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Author Topic: what is Mass?  (Read 23159 times)

Offline lightarrow

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what is Mass?
« Reply #25 on: 25/05/2007 19:22:19 »
It 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!
 

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« Reply #26 on: 25/05/2007 20:14:01 »
It 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.

Since both of those factors include both rest mass (zero) and the Lorenz factor (infinite when one is at the speed of light), so one inevitable has zero multiplied by infinity.

Clearly, with light, the situation is the converse to normal, in that we can measure energy and momentum, but cannot measure rest mass; so the equations are inverted, but you still have the fundamental mix of a measureable quantity divided by infinity returning an assumed zero.
 

Offline lightarrow

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« Reply #27 on: 26/05/2007 15:00:40 »
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.
 

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« Reply #28 on: 26/05/2007 15:37:23 »
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.

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?
 

Offline lightarrow

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« Reply #29 on: 26/05/2007 20:21:46 »
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.

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.

The equations you wrote:
E = γmc2
p = γmv
are valid only for not massless particles.
 

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« Reply #30 on: 27/05/2007 23:31:20 »
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.

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?
« Last Edit: 27/05/2007 23:33:09 by another_someone »
 

Offline lightarrow

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« Reply #31 on: 28/05/2007 23:38:29 »
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.
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?
Yes, if the tennis balls are moving, but the effect is negligible at non-relativistic speeds, that is, m1 + m2 ≈ mass of the system of the 2 balls.
The effect is so striking with photons because they travel at c.

Let's make some computation.

we have 2 tennis balls, each having:

m = 100g = 0.1Kg; so m1 + m2 = 2m = 0.2Kg.
v = 30m/s

p = m*v = 0.1*30 = 3 Kg*m/s (actually, we should write p =m*v*γ, but it's easy to see that the result wouldn't change, at the level of accuracy we will find);

E = Sqrt[(cp)2 + (mc2)2]

and they travel in opposite directions.

be:
ms Es, ps the mass, energy and momentum of the system of 2 balls. We have:
Es = 2E because energy is additive;
ps = 0 because the 2 balls are travelling in opposite directions;
(msc2)2 = (Es)2 - (cps)2 = (Es)2
so:
ms = Es/c2 = 2E/c2 = 2 Sqrt[(cp)2 + (mc2)2]/c2 =

2 Sqrt[(p/c)2 + m2] = 2 Sqrt[(3/3*108)2 + (0.1)2] = 2 Sqrt(10-16 + 10-2) =

= 2*10-1 Sqrt( 1 + 10-14) ≈ 2*10-1 = 0.2Kg.
 

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« Reply #32 on: 29/05/2007 01:56:02 »
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.
 

Offline lightarrow

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« Reply #33 on: 29/05/2007 11:51:48 »
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.
Exactly.
Quote
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.
 

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« Reply #34 on: 29/05/2007 16:02:51 »
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.

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?
 

Offline lightarrow

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« Reply #35 on: 30/05/2007 14:17:08 »
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.
In what way is it an invariant mass?
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
In fact, no mass is actually invariant;
What do you mean? The m defined above is invariant.
Quote
but the notion of 'rest mass' derives from the mass that an object would have in the absence of any kinetic energy,
But what do you mean with "kinetic energy"? For a single particle, kinetic energy is the energy due to the particle's movement, but for a system of two particles travelling in opposite directions with total p = 0, it is not: the system has no kinetic energy, just because total p = 0. Kinetic energy is cp.
Quote
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?
But m is the mass of the system in the ref. frame in which the system's kinetic energy is 0; it's only that it's difficult (for me, at least!) to call it "rest mass" for a system of two balls or particles that travels at high speed in opposite directions, you know what I mean?
 

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« Reply #36 on: 31/05/2007 00:52:28 »
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"

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.

If you are in a frame of reference where you have two particles, each approaching you at equal velocity from opposite directions, then clearly the sum of the momentum of the two particles is zero.  But equally, you can choose a frame of reference where one particle is stationary in respect to you (and thus has zero momentum), while the other particle has a momentum:

p' = 2mv1/sqrt(1 – 4v12/c2)

Would not your calculated “invariant mass” be different in this reference frame?

What you do seem to be saying is that it is the “rest mass” of the system (i.e. the mass of the system when the sum of the velocity of the component parts of the system are zero, thus making the system as as a whole appear to be at rest relative to the observer), where the component parts of the system are not necessarily at rest (i.e. where all the component parts of the system may be in motion, but the total system itself is not in motion).

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).
« Last Edit: 31/05/2007 00:58:04 by another_someone »
 

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« Reply #37 on: 31/05/2007 01:03:08 »
Ofcourse, if one follows the logic of the above, could all mass that we see merely be composed of “systemic rest masses” - i.e. could non-massless particles, such as electrons, really simply be systems composed of massless particles, and what we perceive as the rest mass of an electron, merely be the rest mass of the system of massless particles of which the electron is composed, and the component sub-particles within that system remain in motion, even as the system appears to be at rest?
 

Offline lightarrow

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« Reply #38 on: 31/05/2007 13:39:46 »
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"
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 .

Absolutely! But E also changes, and in such a way that (E/c)2 - |p|2 is invariant.
Quote
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.
Yes.
Quote
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).
No, I talk about rest masses of the components, and rest mass of the system. (This is the only universally recognized useful concept of "mass").
 

Offline Mr. Data

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« Reply #39 on: 02/07/2011 15:47:19 »
What is mass?

