Aer
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I am talking about QM and Relativity, not QM and 'special relativity'. Relativity is in fact 'general relativity' because 'special relativity' is nothing but a special case of 'general relativity'. Now let's be thorough here. The special relativity equations are recovered in general relativity when spacetime is considered to be flat. So, the special case that is 'special relativity' is flat spacetime. Any unification of special relativity and QM should inherently include the general case which is general relativity. Do you disagree with the fact that special relativity is a special case of general relativity?Tom Mattson said:I said that quantum theory and special relativity have been completely unified, and that is true. The status of quantum theories of gravity is irrelevant to that statement.
Very good, this is true - the energy would be undefined according to the relativity equation E=\gamma m c^2Tom Mattson said:No, it isn't. If that were true then the energy of all photons would be undefined, which it isn't.
Yes, from the equation I gave below - I already mentioned this, why did you need to pretend like I didn't know?Tom Mattson said:In relativity the energy of a photon is given by E=pc.
Oh yeah? You'll have to forgive me then for providing the derivation of E^2 = (pc)^2 + (mc^2)^2Tom Mattson said:The first equation is not derived from the second and third equations at all.Aer said:or
E^2 = (pc)^2 + (mc^2)^2
which consequentyly is derived from E = \gamma m c^2 and p = \gamma m v.
Start with the definition of p:
p = \gamma m v
Multiply by c and square both sides:
(pc)^2 = \frac{(m v c)^2}{1-v^2/c^2}
Divide and multiply the right side by c2
(pc)^2 = \frac{m^2 \frac{v^2}{c^2} c^4}{1-v^2/c^2}
Subtract and add the quantity \frac{m^2 c^4}{1-v^2/c^2}:
(pc)^2 = \frac{m^2 c^4 (\frac{v^2}{c^2}-1) }{1-v^2/c^2} + \frac{m^2 c^4}{1-v^2/c^2}
Simplify:
(pc)^2 = -m^2 c^4 + \frac{m^2 c^4}{1-v^2/c^2}
Plug in \gamma^2 = 1/(1-v^2/c^2):<br /> (pc)^2 = -m^2 c^4 + \gamma^2 m^2 c^4<br /> Using the definition of E=\gamma m c^2:<br /> (pc)^2 = -m^2 c^4 + E^2<br /> And viola:<br /> E^2 = (pc)^2 + m^2 c^4<br /> <br /> You were saying?<br /> <br /> <br /> <br /> <blockquote data-attributes="" data-quote="Tom Mattson" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> Tom Mattson said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> The second and third equations only applie to massive particles, while the first applies to all particles, even those with zero mass. </div> </div> </blockquote> That is the ad-hoc nature of taking the result from QM of pc=hf for a photon and throwing it into the relativity equation. It works only if you forget the definition of p = \gamma m v used to derive the equation. That is what I am saying - it is very ad-hoc in nature. You call it unify... I call it ad-hoc. Unification <i>should</i> be all the results from relativity being described in QM.<br /> <br /> Forgive my ignorance, but how does QM explain the relativity of simultanteity? I will admit that I do not know the answer to this and would like for you to tell me since you said special relativity and QM are completely unified.<br /> <br /> <blockquote data-attributes="" data-quote="Tom Mattson" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> Tom Mattson said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Right, the explanation comes from the unification of special relativity and quantum theory, which has been done, despite the fact that you claim otherwise. </div> </div> </blockquote> I told you I already know how the result is arrived at, but it is not from "unification". True, you use both theories, but that is not what I meant by unification and is not what I think most people mean by unification (though I could be wrong - in which case, I need another word for what I am talking about).<br /> <br /> <br /> <br /> <blockquote data-attributes="" data-quote="Tom Mattson" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> Tom Mattson said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Why would I need to explain quantum gravity, in order to defend my point? </div> </div> </blockquote> Well, my point was the original point of contention. And my point was: QM and Relativity are not unified. When I say "Relativity" with no prefix, I mean General Relativity. Sorry for the confusion. And it is for that reason that I believe my point about gravity is relevant.<br /> <br /> <br /> <blockquote data-attributes="" data-quote="Tom Mattson" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> Tom Mattson said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> I didn't forget to do anything. You can, in fact, write force as F=dM/dτ. </div> </div> </blockquote> I didn't say you couldn't do it (actually, I believe you mean F=d(Mv)/dτ but that is beside the point), but if you actually want to take that derivative, you must expand it out, no?<br /> <br /> <blockquote data-attributes="" data-quote="Tom Mattson" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> Tom Mattson said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Obviously, the use of M in that particular equation was to make the relationship more compact. Is that really too difficult to see? </div> </div> </blockquote> What is wrong with just leaving it in the compact form F=dp/dτ?<br /> <br /> <br /> <blockquote data-attributes="" data-quote="Tom Mattson" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> Tom Mattson said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> If by a "non-gravitational energy" term you mean a term that describes "energy that does not contribute to gravitation", then kinetic energy is not one of them. </div> </div> </blockquote> So a particle with kinetic energy has a greater effect on the curvature of spacetime? That is what you mean do you not? In which frame must the kinetic energy be calculated so that the curvature of spacetime that our particle creates can be known?<br /> <br /> <br /> <blockquote data-attributes="" data-quote="Tom Mattson" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> Tom Mattson said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> I said that all energies are put into the energy-momentum tensor, and that is a fact. Nothing you have referred to disagrees with anything that I said. </div> </div> </blockquote> I did not raise any disagreement with that part of your statement.<br /> <br /> <blockquote data-attributes="" data-quote="Tom Mattson" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> Tom Mattson said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Unfortunately, you don't have a single good reason for holding that opinion. </div> </div> </blockquote> Name one good reason you have for holding the opinion that <a href="http://en.wikipedia.org/wiki/Relativistic_mass" target="_blank" class="link link--external" rel="nofollow ugc noopener">relativistic mass is useful</a>. I have one, it is highlighted in red:<br /> <br /> In the earlier years of relativity, it was the relativistic mass that was taken to be the "correct" notion of mass, and the invariant mass was referred to as the rest mass. Gradually, as special relativity gave way to general relativity and found application in quantum field theory, it was realized that the invariant mass was the more useful quantity and scientists stopped referring to the relativistic mass altogether.<br /> <br /> The accepted usage in the scientific community today (at least in the context of special relativity) considers the invariant mass to be the only "mass", while the concept of energy has replaced the relativistic mass. In popular science and basic relativity courses, however, the relativistic mass is usually presented, most likely due to its conceptual simplicity.<br /> <br /> However, I don't think it is a good idea to use it at all because of the confusion it generates.<br /> <br /> <br /> <blockquote data-attributes="" data-quote="Tom Mattson" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> Tom Mattson said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Aer, you have been trying the patience of the staff with your insulting tone to others, and we have been watching. </div> </div> </blockquote> I know that you have been watching and that you have been called here.<br /> <br /> <br /> <blockquote data-attributes="" data-quote="Tom Mattson" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> Tom Mattson said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> But in this post not only have you been patronizing, but you have descended into abject crackpottery. </div> </div> </blockquote> Name one thing I've said that is crackpottery? I can tell you one thing, the only person I agree with in this thread is the thread starter whom just happens to be a qualified physics instructor. Ho-hum, and I am a crackpot?<br /> <br /> <blockquote data-attributes="" data-quote="Tom Mattson" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> Tom Mattson said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> I must insist that you stop doing it. </div> </div> </blockquote> I will agree to stop pushing any crackpot ideas because I've yet to do so.<br /> <br /> Now I am just curious, are you just as qualified or more qualified in physics than the thread starter?
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