It's going for $1,000,000,000 per kilo

[FlFishOn]

That's right, $1 billion per kilo, and most everyone in the modern world buys it every day. Guess away, answer later.

[Nottom]

Electricity?

[ScottHoward]

[ QUOTE ]
Electricity?

[/ QUOTE ]
this is a good answer

[FlFishOn]

Yup. The mass equivalent of $1 billion dollars of electrical energy is ~ 1 kilo.

[hmkpoker]

*whew*

I thought the price of coke had REALLY gone up for a second.

[Nielsio]

[ QUOTE ]
*whew*

I thought the price of coke had REALLY gone up for a second.

[/ QUOTE ]

..and the usage

[r3vbr]

lot's of things cost way more than that per kilo
think nano tech
:O

[StepBangin]

Im ready for an answer

[Silent A]

[ QUOTE ]
lot's of things cost way more than that per kilo
think nano tech
:O

[/ QUOTE ]

Who buys this every day?

[FlFishOn]

No doubt. Most of the trans-uranic elements are more expensive to synthesize. I'd imagine LSD is close.

[aeest400]

LSD is not close to close.

[aeest400]

In fact, nothing with protons is likely close.

[Silent A]

[ QUOTE ]
No doubt. Most of the trans-uranic elements are more expensive to synthesize. I'd imagine LSD is close.

[/ QUOTE ]

Standard dose of LSD is roughly 100 micrograms (googled it, not from experience).

$1B/kg = $1M/g = $1/microg

This works out to about $100/dose

While I doubt LSD costs this much, I assume it's at least $5, which works out to $50M/kg.

Of course, that's retail in an industry with massive markup. If LSD were leagal it would be much, much cheaper.

[FlFishOn]

[ QUOTE ]
LSD is not close to close.

[/ QUOTE ]

Doseage is 1/10000 of a gram at $10 a hit gets me to about $.1 billion/KG. Check my math. One order of magnitude is close in my world.

[Rhett]

I just did the calculations, and $1 Billion worth of electricity at 120 volts has a mass of about 1685 Kg, so nice try, but even electrons are too heavy.

It matters what voltage you buy the power at, because you pay for the energy in watt-hours, even though, physically, you are getting couloumbs of electrons.

I assumed 1 Kw-h to be $0.10, which may be on the cheap side for consumers, but multiply that by 10, and you still have 169Kg of electrons, not 1 Kg for a billion dollars.

Some other usefull conversion factors, if anyone cares to check my work:

1000 w-h = 0.10 $
120v*x A-h = w-h
1 A = 1 couloumb / sec
1 hour = 3600 sec

8.3 A-h = 0.10 $

298,800 coulombs / $

electron mass = 9.1 x 10-31 kg
1 coulomb 6.2E+18


coulombs / dollar 298800
1 coulomb 6.2E+18
electrons / dollar 1.85256E+24
electrons / billion dollars 1.85256E+33
mass / billion dollars 1685.8296 kg


For those still wondering, I estimate you're paying about $590,000 per kilo.

[FlFishOn]

You failed to use E = M C^2. I'm talking about the mass equivalent of the energy.

[Rhett]

If you're talking about that, you're wrong, and clueless.

[ QUOTE ]
You failed to use E = M C^2. I'm talking about the mass equivalent of the energy.

[/ QUOTE ]

[aeest400]

I guess I underestimated the number of electrons we get for the buck. As far as LSD, I'd be shocked if a can of cola (354g) full of LSD were worth $33 million dollars. Maybe it usually comes in a 10000x aqueous solution and the price you are quoting is based on some pure crystalline form.

Regardless, if you are caught with it the federal sentencing guidelines tie the sentence to the mass of the substance it's in [which, incidentally, the Supreme Court recently ruled had to be found by a jury, not a judge]. So keep your LSD away from the orange juice and sugar cubes or you'll be looking at 20 years in the federal pen. (My original response was based on this fact, which the LA Times ran an article on a few years ago, as well as the fact that protons weigh a hell of a lot more than electrons).

