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'Edge' of the Universe
19 years 6 months ago #13302
by Jim
Replied by Jim on topic Reply from
I hate always being a killjoy but if you throw a gravatron from mass "A" to mass "B" you are doing work are you not? That work will cause mass "A" to cool and mass "B" to heat. That does not happen in the real gravity field as near as I can determine at this time. I have a related question posted with no answer so being the dummy here I don't yet know for sure.
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- Larry Burford
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19 years 6 months ago #14169
by Larry Burford
Replied by Larry Burford on topic Reply from Larry Burford
I guess there is a fairly obvious candidate for a name for this model of the atom. "Hovering Electron Model".
New ideas like this rarely survive for long once they are shopped around to other minds. It is easy for others to see what the originator can't see. I haven't provided a lot of detail, but before I spend too much more time on it I was hoping that another brain would spot something about it that is obviously not going to work in the real world.
??,
LB
New ideas like this rarely survive for long once they are shopped around to other minds. It is easy for others to see what the originator can't see. I haven't provided a lot of detail, but before I spend too much more time on it I was hoping that another brain would spot something about it that is obviously not going to work in the real world.
??,
LB
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- tvanflandern
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19 years 6 months ago #14170
by tvanflandern
Replied by tvanflandern on topic Reply from Tom Van Flandern
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Jim</i>
<br />I hate always being a killjoy but if you throw a gravatron from mass "A" to mass "B" you are doing work are you not? That work will cause mass "A" to cool and mass "B" to heat.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">In PG, there is no mass "A". Looking at mass "B", gravitons are absorbed asymmetrically because a source mass shadows some gravitons in that direction. But they are re-emitted isotropically. So the energy books are balanced, but the momentum books are not (at least, not when we confine our attention to mass "B" only). -|Tom|-
<br />I hate always being a killjoy but if you throw a gravatron from mass "A" to mass "B" you are doing work are you not? That work will cause mass "A" to cool and mass "B" to heat.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">In PG, there is no mass "A". Looking at mass "B", gravitons are absorbed asymmetrically because a source mass shadows some gravitons in that direction. But they are re-emitted isotropically. So the energy books are balanced, but the momentum books are not (at least, not when we confine our attention to mass "B" only). -|Tom|-
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19 years 6 months ago #13245
by Jim
Replied by Jim on topic Reply from
Energy and momentum seem to me like water and rain. The two are different and not really interchangable but when you get to the meat of this it is only the math that makes momentum without energy. How do you have a mass with momentum and not energy?
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19 years 6 months ago #13342
by tvanflandern
Replied by tvanflandern on topic Reply from Tom Van Flandern
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Jim</i>
<br />How do you have a mass with momentum and not energy?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">You don't. But both energy and momentum are reference-frame-dependent numbers that depend on your point of view. Take this example from Winkler's article in the 2005 March 15 issue of the <i>Meta Research Bulletin</i>:
"Remember now, that these measures are relative to the reference system! Assume we throw a rock of, say, 1 kg, by giving it an acceleration of 200 m/s/s for 0.1 second, so that it reaches a speed of 20 m/s. We have to use a force of 200 N to do this; it gives this rock an energy of 200 J. However, suppose the rock was thrown in the direction of the momentary velocity of the earth around the Sun, 30,000 m/s. Before being thrown, the rock had a kinetic energy of 30,000^2/2 J or 4.5×10^8 J. After our throw, the rock had a kinetic energy of 30,020^2/2 or 600000 J (3000 times) more than it had before! Obviously, the kinetic energy is a quantity that we must compute in the inertial system where it will be used. It is a quantity relative to this system and not an absolute that is anchored in the distant masses of the universe. Of course, the larger energy will only be released if the rock hits a body that is at rest relative to the Sun. If it hits a body at rest in our local system, only 200 joules will be released in the impact."
-|Tom|-
<br />How do you have a mass with momentum and not energy?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">You don't. But both energy and momentum are reference-frame-dependent numbers that depend on your point of view. Take this example from Winkler's article in the 2005 March 15 issue of the <i>Meta Research Bulletin</i>:
"Remember now, that these measures are relative to the reference system! Assume we throw a rock of, say, 1 kg, by giving it an acceleration of 200 m/s/s for 0.1 second, so that it reaches a speed of 20 m/s. We have to use a force of 200 N to do this; it gives this rock an energy of 200 J. However, suppose the rock was thrown in the direction of the momentary velocity of the earth around the Sun, 30,000 m/s. Before being thrown, the rock had a kinetic energy of 30,000^2/2 J or 4.5×10^8 J. After our throw, the rock had a kinetic energy of 30,020^2/2 or 600000 J (3000 times) more than it had before! Obviously, the kinetic energy is a quantity that we must compute in the inertial system where it will be used. It is a quantity relative to this system and not an absolute that is anchored in the distant masses of the universe. Of course, the larger energy will only be released if the rock hits a body that is at rest relative to the Sun. If it hits a body at rest in our local system, only 200 joules will be released in the impact."
-|Tom|-
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19 years 6 months ago #11242
by Jim
Replied by Jim on topic Reply from
The above example is just the kind of stuff I object to most of the time. This is what math can do when subtile tricks are introduced in the process. In this example the trick is using two rest frames to set the groundwork and only one rest frame is used to make the calculation. But, the same kind of trickery is more or less common practice in science.
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