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 Dec 1 comment The virtual particles are only a fictive tool in equations? DO they exist or DON'T? And if they exist, why do we call them VIRTUAL? @PeterShor An external line doesn't necessitate observation, as you seem to imply. That's just bad logic. External lines correspond to particles which can be observed, but they don't need to be. Dec 1 comment The virtual particles are only a fictive tool in equations? DO they exist or DON'T? And if they exist, why do we call them VIRTUAL? @PeterShor It seems like you're going out of your way to make it sound absurd. Virtual particles are internal lines by definition. Internal lines are not observed by definition. Therefore, virtual particles are not observed by definition. It's not complicated, and it's not circular. Also notice that I never said, "particles are virtual because they're not observed." That simply does not follow from the logic. Dec 1 comment The virtual particles are only a fictive tool in equations? DO they exist or DON'T? And if they exist, why do we call them VIRTUAL? @PeterShor How is it a tautology to say something is true by definition? Nov 30 comment Schwarzschild Radius vs Gravitational Pull? Error in Model? Newtonian gravity is a weak-field static limit approximation of General Relativity. That means it works well when the gravitational field is relatively small and everything is moving slowly compared to light. When you want to consider extreme objects like black holes, Newtonian gravity will simply fail to be accurate in their vicinity. You need to do full-fledged GR to get the correct predictions. Nov 28 comment Why do we use functional integration in QFT? @AlexNelson But the path integral doesn't say that "$x^0$ is the $t$-coordinate." You integrate over four-volume $\sqrt{-g} d^4 x$, which is a coordinate-invariant object. Nov 25 answered A true singularity at $t=0$, coordinate independent Big Bang Nov 25 comment On which basis we think there is something more than just super dense matter in black holes? Peter should also be aware the Schwarzschild coordinate $r$ does not actually correspond to "distance from the center." The distance between two values of $r$ is determined by an integral over the line element. Nov 25 comment Gravity at event horizon Except the quantity you've defined as "g" is not particularly meaningful. It does not correspond to any type of measurable acceleration. The proper acceleration required to hover near the event horizon asymptotically approaches infinity as $r \to 2GM/c^2$. Nov 24 revised Gravitational Time Dilation Formula? Clarified the post. Nov 24 comment Gravitational Time Dilation Formula? @Hypnosifl Indeed. I'll edit my post accordingly. Nov 24 answered Gravitational Time Dilation Formula? Nov 24 comment What does this depiction of a black hole in the movie Interstellar mean? I remember reading somewhere that Chris Nolan decided to cut out Doppler effects, and he also had them drop the angular momentum well below what would be required for the extreme time dilation the crew experiences. He apparently thought that these would confuse the audience too much. Nov 24 comment Movie Interstellar - Question about Escape Velocity @ChristianRau I replied to the post :). Nov 23 comment Deriving a Useful Solution of the free Schrödinger equation @Ruslan "Schrodinger" is a common spelling, whether originally correct or not, and is listed as one of the spellings on Erwin's wiki page. Nov 23 comment Can we explain Newton's first law mathematically? You might be interested in the answers here: physics.stackexchange.com/q/66057 Nov 22 answered What is the most agreed upon quantum mechanical equation of motion? Nov 22 comment Confused about equations for the Big Bang in general relativity ad loop quantum gravity? @21joanna12 $H^2 \to \infty$ means that space is either expanding or contracting at an infinite rate. Nov 22 answered Confused about equations for the Big Bang in general relativity ad loop quantum gravity? Nov 21 comment Derivation of the Riemann tensor confusion Indeed. As ACuriousMind mentioned, this is the whole point of defining the covariant derivative: so that things remain covariant. I.e. tensors go to tensors under the operation. Nov 21 comment Derivation of the Riemann tensor confusion $\lambda_{a;b}$ is a rank-2 tensor. If you take its covariant derivative you'll get two connection terms by definition of the covariant derivative.