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 Question about Special Relativity pythagorean Nerd of Disrepute Posts: 291 Threads: 11 Likes Received: 35 in 27 posts Likes Given: 68 Joined: Nov 2023 Reputation: -6 11-04-2023, 07:01 AM Question about Special Relativity (11-04-2023, 05:42 AM)Cavebear Wrote: (11-04-2023, 05:26 AM)pythagorean Wrote: Massive objects (in special relativity) are at rest with respect to a coordinate system that moves along with them. Massless objects (like photons) are moving too fast for any coordinate system to keep up with them. That is, somebody thinks that the massive object is at rest, but nobody thinks that the photon is at rest.  Poincaré group - Wikipedia OK, so a black hole basically is "at rest" as it relates to itself and its gravitational influence. Massless particles are never at rest (relative to anything)? Why is C squared? I don't think Black Holes exist in Special Relativity. Their mass would curve spacetime.  And special relativity is flat. Massless particles are not at rest with respect to anything. Not sure about c^2. General Relativity is a whole other ball of wax. LIGO Feel That Space (The Weeknd Parody) | A Capella Science test The following 1 user Likes pythagorean's post: polymath257 Junior Member Posts: 232 Threads: 1 Likes Received: 553 in 181 posts Likes Given: 332 Joined: Apr 2022 Reputation: 7 11-04-2023, 08:11 PM Question about Special Relativity (11-04-2023, 05:29 AM)Cavebear Wrote: (11-04-2023, 03:59 AM)Chas Wrote: An object is only at rest relative to other objects and defines its frame of reference. The c is squared for the same reason v is squared in   K.E. = 1/2 m v² OK, I expected "at rest" would be relative to the immediate location. But, as I haven't the slightest idea why C is squared in E=MC^2, I also have no idea why v is squared in K.E. = 1/2 m v².  And at least I understand what kinetic energy is! I'm not asking for a math class here.  Just something a bit more in plain English.  I am better with geometry than physics formulas.  I can intuitively grasp why light diminishes at the square of the distance, that F=MA, and that a spaceship can gain speed sling-shotting around a planet (time in the gravity-well).   But I (and maybe others) don't know "why" kinetic energy = 1/2 momentum times velocity squared".  Why "1/2"?  Why is "velocity" squared?  Give us a little help here.   First, why is v (or c) squared? The answer is that it has to do with the units of the variables used. Let distance (length) have units of L, time have units of T and mass units of M. Then, velocity is distance divided by time, so has units of L/T. Acceleration is a rate of change of velocity, so has units of (L/T)/T=L/T^2. Force: we have F=MA, so the units of force are ML/T^2. (Be careful, though, F=ma is no longer true in special relativity). Energy: work is a form of energy and we have W=FL, work is force times distance, so energy has units of (ML/T^2)*L, so ML^2/T^2. This means that if we have energy as mass times *something*, then something has to have units of L^2/T^2=(L/T)^2, which is the square of a velocity. Hence, in both K=(1/2)mv^2 and E=mc^2, we need the square of a velocity to make the units come out right. Now, why one has a 1/2 and the other does not is trickier, in part because the answer is slightly different for classical (Newtonian) mechanics and for special relativity. And, in fact, the formula for kinetic energy is different for classical mechanics and for special relativity. In both, the kinetic energy is the amount of work that a force needs to do to bring you from rest (in some frame) to that velocity. In classical mechanics, F=ma and a is the rate of change of velocity. When that is multiplied by a small change in distance (to get the work over  small time interval), that distance is the velocity times the time interval. But the acceleration is a velocity divided by that same time interval. So, we are really looking for the accumulated square velocity. But, if we start from zero velocity, the *average* velocity is half of the final velocity, so we get (1/2)v multiplied by v, giving the (1/2)v^2. (The full derivation is helped with some calculus--the integral of v dv is (1/2)v^2 ). In special relativity, the actual formula is no longer