Can we measure with a balance the relativistic increase of mass with velocity?

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Offline flr

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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?

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In other words. The energy that goes into the heated object is:
E=sqrt [ (pc)^2+(m_rest*c^2)^2 ] = m_relativistic*c^2

What will the balance measure: the m_rest or m_relativistic ?

« Last Edit: 18/03/2013 19:57:22 by flr »

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Offline lightarrow

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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^2

What will the balance measure: the m_rest or m_relativistic ?


(If I remember well, the correct term should be "scale").
If the object is stationary in the scale's frame of reference, it's simply mass (not relativistic mass) which increases.
You could say that it should be relativistic mass because the particles move. But the particles' masses don't vary, actually (their relativistic masses do); the system's mass, however, *is not* the sum of the masses of the particles (mass is not additive).
« Last Edit: 18/03/2013 20:38:57 by lightarrow »

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Offline flr

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If 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 ? 

Quote
the 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?

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Offline yor_on

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The kinetic energy of the particles increase, and as energy is a equivalence to mass you should see the (locally measured, also being 'at rest' with what you measure) weight increasing. When it comes to relativistic mass I think I can see how you think, and myself I'm wondering what 'potential energy' means in a wider perspective, and if one can use that description too, to describe a relativistic mass?

Because a 'potential energy' is a energy, not found/measured locally, although existing when colliding two objects for example. A potential energy describes a relation, does a relativistic mass do the same? Or is there a circumstance in where I can measure it locally? It becomes a quagmire when you ask yourself how to define being 'at rest' relative scales. Also relative uniform motion. I prefer Plank scale for it, but? That's a ideal definition of a 'frame of reference' relative any other. And if you use that one, how would you define something as being 'at rest' with something else?
=

Thinking of it. You have 'gravity' of course. That's a real difference, but it doesn't answer what scale I should use to ideally define a 'frame of reference' relative another. Or maybe it does?
=

Like this :)
At what scale do a gravity cease to exist for a mass, locally measured?
Does it ever?

ahem, back to your question ::))
*Sorry.
« Last Edit: 19/03/2013 19:04:35 by yor_on »
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Offline lightarrow

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If 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.
If the object (centre of mass) is stationary with respect to the scale's frame, it doesn't matter if the atoms inside move or not: any increase in the object's energy will be measured as an increase in its mass (the one you call "rest" mass but it's simply called "mass"; if you want to compare it to relativistic mass, I prefer the term "invariant mass" instead of "rest mass", because, for example, photons don't have any frame of reference in which they are "at rest", but they however have an invariant mass).
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Will the scale indicate the same value at 200K and at 900K ? 
No, as I wrote, the scale will indicate a greater mass at 900 K
Quote

Quote
the 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?
I wrote it. It's *not* additive.
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2) is there a simple intuitive reason why masses are not additive?
Example: you have an electron and a proton in a stationary hydrogen atom at the fundamental level. You measure the atom's mass and you find the value Ma. Then you separate the electron from the proton, you measure the electron's mass when it's at rest and you find the value me, the same with the proton and you find Mp, but Ma is not equal to Mp + me, because in the atom there is also the binding energy (in this context it's not important if this binding energy is positive or negative).

The total mass of the atom Ma is its total energy Ea divided c2: Ma = Ea/c2, but in the compute of its total energy Ea there are *not only* the rest energies of the masses me and Mp, there also is the binding energy.

Every time two particles (or bodies, or objects) A and B can form a bound state AB, the mass of the system AB is not equal to mass(A) + mass(B).

In chemistry these differences are actually very small and irrelevant; in nuclear and subatomic physics, not, because the binding energies are greater.

What is additive is energy.
« Last Edit: 19/03/2013 19:18:04 by lightarrow »

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Offline yor_on

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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?
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Offline Pmb

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Quote from: flr
The object is stationary relative to scale's frame.
The atoms making up the whole object can move due to thermal motion.
The increase in energy will cause an increase in the rest mass of the body. If the body was moving then the weight of the body would be W = Mg where M = relativistic mass.