[FlFishOn]

[ QUOTE ]
If you're talking about that, you're wrong, and clueless.

[ QUOTE ]
You failed to use E = M C^2. I'm talking about the mass equivalent of the energy.

[/ QUOTE ]

[/ QUOTE ]

I got it from a theoretical physicist. Argue with him.

[Rhett]

I have degrees in physics and nuclear engineering. You're the one spreading misinformation, and arguing with my facts.

[ QUOTE ]
[ QUOTE ]
If you're talking about that, you're wrong, and clueless.

[ QUOTE ]
You failed to use E = M C^2. I'm talking about the mass equivalent of the energy.

[/ QUOTE ]

[/ QUOTE ]

I got it from a theoretical physicist. Argue with him.

[/ QUOTE ]

[FlFishOn]

I just did the math myself. At 10 cent / KWH, 1 kilo mass turns into $2.5 B energy, totally converted. It's simple equations. I'm sure you'll figure it out.

[Rhett]

Wow, you did the math yourself! Too bad that only proves it's true in ... nowhere. Being a gambling site, I'm willing to bet you're wrong. I do make mistakes sometimes with calculating things, and I posted my reasoning here in case someone wants to try to correct my errors.


1000 A-V-hr = 0.1 dollars
@
120 volts
That's:
83.33333333 a-hour / dollar
1.86E+24 electrons per dollar
1.69446E-06 kg /dollar
1694.46 kg / billion dollars

[CallMeIshmael]

I know little about physics, but I believe this is how he got it:

1 kw-H = 3,600,000 J

3,600,000J/(c^2) = 4*10-11kg

$0.1/(4*10-11kg) = $2.5 bill / kg


Im not sure if any of these steps are not allowed, since this is a field I will admit I dont know a lot about. Thats just how I think he got it.

[Rhett]

Ok, well that reasoning is fantasy. You are assuming that when you buy electricity, something takes exactly the amount of mass and converts all the mass directly to the energy you need. At which point, this pure energy of course has no mass, so the original question doesn't make any sense either.

[FlFishOn]

The issue is not weighing electrons. Get it? Obviously not.

[FlFishOn]

Just so. Sub in 4 cents/KWH and it works out.

[FlFishOn]

You seem to be the only one that doesn't get it.

[Rhett]

What do you think electricity is made out of?

[ QUOTE ]
The issue is not weighing electrons. Get it? Obviously not.

[/ QUOTE ]

[FlFishOn]

"What do you think electricity is made out of?"

In this context, missing mass.

[OrigamiSensei]

It makes far more sense to look at it the way Rhett does the calculation. I can't vouch for all the numbers themselves (haven't looked at this stuff in years) but from a common sense view the methodology sure makes much more sense than E equals mc squared. Once a kilogram of electrons has crossed some arbitrary border between your house and the electric company much did you pay for that electricity? Simple and sensible.

[CallMeIshmael]

[ QUOTE ]
"What do you think electricity is made out of?"

In this context, missing mass.

[/ QUOTE ]


Again, Im balls at physics, but I thought was wrong.


I thought electricty was simply the movement of electrons (current) through resistance, which caused work. I didnt think this involved the loss of mass.

[Silent A]

Sigh. When you pay for electricity, you don't buy whole electrons. Every single electron that enters your house leaves it (from a nonquantum physics perspective - you could also say that for every electron that enters another leaves). The difference is that the electrons that leave your house have slightly less mass than when they entered it. It's this highly valuable loss of mass that you're actually consuming (thus buying).

FlFishOn's right.

[Silent A]

[ QUOTE ]
I thought electricty was simply the movement of electrons (current) through resistance, which caused work. I didnt think this involved the loss of mass.

[/ QUOTE ]

When an electron passes through a resistor it loses a little bit of mass. The resistor converts this mass energy into heat plus whatever else it was designed to do. If it's a recharging battery, then the battery actually gains a tiny bit of mass equal to the extra potential energy of the battery plus the slightly higher temperature of the battery.