F=ma, but F=dp/dt. Here, p is the momentum of the particle. Classically, p=mv, but that has to be modified in special relativity. The derivation is more than I want to do here, but the total energy of a particle ends up being E=mc^2/sqrt(1-v^2/c^2) The square root can be approximated and it is found that 1/sqrt(1-v^2/c^2) is approximately 1+(1/2)v^2/c^2. The (1/2) comes from the square root. When this is multiplied by mc^2, the first term is mc^2 and represents the energy at rest. The next term is (1/2)mv^2 and represents an approximation to the kinetic energy. The following 3 users Like polymath257's post: Cavebear Paragon Posts: 10,396 Threads: 33 Likes Received: 5,818 in 3,862 posts Likes Given: 7,970 Joined: Sep 2019 Reputation: 24 11-05-2023, 01:49 AM Question about Special Relativity (11-04-2023, 08:11 PM)polymath257 Wrote: (11-04-2023, 05:29 AM)Cavebear Wrote: OK, I expected "at rest" would be relative to the immediate location. But, as I haven't the slightest idea why C is squared in E=MC^2, I also have no idea why v is squared in K.E. = 1/2 m v².  And at least I understand what kinetic energy is! I'm not asking for a math class here.  Just something a bit more in plain English.  I am better with geometry than physics formulas.  I can intuitively grasp why light diminishes at the square of the distance, that F=MA, and that a spaceship can gain speed sling-shotting around a planet (time in the gravity-well).   But I (and maybe others) don't know "why" kinetic energy = 1/2 momentum times velocity squared".  Why "1/2"?  Why is "velocity" squared?  Give us a little help here.   First, why is v (or c) squared? The answer is that it has to do with the units of the variables used. Let distance (length) have units of L, time have units of T and mass units of M. Then, velocity is distance divided by time, so has units of L/T. Acceleration is a rate of change of velocity, so has units of (L/T)/T=L/T^2. Force: we have F=MA, so the units of force are ML/T^2. (Be careful, though, F=ma is no longer true in special relativity). Energy: work is a form of energy and we have W=FL, work is force times distance, so energy has units of (ML/T^2)*L, so ML^2/T^2. This means that if we have energy as mass times *something*, then something has to have units of L^2/T^2=(L/T)^2, which is the square of a velocity. Hence, in both K=(1/2)mv^2 and E=mc^2, we need the square of a velocity to make the units come out right. Now, why one has a 1/2 and the other does not is trickier, in part because the answer is slightly different for classical (Newtonian) mechanics and for special relativity. And, in fact, the formula for kinetic energy is different for classical mechanics and for special relativity. In both, the kinetic energy is the amount of work that a force needs to do to bring you from rest (in some frame) to that velocity. In classical mechanics, F=ma and a is the rate of change of velocity. When that is multiplied by a small change in distance (to get the work over  small time interval), that distance is the velocity times the time interval. But the acceleration is a velocity divided by that same time interval. So, we are really looking for the accumulated square velocity. But, if we start from zero velocity, the *average* velocity is half of the final velocity, so we get (1/2)v multiplied by v, giving the (1/2)v^2. (The full derivation is helped with some calculus--the integral of v dv is (1/2)v^2 ). In special relativity, the actual formula is no longer F=ma, but F=dp/dt. Here, p is the momentum of the particle. Classically, p=mv, but that has to be modified in special relativity. The derivation is more than I want to do here, but the total energy of a particle ends up being E=mc^2/sqrt(1-v^2/c^2) The square root can be approximated and it is found that 1/sqrt(1-v^2/c^2) is approximately 1+(1/2)v^2/c^2. The (1/2) comes from the square root. When this is multiplied by mc^2, the first term is mc^2 and represents the energy at rest. The next term is (1/2)mv^2 and represents an approximation to the kinetic energy. Never argue with people who type fast and have too much time on their hands...   • Cavebear Paragon Posts: 10,396 Threads: 33 Likes Received: 5,818 in 3,862 posts Likes Given: 7,970 Joined: Sep 2019 Reputation: 24 11-05-2023, 02:11 AM Question about Special Relativity (11-05-2023, 01:49 AM)Cavebear Wrote: (11-04-2023, 08:11 PM)polymath257 Wrote: First, why is v (or c) squared? The answer is that it has to do with the units of the variables used. Let distance (length) have units of L, time have units of T and mass units of M. Then, velocity is distance divided by time, so has units of L/T. Acceleration is a rate of change of velocity, so has units of (L/T)/T=L/T^2. Force: we have F=MA, so the units of force are ML/T^2. (Be careful, though, F=ma is no longer true in special relativity). Energy: work is a form of energy and we have W=FL, work is force times distance, so energy has units of (ML/T^2)*L, so ML^2/T^2. This means that if we have energy as mass times *something*, then something has to have units of L^2/T^2=(L/T)^2, which is the square of a velocity. Hence, in both K=(1/2)mv^2 and E=mc^2, we need the square of a velocity to make the units come out right. Now, why one has a 1/2 and the other does not is trickier, in part because the answer is slightly different for classical (Newtonian) mechanics and for special relativity. And, in fact, the formula for kinetic energy is different for classical mechanics and for special relativity. In both, the kinetic energy is the amount of work that a force needs to do to bring you from rest (in some frame) to that velocity. In classical mechanics, F=ma and a is the rate of change of velocity. When that is multiplied by a small change in distance (to get the work over  small time interval), that distance is the velocity times the time interval. But the acceleration is a velocity divided by that same time interval. So, we are really looking for the accumulated square velocity. But, if we start from zero velocity, the *average* velocity is half of the final velocity, so we get (1/2)v multiplied by v, giving the (1/2)v^2. (The full derivation is helped with some calculus--the integral of v dv is (1/2)v^2 ). In special relativity, the actual formula is no longer F=ma, but F=dp/dt. Here, p is the momentum of the particle. Classically, p=mv, but that has to be modified in special relativity. The derivation is more than I want to do here, but the total energy of a particle ends up being E=mc^2/sqrt(1-v^2/c^2) The square root can be approximated and it is found that 1/sqrt(1-v^2/c^2) is approximately 1+(1/2)v^2/c^2. The (1/2) comes from the square root. When this is multiplied by mc^2, the first term is mc^2 and represents the energy at rest. The next term is (1/2)mv^2 and represents an approximation to the kinetic energy. First, thank you for some "plain language" as much as math permits... I actually understood parts of it. My high school math teachers would be surprised at that (I struggled through Algebra I and II. But over the years, I have learned to "visualize" some equations. Not that I understand half of what you are describing, but the parts I did helped. Second, in " F=dp/dt", what is "d"? Is it delta? Distance? Third, if I get the "But, if we start from zero velocity, the *average* velocity is half of the final velocity", correctly, that explains the "1/2" value (assuming a constant velocity). That might have occurred to me had I given it more thought, as "1/2" sort of defines average. Fourth, what is "ML"? It seems to mean "mass-length" and I can't even conceive of what that means. LOL! Even if it means M*L, I have trouble with the concept. Again, thanks for the explanations. I appreciate the time it took to write all that. But even if I gain just a little understanding, it is a benefit. Never argue with people who type fast and have too much time on their hands...   • Cavebear Paragon Posts: 10,396 Threads: 33 Likes Received: 5,818 in 3,862 posts Likes Given: 7,970 Joined: Sep 2019 Reputation: 24 11-05-2023, 02:32 AM Question about Special Relativity (11-04-2023, 07:01 AM)pythagorean Wrote: (11-04-2023, 05:42 AM)Cavebear Wrote: OK, so a black hole basically is "at rest" as it relates to itself and its gravitational influence. Massless particles are never at rest (relative to anything)? Why is C squared? I don't think Black Holes exist in Special Relativity. Their mass would curve spacetime.  