Jump to content

Take a look at this paper about General Relativity


jcarlson
 Share

Recommended Posts

http://zanket.home.att.net/

 

This guy claims to have found a small flaw in GR with the way escape velocity is described. I'm usually very skeptical about this sort of thing, and while I am of course still skeptical, all his math seems to be in order, and he claims that the expirimental evidence for General Relativity with regard to the Schwarzschild metric is valid for his idea as well. To summarize, He is basically claiming that escape velocity at a certain radius is described in the same way by both Newtonian physics and General Relativity, when it should be described relativistically in GR. The implications of this are that Escape velocity never meets or exceeds c, and thus a black hole cannot exist.

 

Just looking for your opinions, especially those of you knowledgeable in cosmology and Relativity.

 

Also, there is an ongoing discussion about the paper here, to which I have posted a few times towards the end:

http://www.sciforums.com/showthread.php?t=47434

Link to comment
Share on other sites

  • Replies 111
  • Created
  • Last Reply

Top Posters In This Topic

It´s totally unreadable. Let alone that in the very first line of section 3 the author starts with "Newton’s equation for free-fall velocity in a uniform gravitational field ..." combined with the fact that I fail to find the point at which he realizes that he´s not talking about a uniform gravitational field (and no: I did not spent hours trying to find it) took away all the fun of trying to find an error in it.

The next possible one would be eq [3] which seems rather unjustified at first glance. But I lost my interest in the lines before that, already.

Link to comment
Share on other sites

I am not a good enough mathmatician to evaluate his math, but I do know that black holes are no longer a thoery, they are fact. This being said, I can't see how his theory could be correct.

 

As far as I know, there is no direct evidence of Black holes. Seeing as how no light, or any other information can escape the event horizon of a black hole, Its practically impossible to observe one directly. The closest we have come is determining several Black hole "candidates", or systems where a black hole is thought to exist based on the behavior of visible objects in the area.

Link to comment
Share on other sites

As far as I know, there is no direct evidence of Black holes. Seeing as how no light, or any other information can escape the event horizon of a black hole, Its practically impossible to observe one directly. The closest we have come is determining several Black hole "candidates", or systems where a black hole is thought to exist based on the behavior of visible objects in the area.

a black hole per se might not actually exist, but it is true that something like it does exist, which can pull light waves towards it in such a way that they apparently cannot escape. this is revealed by observations of how light is distorted around a canditate black hole in that there is intense light around it that is displaced from where it normally should be in space. at first it looks like a bright star or a quasar. its this intense gravitational lensing that provides evidence for a force that is explained by a black hole.

Link to comment
Share on other sites

Another one:

 

US.$50,000.00 to defend Relativity

http://members.aol.com/crebigsol/awards.htm

 

Awards of up to US.$50,000.00 each are hereby offered by the author, Cameron Y. Rebigsol, of this web site to people who can successfully defend Relativity - the most dominating theory in physics - in mathematical terms, and thus disprove the mathematical arguments made against relativity shown in Rebigsol's text MATHEMATICAL INVALIDITY OF RELATIVITY published at this web site.

Link to comment
Share on other sites

  • 2 weeks later...
I am not a good enough mathmatician to evaluate his math, but I do know that black holes are no longer a thoery, they are fact. This being said, I can't see how his theory could be correct.

 

Black holes are not a fact. The "black holes" found are just objects that—if general relativity (GR) is correct—would be black holes because their r-coordinate radius is less than their Schwarzschild radius, in which case GR predicts a black hole. In other words, the evidence shows a black hole only if the theory is valid; the evidence does not validate the theory. There is no direct evidence of a black hole.

Link to comment
Share on other sites

It´s totally unreadable. Let alone that in the very first line of section 3 the author starts with "Newton’s equation for free-fall velocity in a uniform gravitational field ..." combined with the fact that I fail to find the point at which he realizes that he´s not talking about a uniform gravitational field (and no: I did not spent hours trying to find it) took away all the fun of trying to find an error in it.