[Silent A]

[ QUOTE ]
It makes far more sense to look at it the way Rhett does the calculation. I can't vouch for all the numbers themselves (haven't looked at this stuff in years) but from a common sense view the methodology sure makes much more sense than E equals mc squared. Once a kilogram of electrons has crossed some arbitrary border between your house and the electric company much did you pay for that electricity? Simple and sensible.

[/ QUOTE ]

The electrons also leave your house and return to the power company. Every single last one of them.

[Rhett]

Hahahahahahahahahahahhaha

No.

[ QUOTE ]

When an electron passes through a resistor it loses a little bit of mass.

[/ QUOTE ]

[Borodog]

Hahahahahahahahahahahhaha

Yes. It doesn't, of course, change the rest mass, but that isn't what we're interested in, is it?

You don't buy the electrons. Because the current on the line is AC, the electrons themselves simply oscillate back and forth more or less in one spot on the line. You are paying for the energy that you take out of these electrons, and the only meaningful way to measure the mass of that used energy is in the way the OP suggests.

[ QUOTE ]
Hahahahahahahahahahahhaha

No.

[ QUOTE ]

When an electron passes through a resistor it loses a little bit of mass.

[/ QUOTE ]

[/ QUOTE ]

[Rhett]

Wow. Not even close.


[ QUOTE ]
Hahahahahahahahahahahhaha

Yes. It doesn't, of course, change the rest mass, but that isn't what we're interested in, is it?

You don't buy the electrons. Because the current on the line is AC, the electrons themselves simply oscillate back and forth more or less in one spot on the line. You are paying for the energy that you take out of these electrons, and the only meaningful way to measure the mass of that used energy is in the way the OP suggests.

[ QUOTE ]
Hahahahahahahahahahahhaha

No.

[ QUOTE ]

When an electron passes through a resistor it loses a little bit of mass.

[/ QUOTE ]

[/ QUOTE ]

[/ QUOTE ]

[Silent A]

[ QUOTE ]
Hahahahahahahahahahahhaha

No.

[ QUOTE ]

When an electron passes through a resistor it loses a little bit of mass.

[/ QUOTE ]

[/ QUOTE ]

All right then, what does happen to an individual electron when it passes through a resistor? Feel free to go into as much detail as you like.

My understanding (in more detail - assuming DC just for simplicity) :

Assume you have a classic constant voltage source (V) providing current to a resistor (R). Assume the wire has zero resistence. The current (I) in the wire is V/R. If v = 1 Volt and R = 1 Ohm, then 1 Coulomb will pass through the resistor every second. That's 6.25e18 electrons every second. The power lost in the resistor is V^2/R = 1 Watt.

Now, electrons are not consumed in such a circuit. Electrons are not absorbed by the resistor and new ones added back by the voltage source. If they were, then the current "upstream" of the resistor whould be larger than the current downstream. But this doesn't happen. Instead, as the electrons pass through the resistor, each one drops 1 Volt of electric potential and in the process they pass from a higher to a lower energy state (they lose 1.6e-19 J each which is added back at the voltage source).

Where does this energy come from? Basic relativity says that when you add energy to a system you increase its mass by E/C^2. A rolling bowling ball has slightly more mass than when it's stationary and I have slightly more mass when I get off an elevator at the top floor then when I stepped on at the ground floor. The same is true for all matter/systems. Each individual electron loses a little mass as it passes through the resistor and gains it back as it passes through the voltage source.

I realize that I've simplified the internal workings of the voltage source and resistor somewhat but this doesn't change the fact that individual electrons are not consumed. At every point electrons in = electrons out.

[Rhett]

The energy comes from the plant that sold it to you. They might get it from kinetic energy in the wind, chemical energy in coal, or a bunch of other ways. The power company uses their energy to spin a turbine and push electrons, that in turn, push other electrons on the power grid.

You guys are thinking to hard. It is really dumb to bring relativity into this because when someone buys electricity, it is impossible for them to get that energy by pure destruction of an equivilant amount of mass. For one thing, electrical energy has low entropy so it can be used for usable work for things like turning motors. Matter anihilation would create explosive, high entropy energy, and some process must be done to create usable work.

