does one kg convert to eV? Or is that even a plausible conversion?
I made a thread about the mathematical formula for that, since most people think they know the right formula but it's not quite the complete formula, especially if talking about 14 TeV collisions at the LHC:

Is E=mc² right or wrong?

So the math for the conversion of mass to energy isn't hard, but as Eros said, doing the actual conversion of mass to energy is harder, at least for us Earthlings. The sun makes fusion look easy and it converts mass to energy at the rate of 4 million TONS a second. (600 million tons of hydrogen get converted to 596 million tons of helium per second, so most of the mass doesn't get converted into energy).



posted on Mar, 24 2015 @ 05:09 PM
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There is a famous cartoon of a young Einstein writing E=MC^2 on a chalkboard and a teacher scolding him to show his work.

Well; the work just for that one equation is six pages of specialized forms of Algebra calculus and other forms of higher math. Einstein when developing relativity trained under the tutlage of another scientist who had developed as special form of algebra for a year and then practiced on his own for another year or two just to be able to master the math needed to complete Relativity.

there can actually be slight errors within those math and still conclude with E=MC^2. signs that cancel out under dimensional analysis and things of a similar nature. if i remember correctly i have read articles that show a couple of cases where that is know to have happened.



posted on Mar, 25 2015 @ 11:25 AM
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originally posted by: ErosA433
a reply to: pfishy

The only known way thus far is annihilation using an anti-matter + matter interaction. The issue here is that you use a lot of energy in the first place to create anti-matter.

Now thats not the worse of it. The worse of it is in order to ' process' matter in this interaction, it is kind of tricky because you would really want to convert everything to protons, all i guess i am saying is that, its tricky and there is no magical switch or known process that can allow you to do that conversion on enormous scales.

in essence fusion and fission are our current ways of harnessing that power.

I think I phrased my question improperly. I am not actually looking to convert mass to energy. What I am looking for is a way to calculate the equivalent energy contained in a given mass. For example, the mass of the Higgs Boson is given in eV. What would be the eV value for, say, a 50kg human being? Or can kg be calculated and expressed in eV to begin with? I honestly don't know if electron volts is even an applicable measure of mass beyond the quantum scale, but if there is a calculation for determining the total energy of a given mass, perhaps in joules, mAh, or some obscure unit I haven't heard of, I would very much like to know it.
I hope that makes my request a bit more understandable.

Thanks,
pfishy



posted on Mar, 25 2015 @ 11:46 AM
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Oh i See

Well we start with E=mc^2

It is then simply unit conversion, the issue is really people cutting off the units.

Unit of energy = eV or electron volt, which is the energy gained by an electron accelerated across a potential of 1 volt

units of mass in this equation is actually eV/c^2

For conversion say of kg,

E = Joule = Kg m^2 / s^2
m = kg
c = Speed of light, units m/s

RIGHT so, how much mass is that? well the energy gained by an electron accelerated across a potential of 1 volt is

E = QV = 1.6x10^-19 C x 1 V

Thus for energy conversion, 1eV = 1.6x10^-19 J

So what about mass?

1.6x10^-19 = mc^2, solve for m an you get 1eV/c^2 = 1.782×10^−36 kg

So this is where you have for example an electron having a mass of 9.11x10-31 kg and if you do the conversion you get 511keV

Hopefully that makes sense?

50kg human would be 2.8x10^36 eV... or in comparison to the TeV level, 2.8x10^25 TeV... which is a lot compared to the 14TeV single collision of the LHC



posted on Mar, 25 2015 @ 11:47 AM
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originally posted by: Arbitrageur

originally posted by: pfishy
Ok, I actually have a real question for you. Being that mass and energy are interchangeable in GR (loosely speaking), how would I convert the mass of an object to an equivalent energy quantity? Say, for instance, how does one kg convert to eV? Or is that even a plausible conversion?
I made a thread about the mathematical formula for that, since most people think they know the right formula but it's not quite the complete formula, especially if talking about 14 TeV collisions at the LHC:

Is E=mc² right or wrong?

So the math for the conversion of mass to energy isn't hard, but as Eros said, doing the actual conversion of mass to energy is harder, at least for us Earthlings. The sun makes fusion look easy and it converts mass to energy at the rate of 4 million TONS a second. (600 million tons of hydrogen get converted to 596 million tons of helium per second, so most of the mass doesn't get converted into energy).