And special relativity is flat. Massless particles are not at rest with respect to anything. Not sure about c^2. General Relativity is a whole other ball of wax. I poorly understand General Relativity, but I get the concept at least. As in, light is the basis for all measurements, and anything we see involves time for light to travel to our senses or equipment. Einstein's observation that the light from the village tower clock took "some" time to reach his eyes makes perfect sense to me. I will admit that light from a moving object it slightly less intuitive. But Special Relatively drives me crazy. The Wiki article I read about it just gives me a headache. It seems to involve both "frames of reference with no acceleration" and moving objects and "the most accurate model of motion at any speed when gravitational and quantum effects are negligible". You don't need to try to explain it; I won't understand it. But I do wonder about "I don't think Black Holes exist in Special Relativity.". They do certainly seem to exist! A brief explanation of that might be educational. Never argue with people who type fast and have too much time on their hands...   • pythagorean Nerd of Disrepute Posts: 291 Threads: 11 Likes Received: 35 in 27 posts Likes Given: 68 Joined: Nov 2023 Reputation: -6 11-05-2023, 11:34 AM Question about Special Relativity (11-05-2023, 02:32 AM)Cavebear Wrote: (11-04-2023, 07:01 AM)pythagorean Wrote: I don't think Black Holes exist in Special Relativity. Their mass would curve spacetime.  And special relativity is flat. Massless particles are not at rest with respect to anything. Not sure about c^2. General Relativity is a whole other ball of wax. I poorly understand General Relativity, but I get the concept at least.  As in, light is the basis for all measurements, and anything we see involves time for light to travel to our senses or equipment.  Einstein's observation that the light from the village tower clock took "some" time to reach his eyes makes perfect sense to me.  I will admit that light from a moving object it slightly less intuitive. But Special Relatively drives me crazy.  The Wiki article I read about it just gives me a headache.  It seems to involve both "frames of reference with no acceleration" and moving objects and "the most accurate model of motion at any speed when gravitational and quantum effects are negligible".  You don't need to try to explain it; I won't understand it.   But I do wonder about "I don't think Black Holes exist in Special Relativity.".  They do certainly seem to exist!  A brief explanation of that might be educational. Not interested. test   • polymath257 Junior Member Posts: 232 Threads: 1 Likes Received: 553 in 181 posts Likes Given: 332 Joined: Apr 2022 Reputation: 7 11-05-2023, 04:48 PM Question about Special Relativity (11-05-2023, 02:11 AM)Cavebear Wrote: (11-05-2023, 01:49 AM)Cavebear Wrote: First, thank you for some "plain language" as much as math permits...  I actually understood parts of it.  My high school math teachers would be surprised at that (I struggled through Algebra I and II.  But over the years, I have learned to "visualize" some equations.  Not that I understand half of what you are describing, but the parts I did helped. Second, in " F=dp/dt", what is "d"?  Is it delta?  Distance? dp/dt is the derivative of momentum with respect to proper time. A bit of calculus. You can approximate it with delta=change in. Quote:Third, if I get the "But, if we start from zero velocity, the *average* velocity is half of the final velocity", correctly, that explains the "1/2" value (assuming a constant velocity).  That might have occurred to me had I given it more thought, as "1/2" sort of defines average. Fourth, what is "ML"?  It seems to mean "mass-length" and I can't even conceive of what that means. LOL!  Even if it means M*L, I have trouble with the concept. Mass times distance. I can't think of a case off the top of my head where the final units are ML. But momentum has units of ML/T, so mass times distance divided by time. Since distance divided by time is a velocity, this is mass times velocity. Quote:Again, thanks for the explanations. I appreciate the time it took to write all that.  But even if I gain just a little understanding, it is a benefit.   • « Next Oldest | Next Newest »

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