What is the experimentally measurable difference between rest mass and the 'relativistic mass' of the photon ??!!
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From: Koobee Wublee on 26 Apr 2010 20:06 On Apr 26, 12:04 pm, PD <thedraperfam...(a)gmail.com> wrote: > On Apr 26, 1:32 pm, Koobee Wublee wrote: > > It sounds like this so-called invariant mass is indeed your rest mass > > as indicated by your own mathematics above. <shrug> Yet, you are > > continuing to use twisted logics to spew up mysticism. > > > Lets see. The following represent indeed mathematical equivalence. > > The self-styled physicists want to promote mysticism. They are > > claiming the first equation is valid while the second one is out of > > favor. > > > ** E^2 = m^2 c^4 + p^2 c^2 > > ** E = m c^2 > > > Where > > > ** E = Observed energy > > ** m = Rest mass > > ** m = observed mass > > ** p = observed momentum > > > Clearly, both equations are indeed the same. Why do the self-styled > > physicists continue chanting about the nonsense of them being > > different? It makes one wonder if they understand the simple > > mathematics behind all that at all. <shrug> > > They're not different. However, you'll notice that m is *invariant*, > which means it has the same value in all frames, not just the frame in > which it is at rest. Numerically, invariant mass has the same value as > the rest mass. But the latter is the mass when measured at rest, by > definition. The former does not require it to be at rest to apply. > You'll also notice that m' is not invariant, and so yes, there are > reasons why m' is not so much used anymore, where m is. So, both equations are correct. Why do you insist that only the first one is correct? So, if your structural engineer is able to apply the second equation to solve the problem, why are you making fun of that engineer using the second equation instead of the first? The concept of the invariant mass is stupid and unnecessary since mass is always an observed quantity. The rest mass makes much more sense in any applications. <shrug> > > OK? How do the self-styled physicists cope with the energy > > conservation in (E = m c^2) applied to gravitation? They dont. In > > desperation, they tossed around this gravitational wave nonsense to > > distract attentions. The answer to that question is indeed to claim > > the supposedly invariant rest mass not so invariant after all. This > > so-called invariant rest mass will now vary with the amount of > > curvature in spacetime. The more curved up spacetime is the less > > invariant mass it has as represented by the following equation > > derived through the geodesic equations of the Schwarzschild metric. > > > ** E^2 = m^2 c^4 (1 2 U) / (1 B^2) > > > Where > > > ** m = Rest mass in flat spacetime > > ** m sqrt(1 2 U) = Rest mass or invariant mass > > ** U = G M / c^2 / r > > ** B^2 c^2 = (dr/dt)^2 / (1 2 U)^2 + r^2 (dO/dt)^2 / (1 2 U) > > ** dO^2 = dLongitude^2 cos^2(Laitude) + dLatitude^2 > > > The above equation can simpler be written as the following. > > > ** E = m c^2 > > > Where > > > ** m = m sqrt(1 2 U) / sqrt(1 B^2) > > > Which one of the first two equations is more elegantly simpler? It > > should take no brainer to answer that question unless you are the one > > promoting mysticism in post-Aether physics where mysticism is the rule > > of the game. <shrug> > > Gee, why do you think that the equation that has fewer symbols in it > is the one that is fundamentally better? Well, the second equation deals with the so-called invariant/rest mass. Although the first equation would lead to the same thing as the second one if the rest mass in variant under the curvature of spacetime, the second equation allows much easier concept of mass in which as Tony M pointed out that there is only one mass that is the observed mass. That observed mass in increased through the time dilation due to motion but decreased through time dilation due to gravitational time dilation. Oops! Did the cat get out of the bag? Does someone smell contradiction in GR?