Thank you for that. But I must admit that my knowledge of advanced mathematics is rather limited. And to be honest, I'm also not sure of the appropriate units of measurement to use for mass to energy calculation. If you could possibly explain it to me in a more verbose fashion, it would be extremely helpful and very appreciated.
And yes, I did intend to use verbose in that context. Less ego damaging than asking you to dumb it down for me.



posted on Mar, 25 2015 @ 11:56 AM
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a reply to: pfishy
"What would be the eV value for, say, a 50kg human being?"

You need to define the question better. If you mean what is the equivalent energy content of the mass, why do you want to know that because you can't readily use or extract it, so it's not a meaningful number to me? If that's what you're after, just take the formula E=mc² and plug in 50kg for the mass, multiply by c squared and you have the equivalent energy.

But if you're thinking of say the mosquito in flight analogy, or maybe a person walking, that is related to momentum and not mass-energy equivalence and would be a more meaningful measure of energy to me. The mosquito in flight analogy link posted at the top of this page (86) shows how that conversion is done. Just plug in the mass and velocity of a human instead of a mosquito.

You can convert between eV and Joules using this:

1 eV = 1.6021773 x 10^-19 Joules

You should probably review the mosquito in flight analogy link.

edit on 25-3-2015 by Arbitrageur because: clarification



posted on Mar, 25 2015 @ 12:09 PM
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originally posted by: Arbitrageur
a reply to: pfishy
"What would be the eV value for, say, a 50kg human being?"

You need to define the question better. If you mean what is the equivalent energy content of the mass, why do you want to know that because you can't extract it, so it's not a meaningful number to me? If that's what you're after, just take the formula Is E=mc² and plug in 50kg for the mass, multiply by c squared and you have the equivalent energy.

But if you're thinking of say the mosquito in flight analogy, or maybe a person walking, that is related to momentum and not mass-energy equivalence. The mosquito in flight analogy link posted at the top of this page (86) shows how that conversion is done. Just plug in the mass and velocity of a human instead of a mosquito.

You can convert between eV and Joules using this:

1 eV = 1.6021773 x 10^-19 Joules

You should probably review the mosquito in flight analogy link.

I know the mass cannot be completely converted to energy, and in general this value has no practical use. The reason I asked is purely out of curiosity. The stuff that has been driving science since time immemorial. And thank you for explaining that more plainly. Oh, I did read your mosquito in flight analogy, and grasp it rather well, as it pertains to our initial discussion. I just was unsure how to scale that to the macro level, and eV was an applicable value for the anything beyond particles. And you have answered that rather well. Thanks.



posted on Mar, 25 2015 @ 12:14 PM
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originally posted by: ErosA433
Oh i See

Well we start with E=mc^2

It is then simply unit conversion, the issue is really people cutting off the units.

Unit of energy = eV or electron volt, which is the energy gained by an electron accelerated across a potential of 1 volt

units of mass in this equation is actually eV/c^2

For conversion say of kg,

E = Joule = Kg m^2 / s^2
m = kg
c = Speed of light, units m/s

RIGHT so, how much mass is that? well the energy gained by an electron accelerated across a potential of 1 volt is

E = QV = 1.6x10^-19 C x 1 V

Thus for energy conversion, 1eV = 1.6x10^-19 J

So what about mass?

1.6x10^-19 = mc^2, solve for m an you get 1eV/c^2 = 1.782×10^−36 kg

So this is where you have for example an electron having a mass of 9.11x10-31 kg and if you do the conversion you get 511keV

Hopefully that makes sense?

50kg human would be 2.8x10^36 eV... or in comparison to the TeV level, 2.8x10^25 TeV... which is a lot compared to the 14TeV single collision of the LHC

Thanks. Mathematics and I have a long standing agreement to avoid each other as much as possible to maintain civility. But between you and Arbitrageur providing relatively simple explanations, I am fairly confident that I can grasp it well enough now to get the values I'm looking for. I might even drum up a spreadsheet to calculate for either end of the equation and in different units. Just to have handy.
Thanks again.
pfishy



posted on Mar, 25 2015 @ 12:37 PM
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a reply to: pfishy
Ok here's a question for you to see if you understand it now, and the math is pretty easy.

The energy released from the largest nuclear explosion ever was about 210 x 10^15 Joules.
Is that bigger, smaller, or about the same as energy equivalent of 50kg of mass?

For E=mc² to come out in Joules you have to use mass in kg, use the speed of light in meters per second (3x10^8 m/s is close enough).



posted on Mar, 25 2015 @ 01:16 PM
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a reply to: ErosA433

That's right. People see E=mc^2 and think "energy is equivalent to matter". It isn't.

Relativistic energy conservation tells you a constraint, "how much might you get", but it doesn't say how. It's clean and simple.

