(Alpha) A new equation for escape velocity

I always agreed in this thread that GR violates the inverse square law. That explains why I gave you a link about it when you brought it up the first time.

I didn’t suggest otherwise. In fact I suggested this, by saying that the violations may be deemed negligible in weak gravity.

You didn’t show that, nor could you. And in the OP I have shown full experimental confirmation to all significant digits for the Sun. GR predicts the same escape velocity for escape from the Sun as my equation does, to all significant digits, for all of the planets in the Solar System.

It is universal.

Sure, with precise-enough measuring equipment. I doubt such equipment exists for the only relatively extremely weak gravity in which GR has been experimentally tested to date.

I do know this; I don’t need a derivation to tell me that. My theory of gravity is yet to come. In its further development I will show anyone how my escape velocity equation (or GR’s) implies a deviation from the inverse square law.

 

Log in or Sign up to hide all adverts.

I am too lazy, lacking enough interest, to really come up to speed here, but want to throw out idea that may (just guessing) help with experimental tests:

I seem to recall that the tests of inertial = gravitational mass have been pushed out to many decimal places. Is there something here that might be useful? (Perhaps that is what Ben TM is efectivelly already doing with his "Venus Year" analysis?)

 

Log in or Sign up to hide all adverts.

Hi Billy---

Yeah---the problem with zanket's idea is that he has offered no evidence that it is correct, other than it agrees with all known experiments. This may or may not be the case---no one can know for sure because no one (read: zanket) has calculated anything with it. I tried, and found a deviation from the orbital period predicted by Kepler's formula at a part per million level or so---testable I think but suspect because G isn't known that well. But because the Kepler formula doesn't hold for Venus, there is no way to know for sure.

If he was serious about his idea being right, then zanket would calculate the perhilion of Mercury with his formula. Zanket claims that his equation agrees with experiment, but (and I put this in bold to stress it) there is absolutely no calculations that should convince anyone reading this post that this is true. Either way, I am not convinced that zanket has ever really calculated anything---it is much easier to criticize physics this way.

Yet another way to disprove Zanket's formula is to look at a Schwarzchild-like geometry. If zanket's formula is correct, there are no objects in the universe with horizons. This means that if there are singularities (note that zanket's formula holds for R = 0), they are naked singularities, which seem to be generically frowned upon by smart people who are working in Quantum Gravity (cf Penrose, Hawking, et al.).

The other idea I had was about the disks that surround the black holes that we've found in the universe. Since we can measure the velocity of the infalling gas by the red/blue shift of the xray spectra, one could possibly show that the gas is falling at FASTER than the escape velocity that zanket has written down. If you're interested in plugging in the numbers, then check this website: http://csep10.phys.utk.edu/astr162/lect/active/smblack.html.

Zanket will no doubt cry to Pete because I have strayed from the alpha rules, but that's ok.

 

Log in or Sign up to hide all adverts.

There’s no way to prove that any predictive equation is correct. Agreement with experiments, self-consistency, and consistency with other theories is the best that can be achieved.

An orbit, or relativistic orbital precession, can’t be calculated with just an escape velocity equation.

The proof of experimental confirmation is given in the OP. You’ve shown nothing wrong with it. Lack of the calculations you expect is not a problem; can you prove otherwise?

This doesn’t disprove my formula, for the simple reason that there need not be singularities when there are no horizons, as I explain further in the OP. When the escape velocity is always less than c, physics-busting objects like singularities need not be predicted. Notice they aren’t predicted by Newtonian mechanics, which also doesn’t predict horizons. Why doesn’t Newton's escape velocity equation hold for r=0?

Such observations are evidence of a black hole only when GR is assumed to be the correct theory of gravity. The astronomers plug the value of the observation (like the orbital velocity of the gas) into GR (or Newtonian mechanics, which has the same orbital velocity equation as GR) to get the mass. Then they plug the mass back into GR to see if it predicts a black hole. But I have shown that GR is self-inconsistent due to its prediction of black holes, in particular their horizons.

