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Author Topic: Can we measure with a balance the relativistic increase of mass with velocity?  (Read 9698 times)

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.
 

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.
 

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.
 

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?).
 

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?
 

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.
 

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.

 

Offline flr

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

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).
 

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.
 

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?


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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?
 

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?
 

Offline yor_on

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

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.

 
 

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.
Quote
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.
Quote
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.
 

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.
 

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.
 

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?
 

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.
 

Offline flr

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

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.

Quote
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.
 

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