Sure, you can buy energy from nuclear sources. Fission power plants, or solar energy that originated from fusion in the sun are some examples. But fission plants use a fuel mixture where only 3% is fissile material. And even when that U235 splits apart into many isotopes, only a tiny fraction of the original fissile mass is converted to energy. Of couse that energy goes through a thermal cycle, capturing only 30% for usable work. But all of this is moot anyway because only 20% of the nuclear plant's cost to sell you energy were fuel related to begin with. Most of the costs were related to building the plant.

Why isn't the answer to this question gasoline? If I buy a billion dollars worth of gasoline, I expect to be able to use some amount of horsepower-hours energy from it. If I converted that energy directly from matter, it would be nearly a kilogram. The reason the answer isn't gasoline is because a billion dollars worth of gasoline weighs more than 1 kg also, a billion dollars worth of electrons delivered to me by the electric company weighs 1695 kg.

Sure, many people give those electrons back, but I don't have to. It is perfectly fine for me to store them all in a capacitor array if I want. Regardless, when I buy electricity, I am recieving electrons.

If you really want your electrons to be that expensive, just plug in a 100KV transformer.



[ QUOTE ]


Where does this energy come from? Basic relativity says that when you add energy to a system you increase its mass by E/C^2. A rolling bowling ball has slightly more mass than when it's stationary and I have slightly more mass when I get off an elevator at the top floor then when I stepped on at the ground floor. The same is true for all matter/systems. Each individual electron loses a little mass as it passes through the resistor and gains it back as it passes through the voltage source.

I realize that I've simplified the internal workings of the voltage source and resistor somewhat but this doesn't change the fact that individual electrons are not consumed. At every point electrons in = electrons out.

[/ QUOTE ]

[Silent A]

[ QUOTE ]
The energy comes from the plant that sold it to you. They might get it from kinetic energy in the wind, chemical energy in coal, or a bunch of other ways. The power company uses their energy to spin a turbine and push electrons, that in turn, push other electrons on the power grid.

[/ QUOTE ]

No one is denying this.

[ QUOTE ]
You guys are thinking to hard. It is really dumb to bring relativity into this because when someone buys electricity, it is impossible for them to get that energy by pure destruction of an equivilant amount of mass.

[/ QUOTE ]

Relativity is always involved. Matter annialation isn't the only way to get energy from mass. You are missing one of the basic truths of relativity. When a turbine moves it gains mass equal to what a classical physicist would call its kintic energy divided by the speed of light squared. This is what happens according to relativity. The change in mass is obviously small (too small to measure conventionally) but it's real, assuming relativity is real.

[ QUOTE ]
Why isn't the answer to this question gasoline?

[/ QUOTE ]

The octane you burn has slightly more mass than the water vapour and CO2 your car produces. This difference represents the lower energy of C-O and H-O bonds relative to C-C and C-H bonds. It's an extremely small change in mass, but it's non-zero and it's where the energy comes from to power your car.

The difference is that for electricity the particles involved actually return to the plant where they're given more mass (kicked back up to a higher energy state) and returned to you and me. I admit this is neither intuitive nor in line with classical physics, but it still what actually happens.

[ QUOTE ]
Sure, many people give those electrons back, but I don't have to. It is perfectly fine for me to store them all in a capacitor array if I want. Regardless, when I buy electricity, I am recieving electrons.

If you really want your electrons to be that expensive, just plug in a 100KV transformer.

[/ QUOTE ]

Correct me if I'm wrong but I don't think that's quite what happens in a capacitor. When a capcitor charges up, electrons get built up on one side and are removed from the other. Electrons still return to the plant (just not the same ones). The potential energy added to the capacitor results in it gaining mass. Basically, the capacitor gains X high mass electrons and loses X low mass electrons.

[Rhett]

You are grossly missinformed. Seriously, did you guys all just pull the same screwed up ideas about physics out of your ass, or was there some big retard physics convention?

