[math-fun] Another physics question
I also had a physics question recently. I was trying to explain e=mc^2 to my family, and was going to try to give an intuitive example of just how much energy there is in 300,000,000^2 kg*m^2/s^2. I realized that impulse (or momentum), measured in kg*m/s, is much easier for me to conceptualize and explain: a baseball travelling at 30mph has a certain amount of kinetic energy that anyone can experience personally just by catching it. It was much more awkward to try to explain pushing a 1kg weight exactly hard enough to accelerate it 1m/s^2, but only until it had travelled 1m. So if both units can be used to express energy, why do we speak of impulse as so different from energy, and is there an intuitive way to explain the conversion from the 30mph baseball's energy into kg*m^2/s^2 without calculus? I still remember the test question in 10th grade physical science in which I was asked to calculate the work required to carry a 10kg weight across a 10m room. I argued that, at least in the ideal sense, it required no work (if you're infinitely patient), since there was no change in gravitaional potential energy, but the teacher wouldn't budge. I suppose I've resented work ever since. Oh yeah, and maybe Mike Stay could also briefly summarize quantum gravity for us in a way that makes it clear just how that principle works. I was wondering about that, too. -J
Hi Jason, Impulse is a change in momentum, which is distinctly different from energy, although anything in practical experience (like your baseball) for which there is an impulse, there is a change in kinetic energy, too. Impulse is a vector quantity, so it has direction, whereas kinetic energy is a scalar quantity, so it has no direction. There can be an impulse with no (net) change in kinetic energy, such as the elastic collision of a ball and a wall. And kinetic energy is directly related to work through the work-energy principle, whereas impulse is not. That may be one more intuitive way to explain the baseball's kinetic energy: if the baseball were slowed to a stop by compressing a spring, the kinetic energy in the ball is exactly the amount of work done in compressing the spring. Kerry
On 7/17/06, Jason Holt <jason@lunkwill.org> wrote:
Oh yeah, and maybe Mike Stay could also briefly summarize quantum gravity for us in a way that makes it clear just how that principle works. I was wondering about that, too.
I could explain topological quantum field theories, which are a toy version; they give quantum gravity in 2D. In ~2001 the guys at Microsoft's Project Q showed that they can be efficiently simulated by the quantum circuit model, and conversely that there's a TQFT universal for quantum computation. Right now they're working on braiding anyons. A TQFT is a tensor-product-preserving functor from 2Cob to Hilb. 2Cob is the category of 1-dimensional manifolds (tensor products of circles) and 2-d cobordisms between them (caps, cups, pants, and upside-down pants). Hilb has hilbert spaces and linear transformations between them. A functor maps objects to objects and morphisms to morphisms such that composition is preserved. See http://www.math.ucr.edu/home/baez/qg-winter2001/qg15.1.html for a very gentle intro. You build the functor by triangulating the cobordism, taking the poincare dual to get a bunch of Y's or inverted Y's, and then interpreting those as multiplication or comultiplication, repsectively; the centre of this algebra is the part that's immune to changes by Pachner moves, and since Pachner moves can take you between any two triangulations, the functor is triangulation independent. See http://www.math.ucr.edu/home/baez/qg-fall2004/ for lecture notes on how that all works. -- Mike Stay metaweta@gmail.com http://math.ucr.edu/~mike
On 7/17/06, Mike Stay <mike@math.ucr.edu> wrote:
Right now they're working on braiding anyons.
By the way, the reason this has something to do with gravity is that space curls up like a cone around a point particle in 2DQG, i.e. flat except at the point. So there's a "phase difference" as you move a vector around a point mass. The wierd thing is that masses get larger and larger until they behave like negative masses and then go back to zero--they wrap around at 2pi. Anyons are 2-d quasiparticles with arbitrary phase as you move around them. In 3D and up, the only differences you can get are 1 and -1, bosons and fermions. -- Mike Stay metaweta@gmail.com http://math.ucr.edu/~mike
--- Jason Holt <jason@lunkwill.org> wrote:
I also had a physics question recently. I was trying to explain e=mc^2 to my family, and was going to try to give an intuitive example of just how much energy there is in 300,000,000^2 kg*m^2/s^2.
