Abstract: A flaw of general relativity is exposed and is shown to source from a misapplication of the equivalence principle, the theory’s core postulate. A replacement for the Schwarzschild metric is simply derived. (The vast majority of experimental tests of general relativity have been tests of the Schwarzschild metric.) The new metric is shown to be confirmed by experiments of the four classical tests of general relativity. The predictions of the new metric are shown to diverge from those of the Schwarzschild metric as gravity strengthens. The cosmological implications explain some observations simpler than do alternative explanations.

A Flaw of General Relativity, a Fix, and Cosmological Implications (http://zanket.home.att.net/)

Edit: Changed abstract to reflect that the new metric is now confirmed by all four classical tests of general relativity.

Have you submitted this to a peer-reviewed journal?

Impressive, Zanket. From what I can understand of your paper so far, it seems to
parallel my own thoughts. I like it, not that my opinion carries any weight.

I am not going to pretend I understand everthing you have writen, honestly I have never even heard of Schwarzschild metric or before. From the proof's that you provided it all looks good.

I have a question though. What practical applications does your new equations have? For example, does it lead to new measurements of distances within the universe? You stated within the paper that you derive the same values that Newton's and Einstein's equations derive...

Have you submitted this to a peer-reviewed journal?

We could do a whole thread on this question. The answer is yes, and it was rejected out of hand. I subsequently corroborated through good sources that the journal I submitted it to has an unwritten policy to summarily reject papers from noncredentialed people; e.g. me.

Given that experience, I did some research and determined to my satisfaction that the odds of a “respected” journal reviewing my paper are nil. If such wasn’t for my lack of credentials alone, it would also be that the paper purports to overturn one of the most revered theories in physics. The days of an “unknown patent office clerk” getting a purportedly groundbreaking paper reviewed by a journal are over.

Impressive, Zanket.

Thanks!

From the proof's that you provided it all looks good.

That’s high praise given your moniker. :) Thanks!

What practical applications does your new equations have? For example, does it lead to new measurements of distances within the universe? You stated within the paper that you derive the same values that Newton's and Einstein's equations derive...

The abstract says, “The predictions of the new metric are shown to diverge from those of the Schwarzschild metric as gravity strengthens. The cosmological implications explain some observations simpler than do alternative explanations.”

The new metric approximates the Schwarzschild metric (Einstein) only in weak gravity. Fig. 10 is an example of the difference between the metrics in strong gravity (in which the Schwarzschild metric has not yet been experimentally tested).

Section 5 notes that black holes are not predicted by the new metric. (The singularity of a black hole is where attempts to merge GR and quantum mechanics blow up with infinities.) Section 8 resolves the flatness and horizon problems of cosmology using gravity alone (whereas, for example, the inflationary theory invokes ad hoc, undiscovered “scalar energy” to resolve those problems), and explains that the accelerating expansion of the universe is due to gravity alone (no ad hoc, undiscovered dark energy is required).

Don't forget, Einstein had a Ph.D. before he became a patent office clerk.

Unless someone can find an objection, this is clearly an important find! Well done.

http://www.the-highway.com/forum/images/icons/clapping.gif

The Crackpot Index (http://math.ucr.edu/home/baez/crackpot.html)

:D <P>

To get to another galaxy within their lifetimes, the crew of a relativistic rocket must effectively exceed the speed of light. This is doable thanks to length contraction. For instance, when the rocket has accelerated to a velocity of sqrt(0.5) ≈ 0.7071c the crew observes length contraction to a percentage given by eq. 9.13, ≈ 0.7071. One light year as they measured before they began accelerating now measures as 0.7071 proper light years along their axis of motion. At 0.7071c a distance of 0.7071 light years is traversed in one year. At this moment in their frame the crew traverses space at the rate of one formerly-measured light year (that is, as they measured before they began accelerating) every year, an effective velocity of c. An effective velocity of c equals an actual velocity of 0.7071c. Whereas the rocket’s actual velocity approaches a limit of c, its effective velocity approaches infinity. The rocket’s effective velocity is inversely proportional to the proper time to reach the galaxy. For example, at double the effective velocity the trip takes half the proper time.

Something doesn't seem right here...maybe you or someone could enlighten me on how this is possible. What reference frame is this in? etc... :confused:

The frame is “their frame”, the crew’s frame. Can you rephrase to be more specific?

Hi Zanket,
In what frame is the local acceleration due to gravity defined?
The frame of the central mass?
Or the frame of an object free-falling from infinity?

To what part are you referring?

OK, you've approached the derivation by using your equation:
v_eff = a * t

This is a sound equation, but only true for a defined in the frame of the accelerating object.

The old equation:
v = a * t
Is also a sound equation, but only true for a defined in a non-accelerating frame.

You've continued your derivation by equating a with the local gravitational acceleration g.