You have a completely screwed up notion about how relativity works. E=mc^2 does not mean a particle's mass increases by it's kinetic energy divided by c^2.

Yes, electrons moving relative to your reference frame have an increased mass by a factor of 1/(1-(v^2/c^2))^(1/2) . So what?

Kinetic energy from steam pushes a turbine, which pushes on electrons. Your resistor at home slows down electrons and increases the kinetic energy of air particles by heating them up.

[ QUOTE ]

Relativity is always involved. Matter annialation isn't the only way to get energy from mass. You are missing one of the basic truths of relativity. When a turbine moves it gains mass equal to what a classical physicist would call its kintic energy divided by the speed of light squared. This is what happens according to relativity. The change in mass is obviously small (too small to measure conventionally) but it's real, assuming relativity is real.

[/ QUOTE ]

[FlFishOn]

"You guys are thinking to hard. It is really dumb to bring relativity into this because when someone buys electricity, it is impossible for them to get that energy by pure destruction of an equivilant amount of mass. "

I know you're not thinking too hard.

There is no other source of energy save mass conversion.

[FlFishOn]

"FlFishOn's right."

Wow. That only get into print once every geologic age. I'll bask in it for a moment.

[mindless]

[ QUOTE ]
"You guys are thinking to hard. It is really dumb to bring relativity into this because when someone buys electricity, it is impossible for them to get that energy by pure destruction of an equivilant amount of mass. "

I know you're not thinking too hard.

There is no other source of energy save mass conversion.

[/ QUOTE ]

If you want, we can get into discussions about the total amount of energy in the universe, but that's not applicable to this discussion. Rhett is completely correct...you're not using the "Power" company to sell you a source of energy, rather they're simply transfering energy to you. The amount of energy is conserved, if we make hte system big enough, but the type of energy is transferred as it proceeds from its source as chemical potential or whatever into your house, where most of it eventually gets disippated as heat into the atmosphere.

Introducing relativity into the equation is also not useful. While it's true that there's something called rest mass, and mass can change as a function of the object's velocity, you're obfuscating matters.

Also, being a nit as a physics major, I also want to point out that we're neglecting to calculate the efficiency of the system here, and thus should up our accounts by a factor of at least 10 here, depending on the system used.

[Silent A]

[ QUOTE ]
Introducing relativity into the equation is also not useful. While it's true that there's something called rest mass, and mass can change as a function of the object's velocity, you're obfuscating matters.

[/ QUOTE ]

Rhett says an electron slows down as it passes through a resistor passing on its kinetic energy to my appliance. Are you arguing that the mass of a fast moving electron is not greater than the mass of a slow moving electron?

The fact is that if you did an extremely careful mass budget on the electric cuircuits entering and exiting my house you would find that more mass was entering than leaving. It might not be the most usefull way to look at what's happening, but it IS happening.

[ QUOTE ]
Also, being a nit as a physics major, I also want to point out that we're neglecting to calculate the efficiency of the system here, and thus should up our accounts by a factor of at least 10 here, depending on the system used.

[/ QUOTE ]

Efficientcy doesn't change the fundamental nature of what happens when an electron passes though a resistor. It only effects how much of the energy I pay for goes where I would like it to.

[Silent A]

[ QUOTE ]
You are grossly missinformed. Seriously, did you guys all just pull the same screwed up ideas about physics out of your ass, or was there some big retard physics convention?

You have a completely screwed up notion about how relativity works. E=mc^2 does not mean a particle's mass increases by it's kinetic energy divided by c^2.

Yes, electrons moving relative to your reference frame have an increased mass by a factor of 1/(1-(v^2/c^2))^(1/2) . So what?

[/ QUOTE ]

I'm approaching the limit of my knowledge of relativity here. The issue isn't relative velocity of inertial frames, it's the acceleration of the electron at the plant and its decelleration in my toaster (i.e. we're talking General not Special relativity).