Here is one illustration. Ocean water contains about 3 micrograms per liter of dissolved uranium. Extract the uranium from a volume of ocean, and fission all of it (using a breeder reactor), yielding about 200 MeV per atom. Use the released energy to heat that same volume of water. Water has a specific heat of 4800 J K^(-1) kg^(-1). What is the temperature rise? Answer: about 110 deg F.
I realized that impulse (or momentum), measured in kg*m/s, is much easier for me to conceptualize and explain: a baseball travelling at 30mph has a certain amount of kinetic energy that anyone can experience personally just by catching it. It was much more awkward to try to explain pushing a 1kg weight exactly hard enough to
accelerate it 1m/s^2, but only until it had travelled 1m. So if both units can be used to express energy, why do we speak of impulse as so different from energy,
Because energy and momentum are different physical quantities.
and is there an intuitive way to explain the conversion from the 30mph baseball's energy into kg*m^2/s^2 without calculus?
Perhaps, think of measuring the energy calorimetrically, by converting the energy into heat. Let a tank of water bring the baseball to a stop, and measure the temperature rise. This doesn't explain what energy actually IS; I can't think of a simple answer to that question.
I still remember the test question in 10th grade physical science in which I was asked to calculate the work required to carry a 10kg weight across a 10m room. I argued that, at least in the ideal sense, it required no work (if you're infinitely patient), since there was no change in gravitaional
potential energy, but the teacher wouldn't budge. I suppose I've resented work ever since.
This is one more example of how incompetent teachers drive students away from the phyical sciences. Was this public school?
Oh yeah, and maybe Mike Stay could also briefly summarize quantum gravity for us in a way that makes it clear just how that principle works. I was
wondering about that, too.
Yeah Mike, do that, would you please? Gene __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com
On Mon, 17 Jul 2006, Eugene Salamin wrote:
and is there an intuitive way to explain the conversion from the 30mph baseball's energy into kg*m^2/s^2 without calculus?
Perhaps, think of measuring the energy calorimetrically, by converting the energy into heat. Let a tank of water bring the baseball to a stop, and measure the temperature rise. This doesn't explain what energy actually IS; I can't think of a simple answer to that question.
The 30mph baseball is my favorite intuitive notion of energy. Batteries are another one I like to use when explaining it; compressed springs are good, too. I've used calorimetry examples as well, although those can be difficult since specific heat isn't as intuitive for people. What I was really trying to get at is that since momentum is, at least to me, a more intuitive way of understanding energy, why do we teach energy in terms of kilograms accelerated for a distance? And secondarily, what's the formula for converting a moving baseball into an accelerating kilogram, and how can I explain it to people who don't know calculus? -J
On 7/17/06, Jason Holt <jason@lunkwill.org> wrote:
What I was really trying to get at is that since momentum is, at least to me, a more intuitive way of understanding energy, why do we teach energy in terms of kilograms accelerated for a distance? And secondarily, what's the formula for converting a moving baseball into an accelerating kilogram, and how can I explain it to people who don't know calculus?
Your accelerating kilogram happens when the pitcher throws the ball: Energy=force applied over a distance and F=ma, so E=m*a*d. So the energy in a baseball is the mass of the baseball times the acceleration (0 to 30 in .5 seconds) times the distance (about one arm's length). -- Mike Stay metaweta@gmail.com http://math.ucr.edu/~mike
I think that beginners won't see why momentum and energy are both needed and why they are conserved under different conditions. Energy is best explained without mentioning momentum. I used friction to explain energy in another post. Steve Gray ----- Original Message ----- From: "Mike Stay" <mike@math.ucr.edu> To: "math-fun" <math-fun@mailman.xmission.com> Sent: Monday, July 17, 2006 2:04 PM Subject: Re: [math-fun] Another physics question
On 7/17/06, Jason Holt <jason@lunkwill.org> wrote:
What I was really trying to get at is that since momentum is, at least to me, a more intuitive way of understanding energy, why do we teach energy in terms of kilograms accelerated for a distance? And secondarily, what's the formula for converting a moving baseball into an accelerating kilogram, and how can I explain it to people who don't know calculus?
