elfmotat

The problem with post quotas is that it incentivizes new users to make lots of useless posts in other sections to get access quicker.

Please quote any and all mistakes I made, and explain why they are mistakes. I have already given you the same courtesy, as can be found here. Although you chose to ignore your mistake and pretend it didn't happen, I promise I won't do the same.

Read MigL's post. Then read the "this is right" reply he received. Do you disagree with anything I said in my previous post, or do you just enjoy bickering?

If you drop an elevator into free-fall, as long as the elevator is small compared to the gravitating planet/star/etc., any light inside will behave completely normally as if you were in an inertial reference frame. So as long as the elevator isn't a significant fraction of the size of the gravitating body, you won't be able to tell that you're falling by measuring light deflection. If the elevator was very very big, then you would be able see deviations in the path of the light. Sorry it took so long for you to receive an answer. No, he's comparing free-fall frames with inertial frames, which is a perfectly valid thing to do locally.

Well, since there's apparently no content left in this discussion, I think I'll be on my way.

Except that was the cause of your disagreement, and I wasn't wrong. Nothing I've said so far is false. On the other hand, you've given us a meaningless equation and barrels of insults.

It was the subject your and MigL's disagreement, which is why I brought it up. Ignoring the part about your nonsense equation, I completely agree and I never said differently. If you had been less reactionary from the get go this would have been obvious. I said they "fall the same way" over small enough times/distances.

Okay, this time I'll be precise: for a short enough time over short enough distances, free-fall reference frames are equivalent to inertial reference frames. Which means that over small enough times/distances if you were to shine a laser at a detector to try to measure some light deflection, you won't be able to. Because it's locally an inertial frame. I can't argue with an argument with no content. Are you really going to pretend you don't know how the forum works? I pressed "quote" a few minutes before I responded. I responded a minute after you edited that post, meaning I had no knowledge of your edit at the time I posted. Now you're going to pretend I'm intentionally ignoring things? Very dishonest. Please derive the equation [math]\Gamma^{\mu}_{\nu \sigma}\frac{dx^\nu}{d \lambda}\frac{dx^\sigma}{d \lambda} =0[/math] from the following: [math]ds^2=0[/math] [math]\frac{d^2 x^{\mu}}{d \lambda^2} + \Gamma^{\mu}_{\nu \sigma}\frac{dx^\nu}{d \lambda}\frac{dx^\sigma}{d \lambda}=0[/math] I'll give you a hint: it can't be done. Because it's nonsense. That equation is meaningless. It's coordinate-dependent, which means it's not generally covariant, so it means absolutely nothing in the context of GR.

"Free-fall" is by definition geodesic motion. I assumed this was understood, because it commonly is. You have quite a nasty attitude, which makes it difficult to talk to you. I was trying to explain something, and you chose to ignore it. "First order" is also a well-defined term which you have, out of ignorance, labeled as "psychobabble." You also ignored the part where I called you out for using a nonsensical equation. I'm well aware of the everything else you just took the unnecessary time write, and have been for quite a while. The sheer condescension is making your posts a bit unbearable to read, so I recommend toning that down a bit as well.

Instead of trying to argue and hurl petty insults, you may find my post enlightening if you actually read it. I was attempting to end the argument between yourself and MigL by explaining where the disagreement is coming from. To first order light and massive particles will fall identically. Globally they won't. MigL was considering the situation locally and you were considering it globally. That's where the disagreement was coming from, and you're both right in a way. For someone so intent on making sure others "understand the math," the irony here is palpable. This is just completely false. This is not even a generally covariant equation; it is completely coordinate-dependent. I don't know where you got it from, but it doesn't even make sense. I believe you're mixing the geodesic equation up with the null-condition for massless particles.

I quoted you directly. You said they have different equations of motions. They don't. The worldline of a photon satisfies the geodesic equation. "Constraints" are not usually considered to be "equations of motion." There's no need to get snippy about it, though I do find the irony of the sentiment amusing. That's just semantics anyway. The argument here stems from whether the question of "how does light fall" is being asked locally or globally. Massive particles and massless particles will locally "fall" identically to first order. Globally, second-order effects like curvature must be considered. That is, the answer depends on the distance scales you're working with and how accurate you want to be.

