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all things have mass, but what is mass and can it be manipulated? IE, reduced or increased.

Most simply, mass is just a measure of how difficult (how much force is required) to cause an object to change speed or direction - i.e. a measure of just how stubborn an object is going to be about continuing to do what it is doing.

What you have pointed out is that the everyday concept of a solid object is flawed. What we precieve as a solid is , as you quite rightly say, merely electrostic repulsion between its atoms and ours.But I don't think this invalidates our concept of inertial mass. A 'solid' ball of one isotope of lead could be constructed with the same number of atoms as one of a different isotope. The balls would appear physically identical and would certainly have the same number (rougghly!) of electrons and protons - both being lead - so the electrostatic forces between it and us would be the same when we pushed them. But we would be able to percieve/measure the difference in force that would have to be applied in order to give a specific acceleration.

Strong Force - Short range .. but eh the strongest!Electromagnetic about 1% as strong as the 'Strong' force but much longer range Weak Force - 1/1,000,000,000 as strong and very Short range Gravitational Force - Long range but very, very weak (10^{-38} as strong as the stronng force).

More technically, mass is just another form of energy (E = mc^{2}),

Quote from: another_someone on 11/05/2007 04:19:40More technically, mass is just another form of energy (E = mc^{2}),Sorry to be annoying but the maths Of E=MC2 is designed to give you the answer for the energy content of an object or whatever- not the mass!!!! To do the equation you need- the mass!

Quote from: jolly on 14/05/2007 01:30:50Quote from: another_someone on 11/05/2007 04:19:40More technically, mass is just another form of energy (E = mc^{2}),Sorry to be annoying but the maths Of E=MC2 is designed to give you the answer for the energy content of an object or whatever- not the mass!!!! To do the equation you need- the mass! The point is that the equation provides an equivalence between mass and energy, such that you can create energy by reducing mass (as is done in an atomic bomb) and by inference can create mass from energy - thus the two are interchangeable.

Well this is just not true mass is another form of energy!The equation is E=MC2! right so under that equation mass and energy are considered distint and seperate! to work out the Mass you would have to do the equationM=E/DIVIDED BY C2

The best bit is really... That you still dont know how to work out the mass or the energy of an object! and until you do you wont be able to check the maths in this equation... but when you do you will realise it is utterly incorrect!!!

m = ^{h}/_{λV}It gets more complex for relativistic speeds, but it still boils down to the mass being a function of the quantum wavelength of the particle.

Well magnetism and electricity are clearly linked! but I do not agree that mass and energy are! As a mass can hold varying amounts of energy, which flux; so under those circumstances how can you relate them! Mass I do not believe realitively afects the amount of energy an object can hold!

The problem really is we still dont know what a mass is; or rather we need better understanding of what is classified/constiutues a mass!

My priest says its where we celebrate the Eucharist

Quote from: jolly on 22/05/2007 11:33:11Well magnetism and electricity are clearly linked! but I do not agree that mass and energy are! As a mass can hold varying amounts of energy, which flux; so under those circumstances how can you relate them! Mass I do not believe realitively afects the amount of energy an object can hold! We are not actually saying that the mass in any way limits the amount of energy matter can hold, but rather the contrary, that the amount of energy a particle contains is manifest in the apparent mass of the particle, and that even at rest, that mass contains some energy. Neither the mass nor the energy are limited (just as neither electricity nor magnetism limit each other, but they are nonetheless directly linked to each other).

Mass is a place!

An interesting feature of mass is this:one photon has exactly zero mass.Two photons, no!(It's not a joke!).

Quote from: lightarrow on 24/05/2007 21:18:45An interesting feature of mass is this:one photon has exactly zero mass.Two photons, no!(It's not a joke!).First I have ever heard about this. Would you like to say (preferably in words of one, maybe two at most, syllables) how this comes about, and what it means?

Do you mean that two photons that are somehow bound, and it is the binding energy that has mass; or do you mean that even two independent photons are measured to have mass?

1. Simple explanation.If you know that a system has energy but no momentum, using simple relativistic considerations, you should easily conclude that it has rest mass, isnt'it?Ok. Now, consider the system of two photons of exactly the same energy, travelling in opposite directions. This system has clearly non-zero energy but zero momentum (total p = p1 - p1 = 0). So, the system mast have a rest mass!2. Just a bit less simple explanation.The exact formula relating rest mass m, energy E and momentum p of a relativistic system of any kind, is:(E/c)^{2} - |p|^{2} = (mc)^{2}So, if p = 0, then (E/c) = mc --> m = E/c^{2} ≠ 0 because E ≠ 0 (in this case, E = 2hν).

It sounds to me simply to be a consequence of the notion that light can have energy, and momentum, and thus mass,

Quote from: another_someone on 25/05/2007 16:29:34It sounds to me simply to be a consequence of the notion that light can have energy, and momentum, and thus mass,Here you are talking about relativistic mass, which is a concept physicists advice not to use anylonger. I was talking about rest mass of photons, which is 0 for one photon and non 0 for 2 of them!

Yes, but as I understand what you are doing is that you are taking the relativistic notions of energy and momentum, which implicitly include relativistic mass, and then back propagating to an implied rest mass.Energy and momentum are thus:where γ is the Lorentz factor.

With these formulas you would certainly have strange results, since they are simply wrong for photons (and for 0 mass objects in general). Those are only valid for non 0 mass objects and travelling at v ≠ c.The one valid for every object is (E/c)^{2} - |p|^{2} = (mc)^{2} as I've already written in the other post.