Your accelerating kilogram happens when the pitcher throws the ball: Energy=force applied over a distance and F=ma, so E=m*a*d. So the energy in a baseball is the mass of the baseball times the acceleration (0 to 30 in .5 seconds) times the distance (about one arm's length). -- Mike Stay metaweta@gmail.com http://math.ucr.edu/~mike
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Try this. Energy is the capacity to do work. Work is force*distance. Let the work not accelerate anything but overcome friction at low speed. Let a 10 kg block sit on a table and have a coefficient of friction of 0.1, so the constant force to move it slowly (without acceleration) is 1 kg. If the energy in question is 4 kg-meters (converted suitably to common units), the block can be pushed 4 meters. This avoids time, velocity, and specific heat, keeping the whole thing purely mechanical for intuitive appeal. Steve Gray ----- Original Message ----- From: "Jason Holt" <jason@lunkwill.org> To: "math-fun" <math-fun@mailman.xmission.com> Sent: Monday, July 17, 2006 1:52 PM Subject: Re: [math-fun] Another physics question
On Mon, 17 Jul 2006, Eugene Salamin wrote:
and is there an intuitive way to explain the conversion from the 30mph baseball's energy into kg*m^2/s^2 without calculus?
Perhaps, think of measuring the energy calorimetrically, by converting the energy into heat. Let a tank of water bring the baseball to a stop, and measure the temperature rise. This doesn't explain what energy actually IS; I can't think of a simple answer to that question.
The 30mph baseball is my favorite intuitive notion of energy. Batteries are another one I like to use when explaining it; compressed springs are good, too. I've used calorimetry examples as well, although those can be difficult since specific heat isn't as intuitive for people.
What I was really trying to get at is that since momentum is, at least to me, a more intuitive way of understanding energy, why do we teach energy in terms of kilograms accelerated for a distance? And secondarily, what's the formula for converting a moving baseball into an accelerating kilogram, and how can I explain it to people who don't know calculus?
-J
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--- Steve Gray <stevebg@adelphia.net> wrote:
Try this. Energy is the capacity to do work. Work is force*distance. Let the work not accelerate anything but overcome friction at low speed. Let a 10 kg block sit on a table and have a coefficient of friction of 0.1, so the constant force to move it slowly (without acceleration) is 1 kg. If the energy in question is 4 kg-meters (converted suitably to common units), the block can be pushed 4 meters. This avoids time, velocity, and specific heat, keeping the whole thing purely mechanical for intuitive appeal.
Steve Gray
There seems to be a bit of confusion here between mass (a quantity of matter) and weight (the force due to gravity). A mass m (located on Earth's surface where g = 9.8 m/s^2 is the acceleration of gravity) has weight mg. So the 10 kg mass weighs 98 N (Newton = SI unit of force), and the work done in your example is 39.2 N m. Gene __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com
You're quite right, of course. The idea was to give a time-independent example of energy, but explaining the difference between mass and weight may not be clear to beginners, especially why acceleration has to be involved. So we should measure the block being pushed in terms of its weight, avoiding mass entirely. Beginners may not be happy with newtons, but my example may still be the simplest example. Steve ----- Original Message ----- From: "Eugene Salamin" <gene_salamin@yahoo.com> To: "math-fun" <math-fun@mailman.xmission.com> Sent: Tuesday, July 18, 2006 7:50 AM Subject: Re: [math-fun] Another physics question
--- Steve Gray <stevebg@adelphia.net> wrote:
Try this. Energy is the capacity to do work. Work is force*distance. Let the work not accelerate anything but overcome friction at low speed. Let a 10 kg block sit on a table and have a coefficient of friction of 0.1, so the constant force to move it slowly (without acceleration) is 1 kg. If the energy in question is 4 kg-meters (converted suitably to common units), the block can be pushed 4 meters. This avoids time, velocity, and specific heat, keeping the whole thing purely mechanical for intuitive appeal.