Mass is a charge, fundamentally speaking, as a Higgs field interacts with a massless particle. Through a process of spontaneous symmetry-breaking, the photon (as an example of a massless particle) would obtain a mass by gobbling up a Goldstone Boson.

But what really is mass?

Mass is a concentrated form of energy. You can measure how compact the energy is if you apply the equation E=Mc2. This is non-relativistic, however, but the rest mass assures us that by using c2 as a scale factor, we can certainly get a lot of energy from a tiny bit of mass: because of this, energy must be a diffused state of matter.

So simply, mass the compact structure of a more concentrated energy - and this energy occupies the intrinsic structure of the particle, providing it with other interesting things, like charge, (or if absent of electric charge) some kind of magnetic dipole.
« Last Edit: 02/07/2011 15:49:32 by Mr. Data »
 

Offline lightarrow

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« Reply #40 on: 05/07/2011 20:46:07 »
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.
 

Offline Mr. Data

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« Reply #41 on: 05/07/2011 23:41:30 »
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.

No, you apply E=Mc2 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=Mc2 is non-relativistic, meaning it does not apply to particles travelling at or near light speed.
 

Offline JP

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« Reply #42 on: 05/07/2011 23:47:03 »
But you would not have said what you did, if you had read my post and understood also what was said. I said E=Mc2 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.
 

Offline Mr. Data

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« Reply #43 on: 05/07/2011 23:51:45 »
But you would not have said what you did, if you had read my post and understood also what was said. I said E=Mc2 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.

This is an overstatement.

We are not talking about light. Light does not even have a mass - E=Mc2 is a relatively good estimate for particles travelling at non-relativistic speeds. In fact, the nature of speed, should not even be an issue here. The question is about mass, what mass is, and E=Mc2 was used to describe it. Now if we had been talking about photons (which I had not been, nor the OP) then the equation to worry about would have been E2=p2c2 + M2c4. But it wasn't needed.
 

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« Reply #44 on: 06/07/2011 00:24:42 »
Uh huh.  It's a good estimate. 

However, E2=p2c2 + M2c4 tells you a more complete story and includes the cases where the particle is moving or the particle happens to be a photon.
 

Offline Mr. Data

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« Reply #45 on: 06/07/2011 00:41:58 »
Why would we want to talk about photons? They contain none of the relevent information to answer the question of what mass is...
 

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« Reply #46 on: 06/07/2011 03:14:00 »
If you want a technical, mathematical explanation, then I will most happily write out some math tomorrow, or the next day and explain how photons and their fields predict the presence of a Higgs field in the presence of some symmetry-breaking. But you will need to know just some basics of partial differentiation for quantum fields, and whilst that part sounds difficult, it's really not.
 

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« Reply #47 on: 06/07/2011 03:15:46 »
Whilst a Higgs field explains the presence of a mass, which is synonymous to a symmetry-breaking in something like a photon field, the photon alone does not answer what mass is - it, and it's field can answer for the presence of the Higgs Boson if you satisfy breaking the symmetries. Of course, however, the Higgs field does not exactly explain why inertia is experienced; all we can surmise from relativity is that inertial mass and gravitational mass are equal. The Weak equivalence.
 

Offline lightarrow

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« Reply #48 on: 06/07/2011 10:09:40 »
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.

No, you apply E=Mc2 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=Mc2 is non-relativistic, meaning it does not apply to particles travelling at or near light speed.
About the concept of mass, I have understood to be quite careful and to make very precise statements, despite its seeming simplicity. You wrote:

"Mass is a concentrated form of energy"

and I reply:
mass is not "a form of energy", mass "is" energy. But that statement is not complete, because we have to add: "fixed a system in space". If it moves, then it's not true anylonger.
examples:
1) I heat a piece of iron which is "not moving in the frame of reference I have chosen", giving it an amount of thermal energy equal to ΔE; its mass increases of an amount Δm. It comes out that Δm = ΔE/c2.
2) That piece of iron has the shape of a spring. I give it the energy ΔE as elastic potential energy by compressing it of the appropriate amount. Its mass increases of Δm.
3) I give it ΔE as any other form of energy. Its mass increases of Δm.
4) I choose to consider a void volume of space. Its mass is zero. Then a beam of light goes through it, for a brief instant. During that instant, the mass of the system increases of Δm = ΔE/c2, if ΔE is the average light beam's energy in that volume of space during that brief instant of time.

It doesn't matter what form of energy ΔE I give to the piece of iron,
It doesn't matter what form of energy ΔE I give to the fixed system I am considering, its mass increases of Δm.
 
Conclusion: mass is not "a form of energy", mass "is" energy, provided the piece of iron or the system I'm considering doesn't move in space.

So, you don't have to explain what mass is, you already have an explanation.

Just to precise your statement.
« Last Edit: 06/07/2011 10:19:38 by lightarrow »
 

Offline Mr. Data

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« Reply #49 on: 06/07/2011 10:21:45 »
Huh?

The word ''form'' does not imply a direct difference. The statement mass is a concentrated form of energy does not imply a difference either. ''Form'' is simply another word for ''structure''. Replace that with ''form'' in my sentence:

Mass is just a concentrated structure of energy.

The use of the word is not implying a difference between mass or energy. It is just as right as saying directly mass is energy and energy is mass. But saying it like the latter has less meaning than my statement, and I will leave it to you to figure out why.
 

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