From: Inertial on 26 Apr 2010 21:45 "Koobee Wublee" <koobee.wublee(a)gmail.com> wrote in message news:bfe9f26d-914f-4679-84e1-d4b6a2bded8d(a)y38g2000prb.googlegroups.com... > On Apr 26, 12:04 pm, PD <thedraperfam...(a)gmail.com> wrote: >> On Apr 26, 1:32 pm, Koobee Wublee wrote: > >> > It sounds like this so-called invariant mass is indeed your rest mass >> > as indicated by your own mathematics above. <shrug> Yet, you are >> > continuing to use twisted logics to spew up mysticism. >> >> > Let�s see. The following represent indeed mathematical equivalence. >> > The self-styled physicists want to promote mysticism. They are >> > claiming the first equation is valid while the second one is �out of >> > favor�. >> >> > ** E^2 = m^2 c^4 + p^2 c^2 >> > ** E = m� c^2 >> >> > Where >> >> > ** E = Observed energy >> > ** m = Rest mass >> > ** m� = observed mass >> > ** p = observed momentum >> >> > Clearly, both equations are indeed the same. Why do the self-styled >> > physicists continue chanting about the nonsense of them being >> > different? It makes one wonder if they understand the simple >> > mathematics behind all that at all. <shrug> >> >> They're not different. However, you'll notice that m is *invariant*, >> which means it has the same value in all frames, not just the frame in >> which it is at rest. Numerically, invariant mass has the same value as >> the rest mass. But the latter is the mass when measured at rest, by >> definition. The former does not require it to be at rest to apply. >> You'll also notice that m' is not invariant, and so yes, there are >> reasons why m' is not so much used anymore, where m is. > > So, both equations are correct. Why do you insist that only the first > one is correct? He didn't > So, if your structural engineer is able to apply the > second equation to solve the problem, why are you making fun of that > engineer using the second equation instead of the first? > > The concept of the invariant mass is stupid and unnecessary since mass > is always an observed quantity. The rest mass makes much more sense > in any applications. <shrug> Using 'rest mass' is only really a 'valid' concept, by definition, when an object is at rest .. so has limited applicability. However, we know that the rest mass is the same as the invariant mass. And so invariant mass is the same regardless of motion .. so is conceptually more useful. Why do you have a problem with that? >> > OK? How do the self-styled physicists cope with the energy >> > conservation in (E = m c^2) applied to gravitation? They don�t. In >> > desperation, they tossed around this gravitational wave nonsense to >> > distract attentions. The answer to that question is indeed to claim >> > the supposedly invariant rest mass not so invariant after all. This >> > so-called invariant rest mass will now vary with the amount of >> > curvature in spacetime. The more curved up spacetime is the less >> > �invariant� mass it has as represented by the following equation >> > derived through the geodesic equations of the Schwarzschild metric. >> >> > ** E^2 = m^2 c^4 (1 � 2 U) / (1 � B^2) >> >> > Where >> >> > ** m = Rest mass in flat spacetime >> > ** m sqrt(1 � 2 U) = Rest mass or �invariant� mass >> > ** U = G M / c^2 / r >> > ** B^2 c^2 = (dr/dt)^2 / (1 � 2 U)^2 + r^2 (dO/dt)^2 / (1 � 2 U) >> > ** dO^2 = dLongitude^2 cos^2(Laitude) + dLatitude^2 >> >> > The above equation can simpler be written as the following. >> >> > ** E = m� c^2 >> >> > Where >> >> > ** m� = m sqrt(1 � 2 U) / sqrt(1 � B^2) >> >> > Which one of the first two equations is more elegantly simpler? It >> > should take no brainer to answer that question unless you are the one >> > promoting mysticism in post-Aether physics where mysticism is the rule >> > of the game. <shrug> >> >> Gee, why do you think that the equation that has fewer symbols in it >> is the one that is fundamentally better? > > Well, the second equation deals with the so-called invariant/rest > mass. Why is that a problem? > Although the first equation would lead to the same thing as the > second one if the rest mass in variant under the curvature of > spacetime, the second equation allows much easier concept of mass in > which as Tony M pointed out that there is only one mass that is the > observed mass. It depends on how you measure / calculate it > That observed mass in increased through the time > dilation due to motion but decreased through time dilation due to > gravitational time dilation. Oops! Did the cat get out of the bag? > Does someone smell contradiction in GR? No
