how can mass be converted into energy?

Of course not. AND not what I said was your strange POV.

You don't want to consider anything other than particles as having mass - that is what is so strange and it is wrong.

No that released energy does not come from "extra baggage." It comes from the mass of the nuclear assembly not associated with the particles. The amount released is that non-particle mass times C^2. Keep up this very strange POV and I will be forced to instead state you are posting non-sense again. I have been kindly just calling it a "strange POV."

 

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Yes the binding energy is also mass.

MetaKrons's problem {and many others also} is that he only wants particles to be considered as having mass. (Why in face of the facts to the contrary, I do not know.) At best he will only allow that it is "technical possible to call the binding energy mass."

Physics does not, however, conform to "MetaKron's wants." Mass is mass in any of its various forms. I.e. if there is gravity produced by what we often call energy then that form of energy is mass as well as energy. Nuclear binding energy is but one of many examples. But it is important to understand that not all energy is mass or making gravity. - See post 15 to better understand via by my simple rule.

 

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I think you picked up a false perception somewhere. Maybe I should have said something like "invariant mass." I am painfully aware that photons have relativistic mass and that nuclear mass defect in particular registers on the scales as having mass.

 

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It's always "someone's problem" with you, Billy.

I meant like converting atomic matter into energy, other than the nuclear mass defect that accounts for the energy generated by fission and fusion.

 

You really do not understand this at all. Photons have energy. They do not have mass of any form. They make no gravity. Go back and read my post 15 until you have some slight understanding of all this.

 

You have just made it clear that you don't understand anything that you have been saying, and it is now considered scientifically correct to say that photons have "relativistic mass."

 

Actually most modern physicists avoid using the "Relavistic Mass" term. They speak instead of momentum only.

As to the root question here I think it is easiest to think of energy as unbound and flowing, producing heat or becoming kenetic. Mass can be thought of as bound energy or space compacted by the ratio of E=mc^2 or m = E/c^2 where energy is confined to relavistic swirls or harmonic standing waves that we call particles.

This of course is strictly the McCoin view and is not one recognized or advocated by modern science but that I find useful myself and perhaps others might also..

 

Hi MacM. Good to see you here again. While you have a "crackpot's POV, you do know what you are speaking about and defend it well, unlike MetaKron

You are correct on this at least. Only a few of us "old timers" are still addicted to our "relativistic mass" and even we are not very confortable with mass then being a vector instead of a scalar. I do not think anyone claims photons have relatistic mass - That is silly nonsense as then that mass would be different in each reference frame - You can not take many things MetaKron states very seriously - I just try to keep the facts straight.

 

Now you are arguing convention rather than fact. Are you drunk on sugar cane alcohol, Billy? I guess that would be rum.

When you want to accuse someone of a crackpot view be sure that it is actually a crackpot view instead of the use of a naming convention that is still used by mainstream science for mainstream ideas. Also, grow up.

 

No, not entirely as I cannot do the GR needed to be sure. Because photons have the same speed in all frames and yet have different energy in different frames (red shift etc.) they cannot have any single "mass" by the normal E = MC^2 = h F where F is the frequency, different in different frames, and the h needs a bar thru it I think, but you understand what my point is I am sure. (I am too lazy to learn how to write it properly with mathtext etc.) Thus it seems non-sense to me to assign any mass to a photon.

In my POV and with my simple rule for deciding when energy makes gravity (see post 15) photon's energy is like the KE of a bullet. - It is “frame dependent” and thus NOT a contributor to mass as are rest mass and thermal energy, both of which are the same in all frames. (In photon case rest mass = 0 so there is no mutual gravity attraction. – Also see, in post 15 I think, my “stars seen by both eyes are the same” argument suggesting that photon rays from distant stars have not mutually attracted to coalesce into fine filament beams that can only enter one eye, not both.)

If you think this logic is wrong, please tell where it errors and how to describe mass that is frame dependent. - That is one of the main reasons why linear KE's "relativistic mass" is not much accepted anymore. – Not even by me, but I still like my "relativistic mass" when it is the rotational energy as in a cyclotron - that I think does look the same in all frames.
Also as qvB = ma and the acceleration, a, in a fixed circular path with circumferential speed being essentially constant at C and q also fixed, there are only two variables in this equation; (B and m).

It is a fact that as the cyclotron gives energy to the particle it is accelerating that B must be increased "very much" to keep the same radius circle. Ergo m must be also increasing “very much” to keep the equation true and fact that v is not exactly the constant C is unimportant compared to “very much”. This is why, at least in the cyclotron case, this "old timer" still likes his "relativistic mass." No doubt there is a more complex set of equations describing the way a cyclotron works that can kill relativistic mass concept even in the cyclotron - but if so, it is beyond me and not very satisfying compared to my beloved qvB = ma. (The vB is of course VxB in the old "bold for vectors" notation we "old timers" used.)*

Let me turn the tables:
Are you sure that ALL energy makes gravity?