But which a is the right a to equate with g?
You have assumed that a in the free-falling object's rest frame equates to g.
Is this assumption valid?
Perhaps a in the central mass's rest frame is the one that equates to g?

I don't know which (if either) is correct... I only know that your derivation includes an unsupported assumption. This doesn't make it wrong - just not definitely right.

Hi Pete.
Do you understand what the terms 'a clock fixed at infinity' or 'a meterstick fixed at
infinity' mean? They are descriptions of the '0' frame with no gravitational influence.
This '0' frame can also be a global reference frame, with the 'central mass' in its own
frame located WITHIN the global frame, but not at the origin of the global frame. The free-falling object would also be in its
own frame located within the global frame. The gravitational acceleration of the central
mass would be calculated by its variation wrt the '0' frame, the global frame. The local
gravitational time dilation of the free-falling object be defined by its location in the
central masses' gravitational potential, again in reference to the overall global frame.
Remember the International Celestial Reference Frame we discussed earlier? The spatial
coordinates of that frame had their origin at the barycenter of the solar system. The
time component of that frame corresponds to Zanket's '0' frame. Clock tick rates of
both the central mass clock and the free-falling objects clock are both referenced to
the '0' baseline of the ICRF. Same with the gravitational potential, both of the other frames are referenced to the '0' gravity of the ICRF's baseline origin.

This is MY interpretation
of Zanket's paper, I feel he will not use the same description I do, but I believe mine
is compatible with his paper. That is why I said his paper paralleled my thoughts.

I don't mean to speak for zanket, it is his paper and his thread, I just thought I would give another way of explaining it.

Hi 2inquisitive,
You are still very confused about frames, I see. I haven't forgotten our thread, just busy.

Have you worked through zanket's deriviation?
Do you understand the point I make about whether a is defined relative to the free-falling mass or the central mass?

OK, you've approached the derivation by using your equation:
v_eff = a * t

This is a sound equation, but only true for a defined in the frame of the accelerating object.

The old equation:
v = a * t
Is also a sound equation, but only true for a defined in a non-accelerating frame.

See the top of section 9. In both equations, a is measured in an inertial frame. In both equations, a is the uniform acceleration felt by those in the frame of the noninertially accelerating object (e.g. a rocket), and felt to the same degree. In section 9, a is measured differently than it was for Galileo. But for both equations, when a = 1g, say, 1 Earth gravity is felt.

You've continued your derivation by equating a with the local gravitational acceleration g.

But which a is the right a to equate with g?

For both Newton and Einstein, 1g is the acceleration felt by those at sea level, and felt to the same degree. The equivalence principle implies that g must be measured in the same way that a is measured in section 9.

In both equations, a is measured in an inertial frame. In both equations, a is the uniform acceleration felt by those in the frame of the noninertially accelerating object (e.g. a rocket), and felt to the same degree.

That doesn't work...
a as measured in an inertial frame is not the same as the uniform acceleration felt by those in the frame of the noninertially accelerating object.

Say the rocket has constant acceleration of 1g as felt by those in the rocket's frame. As the rocket approaches light speed in some inertial frame, the rocket's acceleration as measured in that frame falls away, doesn't it?

The two accelerations are obviously not the same.
So, which acceleration equates to g, and why?

For both Newton and Einstein, 1g is the acceleration felt by those at sea level, and felt to the same degree. The equivalence principle implies that g must be measured in the same way that a is measured in section 9.
Correct, I think.

This implies that g must always be measured in the rest frame of the central mass, right?

If you were free-falling, how would you measure your acceleration?

I've got to say I'm tremendously impressed with zanket's work.
That's a hell of a lot more effort than most forumites are willing to go to.

That doesn't work...
a as measured in an inertial frame is not the same as the uniform acceleration felt by those in the frame of the noninertially accelerating object.

It is the same. Although you feel a uniform acceleration now, you are not accelerating in your frame. You accelerate relative to an object you drop; your acceleration is measurable in its frame, at least for the first moment. In Newton’s viewpoint your acceleration can continue to be measured in the object’s frame. In Einstein’s viewpoint, as implied by the equivalence principle, your acceleration must be measured in successive inertial (free-falling) frames momentarily comoving (stationary) with you, so only in the first moment can your acceleration be measured in the object’s frame.

Say the rocket has constant acceleration of 1g as felt by those in the rocket's frame. As the rocket approaches light speed in some inertial frame, the rocket's acceleration as measured in that frame falls away, doesn't it?

It does not fall away when the acceleration is measured in successive inertial frames momentarily comoving with the rocket. Each of such frames is like a launch pad, the rocket taking off from it at the same acceleration felt by the crew. See also the link in note 15 in the doc.

The two accelerations are obviously not the same.