From: glird on 26 Apr 2010 21:50 On Apr 22, 1:48 am, "Y.Porat" <y.y.po...(a)gmail.com> wrote: > > THERE IS JUST ONE KIND OF MASS!! Although "mass" is "a quantity of matter"; there are two different forms of this on kind of matter: partculate and non-particulate. The non-particulate form is continuous and has no weight, though its density (quantity of matter per unit volume) does change. Because it has no surface it conducts rather than reflects light, thus is invisible. (It is called "dark matter".) Even so, since gravity is an effect caused by a density gradient permeating embedded particles, as the local density of this dark matter increases so will the steepness of the gradient (g-field) and so will the strength of the reaction (g- force) in those particles. Though that's WHY partles have weight and why the g-force is proportional (NOT "equal") to the mass of reacting particles, since present theory denies the existence of this luminiferous matter, there is no way that those who believe everything they are taught can understand these things. glird
From: PD on 26 Apr 2010 22:18 On Apr 26, 7:06 pm, Koobee Wublee <koobee.wub...(a)gmail.com> wrote: > On Apr 26, 12:04 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > On Apr 26, 1:32 pm, Koobee Wublee wrote: > > > It sounds like this so-called invariant mass is indeed your rest mass > > > as indicated by your own mathematics above. <shrug> Yet, you are > > > continuing to use twisted logics to spew up mysticism. > > > > Lets see. The following represent indeed mathematical equivalence. > > > The self-styled physicists want to promote mysticism. They are > > > claiming the first equation is valid while the second one is out of > > > favor. > > > > ** E^2 = m^2 c^4 + p^2 c^2 > > > ** E = m c^2 > > > > Where > > > > ** E = Observed energy > > > ** m = Rest mass > > > ** m = observed mass > > > ** p = observed momentum > > > > Clearly, both equations are indeed the same. Why do the self-styled > > > physicists continue chanting about the nonsense of them being > > > different? It makes one wonder if they understand the simple > > > mathematics behind all that at all. <shrug> > > > They're not different. However, you'll notice that m is *invariant*, > > which means it has the same value in all frames, not just the frame in > > which it is at rest. Numerically, invariant mass has the same value as > > the rest mass. But the latter is the mass when measured at rest, by > > definition. The former does not require it to be at rest to apply. > > You'll also notice that m' is not invariant, and so yes, there are > > reasons why m' is not so much used anymore, where m is. > > So, both equations are correct. Why do you insist that only the first > one is correct? So, if your structural engineer is able to apply the > second equation to solve the problem, why are you making fun of that > engineer using the second equation instead of the first? 1. What I said is that the second equation is used because it makes a more consistent use of m. 2. The structural engineer is not able to apply either equation to solve the problem. 3. The structural engineer is confusing himself with the meaning of mass, which is different in each case. > > The concept of the invariant mass is stupid and unnecessary since mass > is always an observed quantity. The rest mass makes much more sense > in any applications. <shrug> And is it your contention that mass is always observed at rest? Particle physicists would strenuously disagree, and they observe mass all the time. > > > > > > OK? How do the self-styled physicists cope with the energy > > > conservation in (E = m c^2) applied to gravitation? They dont.. In > > > desperation, they tossed around this gravitational wave nonsense to > > > distract attentions. The answer to that question is indeed to claim > > > the supposedly invariant rest mass not so invariant after all. This > > > so-called invariant rest mass will now vary with the amount of > > > curvature in spacetime. The more curved up spacetime