juanrga

I said that K is the kinetic energy. The symbol K is also standard. Kinetic energy and total energy are two different concepts. For matter one can split the total energy and the total momentum into a kinetic plus an interaction term as correspond to the structure of the Hamiltonian. There is no such split for electromagnetic radiation. It is difficult to believe because in the message that you are replying I wrote: Are you aware that my definition of mass uses components of [math]T^{\mu\nu}[/math]? I have not used the stress components of the tensor, because they do not define the mass of a system. I am using a standard concept of mass which is denoted by m (this is not your m below). I wrote something different: It is only now that you give a precise definition of what you mean by inertial mass. Anyone can ask for clarifications, but attributing to me stuff that I never said is something that I am not going to accept. Thank you very much for your advice and the article, although I doubt that it can be useful for me. Time ago I abandoned the continuum limit approximation and I almost always work with more accurate and fundamental descriptions of nature. Precisely in one or two weeks I will plan to start to write a paper on a generalized energy-momentum-stress tensor [math]\Theta^{\mu\nu}[/math] which includes gravitational effects that are not included in the ordinary [math]T^{\mu\nu}[/math] used in general relativity. But thanks in any case.

You asked me in #20 if I could prove that the mainstream concept of matter used by ajb and me was the usual in the Standard Model. And I proved it in #21 with a link to CERN page devoted to the Standard Model. What is your problem?

Many textbooks give the Hamiltonian for matter plus radiation. CERN website also explain to broad audiences why an electron is a matter particle, but a photon is not a matter particle http://public.web.ce...rdmodel-en.html

Which confirms my point of that the constancy of light speed cannot be "the result of the existence of Lorentz invariance". We can select two observers whose frames are not related by a Lorentz transformation and still the speed of a light signal sent from one to the other is constant and equal to c in general relativity. But "in the limit as an inertial reference frame (IRF) approaches the velocity of a photon" the frame ceases to be localizable and cannot be a reliable reference frame (you cannot define/measure distances in it).

This is not the momentum of an electron in an electromagnetic field. Neither the momentum of Mercury near the Sun (due to general relativistic effects). More of the same. The term "proper mass" for m is not broadly used because it is confusing, and because there is no utility in the definition of two or three different concepts of mass. One is enough and this is named just "mass". I do not know any application/utility of the concept of inertial mass M and both particle physics and general relativity use m. For instance four-momentum is defined using m in modern textbooks on general relativity.

For interacting particles one merely substitutes the total energy and momentum by the kinetic energy and kinetic momentum [math]K^2 - (\pi c)^2 = m^2c^4[/math] For fields there is no kinetic vs. non-kinetic distinction, only the total energy and momentum are defined locally for fields. Therefore don't worry about that. I have not assumed what you say. What I did was to apply the transition to the continuum limit as is done in mechanics to derive the equations of continuum systems. In fact a field is merely a collection of particles (harmonic oscillators). I do not know what you mean by "you assume that inertial mass density equals mass density", because I do not know exactly what you mean by inertial mass (I only can guess). I do not know why you think that I am trying to obtain an "invariant mass density". Where I said or even suggested that a density is an invariant? Nowhere I said such one thing about the density.

My use of the term mass is standard. The term "proper mass" must be a term found in old classical relativistic textbooks, but is not found in modern literature dealing with QFT, particle physics, and the like. Also I do not find any advantage on introducing two or three definitions of mass, therefore, when I use the term mass it has a well-defined standard meaning. I think that you misunderstood me. I know that my m is invariant, this is a reason which I use this concept of mass. But I was replying to qft123. When I replied to him, I was replying to what he wrote about Einstein. See the post #6 where he writes about Einstein (relativistic) mass and about how a massive particle would obtain an infinite (relativistic) mass when moving at c.

It can be shown that massless systems travel at c, whereas massive ones cannot.

No. There systems/processes which do not verify Lorentz invariance and still the speed of light is a constant. General relativity was developed for those systems/processes. Therefore the constancy of light speed cannot be "the result of the existence of Lorentz invariance". There is not such thing as "the frame of reference of a photon".

I forgot to ask before, why do you think that energy flux would contribute to the energy density? In my expression [math]T^{00}[/math], [math]T^{i0}[/math], and [math]T^{j0}[/math] are densities (recall I am using c=1). The energy density consists of two terms: an 'inertial' term (given by mass) plus a 'kinetic' term (associated to motion). This is the same structure that energy has for particles as electrons. When there is not electric field the mass density of the EM field is [math]\rho \equiv \sqrt{(T^{00})^2 - (T^{i0})^2} = T^{00}[/math] which coincides with yours (when reintroducing c).

Precisely photons travel to the speed of light because have not mass (m=0). As you said above Einstein showed that a massive particle with nonzero mass cannot travel at the speed of light because its mass would become infinite (which means that you need an infinite energy to accelerate one electron up to c, and this is impossible).

Yes. Thank you.

The concept of matter given by ajb and myself is the usual in the modern Standard Model. Therefore the context is relativistic. Einstein quote is in the context of classical EMTs and their role in the Hilbert & Einstein equations, which only describe matter superficially, as a continuum.