Oh, and I don't think electrons slow down as they pass through a resistor. If they did the current would have to decline. So electrons don't accelerate at the plant but rather gain electric potential energy. Either way though, the energy they gain increases their mass.

[FlFishOn]

"Also, being a nit..."

A single point of agreement. A start.

[Phil153]

How can someone have a physics degree and be so [censored] stupid?

Mass and energy are interchangeable. The fact that you don't know that is astounding. For example, when a chemical reaction occurs that releases heat or light, the mass of the final substance is less than the mass of the ingredients, by an amount m = E/c^2, where E is the energy released in the reaction. It has nothing to do with kinetic energy or relativistic speeds.

FlF, great post.

[mindless]

[ QUOTE ]
How can someone has a physics degree and be so [censored] stupid?

Mass and energy are interchangeable. The fact that you don't know that is astounding. For example, when a chemical reaction occurs that releases heat or light, the mass of the final substance is less than the mass of the ingredients, by an amount m = E/c^2, where E is the energy released in the reaction. It has nothing to do with kinetic energy or relativistic speeds.

FlF, great post.

[/ QUOTE ]

A little knowledge is a dangerous thing. Yes, mass and energy can be related, but they're only interchangeable if you're particularly careful, and it's rather confusing to say that. Do you play the 20/40 game on party, or do you play the 5Big Mac/10 Big Mac game? It's not simply unit conversion.

[FlFishOn]

"How can someone have a physics degree and be so [censored] stupid?"

This is a better puzzle than my OP.

[Phil153]

[ QUOTE ]
they're simply transfering energy to you

[/ QUOTE ]
EXACTLY. And this is what you are paying for, at the rate of $1,000,000,000 per kilo's worth of energy they transfer to you.

[ QUOTE ]
A little knowledge is a dangerous thing.

[/ QUOTE ]
So are unfounded assumptions.

[GMontag]

[ QUOTE ]
[ QUOTE ]
Introducing relativity into the equation is also not useful. While it's true that there's something called rest mass, and mass can change as a function of the object's velocity, you're obfuscating matters.

[/ QUOTE ]

Rhett says an electron slows down as it passes through a resistor passing on its kinetic energy to my appliance. Are you arguing that the mass of a fast moving electron is not greater than the mass of a slow moving electron?

[/ QUOTE ]

Yes. That's exactly it. The mass of the electron does not change. The whole "energy adds mass" thing is a popular misconception. E=mc^2 is only valid for particles with 0 kinetic energy. The total energy formula is E^2 = p^2c^2 + m^2c^4. When you add kinetic energy to a particle, it is the p^2c^2 term that increases, the m^2c^4 term stays constant.

And Borodog, don't bring relativistic mass into it. Relativistic mass is a supremely silly and fundamentally flawed concept that isn't used in physics at all except in bad Intro to Relativity texts.

[ QUOTE ]
The fact is that if you did an extremely careful mass budget on the electric cuircuits entering and exiting my house you would find that more mass was entering than leaving. It might not be the most usefull way to look at what's happening, but it IS happening.

[/ QUOTE ]

No you wouldn't. You'd find that more energy is entering than leaving, not mass.

[ QUOTE ]
I'm approaching the limit of my knowledge of relativity here. The issue isn't relative velocity of inertial frames, it's the acceleration of the electron at the plant and its decelleration in my toaster (i.e. we're talking General not Special relativity).

[/ QUOTE ]

Special Relativity can deal with acceleration and deceleration just fine. Its only gravity that makes GR necessary.

[ QUOTE ]
Oh, and I don't think electrons slow down as they pass through a resistor. If they did the current would have to decline. So electrons don't accelerate at the plant but rather gain electric potential energy. Either way though, the energy they gain increases their mass.

[/ QUOTE ]

Electrons do slow down through a resistor. Current is a measure of movement of electric charge, not electrons themselves. Current depends on the drift velocity, not the actual velocity of the individual electrons. In a resistor, the electrons on average lose kinetic energy (and therefore velocity), but they still keep the same drift velocity.