Now consider an air hockey table. First let the pucks be at rest and measure the weight of the table. Call that weight w = mg where m = intial rest mass of table. Now add energy to the table such that the energy goes into the kinetic energy of the now moving hockey pucks which we'll consider to be part of the table. The moving pucks weigh more than the same pucks at rest so that the entire weight of the table increases. The new weight will be W = Mg where M is the new mass of the table/hockey puck system. M > m.

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Offline simplified

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Kinetic energy without counteraction doesn't create additional gravitation.

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Offline lightarrow

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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?
What you can do is to prove that the quantity:

sqrt[(E/c2)2 - (p/c)2]

is invariant...

...and your problem is solved.

The answer is positive  [;)].

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Offline yor_on

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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.

Mass in special relativity 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)."

But I'm not sure, as always :) Because, although a definition using a idea of invariant mass, as stating that it must be invariant under time, works splendidly for me it then seems to come down to QM.s definitions of particles and their 'life times' too?

"Updated by Don Koks, 2012.
Original by Philip Gibbs and Jim Carr, late 1990s."

 comment on it this way.

"When particles are moving, relativistic mass provides a very economical description that absorbs the particles' motion naturally.  For example, suppose we put an object on a set of scales that are capable of measuring incredibly small increases in weight.  Now heat the object.  As its temperature rises causing its constituents' thermal motion to increase, the reading on the scales will increase.  If we prefer to maintain the usual idea that mass is proportional to weight—assuming we don't step into an elevator or change planets midway through the experiment—then it follows that the object's mass has increased.  If we define mass in such a way that the object's mass does not increase as it heats up, then we will have to give up the idea that mass is proportional to weight."

What is relativistic mass?
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Offline yor_on

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This one "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)"

Seems as a circular logic, to me? It assumes a invariant mass to exist, and as all particles then by definition have it, we can presume that any 'system', no matter how separated in time and space must, by definition, have it too?

I like invariant mass, it makes sense to assume that a particle have a defined 'rest mass' even under translations by 'gravity' 'motion', as well as 'energy' in general. But 'relativistic mass' do make sense to me too.
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Offline yor_on

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Maybe you could define it such as a 'invariant rest mass' indeed is a invariant, even to time? Although a relativistic 'rest mass' use the notion, indirectly, when for example defining a acceleration, or 'uniform motion'? Einsteins definition isn't one of three spatial dimensions being separated from a temporal. To me it's a whole expression, where the spatial and temporal parameters have a balance relative each other. So for example, imagining a 'space' without a arrow shouldn't be possible from such a definition.
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Offline lightarrow

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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?
No. If it is also "time-invariant" = constant, it means the rest energy is conserved. But this can be or not, in general, depending on the situation; for example, if you heat a stationary piece of iron, its mass is not constant and so it's not time-invariant.
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It's not so much the question of if it really will weight more that confuse me, but how that time component are used.
Mass in special relativity 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)."
What it says here is simply this: if the system's centre of mass (or, more generally, the "centre of momentum") is *always zero*, whatever happens (for example a set of some particles could disintegrate in another set of totally different particles) then the system's energy can always be identified with its invariant mass (do you remember when E = mc2 is valid?).
Now, since the system's energy is conserved, because we assume an isolated system, then its invariant mass is conserved too. "Conserved" means that it doesn't vary after the reaction, so it doesn't vary with time. But "invariant" and "conserved/constant" are different concepts, in general.

Hope to have addressed your question.

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lightarrow

« Last Edit: 21/03/2013 18:54:42 by lightarrow »

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Offline yor_on

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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? If that is how you see it I think I agree. My own thoughts on it is that no systems go free from time in this universe, also that I find it very hard to find any closed systems anywhere in nature, and the universe. But as I said :) I may miss your point there.
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Offline Pmb

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Kinetic 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.
« Last Edit: 22/03/2013 18:48:00 by Pmb »

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Offline lightarrow

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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?
*Which* trasformation? Certainly not Lorentz transformation.
Quote
But not that it is invariant under time?
It would be simpler if you used, as it's usually used in physics, the term "invariant" only for Lorentz transformations and the term "constant" to mean that it doesn't vary with time. As "subset" of the concept "constant", physicists sometimes use the term "conserved" to mean that it doesn't vary afyer a reaction; for exampe: "momentum is conserved in particles collisions".