Consider, if I asked you to tell me the formulas, for both Galileo and Einstein, that return the final velocity of a rocket given 1 Earth gravity felt by the crew, over a given time t in the gantry’s frame, you’d have to say eqs. 2.3 and 9.6 respectively. Then a must be the same thing in these formulas. And since, thanks to the equivalence principle, the crew can suppose that their rocket sits unpowered on the Earth’s surface, then the same formulas apply to g.

This implies that g must always be measured in the rest frame of the central mass, right?

I think I covered this one above.

If you were free-falling, how would you measure your acceleration?

In each moment I’d directly measure it relative to the object toward which I’m falling. For example, in each moment I could directly measure my acceleration relative to the face of the cliff I jumped off from. Suppose the cliff were on some other dense planet, such that I approached c relative to the cliff there. Then the cliff’s acceleration in my frame would fall away, just like you imagined for the rocket above.

Thank you for the kudos!

It does not fall away when the acceleration is measured in successive inertial frames momentarily comoving with the rocket.
Of course... but it does fall away when the acceleration is measured in a constant inertial frame - such as the rest frame of an inertial gravitating mass.
Consider, if I asked you to tell me the formulas, for both Galileo and Einstein, that return the final velocity of a rocket given 1 Earth gravity felt by the crew, over a given time t in the gantry’s frame, you’d have to say eqs. 2.3 and 9.6 respectively.
That's because you've explicitly specified acceleration in the comoving frames.
The acceleration in any single inertial frame is not the same, not a constant 1 g.

But this is really just quibbling, I think... I think the clincher is to come:


In each moment I’d directly measure it relative to the object toward which I’m falling. For example, in each moment I could directly measure my acceleration relative to the face of the cliff I jumped off from. Suppose the cliff were on some other dense planet, such that I approached c relative to the cliff there. Then the cliff’s acceleration in my frame would fall away, just like you imagined for the rocket above.
I agree.

Now, I'm going to go away and do some maths to answer the following questions:

1) According to Special Relativity, is your acceleration in the cliff's frame the same as the cliff's acceleration in your frame (with opposite sign, of course)?
2) If they were different, which one corresponds to g?

In each moment I’d directly measure it relative to the object toward which I’m falling. For example, in each moment I could directly measure my acceleration relative to the face of the cliff I jumped off from.Your example does not follow from your rule for measuring your acceleration. How might one jump from a cliff and fall towards said cliff? But that is neither here nor there, you could have simply said "In each moment I’d directly measure it relative to the object toward which I’m falling or away from which I'm falling."

Suppose the cliff were on some other dense planet, such that I approached c relative to the cliff there. Then the cliff’s acceleration in my frame would fall away, just like you imagined for the rocket above.If you are assuming your acceleration is constant in free fall on any planet, then you are ignoring air resistance which neccessitates a terminal velocity, v, where v << c.

Anyway, I fail to see what is so special about what you are saying. If you are considering your acceleration relative to your reference frame, then your acceleration is always 0. If you are considering your acceleration according to the instantaneous inertial reference frame you are in at any given moment, then you can say that you "feel" a constant acceleration relative to your instananeous velocity at that point. Note that this neccessarily requires another reference frame that is not accelerating to measure what your "instantaneous velocity" is. Now if you consider your acceleration in that reference frame, your acceleration is what I believe you refer to as "falling away" or in other words, your acceleration approaches 0 as your velocity as measured in this frame approaches c.

That's because you've explicitly specified acceleration in the comoving frames.

That doesn’t mean that the a in eq. 2.3 is incompatible with the a in section 9.

For both the equations in section 9 (which are not my equations BTW) and eq. 2.3, a is the uniform acceleration that, say, a rocket’s crew feels. For a given experiment involving a uniform acceleration felt by the crew, for both eq. 2.3 and eq. 9.6 you'd input the same value for a, right? You'd input the same gantry's time t, right? If yes to both, then eq. 9.6 supersedes eq. 2.3, and the fact that acceleration is measured invalidly in Galilean physics for a given uniform acceleration felt by the crew is irrelevant to that finding.

Put differently, just because acceleration is measured invalidly in Galilean physics does not mean that a in in Galilean physics is incomparable to a in Einsteinian physics. They can be compared where they have common ground. Eq. 2.3 has common ground with the equations in section 9 in that the a for both is the acceleration felt by the crew. And that lets me validly say, in section 2, that “the formula that Galileo thought gives velocity (eq. 2.3) instead gives effective velocity (eq. 2.4).” The complete implied statement is, “the formula that Galileo thought gives velocity (eq. 2.3) instead gives effective velocity (eq. 2.4), where a in both equations is the acceleration felt by, say, a rocket’s crew.”

