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imsmooth
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I have seen E=pc used to derive E=MC2. I have seen E=MC2 used to derive E=pc. This is circuitous. Which came first and how is E=pc derived?
imsmooth said:I have seen E=pc used to derive E=MC2. I have seen E=MC2 used to derive E=pc.
imsmooth said:how is E=pc derived?
Thank you.lomidrevo said:Both equations come from relativistic energy-momentum relation. See special cases, 1. and 3., in article:
https://en.m.wikipedia.org/wiki/Energy–momentum_relation
love_42 said:The mass of a moving particle is
m = m0/√(1 - v2)
Therefore all relativistic effects are derivable from, or attributed to the increase of mass with velocity, or in other words, the equivalence between mass and energy. Another way to say it is that the momentum is pc = Ev/c (the "flux of energy"), where E = mc2. In terms of this m, the momentum is p = mv. If you substitute pc = Ev/c in the formula for the mass, you get
(m/E)2(E2 - p2) = m02
Using E = mc2, you get E2 - p2 = m02. So essentially you get E = pc just by substituting v = c in the formula pc = Ev/c.
etotheipi said:Your units are all over the place, and you also lose points for bringing up relativistic mass.
love_42 said:All relativistic effects are contained in the idea that the mass increases with velocity
love_42 said:Consider the lagrangian L = -m√1 - v2. Expanding this out to first order, we get
L = 1/2mv2 - m
So m appears in the potential energy place of the lagrangian L = T - U. I find it difficult to understand such things without thinking about the relativistic mass. Maybe there is another way to understand such things without introducing such an idea. If anyone knows, it would be interesting to hear.
love_42 said:All relativistic effects are contained in the idea that the mass increases with velocity
love_42 said:All relativistic effects are contained in the idea that the mass increases with velocity, or the equivalence between mass and energy.
This approach has the disadvantage that it relies on ##m_0## being non-zero, so can only be generalized to apply to massless particles by using dubious mathematical techniques: proof by emphatic assertion and derivation by energetic handwaving.john t said:The way I rationalize it requires recognition of relativistic mass and thus momentum and starts with E=gamma mc^2.
john t said:My algebra is attached as a screenshot.
Thanks, Nugatory. I would like to know what math or physics principle was broken. There is a derivation of E=gamma mc^2 based on thought experiments and that admittedly requires the assumption of non-zero mass. Is the objection that my approach goes from a specific to a general equation? In that case, would it be legal to go from the general equation to the special one using the reverse of the algebra and, if so, how can my approach be unjustified? Note I did not call it a derivation, rather I said I rationalized it that way. I take your view seriously and, as a non-physicist, would apprediate any further comments. I like the thought-experiment approach (based on several publications I read), because I can understand all the algebraic and calculus steps and do not have the background for Minkowski formalism, and I would like to be able to say to myself that I understand the general equation, even as a simple chemist.Nugatory said:This approach has the disadvantage that it relies on ##m_0## being non-zero, so can only be generalized to apply to massless particles by using dubious mathematical techniques: proof by emphatic assertion and derivation by energetic handwaving.
john t said:The way I rationalize it
john t said:I would like to know what math or physics principle was broken.
john t said:would it be legal to go from the general equation to the special one using the reverse of the algebra
john t said:how can my approach be unjustified?
john t said:I like the thought-experiment approach
It's more the exact way you move from the specific to the general.john t said:There is a derivation of E=gamma mc^2 based on thought experiments and that admittedly requires the assumption of non-zero mass. Is the objection that my approach goes from a specific to a general equation?
Thanks Nugatory. I understand, and I do not think you are quibbling, and your analogy with the Lorenz situation makes the point clear. Can one say that my logic shows the consistency of the specific with the general equation, given the acceptance (naieve?) of relativistic mass?Nugatory said:It's more the exact way you move from the specific to the general.
We start with a formula (##m_R=\gamma m_0##) whose derivation critically depends on ##m_0## being non-zero, work through the algebra from there to derive ##E^2=(m_0c^2)^2+(pc)^2##; this is valid as long as the initial ##m_R=\gamma m_0## is valid, which is to say if ##m_0\ne 0##. Thus, setting ##m_0## to zero and watching ##E=pc## pop out of the ##E^2## equation is basically saying that our conclusion is valid even when the premises it depends on are invalid - and that's pretty strong hint that we've been arguing from poorly chosen premises.
I'm not just completely quibbling here. In this case we end up with the right answer, but consider the similar problem with the Lorentz transformations (and the time dilation and length contraction formulas that follow from them). These are derived under assumptions that are equivalent to ##v<c##, just as your argument is derived under the assumption that ##m_0\ne 0## - but when we try to generalize by dumping that assumption and plugging in other values of ##v## the results are nonsensical in a way that has confused generations of physics students. Why should one procedure any less invalid than the other?