I assert that linear kinetic energy does not. (Perhaps the "linear" restriction is not needed, but let’s keeps it simple.)
For example:
Imagine two bullets traveling side-by-side in perfect vacuum in gravity free space with same vector velocity only 0.1 mm apart. You think they gravitationally attract each other by more than their rest mass gravity? (I do not.)
Do you think that the time until they touch is shortened by this extra "KE gravity"? (I do not.)
Obvious the time until touching in your POV will be function of which frame the clocks are in. Note this motion towards each other is orthogonal to the velocity so I think you will have hard time defending your POV with the fact that time is relative. For example all frame observers have agreed to project the separation of the bullets on to the Earth's frame. In the Earth frame, in your POV, do they touch at any and all times you like? This seems crazy to me.

Again are you sure there is extra "KE gravity"?

----------------
*If had been "king of physics," back when GR was first being taught to all, tensors notation would have been with italics so life would not require so much thinking about what repeated subscript sums etc. are doing to the equations. Everything would still be explicit and easy to follow. Some mathematical geniuses, like D.H., probably like the "beauty" of such compact notations, but us simple people like to see the operations written out explicitly.

 

I do not think MacM minds me calling him a crackpot. - I often have and usually add some praise when doing so. Physic needs more crackpots like him. -Very logical and consistent. -Who knows, he might even be correct and most of us be wrong. No need to worry about that in your case.

Also I think MacM knows that I have called myself a crackpot in the field of cognitive science. I consider being a crackpot with well defended alternative POV consistent with (Even supported by facts not well explained* in my case by the accepted theory - For example why is your visual perception equally sharp everywhere in the field of view, when only the fovea has that high resolution is impossible for conventional theory of perception as the "emergent transform of retinal data to explain, but natural a of my crackpot POV) to be a position to be proud of. MacM is and so am I (proud to wear the crackpot hat).
------
*I do not think MacM has any such strong evidence favoring his POV above the accepted one, but he can fit a lot of observations (Dark energy expanding the universe etc.) into to it very consistently and naturally. He does have many experiments he has done, but as far as I know no-one has independently confirmed his supporting experiments. If you read how he did them, you cannot help but admiring him as clever.

PS I may need to stop replies to you as someone I respect greatly (James R) has entered the discussion and is taking my available time.

 

I think the box would have measurable mass.
You could measure the mass of the box by accelerating it: apply a given force, and measure the resulting acceleration.

A box with photons inside will accelerate less than an empty box, because the photons have momentum. They will bounce harder off the back wall than the front wall during acceleration, slowing the box down.

I suspect that equating the change in the photons' average momentum with \(\gamma m_0v\) should lead to \(E=m_0c^2\)

 

Is the gravity produced naturally or by a g-force centrifugal generator which is the cause of the annihilation of the imagined binary tron's ? An ie. is the spiraling white dwarf which exerts an unknown amount of pressure on external objects. Your mass is altered by disintegration.

 

Applying my simple rule (see post 15) I think "yes" to both because the energy inside the box is the same in all frames and in all frames is the same after the annihilation as before. (The two gammas travel in exactly opposite directions so the red shift of one exactly compensates for the blue shift of the other, keeping the total photon energy 2x0.511MeV if memory is serving me correctly.)

You did not state that the box was in an external gravity field but even if it is this still is true - the energy loss by the photon "climbing up the gravity well" is compensated by the gain of the one falling down in it.

We are however back with your "bird in the truck" problem. (Sorry to bring that up for reasons I am sure you know

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) Unless the CoM of the charged particles was "half way up" inside the box there will be "jitter" in the weight of the box as first one then the other hits the top and bottom of the box.

As usual, Pete is correct. The momentum gain by the by the photon reflecting off the "trailing side" of the accelerating box with each reflection will be greater than the momentum loss when one reflects off the "moving away" front-side mirror. Here two, there will be "jitter" in the observed inertia, but not in the long term average.

The case of neither reflecting off either top or bottom for some time is slightly interesting in that it is independent of how close to one side the annihilation is. I.e. one near one side will not fall as much as the other. (Falling either due to real external gravity or the indistinguishable acceleration.) However, when both have traveled the width of the box and each has had one reflection their cumulated falls are equal. Thus if initially traveling level (in the plane orthogonal to the G force. They hit the bottom simultaneously. I am not sure they need ever hit the top (If G big or box wide) - I think they can just repeatedly hit the bottom to make their inertia externally felt on the long term average.

 

Wouldn't this be because of a weakness (inherent error) in the method you are using to measure mass? Momentum shouldn't have any affect on rest mass, and by extension, gravity. In James R's example, both the electron and the positron would bounce off the same wall if the acceleration was done before annihilation, leading to a different mass measurement than in your example. You could also place an empty box in the vacuum and accelerate it. It would be harder to accelerate the box toward the sun (a source of photons) than away from it, thus measuring two different masses for the same box. Besides, you could theoretically make your box of photons long enough that the photons don't reach a wall during the measurement.

 

The reliability of the method would rely on there being many collisions between the particles/photons inside and the walls of the box during the measurement.
It would also rely on there being no biased external influences (like wind).

That's pretty much the question under discussion.

 

You have to measure momentum from a frame other than the rest frame of the object being measured. By acknowledging the effects of relative velocity, the momentum can be eliminated from the rest mass.

Not as I understand the question. The question is does energy have the same gravity as an equivalent amount of rest mass? The gravity would have to be measured from a distant location, of course.

 

If you know momentum and velocity, you know rest mass.
\(p = \gamma m_0 v\)
\(m_0 = p/{\gamma v}\)

That's the other question, which I haven't addressed. James asked two.