[math] \sum_{ij} \eta_{ij}T^{i0}T^{j0} = \sum_{ij} \delta_{ij}T^{i0}T^{j0} = \sum_{i} \delta_{ii}(T^{i0})^2 [/math] The expression is well-defined and unambiguous. [math] \delta_{ii}[/math] has only one index, because the second i is the same than the first i. You seem to believe that there are two i, but there is only one. Moreover, if you use [math] \delta_{ii} = 1 [/math] and sum you will get that the right hand side is equivalent to the left hand side. [math] \sum_{ij} \eta_{ij}T^{i0}T^{j0}[/math] can mean both the sum over i and j or the ij element. You know what it means from context. The same happens with [math](T^{i0})^2[/math]. It can mean both the sum over i or the i element. From context, it is evident that [math](T^{i0})^2[/math] did mean the sum over 1...3, because otherwise what i you choose? The 1? The 2? The 3? Your expression only agrees with mine when using the trace -2 convention. Mine works for both conventions. Both are subjective terms. But as said before if you do not like my notation use your favourite one! I prefer the mine which in c=1 units and convention summation (i.e. sum over index i) is [math]\rho \equiv \sqrt{(T^{00})^2 - (T^{i0})^2} = \sqrt{(T^{00})^2 - \sum_{ij} \delta_{ij}T^{i0}T^{j0}}[/math] This also gives the energy in the general case when the electric field is not zero and coincides with the fundamental definition used in particle physics when the above is applied to matter in a quantum context.

You seem do not understand what [math] \delta_{ii}[/math] is. Start with [math] \sum_{ij} \eta_{ij}T^{i0}T^{j0} = \sum_{ij} \delta_{ij}T^{i0}T^{j0} = \sum_{i} \delta_{ii}(T^{i0})^2 [/math] Now, using [math] \delta_{ii} = 1 [/math] [math] \sum_{i} \delta_{ii}(T^{i0})^2 = \sum_i 1 (T^{i0})^2 = \sum_i (T^{i0})^2 [/math] If you prefer to work with the summation convention the starting point is [math] \eta_{ij}T^{i0}T^{j0} = \delta_{ij}T^{i0}T^{j0} = \delta_{ii}(T^{i0})^2 [/math] Using again [math] \delta_{ii} = 1 [/math] you obtain what I wrote [math] \delta_{ii}(T^{i0})^2 = 1 (T^{i0})^2 = (T^{i0})^2 [/math] But if you do not like my notation you can use your favourite notation, including your original [math] \sum_{ij} \eta_{ij}T^{i0}T^{j0} [/math] Although, the mine is less ambiguous. The yours only gives the correct value when using the trace -2 convention. Whereas my notation is valid with independence of the trace convention. My notation is also simpler and you would understand that the summation convention was invented to write less stuff and simplify the notation.

MOND goes wrong only when it is not understood or when it misapplied beyond its scope. MOND was born from phenomenological (as most of laws of physics) but MOND is now derived from first principles. Several theories derived MOND. Moreover, MOND does not really use any new constant to fit observations (Milgrom's constant is related to c and Hubble H).

[math] \sum_{ij} \eta_{ij}T^{i0}T^{j0} = \sum_{ij} \delta_{ij}T^{i0}T^{j0} = \sum_{i} \delta_{ii}(T^{i0})^2 = \sum_i (T^{i0})^2 [/math] Using summation convention, this can be rewritten in a simpler form [math]\eta_{ij}T^{i0}T^{j0} = \delta_{ij}T^{i0}T^{j0} = \delta_{ii}(T^{i0})^2 = (T^{i0})^2[/math]

It depends of the definition of matter you are using. Many people distinguish between matter and radiation, with photons belonging to the latter (i.e. they consider that photons are not matter). See e.g. http://www.pd.astro.it/E-MOSTRA/NEW/A4000MAT.HTM Personally I consider that electrons, quarks... are matter but photons, gravitons... are not.

Python seems to a kind of fact 'standard' PL for biology applications. Try this http://osq.cs.berkeley.edu/public/EChang-SystemsBiology.ppt and this http://www.ncbi.nlm.nih.gov/pubmed/18647734 to get perspective.

You would be also familiarized with the associated eigenvalues and their measurement. Yes, but the point was that time is not an operator and, thus, expression [math]\Delta{E}\Delta{t}>\hbar[/math] is not a Heisenberg uncertainty relation, but something with a different physical meaning. I quoted the relevant parts explaining why what you said about the measurement of energy and the time of measurement was false. I can cite other parts. For instance, the phrase just after that last that I cited above explains that is not the measurement time for excited states, as it seems you still believe:

The second law of thermodynamics states that the average entropy of an isolated system cannot decrease over time. Brownian motion is a kind of dissipative motion in complete agreement with entropy and the second law. In fact the Brownian motion equations can be obtained from an entropic analysis.

Search the concept "energy eigenfunction" in some textbook. The Heisenberg uncertainty relation is for non-commuting operators. Time is not an operator.

[math]\eta_{ij}T^{i0}T^{j0} = \delta_{ij}T^{i0}T^{j0} = (T^{i0})^2[/math]

Hint: I'm staying out of this thread for most of it. But consider the following regarding your expression [math]\rho_i \equiv \sqrt{(T^{00})^2 - (T^{i0})^2}[/math] Thank you by the hint, but mass density is a scalar. There is no index i on scalars.

Kruskal & Szekeres coordinates