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Offline flr

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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?
The imaginary box is at rest relative to scale, but the photons inside the box are not because they are going back and forth at c.

In this experiment, the energy of the trapped photons E=pc should be 'seen' by my scale as a rest mass of the box, right?



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Offline Pmb

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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?
Yes. There is an interesting article on this in the American Journal of Physics. The mass of a gas of massless photons, H. Kolbenstvedt, Am. J. Phys. 63(1), January 1995.

Would you like to read it? I can e-mail you a copy. I have permission to do so from the editor of that journal.

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Offline flr

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

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Offline flr

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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?
Yes.

Then my question is: The mass measured by the scale is the rest mass of the box trapping photons, is that so?

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Offline simplified

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Kinetic 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.
You can't explain something without counteraction.Energy of satellite creates gravitation to stranger.You can add much energy to the satellite, but additional energy doesn't create gravitation,because it have no counteraction. :P

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Offline flr

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If 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?

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Offline simplified

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If 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?
Can you create counteraction to the energy?

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Offline CD13

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Can anyone explain what exactly attraction and repulsion are? Whether it's charge or gravity. Why should what we call a positive charge move towards a negative charge? 

 


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Offline yor_on

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No, transformations of energy, inside a 'system' labeled 'closed', or in any way 'constricted' by your premises, limits and simplifications for getting that clean presentation. Lorentz transformations is not what I had in mind there :) I mean, the universe has to build on some simple concepts, to please me at least :) And conservation of energy makes sense from that view, although you can discuss it from GR. But from a view of locality, defining it such as you always have a measurable reality making sense, I think it should hold, although I'm never sure :)
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Offline Pmb

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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?
Yes.

Then my question is: The mass measured by the scale is the rest mass of the box trapping photons, is that so?
Yes.

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Offline Pmb

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The abstract is interesting. Unfortunately, I don't have full access to it.
I can e-mail it to you if you want. If you don't want to give me your e-mail address I can upload it onto my website until you're able to download it for yourself. Again, I have permission to do this from the editor.

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Offline lightarrow

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If 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?
No, because its particles will fly away, over a certain temperature.

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Offline lightarrow

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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?
Yes.

Then my question is: The mass measured by the scale is the rest mass of the box trapping photons, is that so?
More precisely, it's the rest mass of *the system*, which consists in a box and photons trapped inside. You can "see" this electromagnetic field trapped inside the box as the electromagnetic field which binds the electron in the H atom (do you remember the example I wrote some posts ago?).

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Offline flr

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So, 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?

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Offline lightarrow

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So, 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?
They would form a black hole.

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Offline yor_on

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CD13.

"Charge conservation can also be understood as a consequence of symmetry through Noether's theorem, a central result in theoretical physics that asserts that each conservation law is associated with a symmetry of the underlying physics. The symmetry that is associated with charge conservation is the global gauge invariance of the electromagnetic field. This is related to the fact that the electric and magnetic fields are not changed by different choices of the value representing the zero point of electrostatic potential. However the full symmetry is more complicated, and also involves the vector potential. The full statement of gauge invariance is that the physics of an electromagnetic field are unchanged when the scalar and vector potential are shifted by the gradient of an arbitrary scalar field."  Charge conservation   

A Test of the Charge Symmetry Hypothesis (1953)

And it was Einstein that coined that idea for good.

"Symmetry is now recognized as providing the basis for many of the most fundamental laws that determine the physical behaviour of the universe. So, for example, the fact that the universe is symmetric under translation in time (implying that experiments give the same results today as they did yesterday or will do tomorrow) underpins the conservation of energy. Similarly, the breakdown of symmetry accounts for some of the most interesting of physical phenomena. Our own existence, for example, may only be possible because of a breakage of the fundamental symmetry between matter and antimatter in the early universe."