[Edit to add:] I’ve been trying to remember the conclusion I came to long ago about this issue. I just remembered it, and it’s a little twist on the above comments. In the “Conventions Herein” section, for the definition of a, I say, “see definition in section 9.” In this way I’m using the same definition for a for both eq. 2.3 and the equations in section 9. Now, if for eq. 2.3 you measure the acceleration as given in section 9 (using comoving inertial frames), then in Galileo’s viewpoint, in which the speed of light is not a speed limit, you get the same acceleration as you would if you measured the acceleration continuously in the gantry’s frame. You can measure the acceleration either way in Galilean physics; the result of the measurement is the same, hence the result of the equation is the same. This argument applies as well to Newtonian physics for g.

1) According to Special Relativity, is your acceleration in the cliff's frame the same as the cliff's acceleration in your frame (with opposite sign, of course)?

To keep it simpler, I’d stick to a uniform gravitational field, such as that within a relativistic rocket. Imagine that the cliff is within the rocket.

2) If they were different, which one corresponds to g?

Section 1 shows that in SR it must be the acceleration felt by someone affixed to the cliff, which is measurable in the frame of a stone dropped by that someone, for a single moment. The stone’s frame is an inertial frame momentarily comoving with the someone.

Regardless how you measure g in Newtonian mechanics, so long as g is a uniform acceleration felt by someone affixed to the cliff, or the rocket’s crew, etc., then I can validly say, in section 2, that “the formula that Newton thought gives velocity (eq. 2.3) instead gives effective velocity (eq. 2.4).” The complete implied statement is, “the formula that Newton thought gives velocity (eq. 2.3) instead gives effective velocity (eq. 2.4), where g in both equations is the acceleration felt by, say, someone on the Earth’s surface.”

Your example does not follow from your rule for measuring your acceleration. How might one jump from a cliff and fall towards said cliff? But that is neither here nor there, you could have simply said "In each moment I’d directly measure it relative to the object toward which I’m falling or away from which I'm falling."

I’m falling toward the ground at the bottom of a cliff face I’m passing by—I think Pete figured that out.

If you are assuming your acceleration is constant in free fall on any planet, then you are ignoring air resistance which neccessitates a terminal velocity, v, where v << c.

I think ignoring air resistance is implied for this type of discussion.

Anyway, I fail to see what is so special about what you are saying.

I’m just responding to Pete’s points. Nothing special intended.

To get to another galaxy within their lifetimes, the crew of a relativistic rocket must effectively exceed the speed of light. This is doable thanks to length contraction.Interesting take on the concept of length contraction. Let's analyze your statement "the crew of a relativistic rocket must effectively exceed the speed of light." I suppose we first have to define a relativistic rocket as one that has achieved a certain velocity. Since the rocket is travelling between galaxies, I assume it is prudent to measure the velocity of this rocket with respect to the center of the Milky Way. Now I am not going to assume I know what you mean by "effectively exceed the speed of light" because to be truely honest, I haven't the slightest clue. Maybe further reading of your paper will guide me. Anyway, I would like to point out that if you were able to even travel at .999(repeat with 100 9's)c then without the aid of a calculator (I might be slightly off) the trip time according to the crew onboard would be about 1 second neglecting the time it takes to accelerate to said velocity.

Now I haven't even taken on the length contraction issue yet, but I'll see what else your paper has to say before I comment further. But at this point, I don't find your initial arguement credible yet alone anything that follows.

Interesting take on the concept of length contraction.

There’s nothing new there. It’s standard SR.

Let's analyze your statement "the crew of a relativistic rocket must effectively exceed the speed of light." I suppose we first have to define a relativistic rocket as one that has achieved a certain velocity.

The “Conventions Herein” section says that a relativistic rocket is “a rocket accelerating under the conditions specified in section 9.”

Now I am not going to assume I know what you mean by "effectively exceed the speed of light" because to be truely honest, I haven't the slightest clue.

I’m using the dictionary definition of “effective.” You say the crew can traverse the galaxy in 1 second on their clock. I agree. That’s effectively exceeding the speed of light, without actually exceeding it.

Now I haven't even taken on the length contraction issue yet, but I'll see what else your paper has to say before I comment further. But at this point, I don't find your initial arguement credible yet alone anything that follows.

The sentences you debated are standard SR and you’ve shown that you don’t know what “effective” means. Then I wouldn’t expect you to find any part of the paper to be credible.

There’s nothing new there. It’s standard SR. Special Relativity does not say you can exceed the speed of light. In fact it exactly says that you cannot exceed the speed of light and one of these reasons is because of length contraction. However you just said, To get to another galaxy within their lifetimes, the crew of a relativistic rocket must effectively exceed the speed of light. This is doable thanks to length contraction. That you can exceed the speed of light because of length contraction with the quantifier "effectively". I know what the meaning of effective (http://www.google.com/search?q=define+effective&sourceid=mozilla-search&start=0&start=0&ie=utf-8&oe=utf-8&client=firefox-a&rls=org.mozilla:en-US:official) is. My point is, in what context are you saying the speed of light is effectively exceeded? It seems to me you are comparing apples to oranges.