Steve Gray
There seems to be a bit of confusion here between mass (a quantity of matter) and weight (the force due to gravity). A mass m (located on Earth's surface where g = 9.8 m/s^2 is the acceleration of gravity) has weight mg. So the 10 kg mass weighs 98 N (Newton = SI unit of force), and the work done in your example is 39.2 N m.
Gene
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--- Steve Gray <stevebg@adelphia.net> wrote:
You're quite right, of course. The idea was to give a time-independent example of energy, but explaining the difference between mass and weight may not be clear to beginners, especially why acceleration has to be involved. So we should measure the block being pushed in terms of its weight, avoiding mass entirely. Beginners may not be happy with newtons, but my example may still be the simplest example.
Steve
Use a spring scale to drag the weight. The scale measures pounds, which is a unit of force. Then you get energy in foot-pounds. Gene
----- Original Message ----- From: "Eugene Salamin" <gene_salamin@yahoo.com> To: "math-fun" <math-fun@mailman.xmission.com> Sent: Tuesday, July 18, 2006 7:50 AM Subject: Re: [math-fun] Another physics question
--- Steve Gray <stevebg@adelphia.net> wrote:
Try this. Energy is the capacity to do work. Work is
force*distance. Let
the work not accelerate anything but overcome friction at low speed. Let a 10 kg block sit on a table and have a coefficient of friction of 0.1, so the constant force to move it slowly (without acceleration) is 1 kg. If the energy in question is 4 kg-meters (converted suitably to common units), the block can be pushed 4 meters. This avoids time, velocity, and specific heat, keeping the whole thing purely mechanical for intuitive appeal.
Steve Gray
There seems to be a bit of confusion here between mass (a quantity of matter) and weight (the force due to gravity). A mass m (located on Earth's surface where g = 9.8 m/s^2 is the acceleration of gravity) has weight mg. So the 10 kg mass weighs 98 N (Newton = SI unit of force), and the work done in your example is 39.2 N m.
Gene
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(It's getting distinctly toasty here in Utah, which suggested the questions below.) 1) So the Uranium in the ocean is enough to warm it up, but not to boil it. Suppose we also used the Deuterium, supposing fusion worked. Would that be enough? 2) You can't boil the ocean by raising it to 212F. First, you need another 540 calories/gram to convert to vapor. And second, after you've boiled a little bit of ocean, atmospheric pressure rises and you have to get the temperature higher to boil the next bit. (Atmospheric pressure is about 30 feet of water, so boiling the top 30 feet of ocean would double the pressure.) Is there some convergence, or would some critical point phenomenon set in as heat is added? Rich -----Original Message----- From: math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com on behalf of Eugene Salamin Sent: Mon 7/17/2006 1:46 PM To: math-fun Subject: Re: [math-fun] Another physics question --- Jason Holt <jason@lunkwill.org> wrote:
I also had a physics question recently. I was trying to explain e=mc^2 to my family, and was going to try to give an intuitive example of just how much energy there is in 300,000,000^2 kg*m^2/s^2.
Here is one illustration. Ocean water contains about 3 micrograms per liter of dissolved uranium. Extract the uranium from a volume of ocean, and fission all of it (using a breeder reactor), yielding about 200 MeV per atom. Use the released energy to heat that same volume of water. Water has a specific heat of 4800 J K^(-1) kg^(-1). What is the temperature rise? Answer: about 110 deg F.
I realized that impulse (or momentum), measured in kg*m/s, is much easier for me to conceptualize and explain: a baseball travelling at 30mph has a certain amount of kinetic energy that anyone can experience personally just by catching it. It was much more awkward to try to explain pushing a 1kg weight exactly hard enough to
accelerate it 1m/s^2, but only until it had travelled 1m. So if both units can be used to express energy, why do we speak of impulse as so different from energy,
Because energy and momentum are different physical quantities.
and is there an intuitive way to explain the conversion from the 30mph baseball's energy into kg*m^2/s^2 without calculus?