I don’t see how it could be shown what you propose. The gas is falling slower than c, and my equation has a range up to a limit of c. Nor could the observation be used to figure out which equation (GR’s or mine) is correct, because the mass is not known until the observation is “used up” by plugging it into a theory of gravity to get the mass. That the astronomers use up their only observation to get the mass is the reason why all the evidence of black holes to date is based on an assumption that GR is the correct theory of gravity. An observation that is plugged into GR to get the mass cannot be used to experimentally confirm GR.

Your claims have been a good test for me, which I appreciate.

 

Do you deny that your formula gives v = c at R = 0?

 

I asked first: Why doesn’t Newton's escape velocity equation hold for r=0?

 

Because there aren't any naked singularities. Every singularity is behind an horizon, so being at a singularity precludes an escape.

 

So I ask again---do you deny that your formula gives an escape velocity of c at R = 0?

 

This doesn’t begin to answer my question. I asked “Why doesn’t Newton's escape velocity equation hold for r=0?” You gave an answer that depends on that equation holding for r=0 (a singularity). Newtonian mechanics does not predict singularities (or horizons). Why not? Why doesn’t its escape velocity equation hold for r=0? GR was the first widely accepted theory of gravity to predict singularities. If you think I’m wrong that Newtonian mechanics does not predict singularities (that is, if you think its escape velocity equation holds for r=0), then say so.

After you give a decent attempt to answer my question, I will answer yours.

 

zanket I won't play your games. Your theory is the one being challenged. Your debate tacticts are reminiscent of those seeking to further intelligent design. If you don't believe me, ask Skinwalker.

If you won't answer the questions posed, then you are CLEARLY in violation of the alpha rules, which you seem to adore so much.

 

The alpha rules apply to you too. I asked you a question and, while you gave an answer, it didn't really apply to the question.

I’m going to assume that you think that Newtonian mechanics predicts singularities, since your answer implied that. That’s not so; that theory does not predict singularities. So why doesn’t r=0 hold in that theory? The answer is that, if it did, then the theory would predict a breakdown of the laws of physics. A theory need not predict that the laws of physics break down. GR does that voluntarily—it predicts its own demise by predicting that all objects below r=2M must fall to r=0. There is no postulate in Newtonian mechanics that precludes r=0. Postulates are not needed in theories to preclude absurdities. Plugging r=0 into either Newton’s equation for escape velocity or my equation is presumably invalid. In the OP I showed that my equation never predicts that a body must implode to r=0, unlike GR.

 

A valid point, thanks.

Ben was I think asking if it is applicable throughout the universe. It is meant to be. And it is applicable only in Schwarzschild geometry, the geometry of empty spacetime around a spherically symmetric, uncharged, nonrotating body. The theory of gravity I am building applies to only Schwarzschild geometry.

To what other flaw do you refer? I am aware of none.

 

No, the Schwarzschild metric is. The metric describes Schwarzschild geometry. Any theory of gravity can have its own metric that describes that geometry. Experimental tests of the Schwarzschild metric show how well it describes Schwarzschild geometry. Get it?

 

If I could derive the Schwarzschild metric from my new theory, it wouldn't be a new theory for Schwarzschild geometry. To be a new theory it must offer a different metric. The metric need not be derived, for no equation need be derived in any theory. But that doesn't mean that I will present it with no derivation. I will at least show that the new equation for escape velocity is incorporated into it.

 

The metric will have a logical derivation. But the derivation will not look anything like Schwarzschild’s.

At some level every theory incorporates a guess as to how nature works. That must be so, for nobody knows for sure how nature works.

 

Then it's a line of best fit - not an explanation.

I really don't see why you insist on clinging on to this idea or where you got it from. I suppose you could come up with a bunch of equations, collect experimantal evidence for each one seperately, and call the collective set of equations your "theory", but this would be inelegant and doesn't score well on Occam's razor (as each equation is effectively its own postulate). It also increases the risk that some of these equations may contradict one another.

A good theory is a relatively small set of postulates from which a great number of predictions demonstrably follow. By showing that one equation is a consequence of other equations in your theory, you reduce the number of necessary postulates. Not only is this preferable by Occam's razor (you explain as much as possible with as little as possible), but it allows a much stronger case to be built in favour of the whole theory: evidence for a particular equation or prediction is also evidence for all the postulates on which it is based.