And that is what I mean by 'invariant'. If you have something able to transform but keeping a constant 'value' defined as some minimalistic description of what it is, normally 'energy', then it becomes 'invariant' to me. But when you discuss mass it becomes a lot more trickier. The 'invariant' part here is our idea of there being some irreducible part of rest mass, that will exist under all transformations. Lifting in a Lorentz transformation nothing can be 'invariant' as the description of position and time will shift, depending on coordinate system aka 'observer' according to Einstein. But to me that also is building on a presumption of us having a common 'same universe', that we just need to 'correct' under relativistic regimes. And I think that is not the whole truth.

We communicate, so in that matter we're sharing it. We touch each other so we know we exist together, but, there has to be another way to describe it.

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Offline flr

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

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Offline lightarrow

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If all photons move parallel, will the system still have rest mass?
You mean parallel and with the same sense of motion? No, in that case the system will have zero invariant mass (as you know I don't like the term "rest" mass).

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Offline Pmb

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If all photons move parallel, will the system still have rest mass?
Yes. All yhou have to do is figure out whether there is a zero momentum frame of a system. If there is, go to that system and calculate the energy of that "closed" system (if it's not a closed system then it won't work). Then divide by c^2 to get the invariant mass of the system.

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Offline flr

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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.

If the 2 photons move at any non-zero angle then the invariant mass is non-zero.
When the 2 photons are at 180 degree (move in opposite directions) the invariant mass is maximum.
What is the wave representation of 2 photons moving in opposite directions? 2 waves canceling each other? Does the question makes sense?


------------------


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?

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Offline yor_on

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Let us make it simple.

In a black box experiment, do you assume that the angles, as directions,  of it will make a difference?
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Offline yor_on

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How will you define a 'zero momentum frame' to something never at rest?
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Offline flr

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But nonzero c*p of the individual particles can be seen as a rest mass when total p (of all particles) is zero relative to the measuring scale.
That should apply to photons as well if they are boxed and the box does not move (relative to the scale).

A photon moving in a circle (if possible) probably should produce non-zero (rest/invariant mass) reading into a scale that does not move relative to the center of the circle.

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.

 

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Offline lightarrow

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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.
Yes. Notice a subtle concept, however: which is *the system*. In the case you discuss, the system is the set of two (or many) photons.
But the situation would be completely different if you considered the stationary region of space which is traversed by the light beam: it would have non-zero invariant mass.
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2 photons moving parallel in wave representation will simply be a wave like the initial one but of double amplitude.
Unfortunately it's not (usually) possible to give a wave representation for a single photon or two of them.
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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?
Yes.

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Offline lightarrow

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Let us make it simple.
In a black box experiment, do you assume that the angles, as directions,  of it will make a difference?
No because in this case the overall momentum is however zero. It makes a difference if the overall momentum changes.

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Offline lightarrow

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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.
There's essentially no difference between this case and the one with photons in a box: the average overall momentum is still zero.
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I find it interesting and surprising that it could be convinced geometries/frames where p*c actually contributes to the rest mass.
You simply have to realize that mass is nothing else than energy in a stationary frame.

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Offline flr

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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?

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Offline lightarrow

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Don't know, but I'm quite ... excusable in this case, since there isn't any (recognized) quantistic theory of gravity, yet.

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Offline flr

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Let's ignore QM.
What Einstein general relativity (non-QM) would predict in this case?

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Offline Pmb

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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.
You can define it. It's zero.

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?
In each case each photon is deflected. Note that in such a gravitational field the spacetime curvature is zero.

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Equal, or the one for which we can define rest mass will interact stronger?
The same amount of deflection of each photon in both cases. The inter-photon interaction can be ignored. Each photon is deflected by the same amount of the photons are moving perpendicular to the gravitational field lines.