jajrussel Posted December 12, 2012 Posted December 12, 2012 I have question; What exactly does E=MC^2 mean? I thought I had a basic understanding, but the more I learn, well questions just seem to appear out of nowhere. It seems to be nothing more than the formula for momentum, where energy takes the place for momentum. Then there is the understanding that nothing with mass can move at C which would make E an imaginary value. The squaring of C implies that there should be a reason for squaring. So it seems we should have (M+M)C^2 with the result still being an imaginary value. And one more question; if anything with a mass of zero has to move at C, it would seem to make sense that anything with a mass greater than zero would have to move also. This is not exactly worded like a question, but it is a question.
swansont Posted December 12, 2012 Posted December 12, 2012 The total energy of anything at rest is not zero — it is mc^2. The is not a momentum equation, it is an energy equation. If the object is moving, the relevant equation is [math]E^2 = m^2c^4 + p^2c^2[/math] The momentum of a photon, which is massless, is E/c. Conservation of momentum and energy involving a photon and a massive object is a path for deriving E=mc^2 and lets you see where the c^2 comes from. 2
jajrussel Posted December 12, 2012 Author Posted December 12, 2012 So, if something at rest has a total energy of mc^2; then i need to ask... What does it mean to be at rest? Wouldn't something being at rest be at rest, as in it's relation to being part of a system, and wouldn't it's total energy be effected by its interaction with parts of the system that cause it to be viewed as, at rest? And wouldn't the size of the system it is interacting with contribute to its total energy? I thought that c implied a vacuum therefore no interaction? If there is no interaction how can it be viewed as at rest? I confess, I was thinking that mc^2 had to do with momentum.
swansont Posted December 12, 2012 Posted December 12, 2012 If it's interacting, that interaction and the partner in that interaction generally have to be considered part of the system. An object is at rest in its own frame. No interaction required.
juanrga Posted December 13, 2012 Posted December 13, 2012 (edited) I have question; What exactly does E=MC^2 mean? It says, in words, that the energy of a free particle at rest (zero velocity) is the product of its mass and of the speed of light squared. I thought I had a basic understanding, but the more I learn, well questions just seem to appear out of nowhere. It seems to be nothing more than the formula for momentum, where energy takes the place for momentum. Momentum has units of mass x velocity. Energy has units of mass x velocity squared. Then there is the understanding that nothing with mass can move at C which would make E an imaginary value. Energy would not be imaginary but infinity for a massive body moving at c. Or said in another way: you need apply a infinite energy to accelerate the object up to c. And one more question; if anything with a mass of zero has to move at C, it would seem to make sense that anything with a mass greater than zero would have to move also. This is not exactly worded like a question, but it is a question. To "make sense" according to your usual experience in real life? Maybe, but at high velocities the behaviour of objects is very different from the ordinary low-velocity behaviour over which you base your experience. In reality it makes sense why a massless object has to move at c. A massless object has not the traditional concept of inertia and cannot be accelerated as when you accelerate a rock. Massless objects are forced to move always to the same speed, and it seems natural that this speed is c, which is an universal constant. A more detailed and rigorous explanation of why massless objects are forced to travel at c requires the use of the relativistic equations. Edited December 13, 2012 by juanrga 1
eytan_il Posted January 22, 2013 Posted January 22, 2013 The total energy of anything at rest is not zero — it is mc^2. The is not a momentum equation, it is an energy equation. If the object is moving, the relevant equation is [math]E^2 = m^2c^4 + p^2c^2[/math] The momentum of a photon, which is massless, is E/c. Conservation of momentum and energy involving a photon and a massive object is a path for deriving E=mc^2 and lets you see where the c^2 comes from. E^2 = m^2c^4 + p^2c^2 is correct only if you mean m(0), the rest mass. m(0)^2 c^4 + p^2c^2 = ( m(0)^2c^4(1-v^2/c^2) + m(0)^2v^2c^2 ) / (1-v^2/c^2) = m(0)2c^4/(1-v^2/c^2) = E^2. Without writing explicitly m(0), the equation is pedagogically wrong. 1
swansont Posted January 22, 2013 Posted January 22, 2013 E^2 = m^2c^4 + p^2c^2 is correct only if you mean m(0), the rest mass. Considering that I defined the system to be at rest, I'd say I had taken care of that.