 

Immediately after I say "Newton’s equation for free-fall velocity in a uniform gravitational field ... " I give Newton’s equation for free-fall velocity in a uniform gravitational field, eq. 1. I don’t see a problem.

 

The next possible one would be eq [3] which seems rather unjustified at first glance.

 

How so?

Link to comment
Share on other sites

a black hole per se might not actually exist, but it is true that something like it does exist, which can pull light waves towards it in such a way that they apparently cannot escape.

 

Or that redshifts light so much that we cannot detect the object directly. My paper predicts that such an object can exist.

 

this is revealed by observations of how light is distorted around a canditate black hole in that there is intense light around it that is displaced from where it normally should be in space. at first it looks like a bright star or a quasar. its this intense gravitational lensing that provides evidence for a force that is explained by a black hole.

 

Agreed, so long as general relativity (GR) is held to be valid. In my theory the same indirect evidence would provide evidence for a mass that has a relativistic escape velocity at its surface, but an escape velocity still less than c, the speed of light.

Link to comment
Share on other sites

its damn near impossible to read. he changes the SI symbols for equations to other symbols so anyone with even a basic understanding of physics gets confused.

 

The Conventions section has a link that explains the geometric units used in the paper. Geometric units are common in papers about relativity, because the equations are simpler. G (the gravitational constant) = c (the speed of light) = 1, so G and c drop out of the equations. The velocity v becomes just a percentage of c, with no units. The mass M has units of length, and R = 2 * M in the paper. This is the unit system used by Taylor and Wheeler in their book Exploring Black Holes. It is used even in some of my layman’s books.

Link to comment
Share on other sites

- The point is that you are giving a movement equation for a uniform gravitational field. The problem is that you are not dealing with a uniform gravitational field.

 

- eq [3] seems unjustified at first glance because the reason you obtain it is "substituting [2] in [19]" with [19] standing in a block of other uncommented equations. It´s simply not enough for me to understand what you are doing there. It simply makes the impression as if the reason for doing it was that the letters match.

 

- the natural units are definitely not the problem in reading the text. The formatting and the writing style that seems like a patchwork of different internet sources are the problems - at least for me.

Link to comment
Share on other sites

- The point is that you are giving a movement equation for a uniform gravitational field. The problem is that you are not dealing with a uniform gravitational field.

 

Just because the section has equations for both uniform and nonuniform gravitational fields does not mean that the former is used incorrectly. You need to make a better case than that.

 

- eq [3] seems unjustified at first glance because I the reason you obtain it is "substituting [2] in [19]" with [19] standing in a block of other uncommented equations. It´s simply not enough for me to understand what you are doing there.

 

Section 1 shows (and section 2 reiterates) that free-fall velocity in a uniform gravitational field is given by eq. 19.

Link to comment
Share on other sites

It simply makes the impression as if the reason for doing it was that the letters match.

 

When the symbols are the same, including the same meaning, the substitution is valid. The reason for doing the substitution is to derive an equation that converts Newtonian velocity, k, to relativistic velocity, v.

 

The formatting and the writing style that seems like a patchwork of different internet sources are the problems - at least for me.

 

That’s too vague. Can you be more specific? For example, what exactly is the problem with the “internet sources”? Papers often reference sources.

 

If you just don’t like the look or feel of the paper, that’s okay—I can’t please everyone. I’m more interested in mathematical or logical problems of the paper.

Link to comment
Share on other sites

Perhaps it would help if you could simply explain why you are talking about a uniform gravitational field at all. I can´t see any reason for doing so because you mainly seem to consider non-uniform fields. And why do you plug your result for a uniform field [3] into an equation for an escape velocity [4,5] - a quantity inherently bond to non-uniform fields?

 

EDIT: The keyword in my comment about the patchwork of internet sources was "patchwork", not "internet sources". It´s imho not a fluent read. It reads as if you plug things in the equations because the letters are equal. That´s how reading your text makes me feel. No, I´m afraid I can´t be more specific about it.

