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Paradoxes and Dilemmas
21 years 10 months ago #4017
by Samizdat
Replied by Samizdat on topic Reply from Frederick Wilson
Must we start the symphony of the new year on a hysterical note?
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- Larry Burford
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21 years 10 months ago #4679
by Larry Burford
Replied by Larry Burford on topic Reply from Larry Burford
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>If c^2 is not a velocity then what is it?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
It (c^2) is a speed squared. (Speed rather than velocity, because no directional information is included. Velocity is speed in a specific direction. If you are walking towards the North Star, your speed is "about 5 km/hr" but your velocity is "about 5 km/hr at a compass heading of about zero degrees". If you are walking <b>away from</b> the North Star your speed is the same but your velocity is not.)
c is a speed
c^2 is a speed squared
speed has (AND, by definition, MUST HAVE!!!!!) the units distance/time.
speed squared has (AND, by definition, MUST HAVE!!!!!) the units distance^2/time^2.
(BTW, a number without units attached to it has a special name. We refer to it as a "number". But a lot of us get sloppy with correct usage from time to time. And when we do the rest of us pile on and give us a hard time. Just like we're doing with you, now. ITS NOT PERSONAL. [Are you listening, makis?] )
The symbol "c" is, by convention in discussions of Physics, reserved as the shorthand notation for the speed of light. It is not the numerical part by itself, nor is it the units part by itself. It must be both parts together.
And the tricky part is you have to use a different numerical part for each different units part, depending the system of units in which you are working. A screw-up like this is what happened to one of the Martian space craft a few years ago. Some navigational calculations were done using the right numbers but the wrong units. Or maybe it was the other way around. Any way, the numbers didn't go with the units and we lost a space ship.
The above does not mean that the symbol "c" can't be used for something else, even in a Physics discussion. But if you do, and IF you want others to understand what you are saying, then you have to explain in detail what you mean by <b>your use</b> of the symbol "c".
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>How can it not be a velocity if you are multiplying one velocity by another velocity? Specifically, what does "MULTIPLIED BY THE SQUARE OF THE VELOCITY" mean?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
Multiplying one length by another length does not produce a length
Multiplying one time by another time does not produce a time
and
Multiplying one speed by another speed does not produce a speed
(some other stuff) "MULTIPLIED BY THE SQUARE OF THE VELOCITY" means:
step 1) start with a speed (number AND units)
step 2) square that speed (square the number AND square the units)
step 3) use the result of step 2 for multiplication by (some other stuff)
I'm not sure how basic I need to get to make this clear to you. If I've gone too far, forgive me. If it is still not clear, let me know and I'll see if I can come at it from another direction.
Regards,
LB
It (c^2) is a speed squared. (Speed rather than velocity, because no directional information is included. Velocity is speed in a specific direction. If you are walking towards the North Star, your speed is "about 5 km/hr" but your velocity is "about 5 km/hr at a compass heading of about zero degrees". If you are walking <b>away from</b> the North Star your speed is the same but your velocity is not.)
c is a speed
c^2 is a speed squared
speed has (AND, by definition, MUST HAVE!!!!!) the units distance/time.
speed squared has (AND, by definition, MUST HAVE!!!!!) the units distance^2/time^2.
(BTW, a number without units attached to it has a special name. We refer to it as a "number". But a lot of us get sloppy with correct usage from time to time. And when we do the rest of us pile on and give us a hard time. Just like we're doing with you, now. ITS NOT PERSONAL. [Are you listening, makis?] )
The symbol "c" is, by convention in discussions of Physics, reserved as the shorthand notation for the speed of light. It is not the numerical part by itself, nor is it the units part by itself. It must be both parts together.
And the tricky part is you have to use a different numerical part for each different units part, depending the system of units in which you are working. A screw-up like this is what happened to one of the Martian space craft a few years ago. Some navigational calculations were done using the right numbers but the wrong units. Or maybe it was the other way around. Any way, the numbers didn't go with the units and we lost a space ship.
The above does not mean that the symbol "c" can't be used for something else, even in a Physics discussion. But if you do, and IF you want others to understand what you are saying, then you have to explain in detail what you mean by <b>your use</b> of the symbol "c".
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>How can it not be a velocity if you are multiplying one velocity by another velocity? Specifically, what does "MULTIPLIED BY THE SQUARE OF THE VELOCITY" mean?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
Multiplying one length by another length does not produce a length
Multiplying one time by another time does not produce a time
and
Multiplying one speed by another speed does not produce a speed
(some other stuff) "MULTIPLIED BY THE SQUARE OF THE VELOCITY" means:
step 1) start with a speed (number AND units)
step 2) square that speed (square the number AND square the units)
step 3) use the result of step 2 for multiplication by (some other stuff)
I'm not sure how basic I need to get to make this clear to you. If I've gone too far, forgive me. If it is still not clear, let me know and I'll see if I can come at it from another direction.