eytan_il Posted January 22, 2013 Posted January 22, 2013 If it's interacting, that interaction and the partner in that interaction generally have to be considered part of the system. An object is at rest in its own frame. No interaction required. Here you touched a very sensitive problem. We see a system of interacting particles as an object with rest mass regardless of weather there are moving parts in it or not. For the physicist, such a system is a packed black box. However, the laws of physics are totally local and a "system of particles" is not. If we look at the parts of a system of interacting particles, we will see that indeed no part is at rest, ever. That simply means that classical physics only approximates such a system. Only when there are effects that can't be ignored, we look at the parts of the system, e.g. Stern - Gerlach experiment. Objects at rest are nothing more than figments of the imagination and so are objects in general. The amazing fact is that we can represent an entire system of particles as one object by a wave function of that system. The act of putting the pieces together to form an object at rest is purely conscious, not phenomenological !!! The fact that quantum mechanics agrees with that conscious synthesis is simply a miracle. 1
swansont Posted January 22, 2013 Posted January 22, 2013 Here you touched a very sensitive problem. We see a system of interacting particles as an object with rest mass regardless of weather there are moving parts in it or not. For the physicist, such a system is a packed black box. However, the laws of physics are totally local and a "system of particles" is not. If we look at the parts of a system of interacting particles, we will see that indeed no part is at rest, ever. That simply means that classical physics only approximates such a system. Only when there are effects that can't be ignored, we look at the parts of the system, e.g. Stern - Gerlach experiment. Objects at rest are nothing more than figments of the imagination and so are objects in general. The amazing fact is that we can represent an entire system of particles as one object by a wave function of that system. The act of putting the pieces together to form an object at rest is purely conscious, not phenomenological !!! The fact that quantum mechanics agrees with that conscious synthesis is simply a miracle. The subject under discussion here is relativity, not QM.
eytan_il Posted January 23, 2013 Posted January 23, 2013 The subject under discussion here is relativity, not QM. "And one more question; if anything with a mass of zero has to move at C, it would seem to make sense that anything with a mass greater than zero would have to move also. This is not exactly worded like a question, but it is a question". This understanding is correct only in QM i.e. zero state energy. That is why it seems to me correct to make this information available. Considering that I defined the system to be at rest, I'd say I had taken care of that. "if the object is moving" succeeded to confuse me. 1
juanrga Posted January 29, 2013 Posted January 29, 2013 (edited) E^2 = m^2c^4 + p^2c^2 is correct only if you mean m(0), the rest mass. m(0)^2 c^4 + p^2c^2 = ( m(0)^2c^4(1-v^2/c^2) + m(0)^2v^2c^2 ) / (1-v^2/c^2) = m(0)2c^4/(1-v^2/c^2) = E^2. Without writing explicitly m(0), the equation is pedagogically wrong. At contrary, your notation [math]m(0)[/math] is pedagogically misleading and, as a consequence, cannot be found in any textbook. Your [math]m(0)[/math] seems to indicate that mass is a function of speed, when it is not. [math]m[/math] in the formula [math]E^2 = m^2 c^4 + p^2 c^2[/math] denotes a relativistic invariant. It is the mass of the object at any speed, it is the mass when the object is at rest and it is the mass when the object is moving at half the speed of light. Edited January 29, 2013 by juanrga
howlingmadpanda Posted February 5, 2013 Posted February 5, 2013 Well E=MC^2 basically states the energy-mass equivalence, which can be demonstrated in the common fission reactor, my splitting mass energy is released therefore there is an equivalence on some proportion. Basically this could say how much energy would be released if an object just disappeared as an object going at the speed of light... well I'm not going to speak of that due to it's unpleasant nature. Anyways watch some videos and look up a bit more and you should be able to understand this.
maskman` Posted February 10, 2013 Posted February 10, 2013 momentum is force * time. ma*t = mv*t/t =mv which is different from mv2 which has a square on v. momentum does not have a square on it. which is the equation of momentum. this is a equation about energy =force * distance =f*d =ma*d =mv/t*d =mv2 so mass and energy are equivalent and interconvertible. e=mc2 in this e is energy and m is mass and c is the speed of light. here we know that the speed of light is tremendous and that value is very large. that is trying to say that a small amount of mass can be converted to a large amount of energy. and a large amount of energy can be converted to a small amount of mass. now it also has implications in velocity. when a object is moving with certain velocity it has certain amount of kinetic energy. this energy kinetic adds to the mass of the object and makes it heavier. however this value is small at smaller velocities but only significant for the bodies near the speed of light. the more the body moves towards the speed of light the more the energy adds up to his mass and has a rather significant impact on it. this theory i believe has also implications in proving that a accelerating body is space feels similar laws to a body that is within a gravitational field of a big mass.