In the end, it's the properties of the Standard Model and the particular physical interactions which make the difference, and these are not clean and simple.

There's a more familiar analogy. Consider a chemical manufacturing plant and refinery. Basic conservation laws will tell you that to get 'x' product out you need 'y' inputs with certain molecules & elements with 'z' waste products. This is like the E=mc^2.

But it won't tell you how to make the stuff. It won't tell you what kind of catalysts you need or the sequence of reactions or the pressures and temperatures or how to do it efficiently or any of the rest of the physics & chemistry to make it actually happen. This is like understanding details of nuclear fission and how to build a reactor and what it's made out of.


edit on 25-3-2015 by mbkennel because: (no reason given)



posted on Mar, 25 2015 @ 01:25 PM
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exactly, its quite tricky

But if we work through what we know regarding binding energy an fusion...

the mass of a proton is 938 MeV

the mass of a alpha (helium nucleus) is 3727 MeV

so it requires basically 4 protons to make 1 alpha (not exactly, you have neutrons, but you can see where i am going with this hopefully) so 4x938MeV = 3753 MeV

The difference is.... 26 MeV

As i said its not exactly like that but basically fusion gives you about 25 MeV per process... these are the only ways other than antimatter in which we may directly convert energy and matter... other than that it is other things like pair prouction using high energy interactions



posted on Mar, 25 2015 @ 02:52 PM
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I have a question I want to ask (not crackpot!) about magnets!

I am having a hard time justifying the effects of natural or manufactured (ie, not electromagnets) and how they fit into the laws of thermodynamics and mass/energy conservation.

I'll try to break down my question into a couple of sub-questions.

1) where does the energy, the ability of a magnet able to repel another magnet, and the motive force thus generated, even in complete vacuum and zero gravity, come from? Without violating energy conservation?

2) the reverse (attraction) of point 1. Because, when those magnets touch, they are applying their equal and opposite forces on each other, cancelling out? And yet they are still bound by the magnetic field? And require energy to separate?

I thought of these questions while trying to figure out something that has bothered me for some time. Imagine a thin metal rod, made of spring steel, with a magnet attached at the top, fixed to a sturdy base. Now have another base / rod / magnet with the opposite pole facing the first.

The two rods will be bent toward each other, overcoming the resistance of the spring steel, permanently, yet no external energy is being applied to the construction. I JUST DONT GET IT

edit on 25-3-2015 by MasterAtArms because: (no reason given)

edit on 25-3-2015 by MasterAtArms because: (no reason given)



posted on Mar, 25 2015 @ 05:32 PM
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originally posted by: MasterAtArms
I have a question I want to ask (not crackpot!) about magnets!...
2) the reverse (attraction) of point 1. Because, when those magnets touch, they are applying their equal and opposite forces on each other, cancelling out? And yet they are still bound by the magnetic field? And require energy to separate?
Let's start with that one, and see if you might already know the answer but don't realize it.

Ever do the problem in school where you have a ball at the top of a hill with no motion? All its energy is potential, and none is kinetic. Then it rolls to the bottom of the hill and all the energy is kinetic, and none is potential? Usually the teacher says assume friction is negligible, and thus energy is conserved. If you're familiar with that and it works for a gravitational field, is there any reason it can't work for a magnetic field, at least in the attraction mode?

If you see how energy is conserved in the gravity example why wouldn't it be conserved in the magnetism example?



posted on Mar, 25 2015 @ 06:02 PM
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originally posted by: Arbitrageur
a reply to: pfishy
Ok here's a question for you to see if you understand it now, and the math is pretty easy.

The energy released from the largest nuclear explosion ever was about 210 x 10^15 Joules.
Is that bigger, smaller, or about the same as energy equivalent of 50kg of mass?

For E=mc² to come out in Joules you have to use mass in kg, use the speed of light in meters per second (3x10^8 m/s is close enough).
Ok, unless I am calculating this incorrectly, it is phenomenally less.
The mass equivalent to the release would be 7×10^-48 kg.
Of course, I had to break a long-standing treaty with Mathematics and press the on button on my TI-30XIIS. And there's always the very strong possibility that I calculated it completely incorrectly.
EDIT: Yeah, looking back at the thread it would seem that answer cannot be right. I'll figure out where I miffed it and get back to.you.
edit on 25-3-2015 by pfishy because: The Electric Universe made me do it.



posted on Mar, 25 2015 @ 06:05 PM
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a reply to: Arbitrageur

I'm feeling pretty stupid today, and my school physics was a long time ago. I don't mind being spoon fed but I'm either totally not thinking about this right or just down right forgotten everything



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