Regards,
LB
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- tvanflandern
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21 years 10 months ago #4680
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>[Patrick]: If c^2 is not a velocity then what is it?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
Augmenting Larry's fine answer, if v is any speed or velocity in units of meters per second (m/s), then v^2 is energy in units of joules per kilogram.
For example, if an object weighs 10 kg and travels at 5 m/s before it suddenly hits a brick wall, it will release 250 joules of energy (10 x 5^2 = 250). If we doubled the speed, the energy released would go up by a factor of four because kinetic energy is proportional to speed squared.
What is wrong with calling v^2 a speed? Consider this: 60 meters per hour = 1 meter per minute = 1/60 meters per second are all exactly the same physical speed. It can simply be measured in whatever units we please. But when we square those speeds and ignore that the units change, we would get 3600 or 1 or 1/3600 as the square of the original speed, respectively. So which is correct? In each case, we are talking about the same speed, but its square can be larger than, the same as, or smaller than the original speed. So the square of a speed is an arbitrary number that could be anything, if we ignore that the units must be squared also.
If we retain the units with the number, then the three "speed squared" values will be: 3600 m^2/hour^2 = 1 m^2/min^2 = 1/3600 m^2/s^2, respectively, all of which convert to the same amount of energy in joules.
This subject, the meaning of physical units, is not just Physics 101 material; it is first-day Physics 101 material. Pick up any physics book at your local library and read the material on the meaning of physical units, and material on "dimensional analysis". The number and its units cannot be separated or one will arrive at meaningless and contradictory results when plugging numbers into equations.
However, for anyone to be able to learn something new, his/her mind must be open to the possibility that he/she does not already have all the answers. The opposite mindset is inimical to the learning process. Of course, any teaching is fair game to be challenged. But always ensure that you fully understand the position you are challenging and have read and listened to everything the teachers/experts have to say about the matter before issuing any challenge -- unless you enjoy being abused or made to look foolish. There are no shortcuts in science if you wish to acquire wisdom and make a difference in the world. -|Tom|-
Augmenting Larry's fine answer, if v is any speed or velocity in units of meters per second (m/s), then v^2 is energy in units of joules per kilogram.
For example, if an object weighs 10 kg and travels at 5 m/s before it suddenly hits a brick wall, it will release 250 joules of energy (10 x 5^2 = 250). If we doubled the speed, the energy released would go up by a factor of four because kinetic energy is proportional to speed squared.
What is wrong with calling v^2 a speed? Consider this: 60 meters per hour = 1 meter per minute = 1/60 meters per second are all exactly the same physical speed. It can simply be measured in whatever units we please. But when we square those speeds and ignore that the units change, we would get 3600 or 1 or 1/3600 as the square of the original speed, respectively. So which is correct? In each case, we are talking about the same speed, but its square can be larger than, the same as, or smaller than the original speed. So the square of a speed is an arbitrary number that could be anything, if we ignore that the units must be squared also.
If we retain the units with the number, then the three "speed squared" values will be: 3600 m^2/hour^2 = 1 m^2/min^2 = 1/3600 m^2/s^2, respectively, all of which convert to the same amount of energy in joules.
This subject, the meaning of physical units, is not just Physics 101 material; it is first-day Physics 101 material. Pick up any physics book at your local library and read the material on the meaning of physical units, and material on "dimensional analysis". The number and its units cannot be separated or one will arrive at meaningless and contradictory results when plugging numbers into equations.
However, for anyone to be able to learn something new, his/her mind must be open to the possibility that he/she does not already have all the answers. The opposite mindset is inimical to the learning process. Of course, any teaching is fair game to be challenged. But always ensure that you fully understand the position you are challenging and have read and listened to everything the teachers/experts have to say about the matter before issuing any challenge -- unless you enjoy being abused or made to look foolish. There are no shortcuts in science if you wish to acquire wisdom and make a difference in the world. -|Tom|-
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21 years 10 months ago #4019
by Larry Burford
Replied by Larry Burford on topic Reply from Larry Burford
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>(Patrick)
Perhaps you and the others should heed your own advice:
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote><b>(tvf)
However, for anyone to be able to learn something new, his/her mind must be open to the possibility that he/she does not already have all the answers. </b><hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
AhHa! You know, if you look hard enough you can usually find something to agree with. I'm refering to BOTH statements.
Regards,
LB
<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
Perhaps you and the others should heed your own advice:
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote><b>(tvf)
However, for anyone to be able to learn something new, his/her mind must be open to the possibility that he/she does not already have all the answers. </b><hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
AhHa! You know, if you look hard enough you can usually find something to agree with. I'm refering to BOTH statements.