swansont Posted February 10, 2013 Posted February 10, 2013 so mass and energy are equivalent and interconvertible. There's actually no guarantee that you can do the conversion. e=mc2 in this e is energy and m is mass and c is the speed of light. here we know that the speed of light is tremendous and that value is very large. that is trying to say that a small amount of mass can be converted to a large amount of energy. and a large amount of energy can be converted to a small amount of mass. now it also has implications in velocity. when a object is moving with certain velocity it has certain amount of kinetic energy. this energy kinetic adds to the mass of the object and makes it heavier. however this value is small at smaller velocities but only significant for the bodies near the speed of light. the more the body moves towards the speed of light the more the energy adds up to his mass and has a rather significant impact on it. this theory i believe has also implications in proving that a accelerating body is space feels similar laws to a body that is within a gravitational field of a big mass. If you will refer to the equation given by juanrga above, you will see that mass does not change. Mass in motion will have momentum, and that's where the energy appears.
maskman` Posted February 11, 2013 Posted February 11, 2013 (edited) swansont The total energy of anything at rest is not zero — it is mc^2. The is not a momentum equation, it is an energy equation. If the object is moving, the relevant equation is The momentum of a photon, which is massless, is E/c. Conservationof momentum and energy involving a photon and a massive object is apath for deriving E=mc^2 and lets you see where the c^2 comes from. respected swansont, You yourself have quoted that energy is not momentum. a moving body has kinetic energy. but i agree on the fact that energy is dependent on mass and velocity. i think this is what your trying to say to me. that being said i know that the moving body doesnt not get its energy from the equation e=mc^2. momentum it gets it from the equation, kinetic energy= 1/2mv^2 now if you suppose a body is moving at the 90% the speed of light. it has a velocity of 3 * 10^7. say this body has a mass of 5 kg. now its K E is 5/2*(3*10^7)^2. agreed? ke =2.25*10^15 now this energy according to einsteins theory e=mc^2 adds up to the mass. e=mc^2 e= KE m= KE/c^2 about 0.025 which adds up to the mass of the object and is about 0.5 % of an increase in mass. which is not much but the more a person moves towards the speed of light the more the energy adds up to the mass. cheers please correct me if i am wrong. thx for replying Edited February 11, 2013 by maskman`
swansont Posted February 11, 2013 Posted February 11, 2013 swansont The total energy of anything at rest is not zero — it is mc^2. The is not a momentum equation, it is an energy equation. AFAICT, I have not stated anything to the contrary. that being said i know that the moving body doesnt not get its energy from the equation e=mc^2. momentum it gets it from the equation, kinetic energy= 1/2mv^2 now if you suppose a body is moving at the 90% the speed of light. it has a velocity of 3 * 10^7. say this body has a mass of 5 kg. now its K E is 5/2*(3*10^7)^2. agreed? ke =2.25*10^15 now this energy according to einsteins theory e=mc^2 adds up to the mass. No, according to Einstein's theory [math]E^2 = p^2c^2 + m^2c^4[/math] This reduces to [math]E = mc^2[/math] only when the system is at rest. When it's moving, p≠0. m is unchanged
juanrga Posted February 14, 2013 Posted February 14, 2013 The total energy of anything at rest is not zero — it is mc^2. The E in the above formula is not the total energy but is lacking the potential energy. The total energy of a charged particle at rest in an external electric field is not mc2
jajrussel Posted February 27, 2013 Author Posted February 27, 2013 It does seem to be an energy equation, but more of a energy/mass equation, or more seemingly precise a conversion equation. It is still puzzling. I understand the concept mass to energy and energy to mass, and that light is the visible form of electromagnetic energy, and that lights velocity is c. But why c in the equation? How do you convert something that has a mass value of zero into some that has mass, or reverse the question, how do you convert something that has mass into something that has no mass, and still have conservation?
elfmotat Posted February 27, 2013 Posted February 27, 2013 (edited) The whole "conversion" thing is rather contrived. All E=mc2 says is that mass is a form of energy. The equation falls directly out of Special Relativity, so if you really want to understand where it comes from you're probably going to have to read an SR textbook. You could use the following line of reasoning to get an intuition for it: Suppose mass is a form of energy. If this is true, then there should be some universal constant relating mass to mass-energy. By dimensional analysis, this constant should have units of [length]2[time]-2. Since [math]c[/math] is a universal constant with dimensions of [length][time]-1, it seems natural that this constant should be proportional to [math]c^2[/math] This doesn't tell you exactly what the constant should be, i.e. it could be [math]\frac{1}{2} c^2[/math], or [math]2 c^2[/math], etc. The correct constant comes from SR. Edited February 27, 2013 by elfmotat
jajrussel Posted February 27, 2013 Author Posted February 27, 2013 Still, the equation seems to suggest that a specific energy value is equal to a specific mass value. If this is what it suggest then anything that has a energy value would seem to need to have a mass value. Is this not what it is saying?
