CraigD Posted July 29, 2012 Report Posted July 29, 2012 Bosons can be described as “force carriers”, fermions as what they carry the force between.I disagree. Just because some bosons are force carriers it doesn’t mean that all of them are. Alpha particles certainly aren’t force carriers. Bosons and fermions are defined in terms of the symmetry or antismmetry of their wave functions. They are also defined by the eigenvalue of the exchange operator.I agree with your disagreement. :) I was taking substantial liberties with my working in the interest of giving simple, at the expense of very accurate, answers to the OP’s question. In this case, to avoid a complicating adjective, I wrote “bosons”, where I should more correctly have written “elementary “ or “gauge bosons”. The concept I’m trying to convey is of the elementary particle of the Standard Model being somewhat neatly divisible into those that “are matter” - fermions, and those that interact with them and “carry” interaction between fermions – bosons. Even this is substantially over-simplified, in that the gauge bosons interact, and, except for the photon and z boson, self-interact. “Matter is made up of energy” is a rough way of describing mass–energy equivalence,..I disagree. In my opinion one should never think of matter as being made up of energy. There is merely a relationship between mass and energy. That relationship has to do with two things (1) if a body decreases its energy content by [math]\Delta m[/math] then its mass decreases by the amount [math]\Delta E[/math] where [math]\Delta E = \Delta m c^2[/math]. (2) If a particle has proper mass m then there it has proper energy E where [math]E = mc^2[/math]. This comes into place in energy conservation equations where particles are being created and/or destroyed.I agree. By “a rough way”, I meant that “matter is made up of energy” is alluding to mass energy equivalence, not that it’s a very good natural language statement. While we should be careful to suggest egregious misconception like “all matter is made of ‘frozen’ photons” (one I often see go unchallenged in bad science blogs and forums), I think there is nonetheless some sense to the view that “matter is made of energy”, because, if the hypothesis that all the universe’s mass-energy arose from a large scale fluctuation of the “sea” of virtual particles that the vacuum can be described as, virtually, containing, where these particles became real, is correct, it’s reasonable to say these real particles “came from” the vacuum energy. Natural language and metaphors are fraught with peril of formal inaccuracy, but they’re the main tools most, possibly all (as I doubt that even the best educated and smartest physicists think only in theoretical formalism), of us have to steer us toward the best theoretical understanding of physical reality. Coining and promoting the best metaphors is a big part of promoting and popularizing science. It's also funny how you may think of density. Density in our dimensions we use everyday has dimensions of [math]\frac{M}{\ell^3}[/math] which is mass over some length cubed. But you never see an energy density have the form [math]\frac{E}{\ell^3}[/math] because that has the wrong dimensions.As I’m acquainted with the term, and as its defined by the first sentence of its wikipedia article, energy density does have dimension [math]\frac{E}{\mathbf{L}^3} = \mathbf{M}\mathbf{L}^{-1}\mathbf{T}^{-2}[/math]. Practically, though, especially in engineering, energy density is used interchangeably with specific energy, which has dimension [math]\frac{E}{\mathbf{M}} = \mathbf{L}^2\mathbf{T}^{-2}[/math] I don’t see any dimensional problems here, just some mildly imprecise common language. So when physicists talk about an energy density over some volume, we surely don't mean it to have the same kind of Newtonian density.I believe we surely do – though when physicists talk about energy density, they’re often talking about virtual kinds like vacuum energy, which is pretty far divorced from Newtonian physics. Quote
Aethelwulf Posted July 29, 2012 Report Posted July 29, 2012 (edited) As I’m acquainted with the term, and as its defined by the first sentence of its wikipedia article, energy density does have dimension [math]\frac{E}{\mathbf{L}^3} = \mathbf{M}\mathbf{L}^{-1}\mathbf{T}^{-2}[/math]. Practically, though, especially in engineering, energy density is used interchangeably with specific energy, which has dimension [math]\frac{E}{\mathbf{M}} = \mathbf{L}^2\mathbf{T}^{-2}[/math] I don’t see any dimensional problems here, just some mildly imprecise common language. I believe we surely do – though when physicists talk about energy density, they’re often talking about virtual kinds like vacuum energy, which is pretty far divorced from Newtonian physics. Interesting... I never knew energy density was defined as such :) I guess we all learn something every day :) I guess it just seems ''iffy'' how one can associate M/V^3 to a density to one which is E/V^3... Oh well. Edited July 29, 2012 by Aethelwulf Quote
Aethelwulf Posted July 30, 2012 Report Posted July 30, 2012 Have I done something wrong in post 17... just doesn't look right for some reason... Quote
Pmb Posted July 30, 2012 Report Posted July 30, 2012 Here's a chicken egg scenario... is mass the presence of curvature or does curvature create mass?In my humble opinion, mass is the source of spacetime curvature, not the other way around. In any case they co-exist so I don't know how to prove it, at least not now. Taylor and Wheeler: ''The concept of 'relativistic mass' is subject to misunderstanding. That's why we don't use it. First, it applies the name mass - belonging to the magnitude of a 4-vector - to a very different concept, the time component of a 4-vector. Second, it makes increase of energy of an object with velocity or momentum appear to be connected with some change in internal structure of the object. In reality, the increase of energy with velocity originates not in the object but in the geometric properties of space-time itself.''I strongly disagree with this statement. First off, obody who has carefullt studied relativity and the concept of relativistic mass has been confused about it. It's a very simple concept. TWs second point about it being the time component of the 4-momentum is not quite right. It's quite reasonable to define the time component as relativistic mass. Several physicists do in fact. Their next objection about it being a change in internal structure of the object also isn't true. Anybody who carefully studies it carefully knows this. After all those who study relativity caefully never make the mistake of thinking that the slowing of clocks has something to do with changesin its internal structure either. Then again there might just be a few people who get it confused, but that happens with time dilation too. Once someome learns it then there's no real problem. If increase of energy of a system originates from the properties of space and time, then perhaps mass is a type of geometrical property as well; keep in mind that mass is the presence of curvature within the standard equations of relativity. So what does my own equations have to say about this? Let's go back to them: [math]8\pi \rho_0 (\frac{G}{c^2}) = \frac{GM^2}{Mc^2}[/math] Where did you get this from? I don't know what it means. Quote
Aethelwulf Posted July 30, 2012 Report Posted July 30, 2012 Where did you get this from? I don't know what it means. I know how you feel about using the term relativistic mass so I won't debate that. Anyway deriving the above equation is relatively simple. First of all, take [math]\hbar c = GM^2[/math] 1] which is a famous relationship. Divide through by [math]Mc^2[/math] gives [math]\frac{\hbar}{Mc} = \frac{GM^2}{Mc^2}[/math] 2] The left is recognized as the Compton wavelength and the right is recognized as the Schwarzschild radius, however, I also note before any kind of simplification [math]GM^2[/math] is the square of the gravitational charge. What Lloyd Motz has done is equate the proper mass density relationship to the Compton wavelength by setting them equal [math]8\pi \rho (\frac{G}{c^2}) = \frac{\hbar}{Mc}[/math] 3] http://www.gravityresearchfoundation.org/pdf/awarded/1971/motz.pdf If you go back to the relationship derived from the famous quantization condition in 2] you can simply replace the right hand side by the new form [math]8\pi \rho (\frac{G}{c^2}) = \frac{GM^2}{Mc^2}[/math] Now, you won't have seen this equation because I am the one who derived it. Quote
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