This isn't true. The equation of motion is the geodesic equation for both massive and massless particles. Massless particles just have the additional constraint that [math]ds^2=0.[/math] In other words, they both follow geodesics, but massless particles only follow geodesics such that their velocity is locally constant. I agree. A free-falling observer can be locally considered as an inertial frame. (I.e. ignoring all second-order effects due to curvature.) This is one of the core principles of GR: the equivalence principle. If you are in a free-falling elevator and shine a laser, the photons had better fall at the same rate that you do. Otherwise you'd be able to tell you were free-falling by measuring light-deflection. Equivalently, a uniformly accelerating elevator and an elevator sitting on the surface of a planet (with equivalent g-value) should also behave identically: light should be deflected by the same amount in both instances.

That's a false dichotomy. We already know of models with no hidden variables where no information is exchanged: namely quantum mechanics.

That's a different argument. Swansont didn't say that the state of each particle is determined at their creation, he said the correlation between particle states is determined at their creation.

Lots of curiosity and lots of practice. People sometimes like to attribute this kind of stuff to external factors like brain hard-wiring, when in reality I think the vast majority of people who are considered "geniuses" simply got very interested in a subject and pursued that line of research.

You miss the entire purpose of our objections. You haven't really simulated anything. What you've done is use known equations to output examples of those equations. You haven't done anything new. If you say you've shown the Bell inequalities to be true then you're just wrong, because reality does not conform to them. You haven't given us a new theory or a new understanding of QM, you've given us precisely nothing.

Except we know what you're telling us is false. The neutron is comprised of gluons and three quarks, not electrons protons and neutrinos. There are no electrons or neutrinos or protons inside a neutron.

Ha, I don't know. Maybe it's the style in Israel.

Just to make something a bit more clear: the Bell inequalities don't rule out all hidden variable theories, just hidden variables which obey local equations. de Broglie-Bohm Theory is an example of a non-local hidden variable theory which is not ruled out by the Bell inequalities.

It does seem a bit speculative. As I read it the question seems to be about what would happen if gravity were suddenly "turned off." I'm not sure that's even a reasonable physical question to ask. For example, what would happen to black holes? I think situations which clearly violate known physics qualify as speculative.

Fun song to play: https://www.youtube.com/watch?v=ahJCERfeehY

What's your problem?

I'm not really sure why you're trying to use that equation for tidal forces. It's just the leading term from a series expansion of the exact force differential between two radially separated points, and is only valid for small separations. The time for a small mass starting at rest at coordinate [math]r_0[/math] to fall to coordinate [math]r[/math] is given by: [math]t=\sqrt{\frac{r_0^3}{2GM}} \, \left(\sqrt{{\frac{r}{r_0}\left(1-\frac{r}{r_0}\right)}} + \tan^{-1} \sqrt{\frac{r_0}{r}-1} \,\right)[/math] Therefore the two particles, after falling for some time, have r-coordinates which can be related to each other: [math]\sqrt{\frac{(r_i-R)^3}{2GM}} \left(\sqrt{{\frac{r_1}{r_i-R}\left(1-\frac{r_1}{r_i-R}\right)}} + \tan^{-1} \sqrt{\frac{r_i-R}{r_1}-1} \,\right) = \sqrt{\frac{(r_i+R)^3}{2GM}} \left(\sqrt{{\frac{r_2}{r_i+R}\left(1-\frac{r_2}{r_i+R}\right)}} + \tan^{-1} \sqrt{\frac{r_i+R}{r_2}-1} \,\right)[/math] Now, you could in principle solve this for either [math]r_1[/math] or [math]r_2[/math] given the condition [math]r_2-r_1=4R.[/math] Once you've done that you can just plug the value into the time equation. The above looks incredibly tedious, which is why I won't try it. If you're feeling brave I suppose you could try it yourself. EDIT: Now that I read your question again, it seems you're more interested in the case where the separation between the particles is small. You could expand that massive equation above into two series and discard all higher order terms in R.

c is a fundamental constant related to the geometry of spacetime. The fact that light travels at c in a vacuum is just a consequence of photons having zero mass. Other particles have zero mass as well, and they also propagate at c. I'm not quite sure what you mean by a "calculated speed," so perhaps you could expand on that.

Historically, I'm not really sure what prompted Newton to write down his third law. Physically, however, it is just a statement of momentum conservation. Say object 1 pushes on object 2 with force [math]F_{12}[/math]. Then by the third law object 2 pushes on object 1 with force [math]F_{21}=-F_{12}[/math]. Rearranging and using Newton's second law: [math]F_{12}+F_{21} = \frac{d}{dt} \left ( p_1 + p_2 \right ) = 0[/math] This just says that the total momentum of the system must be conserved. Whenever there is spatial translation symmetry present you will get some version of Newton's third law.