juanrga Posted February 27, 2013 Posted February 27, 2013 (edited) Still, the equation seems to suggest that a specific energy value is equal to a specific mass value. The physical interpretation of rest energy [math]E_0=mc^2[/math] is not very different from that for the Newtonian kinetic energy [math]E_\mathrm{N}=1/2 m v^2[/math]. Consider a free massive particle moving at [math]v=c\sqrt{2}[/math]. Then its Newtonian kinetic energy is [math]E_\mathrm{N} = mc^2[/math]. The first expression of above says you that a free mass at rest has an energy [math]E_0[/math]. The second says that a moving free mass has a Newtonian energy [math]E_\mathrm{N}[/math]. In this particular case both are [math]mc^2[/math]. The total energy of a free massive particle is given by [math]E = E_0 + E_\mathrm{N} + \cdots [/math], where the dots denote post-Newtonian corrections. If this is what it suggest then anything that has a energy value would seem to need to have a mass value. Is this not what it is saying? No. The expression [math]E=mc^2[/math] is only valid for a free massive particle at rest. Thus, it is not valid for a massless particle such as the photon, for which [math]E=pc[/math]. Thus, the photon has an energy value but it has not mass: [math]m=0[/math]. Edited February 27, 2013 by juanrga
swansont Posted February 27, 2013 Posted February 27, 2013 Still, the equation seems to suggest that a specific energy value is equal to a specific mass value. If this is what it suggest then anything that has a energy value would seem to need to have a mass value. Is this not what it is saying? It tells you how much mass you could get, if you do the conversion, because energy is conserved. Mass is one form of energy, just like coins are one form of money. It does not mean that all money is in coin form.
imatfaal Posted October 5, 2013 Posted October 5, 2013 ! Moderator Note Hijack with speculative ideas regarding misapprehension of the meaning of the equation have been split off to the trash. It is unacceptable to answer questions in the main forum with speculative ideas. Do not respond to this moderation within the thread.
Delbert Posted October 8, 2013 Posted October 8, 2013 (edited) What does it mean to be at rest? There is no such thing as being at 'rest'. It just means not moving relative to something else - like the Earth perhaps. Being at 'rest' is a nonentity as far as the universe is concerned. Like absolute speed, there is no such thing. For example, ask yourself: what speed am I doing right now? I suggest you'll come up with anything from zero to a tad under the speed of light - if not the speed of light. As I believe Brian Cox remarked: We are travelling through time at the speed of light according to E = MC2. Had a discussion with friend about a ball on a table, in which I asked him how much potential energy it has? Well, he said I haven't lifted it, moved it in any way to used it to store some potential energy, so it hasn't any. I said it has a lot of potential energy because I lifted it out of a deep hole earlier on before you arrived. And what's more, I understand according to Einstein energy is interchangeable with mass, which means, does it not, I've increased its mass a very tiny bit in keeping with the potential energy imparted when I lifted it out of the hole. Whereas my friend appeared to take it's rest mass to be on the table and nothing has happened, let alone a tiny increase in mass. It's all relative, there is no such thing as absolute rest mass. Unless of course, you take rest mass to be the tiny point at the moment of the big bang! With mass, as we experience it, being a property of the things around us nowadays being an exchange of the potential energy from the big bang, according to E=MC2. Edited October 8, 2013 by Delbert
jajrussel Posted October 23, 2013 Author Posted October 23, 2013 (edited) I have been reading physics books, and my own thoughts tend to interrupt the learning process. There was another question about something being at rest that I had while reading (Light and Matter) which can be found from the website of the same name. I am trying to find the exact spot in the text that brought the question forward in my mind, but for the moment I haven't found it, so I am relying on my memory, which is probably not so good. Anyway - I was wondering if it is actually possible to distinguish if something is in motion, or at rest without simply stating that it is, or it is not? The statement that there is no such thing as something being at rest; I have read before, so the ball sitting on the table from one perspective is at rest, from another; it is not. If the ball were rolling across the table at any point it could be said to be at rest. The thing that seemed interesting to me is that there does not seem to be anything physically different about the ball whether it is perceived to be at rest, or if it is perceived to be moving because the ball is always in balance with the forces acting on it. If this thinking is correct then it would seem that the balls potential energy would be the same whether it is perceived to be moving, or is perceived to be at rest, and that any measurable difference between the two perceptions would simply be one of the coins that Swansont was talking about that makes up part of the balls total potential energy. There is the mass gain as the ball moves to consider. Does E=MC2 account for this gain in mass? Maybe I need to think about this question a little bit more? If the ball can not be accelerated to C because it has mass. It would seem that once the ball has been accelerated to is maximum, that the formula says that at this point there should be a measurable difference between the balls maximum acceleration energy and its total potential energy. Maximum acceleration energy would seem to mean maximum mass potential for the ball, and the measurable difference would have to be mass-less energy. This is actually a question and not a statement. Sometimes I feel like I am totally clueless, and for that reason I am not sure the questions have merit. Edited October 23, 2013 by jajrussel
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now