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Let's suppose I heat an object and measure its mass with very accurate balance. As I heat is, the thermal motion of atoms increases. Will the balance indicate an increase of the mass of the heated object, as the temperature increases? -----In other words. The energy that goes into the heated object is: E=sqrt [ (pc)^2+(m_rest*c^2)^2 ] = m_relativistic*c^2What will the balance measure: the m_rest or m_relativistic ?

If the object is stationary in the scale's frame of reference, it's simply mass (not relativistic mass) which increases.

the system's mass, however, *is not* the sum of the masses of the particles (mass is not additive).

QuoteIf the object is stationary in the scale's frame of reference, it's simply mass (not relativistic mass) which increases.The object is stationary relative to scale's frame. The atoms making up the whole object can move due to thermal motion.

Will the scale indicate the same value at 200K and at 900K ?

Quotethe system's mass, however, *is not* the sum of the masses of the particles (mass is not additive).This is counter-intuitive and I have 2 more questions:1) rest masses are additive?

2) is there a simple intuitive reason why masses are not additive?

The object is stationary relative to scale's frame. The atoms making up the whole object can move due to thermal motion.

You know Lightarrow, if I define mass (rest mass) as a property being invariant, intrinsically belonging to a rest mass, unchanging although I may change its frame of reference from where I, for example, measure its weight. Can I really state that I by heating it up change its rest mass? Also photons and rest mass? I think I know what you are referring to, but?

There is a time component to the definition of a invariant mass, to me that is. If it is invariant, it should be so under time too?

It's not so much the question of if it really will weight more that confuse me, but how that time component are used....sorry, you cannot view external links. To see them, please REGISTER or LOGIN uses this definition."The term mass in special relativity usually refers to the rest mass of the object, which is the Newtonian mass as measured by an observer moving along with the object. The invariant mass is another name for the rest mass of single particles. The more general invariant mass (calculated with a more complicated formula) loosely corresponds to the "rest mass" of a "system". Thus, invariant mass is a natural unit of mass used for systems which are being viewed from their center of momentum frame (COM frame), as when any closed system ..... (for example a bottle of hot gas) ...... is weighed, which requires that the measurement be taken in the center of momentum frame where the system has no net momentum. Under such circumstances the invariant mass is equal to the relativistic mass (discussed below), which is the total energy of the system divided by c (the speed of light) squared.The concept of invariant mass does not require bound systems of particles, however. As such, it may also be applied to systems of unbound particles in high-speed relative motion. Because of this, it is often employed in particle physics for systems which consist of widely separated high-energy particles. If such systems were derived from a single particle, then the calculation of the invariant mass of such systems, which is a never-changing quantity, will provide the rest mass of the parent particle (because it is conserved over time)."

Kinetic energy without counteraction doesn't create additional gravitation.

Let me think about it for a time Lightarrow, and I may misstate it. What you seem to say is that the 'energy', of any 'closed' system defined, is invariant under transformations?

But not that it is invariant under time?

Many thanks for so insightful answers. Let me ask this question, because it seems inevitable:In a box with fictitious zero-rest-mass walls we 'trap' some light. The photos can move back and forth inside this fictitious box. They cannot tunnel (assume for the sake or argument they cannot).If I put this imaginary box on a scale, I should get a non-zero reading right?

Quote from: flr on 22/03/2013 21:41:26Many thanks for so insightful answers. Let me ask this question, because it seems inevitable:In a box with fictitious zero-rest-mass walls we 'trap' some light. The photos can move back and forth inside this fictitious box. They cannot tunnel (assume for the sake or argument they cannot).If I put this imaginary box on a scale, I should get a non-zero reading right? Yes.

Quote from: simplified on 20/03/2013 09:36:57Kinetic energy without counteraction doesn't create additional gravitation.I don't understand what you mean by "counteraction." Kinetic energy DOES generate a gravitational field though. That's quite a well-known fact.Regarding the term "invariant." Invariant does sound like it means "does not change with time" but this is not the case. It gets its name from the fact that it remains inchanged upon applying a Lorentz transformation.I want to remind you that invariant mass only applies to closed systems. It's meaningless in more general cases.

Quote from: lightarrow on 18/03/2013 20:34:58If the object is stationary in the scale's frame of reference, it's simply mass (not relativistic mass) which increases.So, if I keep heating it, it will eventually form a black hole?

Quote from: Pmb on 22/03/2013 23:39:47Quote from: flr on 22/03/2013 21:41:26Many thanks for so insightful answers. Let me ask this question, because it seems inevitable:In a box with fictitious zero-rest-mass walls we 'trap' some light. The photos can move back and forth inside this fictitious box. They cannot tunnel (assume for the sake or argument they cannot).If I put this imaginary box on a scale, I should get a non-zero reading right? Yes. Then my question is: The mass measured by the scale is the rest mass of the box trapping photons, is that so?

The abstract is interesting. Unfortunately, I don't have full access to it.

Quote from: flr on 23/03/2013 07:21:19So, if I keep heating it, it will eventually form a black hole?No, because its particles will fly away, over a certain temperature.

So, if I keep heating it, it will eventually form a black hole?

Quote from: lightarrow on 23/03/2013 13:04:54Quote from: flr on 23/03/2013 07:21:19So, if I keep heating it, it will eventually form a black hole?No, because its particles will fly away, over a certain temperature.Let's assume, for the sake of argument that they are constrained NOT to fly away. Will then form black hole?

If all photons move parallel, will the system still have rest mass?

The total momentum depends on the angle between photons, therefore the invariant mass depends on the angle as well.If they move parallel and are unconstrained in that fictitious box, then their invariant mass is zero, just like it is the case for one single photon.

2 photons moving parallel in wave representation will simply be a wave like the initial one but of double amplitude.

Let's assume a photon "moves in a circle". Unless there is a fundamental law that prevent the photon "moving in a circle", let's suppose for argument that it is possible to get a photon trapped such that it moves in circle. Will such a system made of a photon "moving in a circle" have a nonzero invariant mass?

Let us make it simple.In a black box experiment, do you assume that the angles, as directions, of it will make a difference?

If I want to follow the photon around the circle I agree with you that I will not be able to get into a frame from where I see zero momentum.

I find it interesting and surprising that it could be convinced geometries/frames where p*c actually contributes to the rest mass.

I have another question on this issue.Lets consider an ensemble made of 2 two photons having equal energy (and therefore equal momentum), in the following situations:i) the angle between them is zero degrees , in which case I cannot define a frame from which to define the invariant mass of the ensemble (and therefore I make the rest mass zero and all energy goes into p*c term.

ii) the angle between photons is 180 degrees (or Pi radians), and in this case there is a frame from which I can determine that the ensemble of the two photons has a non-zero invariant mass.The question is: how will the above ensemble of 2 photons will interact with an homogeneous gravitational field?

Equal, or the one for which we can define rest mass will interact stronger?