Editing Talk:2007: Brookhaven RHIC
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I think the correct explanation has to do with relativistic mass. Accelerated gold ions will have an increased (relativistic) mass. Therefore, the gold coming out of the accelerator will have a higher value than the input. You just have to sell by mass. --[[Special:Contributions/172.68.78.22|172.68.78.22]] 16:14, 15 June 2018 (UTC) | I think the correct explanation has to do with relativistic mass. Accelerated gold ions will have an increased (relativistic) mass. Therefore, the gold coming out of the accelerator will have a higher value than the input. You just have to sell by mass. --[[Special:Contributions/172.68.78.22|172.68.78.22]] 16:14, 15 June 2018 (UTC) | ||
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:Yeah, we need to calculate the increase of mass at 99.99, 99,999, 99,9999 percent of the speed of light. I didn't check but I'm curious how many 9s we need to reach the mass of the Earth for a single gold nuclei. Nevertheless the speed is zero again when you collect them, or you have to move at the same speed, but than you can't measure that increase because you're in the same reference -- and back on Earth your own mass would have grown out far beyond the mass of our Milky Way. --[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 18:11, 15 June 2018 (UTC) | :Yeah, we need to calculate the increase of mass at 99.99, 99,999, 99,9999 percent of the speed of light. I didn't check but I'm curious how many 9s we need to reach the mass of the Earth for a single gold nuclei. Nevertheless the speed is zero again when you collect them, or you have to move at the same speed, but than you can't measure that increase because you're in the same reference -- and back on Earth your own mass would have grown out far beyond the mass of our Milky Way. --[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 18:11, 15 June 2018 (UTC) | ||
::Mass is an invariant quantity. What is increased is the total energy, E=sqrt(p^2+m^2). As previously mentioned, when they are delivered they would have to be measured in a comoving frame in which case no increase in "mass" would be noticed.-- [[Special:Contributions/162.158.146.70|162.158.146.70]] 20:46, 15 June 2018 (UTC) | ::Mass is an invariant quantity. What is increased is the total energy, E=sqrt(p^2+m^2). As previously mentioned, when they are delivered they would have to be measured in a comoving frame in which case no increase in "mass" would be noticed.-- [[Special:Contributions/162.158.146.70|162.158.146.70]] 20:46, 15 June 2018 (UTC) | ||
:::Sorry, you know what ''invariant'' means in physics? And you can't sum p and m (simply squared both) by using the Einstein conventions c=1, it's in fact E<sup>2</sup>=m<sup>4</sup>+p<sup>2</sup> -- I fear most people still don't understand. --[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 22:02, 15 June 2018 (UTC) | :::Sorry, you know what ''invariant'' means in physics? And you can't sum p and m (simply squared both) by using the Einstein conventions c=1, it's in fact E<sup>2</sup>=m<sup>4</sup>+p<sup>2</sup> -- I fear most people still don't understand. --[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 22:02, 15 June 2018 (UTC) | ||
::::Comes across a bit condescending considering m^2c^4=(mc^2)^2, not (mc)^4. The mass is the rest mass, which is always the same, considering it is measured at rest. It is a misnomer to call the total energy the mass. The physics you're describing is correct, I just take issue with the wording. I'm sorry if I didn't make myself clear enough. --[[Special:Contributions/162.158.146.70|162.158.146.70]] 22:29, 15 June 2018 (UTC) | ::::Comes across a bit condescending considering m^2c^4=(mc^2)^2, not (mc)^4. The mass is the rest mass, which is always the same, considering it is measured at rest. It is a misnomer to call the total energy the mass. The physics you're describing is correct, I just take issue with the wording. I'm sorry if I didn't make myself clear enough. --[[Special:Contributions/162.158.146.70|162.158.146.70]] 22:29, 15 June 2018 (UTC) | ||
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Isn't a typical cash-for-gold store where you would take your gold and walk out with cash for your gold? It sounds like Randall's proposal is gold-for-cash stores instead. [[User:Ianrbibtitlht|Ianrbibtitlht]] ([[User talk:Ianrbibtitlht|talk]]) 19:03, 15 June 2018 (UTC) | Isn't a typical cash-for-gold store where you would take your gold and walk out with cash for your gold? It sounds like Randall's proposal is gold-for-cash stores instead. [[User:Ianrbibtitlht|Ianrbibtitlht]] ([[User talk:Ianrbibtitlht|talk]]) 19:03, 15 June 2018 (UTC) | ||
: It makes sense if his proposal is to sell the gold to the stores, where he would just need the stores to agree to buy the high speed particles. I thought he was suggesting consumers would buy the excess gold particles in the stores, but that's probably not what he meant. [[User:Ianrbibtitlht|Ianrbibtitlht]] ([[User talk:Ianrbibtitlht|talk]]) 19:14, 15 June 2018 (UTC) | : It makes sense if his proposal is to sell the gold to the stores, where he would just need the stores to agree to buy the high speed particles. I thought he was suggesting consumers would buy the excess gold particles in the stores, but that's probably not what he meant. [[User:Ianrbibtitlht|Ianrbibtitlht]] ([[User talk:Ianrbibtitlht|talk]]) 19:14, 15 June 2018 (UTC) | ||
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: Actually all three locations exists in real life. From left to right (north to south): New York Gold Center, Cash for Gold, and Gold Traders Inc. --[[Special:Contributions/162.158.134.142|162.158.134.142]] 07:03, 16 June 2018 (UTC) | : Actually all three locations exists in real life. From left to right (north to south): New York Gold Center, Cash for Gold, and Gold Traders Inc. --[[Special:Contributions/162.158.134.142|162.158.134.142]] 07:03, 16 June 2018 (UTC) | ||
− | If anyone wants to calculate the revenue of the project: particles move at 0.99995c and according to one source they use | + | If anyone wants to calculate the revenue of the project: particles move at 0.99995c and according to one source they use Leeds than 1/1000000 Gramm of gold in 20years. Current gold price is about 1280€ per kg |
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