Regards,
LB
<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
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21 years 10 months ago #4418
by jacques
Replied by jacques on topic Reply from
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>"E is equal to m c-squared, in which energy is put equal to mass, multiplied by the square of the velocity of light" <hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
square of the velocity of light = (velocity of light ) * (velocity of light )= (3 * 10^9 m/s) * (3 * 10^9 m/s)= 9 * 10^18 m^2/s^2= c^2
That what Einstein told and the unit are m^2/s^2 (not a speed)
If you consider c as a vector c^2 is a scalar.
What is the problem?
square of the velocity of light = (velocity of light ) * (velocity of light )= (3 * 10^9 m/s) * (3 * 10^9 m/s)= 9 * 10^18 m^2/s^2= c^2
That what Einstein told and the unit are m^2/s^2 (not a speed)
If you consider c as a vector c^2 is a scalar.
What is the problem?
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21 years 10 months ago #4682
by Larry Burford
Replied by Larry Burford on topic Reply from Larry Burford
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>[Patrick]
If the concept that, c^2 cannot be velocity or speed, and that the disproof is so elementary, then why did Mr. Einstein use it in E=mc^2 and clearly and plainly state:
[AE]
"E is equal to m c-squared, in which energy is put equal to mass, multiplied by the square of the velocity of light"
[Patrick]
If you can then please explain how Einstein intended its use then. I'm sorry, but nobody is answering the question here. I don't think Mr. Einstein was wrong. The real question here is, is c^2 the speed or velocity of gravity?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
It pleases me greatly to be able to tell you that Mr Einstein wasn't wrong.
It pains me *deeply* to have to tell you that you don't understand what he was saying. This quote from Dr. Einstein says nothing about velocity. It says plenty, however, about velocity squared. And guess what? There really is a difference between velocity and velocity squared.
"Nobody is answering the question..."? Several of us have tried. And we will probably continue to try, at least a few more times. A mind really is a terrible thing to waste.
It's possible, of course, that you actually understand these things and are just playing some kind of game with us. I don't pretend to understand such behaviour, but it's the way some people are. I don't think this sort of thing is against board policy, so if it gives you pleasure, by all means have at it.
Just realize that most of what you have said conveys little or no meaning to us. To us it seems wrong. Actually, wrong may be the wrong word here. Your ideas seem not-useful, whether or not they are right. And unless you come up with explanations that we can understand for your ideas, a way <b>for us</b> to see them as useful, we will eventually drift away.
You aren't wasting my time because I'm getting some good practice thinking about the fundamentals of Physics, and good practice trying to commumicate. It's a good idea to reexamine the basics from time to time. And I've also spent some time thinking about alternatives to those basics, as you suggested. Nothing you've said so far has tempted me to think inside your box instead of mine, however.
Do you have anything else?
Regards,
LB
If the concept that, c^2 cannot be velocity or speed, and that the disproof is so elementary, then why did Mr. Einstein use it in E=mc^2 and clearly and plainly state:
[AE]
"E is equal to m c-squared, in which energy is put equal to mass, multiplied by the square of the velocity of light"
[Patrick]
If you can then please explain how Einstein intended its use then. I'm sorry, but nobody is answering the question here. I don't think Mr. Einstein was wrong. The real question here is, is c^2 the speed or velocity of gravity?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
It pleases me greatly to be able to tell you that Mr Einstein wasn't wrong.
It pains me *deeply* to have to tell you that you don't understand what he was saying. This quote from Dr. Einstein says nothing about velocity. It says plenty, however, about velocity squared. And guess what? There really is a difference between velocity and velocity squared.
"Nobody is answering the question..."? Several of us have tried. And we will probably continue to try, at least a few more times. A mind really is a terrible thing to waste.
It's possible, of course, that you actually understand these things and are just playing some kind of game with us. I don't pretend to understand such behaviour, but it's the way some people are. I don't think this sort of thing is against board policy, so if it gives you pleasure, by all means have at it.
Just realize that most of what you have said conveys little or no meaning to us. To us it seems wrong. Actually, wrong may be the wrong word here. Your ideas seem not-useful, whether or not they are right. And unless you come up with explanations that we can understand for your ideas, a way <b>for us</b> to see them as useful, we will eventually drift away.
You aren't wasting my time because I'm getting some good practice thinking about the fundamentals of Physics, and good practice trying to commumicate. It's a good idea to reexamine the basics from time to time. And I've also spent some time thinking about alternatives to those basics, as you suggested. Nothing you've said so far has tempted me to think inside your box instead of mine, however.
Do you have anything else?
Regards,
LB
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