The “Conventions Herein” section says that a relativistic rocket is “a rocket accelerating under the conditions specified in section 9.” This is nothing special that allows the rocket to exceed the speed of light.



I’m using the dictionary definition of “effective.” You say the crew can traverse the galaxy in 1 second on their clock. I agree. That’s effectively exceeding the speed of light, without actually exceeding it. No it isn't. I challenge you to find one person with appropriate credentials that will agree with this statement.

The sentences you debated are standard SR and you’ve shown that you don’t know what “effective” means. Then I wouldn’t expect you to find any part of the paper to be credible. They are not standard special relativity. Length contraction does not allow you to exceed the speed of light.

Aer:
He is saying because they crew travels farther then 1 lightyear is less then one standard year, they are "effectively" surpassing c.

I don't agree with this, and I tried to point it out eariler in this threat. I don't know how to fully explain why he's wrong, but it has to due with the frame of reference and length contraction. Perhaps someone will be able to put forward a better explanation.

Special Relativity does not say you can exceed the speed of light. In fact it exactly says that you cannot exceed the speed of light and one of these reasons is because of length contraction.

I did not say or imply otherwise.

My point is, in what context are you saying the speed of light is effectively exceeded? It seems to me you are comparing apples to oranges.

In the context of the dictionary definition “in practice, even if not officially so”. Given that you can traverse between galaxies in your lifetime, even though you cannot officially exceed c, in practice you can (thanks to length contraction).

This is nothing special that allows the rocket to exceed the speed of light.

I did not say or imply otherwise.

No it isn't. I challenge you to find one person with appropriate credentials that will agree with this statement.

Then you do not understand the definition of “effective.” Here is the example used in my dictionary: “She was the effective leader during the premier's illness”. Was she the leader? No. Was she the leader in practice? Yes. Can you exceed c while traversing between galaxies? No. Can you exceed c in practice, by traversing between galaxies in any time on your clock? Yes.

Length contraction does not allow you to exceed the speed of light.

I did not say or imply otherwise. Please be more careful before you allude that I said or implied something.

He is saying because they crew travels farther then 1 lightyear is less then one standard year, they are "effectively" surpassing c.

Agreed. And I am careful to say that the “1 light year” is as measured before they began accelerating.

I don't agree with this, and I tried to point it out eariler in this threat.

You didn’t say specifically why you disagree with it. Do you disagree that the crew can traverse between galaxies in their lifetime? Or?

Aer:
He is saying because they crew travels farther then 1 lightyear is less then one standard year, they are "effectively" surpassing c.

I don't agree with this, and I tried to point it out eariler in this threat. I don't know how to fully explain why he's wrong, but it has to due with the frame of reference and length contraction. Perhaps someone will be able to put forward a better explanation.So zanket's effective velocity is v = contracted-length / rest-time where:

contracted-length = distance traveled as measured by the ship reference frame moving relative to the Milky Way.

rest-time = time it takes for said ship moving relative to the Milky Way to travel to another galaxy according to a stationary observer in the Milky Way reference frame.

This is not a measure of velocity even if zanket wants to label it as "effective velocity". You cannot take two quanities from two different inertial reference frames and attribute some type of significance to it in either reference frame. That is to say, the crew has an infinite number of "effective velocities" as defined above (If this is what zanket has intended) according to the infinite number of inertial reference frames you can compare the crews velocity to.

It is a meaningless quantity and should not be used in any sequential derivation.

I could easily define a second effective velocity as v = rest-length / time-dilated which would not be equal to the "effective velocity" defined above and be as equally useless.

Then you do not understand the definition of “effective.” Here is the example used in my dictionary: “She was the effective leader during the premier's illness”. Was she the leader? No. Was she the leader in practice? Yes. Can you exceed c while traversing between galaxies? No. Can you exceed c in practice, by traversing between galaxies in any time on your clock? Yes. Very well, do you agree that your "effective velocity" measurement is not a measure of velocity?

I think you don't understand that comparing two quantities in different reference frames is the same as comparing apples and oranges.

So zanket's effective velocity is v = contracted-length / rest-time where:

I did not say or imply this. My equation for effective velocity, eq. 2.1, differs.

Very well, do you agree that your "effective velocity" measurement is not a measure of velocity?

It is a measure of effective velocity, different from actual velocity.

I think you don't understand that comparing two quantities in different reference frames is the same as comparing apples and oranges.

I did no such comparison. Eq. 2.1 differs from what you say it is above.

You didn’t say specifically why you disagree with it. Do you disagree that the crew can traverse between galaxies in their lifetime? Or?