[Borodog]

[ QUOTE ]
Yes. That's exactly it. The mass of the electron does not change. The whole "energy adds mass" thing is a popular misconception. E=mc^2 is only valid for particles with 0 kinetic energy. The total energy formula is E^2 = p^2c^2 + m^2c^4. When you add kinetic energy to a particle, it is the p^2c^2 term that increases, the m^2c^4 term stays constant.

[/ QUOTE ]

You are claiming that one mathematical description for reality is "correct" and the other is "incorrect", when they produce identical results. The "m" in the equation you quote is specifically m_0, the rest mass. While the m in E=mc^2 must be the relativistic mass for it to always be correct. I.e. m^2c^4 = p^2c^2 + m_0^2c^4.

[ QUOTE ]
And Borodog, don't bring relativistic mass into it. Relativistic mass is a supremely silly and fundamentally flawed concept that isn't used in physics at all except in bad Intro to Relativity texts.

[/ QUOTE ]

It is a mathematical description of reality that works. The fact that you don't like it doesn't make it incorrect. The only way that E=mc^2 is always correct is if m is the relativistic mass.

Hey, but what do I know? I only have a Ph.D. in astrophysics. /images/graemlins/tongue.gif

[Silent A]

[ QUOTE ]
Electrons do slow down through a resistor. Current is a measure of movement of electric charge, not electrons themselves. Current depends on the drift velocity, not the actual velocity of the individual electrons. In a resistor, the electrons on average lose kinetic energy (and therefore velocity), but they still keep the same drift velocity.

[/ QUOTE ]

Yeah, I understand these differences. I admit I screwed this up. I just had difficulty imaging how they could maintain their drift velocity when they slow down in what could well be a constant cross-sectional area. I was carrying the water flow analog too far in my head. Water's incompressible, electons obviously aren't.

I'm still sticking to my "mass decreases" argument though. Yeah Borogod! oops dog. <--- actual slip - decided to keep it.

[renodoc]

Lucentis, a new macular degeneration drug, will likely be sold for $2000 per 1 mg. That gets to $2B per kilo if my math isn't off too much....

[srjunkacct]

The electric company doesn't sell you electrons. It sells you energy. You provide the electrons.

[Sack Lodge]

[ QUOTE ]
The electric company doesn't sell you electrons. It sells you energy. You provide the electrons.

[/ QUOTE ]

What if I want to buy some electrons? Where can I get them?

Please help me, I need an electron fix.

[diebitter]

[ QUOTE ]
[ QUOTE ]
The electric company doesn't sell you electrons. It sells you energy. You provide the electrons.

[/ QUOTE ]

What if I want to buy some electrons? Where can I get them?

Please help me, I need an electron fix.

[/ QUOTE ]

Are you positive?

[GMontag]

[ QUOTE ]
[ QUOTE ]
Yes. That's exactly it. The mass of the electron does not change. The whole "energy adds mass" thing is a popular misconception. E=mc^2 is only valid for particles with 0 kinetic energy. The total energy formula is E^2 = p^2c^2 + m^2c^4. When you add kinetic energy to a particle, it is the p^2c^2 term that increases, the m^2c^4 term stays constant.

[/ QUOTE ]

You are claiming that one mathematical description for reality is "correct" and the other is "incorrect", when they produce identical results. The "m" in the equation you quote is specifically m_0, the rest mass. While the m in E=mc^2 must be the relativistic mass for it to always be correct. I.e. m^2c^4 = p^2c^2 + m_0^2c^4.

[ QUOTE ]
And Borodog, don't bring relativistic mass into it. Relativistic mass is a supremely silly and fundamentally flawed concept that isn't used in physics at all except in bad Intro to Relativity texts.

[/ QUOTE ]

It is a mathematical description of reality that works. The fact that you don't like it doesn't make it incorrect. The only way that E=mc^2 is always correct is if m is the relativistic mass.

[/ QUOTE ]

Just because the math works out the same doesn't mean the underlying physical model is sensical. There is absolutely no reason to interpret γm as "relativistic mass" as if it had some kind of physical meaning, and there is a lot of reasons to not interpret it as such.