No. In you attachment to posting #16, you are using the same symbol ##E## for different things in the first and the last equation (total energy vs. rest energy). With ##m## you can't mean relativistic mass, because in the first equation you wrote:john t said:Can one say that my logic shows the consistency of the specific with the general equation, given the acceptance (naieve?) of relativistic mass?
But according to Wikipedia, relativistic mass ##m_R## ist defined as total energy (devided by c²):john t said:##E=\gamma mc^2##
Source:Wikipedia said:The relativistic mass is the sum total quantity of energy in a body or system (divided by c²). Thus, the mass in the formula
##E = m_R c^2##
is the relativistic mass.
...
A so-called massless particle (such as a photon, or a theoretical graviton) moves at the speed of light in every frame of reference. In this case there is no transformation that will bring the particle to rest. ... This property of having no rest mass is what causes these particles to be termed "massless". However, even massless particles have a relativistic mass, which varies with their observed energy in various frames of reference.
Then this disqualifies me from using the forum. I am legally blind, and use of LaTex is beyond my ability. (I installed it on my mac, and see that it is beyond me.) The keystrokes are just too difficult. I am able to use MathType, but you guys do not recognize that apparently. Maybe I can find another forum which allows screenshots of written text I prepare on an ipad.PeterDonis said:This is not allowed per the PF rules. Please use the PF LaTeX feature to enter equations directly in the post. A link to the LaTeX Guide is at the lower left of the post window.
john t said:Then this disqualifies me from using the forum. I am legally blind, and use of LaTex is beyond my ability. (I installed it on my mac, and see that it is beyond me.) The keystrokes are just too difficult. I am able to use MathType, but you guys do not recognize that apparently. Maybe I can find another forum which allows screenshots of written text I prepare on an ipad.
\left. \begin{cases} { E = \gamma m { c }^{ 2 } } \\ { p = \gamma mv } \end{cases} \right.
\left. \begin{array} { l } { E = \gamma m c ^ { 2 } \text { and } p = \gamma mv } \\ { E ^ { 2 } = m ^ { 2 } c ^ { 4 } \frac { 1 } { 1 - \frac { v ^ { 2 } } { c ^ { 2 } } } = m ^ { 2 } c ^ { 4 } [ \frac { 1 - v ^ { 2 } / c ^ { 2 } + v ^ { 2 } / c ^ { 2 } } { 1 - \frac { v ^ { 2 } } { c ^ { 2 } } } ] = } \end{array} \right.
I think you could derive an equation, that is also valid for photons, by avoiding ##\gamma## as a factor, with ##m## being the "invariant mass", which is defined, but zero for photons:john t said:Can one say that my logic shows the consistency of the specific with the general equation
Nugatory said:I'm not just completely quibbling here. In this case we end up with the right answer, but consider the similar problem with the Lorentz transformations (and the time dilation and length contraction formulas that follow from them). These are derived under assumptions that are equivalent to ##v<c##, just as your argument is derived under the assumption that ##m_0\ne 0## - but when we try to generalize by dumping that assumption and plugging in other values of ##v## the results are nonsensical in a way that has confused generations of physics students. Why should one procedure any less invalid than the other?
john t said:Thanks Nugatory. I understand, and I do not think you are quibbling, and your analogy with the Lorenz situation makes the point clear. Can one say that my logic shows the consistency of the specific with the general equation, given the acceptance (naieve?) of relativistic mass?
atyy said:Relativistic mass is not wrong, as it's found in many good books (eg. by Feynman, Purcell, Rindler) just that one has to be careful with how one uses it. In many cases, it is easier to avoid the relativistic mass for the purpose of obtaining a correct calculation.
atyy said:Anyway, maybe to rephrase @john t's question, why does using an argument beyond the validity of its assumptions sometimes work? From the point of view of theory, massless photons are completely different from massive particles, as they are not at rest in any of the usual inertial reference frames, However, given that we don't know for sure that neutrinos and photons are massless, it seems that the equations for massless photons and for massive photons give almost the same results, ie. it would seem that the equations for massive particles should give results "close to" those for massless particles.
Sagittarius A-Star said:I think you could derive an equation, that is also valid for photons, by avoiding ##\gamma## as a factor
I am not using ##m = E / \gamma##, which would be undefined for ##v=c##. What I wrote, is different for the case ##v=c##.PeterDonis said:But you didn't do that. Rearranging ##E = \gamma m## to ##m = E / \gamma## is not avoiding the gamma factor.