Perhaps, think of measuring the energy calorimetrically, by converting the energy into heat. Let a tank of water bring the baseball to a stop, and measure the temperature rise. This doesn't explain what energy actually IS; I can't think of a simple answer to that question.
I still remember the test question in 10th grade physical science in which I was asked to calculate the work required to carry a 10kg weight across a 10m room. I argued that, at least in the ideal sense, it required no work (if you're infinitely patient), since there was no change in gravitaional
potential energy, but the teacher wouldn't budge. I suppose I've resented work ever since.
This is one more example of how incompetent teachers drive students away from the phyical sciences. Was this public school?
Oh yeah, and maybe Mike Stay could also briefly summarize quantum gravity for us in a way that makes it clear just how that principle works. I was
wondering about that, too.
Yeah Mike, do that, would you please? Gene __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
--- "Schroeppel, Richard" <rschroe@sandia.gov> wrote:
(It's getting distinctly toasty here in Utah, which suggested the questions below.)
1) So the Uranium in the ocean is enough to warm it up, but not to boil it. Suppose we also used the Deuterium, supposing fusion worked. Would that be enough?
D + D --> He4 + 24 MeV. Water is 55 molar H2O, and D is 1/6700, so D concentration is 17 mM or 10^22 D atoms/liter. Each D atom yields 12 MeV = 1.9x10^-12 J, and 1 liter yields 1.9x10^10 J. To boil that 1 liter starting from 20 C, you need (80+540) cal/g x 4.8 J/cal x 1000 g = 3x10^6 J.
2) You can't boil the ocean by raising it to 212F. First, you need another 540 calories/gram to convert to vapor. And second, after you've boiled a little bit of ocean, atmospheric pressure rises and you have to get the temperature higher to boil the next bit. (Atmospheric pressure is about 30 feet of water, so boiling the top 30 feet of ocean would double the pressure.) Is there some convergence, or would some critical point phenomenon set in as heat is added?
Rich
You easily have sufficient energy to heat the ocean to the critical point of water: 374 C, 220 atm. At the critical temperature, the liquid and vapor phases become identical. Gene __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com
Work is NOT force*distance, it's force . distance; both are vector quantities. So your answer to your 10th grade physical science question was correct, since the force to carry the weight is perpendicular to the direction of motion, giving a dot product of zero. On the other hand, if you were pushing the weight across the floor, with some appropriate coefficient of friction, then you would indeed be doing work, given by the product of weight, coefficient of friction, and distance. --ms Jason Holt wrote:
I also had a physics question recently. I was trying to explain e=mc^2 to my family, and was going to try to give an intuitive example of just how much energy there is in 300,000,000^2 kg*m^2/s^2. I realized that impulse (or momentum), measured in kg*m/s, is much easier for me to conceptualize and explain: a baseball travelling at 30mph has a certain amount of kinetic energy that anyone can experience personally just by catching it. It was much more awkward to try to explain pushing a 1kg weight exactly hard enough to accelerate it 1m/s^2, but only until it had travelled 1m. So if both units can be used to express energy, why do we speak of impulse as so different from energy, and is there an intuitive way to explain the conversion from the 30mph baseball's energy into kg*m^2/s^2 without calculus?
I still remember the test question in 10th grade physical science in which I was asked to calculate the work required to carry a 10kg weight across a 10m room. I argued that, at least in the ideal sense, it required no work (if you're infinitely patient), since there was no change in gravitaional potential energy, but the teacher wouldn't budge. I suppose I've resented work ever since.
Oh yeah, and maybe Mike Stay could also briefly summarize quantum gravity for us in a way that makes it clear just how that principle works. I was wondering about that, too.
-J
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participants (7)
-
Eugene Salamin -
Jason Holt -
Kerry Mitchell -
Mike Speciner -
Mike Stay -
Schroeppel, Richard -
Steve Gray