I agree that they could travel between galaxies, but I don't think they exceed "c" (effectively or not)

I guess the way I see it the definition of a light year (distance) is depentant on the speed at which your traveling. Therefore if said ship is travelling at or near "c" then the length of a lightyear should be adjusted accordingly.

Shouldn't length be expanded near "c" not contracted.

Am I making sense, or am I just completely backwards here? :confused:

I did not say or imply this. My equation for effective velocity, eq. 2.1, differs.



It is a measure of effective velocity, different from actual velocity.



I did no such comparison. Eq. 2.1 differs from what you say it is above. OK, I'll ignore the fact that your equation 2.1 comes out of thin air (non-formula explaination prior to not included). And I'll reproduce it here:

veff ≡ v / sqrt(1 - v^2) [2.1]

Apparently this equation is a definition. That's fine, I'll work with that.

At this point I want to point out that the 2 equations I gave above should be interchanged with each other. However, I've already said that both equations are meaningless so it doesn't matter which one we are talking about.

The equation I am saying is exactly the same as your 2.1 is:

v = rest-length / time-dilated

OR

veff = L / T'

Now I'll prove it:

L = &gamma; L'
T' = &gamma; T
V = L' / T'

L / T' = L' / T ... (L = &gamma; L') ... (T' = &gamma; T)
L / T' = V T' / T ... (L' = v T')
L / T' = V &gamma; ... (T' / T = &gamma; )
L / T' = V / sqrt(1 - V^2)
L / T' = veff

I agree that they could travel between galaxies, but I don't think they exceed "c" (effectively or not)

I guess the way I see it the definition of a light year (distance) is depentant on the speed at which your traveling. Therefore if said ship is travelling at or near "c" then the length of a lightyear should be adjusted accordingly.

OK, your issue seems to be with the word “effectively.” It seems that you agree that, while you cannot exceed c, the speed of light limit does not prevent you from getting between any given points A and B in any time on your clock.

Let the distance between galaxies be a million light years as measured by the crew before they take off. Let them cross that gulf in 1 year on their clock. Although they never exceed c, they effectively crossed at an average rate of a million light years per year, right? Did they actually cross a million light years? No, they didn’t have to, thanks to length contraction.

There is no problem with saying that their effective rate of travel was 1 million c. It’s just another way of expressing their velocity. Their effective velocity relates to a specific actual velocity less than c.

There is no problem with saying that their effective rate of travel was 1 million c. It’s just another way of expressing their velocity. Their effective velocity relates to a specific actual velocity less than c. As I pointed out initially, your definition of "effective velocity" is not a velocity. Any meaning you might think it has cannot be applied into either frame used to derive it (I derived it above, you mearly defined it) or any other frame for that matter.

OK, I'll ignore the fact that your equation 2.1 comes out of thin air (non-formula explaination prior to not included). And I'll reproduce it here:

veff ≡ v / sqrt(1 - v^2) [2.1]

Apparently this equation is a definition. That's fine, I'll work with that.

Yes, it is a definition, which is why I say “Effective velocity ... is defined herein as ...” and use the ≡ symbol (indicates a definition) in the equation.

At this point I want to point out that the 2 equations I gave above should be interchanged with each other. However, I've already said that both equations are meaningless so it doesn't matter which one we are talking about.

Regardless what you can equate eq. 2.1 to, it is not meaningless. As I have defined it, a given effective velocity relates to a specific actual velocity. They have a one-to-one correlation. That makes the definition perfectly okay. Velocity can be stated in terms of kilometers or miles per hour, as a fraction of c, and lots of other ways. My effective velocity states it in terms of “velocity [as a fraction of c] divided by the special relativistic distortion factor (eq. 9.13) for that velocity”. Nothing wrong with that. It is convertible to other ways of specifying velocity with no ambiguity.

As I pointed out initially, your definition of "effective velocity" is not a velocity. Any meaning you might think it has cannot be applied into either frame used to derive it (I derived it above, you mearly defined it) or any other frame for that matter.

Do you agree that a velocity expressed in terms of kilometers per hour can be expressed in terms of miles per hour? You cannot agree to that, given your logic here.

Look at eq. 2.1. The only variable on the right-hand side is v. If I define “Zanket’s velocity” as 2 * v, is that meaningless? Of course not. I can define it however I want, so long as it is convertible to other ways of specifying velocity with no ambiguity. If I am moving at v relative to something, do I need to apply another frame to calculate Zanket’s velocity? No, there’s a one-to-one correlation between v and Zanket’s velocity. It’s a simple conversion.

Yes, it is a definition, which is why I say “Effective velocity ... is defined herein as ...” and use the ≡ symbol (indicates a definition) in the equation. I have a firm understanding of the symbol ≡, hence my comment. :rolleyes:

Regardless what you can equate eq. 2.1 to, it is not meaningless. Unsupported claim by you.