Link to comment
Share on other sites

Perhaps it would help if you could simply explain why you are talking about a uniform gravitational field at all. I can´t see any reason for doing so because you mainly seem to consider non-uniform fields.

 

Newtonian velocity is denoted with a new symbol, k. Then Newton’s equation for free-fall velocity in a uniform gravitational field is used in a valid substitution to derive eq. 3, which converts k, Newtonian velocity, to v, relativistic velocity. Eq. 3 is not dependent upon a uniform field (k is just Newtonian velocity, it has nothing to do per se with acceleration). So eq. 3 can be used to convert Newtonian velocity to relativistic velocity regardless of how that velocity was obtained (whether in a uniform or nonuniform gravitational field). Eq. 3 is then used to convert Newton’s escape velocity equation to a relativistic (valid) equation.

 

And why do you plug your result for a uniform field [3] into an equation for an escape velocity [4,5] - a quantity inherently bond to non-uniform fields?

 

That eq. 3 is derived from an equation for a uniform field does not make it dependent on a uniform field. Its only input is k, Newtonian velocity. The same Newtonian velocity can be obtained by free-falling through either a uniform or a nonuniform field.

 

EDIT: The keyword in my comment about the patchwork of internet sources was "patchwork", not "internet sources". It´s imho not a fluent read. It reads as if you plug things in the equations because the letters are equal. That´s how reading your text makes me feel. No, I´m afraid I can´t be more specific about it.

 

To “plug things in the equations because the letters are equal” (that is, when the symbols have the same meaning) is substitution, a common algebraic manipulation. You’d have a hard time finding a theory of physics that doesn’t employ substitution.

Link to comment
Share on other sites

Quote from the link

Newton’s (and general relativity’s) equation for escape velocity is:

v = sqrt(R / r)

And earlyer you state

Schwarzschild radius ® The r-coordinate 2 * M

This would mean that you say Newtons equation for escape velocity is

v=sqrt((2Gm/c2)/r)

as the equation for a schwarzchild radius is r=(2Gm/c^2)

However I think you will find that newtons equation for escape velocity is

v=sqrt(2Gm/r). So you appear to have incorrectly stated newtons equation for escape velocity. Is this a major problem......?

 

`Scott

Link to comment
Share on other sites

However I think you will find that newtons equation for escape velocity is

v=sqrt(2Gm/r). So you appear to have incorrectly stated newtons equation for escape velocity. Is this a major problem......?

 

Note that geometric units are used throughout the paper. This means that c = G = 1. Then the equation for the Schwarzschild radius you gave here simplifies to r = 2M. In the paper, R = 2M. Then the escape velocity equation you gave here simplifies to v = sqrt(R / r), the same as in the paper.

Link to comment
Share on other sites

Note that geometric units are used throughout the paper. This means that c = G = 1. Then the equation for the Schwarzschild radius you gave here simplifies to r = 2M. In the paper, R = 2M. Then the escape velocity equation you gave here simplifies to v = sqrt(R / r), the same as in the paper.

I thought geometricized units were exclusively used in generaly reletivity? maybe I'm wrong. So does it work when you use normal units?

Link to comment
Share on other sites

I thought geometricized units were exclusively used in generaly reletivity? maybe I'm wrong. So does it work when you use normal units?

 

Geometric units are not limited to GR. It's a unit system that any theory of gravity can adopt. Geometric units can be converted with no ambiguity to SI units, as described in the link above. So my theory could be put in terms of SI units with no problem.

Link to comment
Share on other sites

  • 7 months later...
  • 4 months later...
I may challenge your comments, but they are always appreciated.

 

I have a comment about a statement from your website, but first I'd like to give an equation I will frequently refer to as Eq. (17) (see reference below), for readers without immediate access to the reference:

 

v = (2M/r)^{1/2} (17)

 

So, I have a comment on the following statement from your website:

 

According to general relativity, where v is less than c it equals the escape velocity there. (7) Then when v always asymptotes to c, so does escape velocity, in which case escape velocity is always less than c and then there are no black holes.