1. Relativistic mass isn't frame invariant, and a change in it doesn't necessarily imply a change in the stress-energy tensor (and therefore the gravitational field). In other words, it doesn't do what we would expect mass to do.

2. The math works out for this formula, but it doesn't for most other formulas. For example, F=ma doesn't work for relativistic mass.

3. Using E=γmc^2 implies that moving particles have no kinetic energy, as all of their total energy is contributed by energy from their mass rather than their motion.

I think it is pretty obvious that calling γm relativistic mass, just so E=mc^2 looks pretty is nothing more than a mathematical hack, and doesn't reflect the actual physical reality.

[ QUOTE ]
Hey, but what do I know? I only have a Ph.D. in astrophysics. /images/graemlins/tongue.gif

[/ QUOTE ]

Then you should know better.

EDIT: Sorry, the γ stuff is supposed to be the greek letter gamma. Is there anyway to get that to show up correctly?

[Borodog]

All mathematical descriptions of reality are "mathematical hacks" that don't affect the underlying physical reality.

Lots of quantities are not frame invariant, like length and time. That doesn't make them useless or wrong.

Your point 2 is a restatement of point 1. As for point 3, who cares? It's a different interpretation. Are you claiming that you have some magic ability to peer behind the veil of reality and tell us that 6 is correct but half a dozen is wrong?

[boo5000]

Seriously, think about what you are claiming.

Do electrons have mass? Yes.

Does "electricity" have mass? No.

I understand that mass and energy can be equated, but it is marginal to propose that we can give some mass to an energy that is not associated with a mass at the instant we want to determine it. We can calculate how much the electrons weigh that are used, and use this as what we pay for - although ina renewable sense. We can also calculate the cost of certain methods that create electrical energy - combustion, etc. But assigning a mass to the electricy itself is marginal at best. And even if you did, it would have no meaning in terms of economics.

[Borodog]

[ QUOTE ]
Seriously, think about what you are claiming.

Do electrons have mass? Yes.

Does "electricity" have mass? No.

I understand that mass and energy can be equated, but it is marginal to propose that we can give some mass to an energy that is not associated with a mass at the instant we want to determine it. We can calculate how much the electrons weigh that are used, and use this as what we pay for - although ina renewable sense. We can also calculate the cost of certain methods that create electrical energy - combustion, etc. But assigning a mass to the electricy itself is marginal at best. And even if you did, it would have no meaning in terms of economics.

[/ QUOTE ]

Sigh.

Just go back and read my initial comment again, because given that:

A) You are NOT buying electrons
B) You are buying ENERGY
C) The question was about buying something for dollars PER KILOGRAM,

The ONLY way to calculate a meaningful result is to use the mass-energy relation.

Jesus H Christ.

[boo5000]

But do we buy batterys for the amount of "electricty-grams" they provide? No. So it is illogical wo create a $ amount that is in terms of electrical mass. This amount HAS NO BEARING on economics. It can't be compared to any other cost/gram numbers. I agree with you point that a mass for a certain energy can be calculated, but consider the outcome and what it applies to. And I also agree that we dont buy electrons, that was stated as an alternative that actually relies on mass.

[Borodog]

Your problem is with the OP. Perhaps it is poorly formed. I'm satisfied with my interpretation of it. Given that I'm asked about a $/mass sale on a product that is actually energy, the mass-energy relation is the only thing that makes any sense.

[hoterdoc]

vnh

[KeysrSoze]

Thread way to nitty. We might as well be saying that we're not buying a gallon of gas, but buying the fractional amount transformed into heat radiation by reaction with oxygen, or you're not buying an apple but the tiny amount of mass lost by breaking up the glucose molecules. I'm paying a billion dollars a pound for gas and apples now!

[KaneKungFu123]

Physics is gay.

[wheeler]

This thread is amusing. But the guy with "degrees in physics and nuclear engineering" needs to start posting again. It's awesome when people who are wrong are so sure they're right that they insult and ridicule everyone who disagrees with them.