As I have defined it, a given effective velocity relates to a specific actual velocity. They have a one-to-one correlation. That makes the definition perfectly okay. This is commonly referred to as scaling. Why are you scaling the velocity? You apparently do not realize if you scale one velocity, you must scale all other velocities accordingly (i.e. c becomes c*&gamma; )

Velocity can be stated in terms of kilometers or miles per hour, as a fraction of c, and lots of other ways. What you are doing does not relate to this in ANY WAY. Conversion to different units is one thing, but you are multiplying by a non-dimension quantity (hence, not a unit conversion). A fraction of c is also a unit as c must be defined in some sort of units.

My effective velocity states it in terms of “velocity [as a fraction of c] divided by the special relativistic distortion factor (eq. 9.13) for that velocity”. I know this, you scaled the velocity with &gamma;

Nothing wrong with that. It is convertible to other ways of specifying velocity with no ambiguity. Obviously you are having problems with the ambiguity caused by your own notation.

Do you agree that a velocity expressed in terms of kilometers per hour can be expressed in terms of miles per hour? You cannot agree to that, given your logic here. If you understood anything I said, you'd realize this statement is asinine.

Look at eq. 2.1. The only variable on the right-hand side is v. If I define “Zanket’s velocity” as 2 * v, is that meaningless? did you multiply all other velocities by 2?

Of course not. You cannot say this until you answer the above.

I can define it however I want, so long as it is convertible to other ways of specifying velocity with no ambiguity. This is a rehash of misguided thoughts you expressed above.

If I am moving at v relative to something, do I need to apply another frame to calculate Zanket’s velocity? Zanket's velocity is a scaled velocity, what is the point?

No, there’s a one-to-one correlation between v and Zanket’s velocity. It’s a simple conversion.
conversion (http://www.google.com/search?q=define+conversion&sourceid=mozilla-search&start=0&start=0&ie=utf-8&oe=utf-8&client=firefox-a&rls=org.mozilla:en-US:official)
scale (http://www.google.com/search?q=define+%22to+scale%22&sourceid=mozilla-search&start=0&start=0&ie=utf-8&oe=utf-8&client=firefox-a&rls=org.mozilla:en-US:official)

I have a firm understanding of the symbol ≡, hence my comment. :rolleyes:

Then why did you say, “Apparently this equation is a definition,” as if it might not be one?

Why are you scaling the velocity?

So I can use eq. 2.1 in subsequent derivations.

You apparently do not realize if you scale one velocity, you must scale all other velocities accordingly (i.e. c becomes c*&gamma; )

Why? When Zanket’s velocity is defined as 2 * v, c stays c. Zanket’s velocity is not an actual velocity. Neither is the effective velocity I defined. Hence the qualifier “effective”.

What you are doing does not relate to this in ANY WAY.

It relates in that it’s a one-to-one conversion, just like a conversion to different units.

Obviously you are having problems with the ambiguity caused by your own notation.

How so? I see no ambiguity.

did you multiply all other velocities by 2?

Zanket’s velocity is not an actual velocity. It’s twice an actual velocity. Why would I multiply the actual velocity by 2 as well?

Zanket's velocity is a scaled velocity, what is the point?

The point of defining effective velocity is so I can use the equation (eq. 2.1) in subsequent derivations, to make other points.

Then why did you say, “Apparently this equation is a definition,” as if it might not be one?



So I can use eq. 2.1 in subsequent derivations.



Why? When Zanket’s velocity is defined as 2 * v, c stays c. Zanket’s velocity is not an actual velocity. Neither is the effective velocity I defined. Hence the qualifier “effective”.



It relates in that it’s a one-to-one conversion, just like a conversion to different units.



How so? I see no ambiguity.



Zanket’s velocity is not an actual velocity. It’s twice an actual velocity. Why would I multiply the actual velocity by 2 as well?



The point of defining effective velocity is so I can use the equation (eq. 2.1) in subsequent derivations, to make other points. I can testify as to exactly why your paper will never be peer review in a scientific journal.

You continue to express the idea that scaling is the same as conversion. It is not!

You are using something that you say is not a velocity, but yet has the units of velocity in unit dependent equations. Absurd!

And you say the point of defining something is to use it in derivations. No! The point of defining something is to express exactly what a symbol represents. Whatever your "effective velocity" represents is unknown by your definition. But we've been able to conclude one thing it is not and that is velocity.

You continue to express the idea that scaling is the same as conversion. It is not!

I did not say or imply that. I said they relate in that they are both one-to-one conversions.

You are using something that you say is not a velocity, but yet has the units of velocity in unit dependent equations. Absurd!

You’re getting hung up on the word “velocity” in “effective velocity.” The symbol v_eff just represents the right-hand side of the equation, v / sqrt(1 - v^2). That’s all, and that’s a perfectly valid definition. Had I used the symbol w instead of v_eff and referred to it only as w, then we could not have had this debate, right?