 

Where (7) refers to the book "Exploring Black Holes", p. 2-22, Eq. (17). I'm looking in the box "Newton Predicts the Horizon of a Black Hole" where Eq. (17) is listed, but I can't seem to find any statements that says that wherever "v is less than c it equals the escape velocity there".

 

Eq. (17) in the book refers to the velocity obtained by a free-falling object starting at a "great distance" from rest. Technically this "great distance" has to be infinity for Eq. (14) to hold (that's where the potential energy as defined in Eq. (13) also is zero). So, while equation (17) refers to the escape velocity of a black hole, an object at an arbitrary radius (at least if it's larger than the Schwarzschild radius) can have either larger or smaller velocity than that given by Eq. (17).

 

Finally, your statement that "the escape velocity is always less than c" litterally contradicts the book you're referring to, which actually says:

 

What is the maximum possible escape velocity? Here we elbow Newton aside and give the relativistic answer: The maximum escape velocity is the speed of light, v = [...] = 1.

 

If you look at Eq. (17) you'll see that for a small enough radius, the velocity will be larger than c, but because the max escape velocity is c, anything inside that radius will not be able to escape.

Link to comment
Share on other sites

Where (7) refers to the book "Exploring Black Holes", p. 2-22, Eq. (17). I'm looking in the box "Newton Predicts the Horizon of a Black Hole" where Eq. (17) is listed, but I can't seem to find any statements that says that wherever "v is less than c it equals the escape velocity there".

That is given by eq. 17, simply by the fact that this is GR’s equation for both escape velocity (as low as the Schwarzschild radius) and free-fall velocity from rest at infinity (directly measurable only above the Schwarzschild radius). Eq. 17 returns v < c above the Schwarzschild radius. Then, where section 2’s v (the velocity of objects fixed at each altitude as they pass directly by the particle) is less than c, it equals the escape velocity there.

 

Eq. (17) in the book refers to the velocity obtained by a free-falling object starting at a "great distance" from rest. Technically this "great distance" has to be infinity for Eq. (14) to hold (that's where the potential energy as defined in Eq. (13) also is zero). So, while equation (17) refers to the escape velocity of a black hole, an object at an arbitrary radius (at least if it's larger than the Schwarzschild radius) can have either larger or smaller velocity than that given by Eq. (17).

I don’t see your point. Eq. 17 is used to show just one example of how GR contradicts itself. I need only one example to refute the theory.

 

Finally, your statement that "the escape velocity is always less than c" litterally contradicts the book you're referring to, which actually says:

 

If you look at Eq. (17) you'll see that for a small enough radius, the velocity will be larger than c, but because the max escape velocity is c, anything inside that radius will not be able to escape.

Well, if I didn’t contradict the book then I’d have nothing new to say, right? There’s nothing wrong with using statements from the book to show a contradiction of GR. The contradiction is shown between the statements of the book and the new information shown in sections 1 and 2, which was inferred by means GR allows.

 

Sections 1 and 2 show that it is deducible from GR that v (the velocity of objects fixed at each altitude as they pass directly by the particle) always asymptotes to c (that is, as long as the particle falls), in which case v is always less than c, which contradicts eq. 17, GR’s equation for that velocity. The theory contradicts itself. According to eq. 17, where v is less than c, it equals the escape velocity there. Sections 1 and 2 show that v is always less than c. Then escape velocity is always less than c, and this was inferred by means GR allows.

Link to comment
Share on other sites

Thanks for your reply Zanket, I now think I know where you're misunderstanding. It appears as if you've forgotten or old pal, the photon. You see, Eq. (17) only applies to massive objects, and as we both know, their velocity is always less than c. For massless objects, however, like the good ol' photon, other rules apply, and, again, as we both know, the speed of light is always constant, so a photon moving radially towards a black hole will always travel at the speed of light.

 

So, if, from Eq. (17) we get a finite, nonzero radius when we plug in v = c (for the photon), there will be areas that require a v > c to escape from, i.e. Black Holes.

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
 Share


×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.