Quote from: lightarrow on 26/05/2007 15:00:40With these formulas you would certainly have strange results, since they are simply wrong for photons (and for 0 mass objects in general). Those are only valid for non 0 mass objects and travelling at v ≠ c.The one valid for every object is (E/c)^{2} - |p|^{2} = (mc)^{2} as I've already written in the other post.Is there not a paradox here.You are saying that you are using an equation that is only valid for massless particles to calculate that the particle is not massless (or, at least that the collections of particles are not massless). If the particles are not massless, then how can you use an equation that is only valid for massless particles?

Probably I didn't express myself clearly.The equation I use: (E/c)^{2} - |p|^{2} = (mc)^{2} is valid for all kinds of particles, massless and not massless, it's the most general one.

Quote from: lightarrow on 26/05/2007 20:21:46Probably I didn't express myself clearly.The equation I use: (E/c)^{2} - |p|^{2} = (mc)^{2} is valid for all kinds of particles, massless and not massless, it's the most general one.Sorry, I did misunderstand that.On the other hand, if the equation is true for both massless and non-massless bodies; and for all such bodies, it can represent a mass for two objects that is different from the sum of the masses of the single objects, then should that not be just as true for tennis balls as for photons?

OK, I see what you are saying.That for any particle or object, if two of them are moving antiparallel directions, they will have a greater calculated rest mass than if they travel in parallel; but that for low speed objects, most of the energy in the particle is bound up in the rest mass anyway, so the amount of kinetic energy is too insignificant to cause the confusion in rest mass.So ant particle, be it massless or non-massless; if it has kinetic energy that is comparable to the zero velocity rest mass, then the calculated rest mass using that formula for particles that are antiparallel will differ from the rest mass of the same particles travelling in parallel.

As such, it does bring into question whether the equation can really be used to measure true rest mass, since true rest mass I would expect to be the mass of the object with zero kinetic energy, and if that equation causes a different answer to that, then the equation sounds wrong.

Infact, calling that m "rest mass" is not appropriate, in this case; some physicist call it, better, "invariant mass" and I agree with them. You could say that, in the case of two objects travelling in opposite directions with total p = 0, however, the system is "at rest" because it is its centre of gravity; however it would be very tricky to keep calling m "rest mass".The term everyone should use is invariant mass simply because it doesn't change from a ref frame to another, and this is what really counts.

Quote from: lightarrow on 29/05/2007 11:51:48Infact, calling that m "rest mass" is not appropriate, in this case; some physicist call it, better, "invariant mass" and I agree with them. You could say that, in the case of two objects travelling in opposite directions with total p = 0, however, the system is "at rest" because it is its centre of gravity; however it would be very tricky to keep calling m "rest mass".The term everyone should use is invariant mass simply because it doesn't change from a ref frame to another, and this is what really counts.In what way is it an invariant mass?

In fact, no mass is actually invariant;

but the notion of 'rest mass' derives from the mass that an object would have in the absence of any kinetic energy,

whereas the equation (^{E}/_{c})^{2} - |p|^{2} = (mc)^{2} must include kinetic energy (otherwise there would not be any momentum, even for a lone particle, and thus the paradox we are talking about disappears).In practical terms, what measurable quantity does this equation provide? It does not show an apparent mass (because that would be the full relativistic mass), and it does not show the mass measured when the system is at rest - so what measurable quantity would relate to this number?

I wrote it:"The term everyone should use is invariant mass simply because it doesn't change from a ref frame to another, and this is what really counts"

Quote from: lightarrow on 30/05/2007 14:17:08I wrote it:"The term everyone should use is invariant mass simply because it doesn't change from a ref frame to another, and this is what really counts"Not sure even this is true.The equation (^{E}/_{c})^{2} - |p|^{2} = (mc)^{2} will assume that |p|^{2} is zero in only one particular frame of reference. If p = γmv for each of the two particles (which it will be is we are talking about non-massless particles at less than the speed of light), then it follows that p changes according to the frame of reference you use .

This would seem more sensibly to be regarded as a systemic rest mass, rather than any sort of reference invariant mass. Thus, you are saying that a system of two photons can have a non-zero systemic rest mass, but that in any system, the rest mass of the system may be different from the sum of the rest masses of its component parts, if the component parts of the system are not at rest.

This is probably not unreasonable, since you are not actually measuring the rest masses of the component parts of the system, but the apparent masses as they occur within the system (i.e. the relativistic masses of the individual component parts).

Wrong. According to your idea, a photon with total energy E and an electron with total energy E should have the same mass, but it's not.

But you would not have said what you did, if you had read my post and understood also what was said. I said E=Mc^{2} is non-relativistic, meaning it does not apply to particles travelling at or near light speed.

Quote from: Mr. Data on 05/07/2011 23:41:30But you would not have said what you did, if you had read my post and understood also what was said. I said E=Mc^{2} is non-relativistic, meaning it does not apply to particles travelling at or near light speed. That's an understatement, since it only applies in the rest frames of particles, which is as far from light speed as they can get. That equation is wrong if the particle is moving in your reference frame--and of course light is always moving.

Quote from: lightarrow on 05/07/2011 20:46:07Wrong. According to your idea, a photon with total energy E and an electron with total energy E should have the same mass, but it's not.No, you apply E=Mc^{2} to an individual particle system. Sure you can work out different values for mass, but for a photon, it does not even have a mass. But that's a whole different ball game involving more difficult field processes I doubt you could fully understand unless you sat down and learned this stuff. But you would not have said what you did, if you had read my post and understood also what was said. I said E=Mc^{2} is non-relativistic, meaning it does not apply to particles travelling at or near light speed.