is the less > > > invariant mass it has as represented by the following equation > > > derived through the geodesic equations of the Schwarzschild metric. > > > > ** E^2 = m^2 c^4 (1 2 U) / (1 B^2) > > > > Where > > > > ** m = Rest mass in flat spacetime > > > ** m sqrt(1 2 U) = Rest mass or invariant mass > > > ** U = G M / c^2 / r > > > ** B^2 c^2 = (dr/dt)^2 / (1 2 U)^2 + r^2 (dO/dt)^2 / (1 2 U) > > > ** dO^2 = dLongitude^2 cos^2(Laitude) + dLatitude^2 > > > > The above equation can simpler be written as the following. > > > > ** E = m c^2 > > > > Where > > > > ** m = m sqrt(1 2 U) / sqrt(1 B^2) > > > > Which one of the first two equations is more elegantly simpler? It > > > should take no brainer to answer that question unless you are the one > > > promoting mysticism in post-Aether physics where mysticism is the rule > > > of the game. <shrug> > > > Gee, why do you think that the equation that has fewer symbols in it > > is the one that is fundamentally better? > > Well, the second equation deals with the so-called invariant/rest > mass. Although the first equation would lead to the same thing as the > second one if the rest mass in variant under the curvature of > spacetime, the second equation allows much easier concept of mass in > which as Tony M pointed out that there is only one mass that is the > observed mass. That observed mass in increased through the time > dilation due to motion but decreased through time dilation due to > gravitational time dilation. Oops! Did the cat get out of the bag? > Does someone smell contradiction in GR? No, sorry, I don't sense any contradiction. Do you?
From: Koobee Wublee on 27 Apr 2010 00:32
On Apr 26, 7:18 pm, PD <thedraperfam...(a)gmail.com> wrote: > On Apr 26, 7:06 pm, Koobee Wublee wrote: > > > > ** E^2 = m^2 c^4 + p^2 c^2 > > > > ** E = m c^2 > > > > > Where > > > > > ** E = Observed energy > > > > ** m = Rest mass > > > > ** m = observed mass > > > > ** p = observed momentum > > > > > Clearly, both equations are indeed the same. Why do the self-styled > > > > physicists continue chanting about the nonsense of them being > > > > different? It makes one wonder if they understand the simple > > > > mathematics behind all that at all. <shrug> > > > > They're not different. However, you'll notice that m is *invariant*, > > > which means it has the same value in all frames, not just the frame in > > > which it is at rest. Numerically, invariant mass has the same value as > > > the rest mass. But the latter is the mass when measured at rest, by > > > definition. The former does not require it to be at rest to apply. > > > You'll also notice that m' is not invariant, and so yes, there are > > > reasons why m' is not so much used anymore, where m is. > > > So, both equations are correct. Why do you insist that only the first > > one is correct? So, if your structural engineer is able to apply the > > second equation to solve the problem, why are you making fun of that > > engineer using the second equation instead of the first? > > 1. What I said is that the second equation is used because it makes a > more consistent use of m. Wait! You have been saying the first equation is valid while the second equation is valid when applying to this rest mass. <shrug> > 2. The structural engineer is not able to apply either equation to > solve the problem. Who is this structural engineer? > 3. The structural engineer is confusing himself with the meaning of > mass, which is different in each case. Tony M said it well. Mass is an observed quantity. Thus, there needs to have only one concept of mass whether it is relativistic or rest mass. <shrug> > > The concept of the invariant mass is stupid and unnecessary since mass > > is always an observed quantity. The rest mass makes much more sense > > in any applications. <shrug> > > And is it your contention that mass is always observed at rest? No, not at all. I have been saying the mass is an observed quantity regardless who the observer is. <shrug> > Particle physicists would strenuously disagree, and they observe mass > all the time. <shrug> Disagree with what? Remember that kid who was shoved your way with vivid, wild ideas about physics? Your complaint was that the kid knew no math to back up his hypotheses. Well, you are in the same boat. <shrug> |