And you say the point of defining something is to use it in derivations.

I did not say that. What was my exact quote?

No! The point of defining something is to express exactly what a symbol represents.

That is the point of defining something in general. You’re nitpicking. Obviously there is a larger point in defining something.

Whatever your "effective velocity" represents is unknown by your definition.

It clearly represents the right-hand side of the equation (eq. 2.1). Notice that Pete was not confused by that.

But we've been able to conclude one thing it is not and that is velocity.

I did not say or imply otherwise.

I did not say or imply that. I said they relate in that they are both one-to-one conversions. This is a moot point compared to the other issues, so I'm not going to argue with you over this.



You’re getting hung up on the word “velocity” in “effective velocity.”

Let's take a look at some of what you say.

Can you exceed c while traversing between galaxies? No. Can you exceed c in practice, by traversing between galaxies in any time on your clock? Yes. You are telling me here that you are effectively exceeding the speed of light. What does it mean to effectively exceed the speed of light when this "effective velocity" is not a velocity. :bugeye: Are you going to now say that your "effective velocity" is a velocity? Or do you wish to retract your entire argument you've been making thus far?

Let the distance between galaxies be a million light years as measured by the crew before they take off. Let them cross that gulf in 1 year on their clock. Although they never exceed c, they effectively crossed at an average rate of a million light years per year, right? Did they actually cross a million light years? No, they didn’t have to, thanks to length contraction. "effectively crossed at an average rate" - average rate of what? based on your calculation, it is an average rate of rest-length/time-dilated. So by your own argument, you are confirming what I said:


The equation I am saying is exactly the same as your 2.1 is:

v = rest-length / time-dilated

OR

veff = L / T'

Now I'll prove it:

L = γ L'
T' = γ T
V = L' / T'

L / T' = L' / T ... (L = γ L') ... (T' = γ T)
L / T' = V T' / T ... (L' = v T')
L / T' = V γ ... (T' / T = γ )
L / T' = V / sqrt(1 - V^2)
L / T' = veff

Now, not only am I telling you this, but you said the exact same thing yourself! So tell me, what meaning are you attaching to rest-length/time-dilated and how is it applied to the rest of your theory. I'll tell you right now that rest-length/time-dilated is a useless quantity, you are taking two quantities (rest-length and time-dilated) from two different inertial reference frames and coming up with a quantity you call "effective velocity" and trying to argue that is significant somehow. I am not going to argue the absurdity of this point any further until you bring to me anyone who agrees with your position. Good Luck!

You are telling me here that you are effectively exceeding the speed of light. What does it mean to effectively exceed the speed of light when this "effective velocity" is not a velocity. :bugeye:

It means what I described at the top of section 2. I could have called it “blobwow” instead of “effective velocity,” but I showed in section 2 how “effective velocity” is an appropriate name. It should not be confused with actual velocity.

"effectively crossed at an average rate" - average rate of what? based on your calculation, it is an average rate of rest-length/time-dilated. So by your own argument, you are confirming what I said:

Whatever you can equate to eq. 2.1 doesn’t change the fact that it is a valid definition.

I am not going to argue the absurdity of this point any further until you bring to me anyone who agrees with your position. Good Luck!

You have essentially argued that I cannot define a symbol to be equivalent to v / sqrt(1 - v^2), and then use that symbol in some other derivation. That’s basic algebraic substitution! There’s no problem with it, which explains why Pete did not contest it. It doesn’t matter how meaningful or not you think v / sqrt(1 - v^2) is. It doesn’t matter what descriptive name I give the symbol.

You have essentially argued that I cannot define a symbol to be equivalent to v / sqrt(1 - v^2), and then use that symbol in some other derivation. I give up... you want to substitute apples for oranges and then refer to the authority of algebra to say its valid. If you want to go down this road, fine - I'll prove you cannot do this either!

First, one comment -
That’s basic algebraic substitution! There’s no problem with it, which explains why Pete did not contest it.I am not going to argue what Pete says or does not say on this issue as he has not made any comment. The mere fact that he said nothing does not mean he would neccessarily agree with you. Sorry, you can not use this to prove that someone else agrees with you!

OK, moving on... I'll will reproduce your derivation here:

v = (a * t) / sqrt(1 + (a * t)^2)

This is the equation you start with, where we define:

v Final velocity, as a fraction of c; no units (Oh! but it does have units, whatever units you define c in, note that if you define c=1, then the units are unit-length/unit-time, where light travels 1 unit-length over the duration of 1 unit-time)
a Acceleration as described above, in units where c = 1 (vague? maybe my explaination above is what you meant)
t Time elapsed in the gantry’s frame, in units where c = 1 (same)

OK, so what do you mean by Final velocity? because everyone here will know velocity to be v = a * t, yet you have v = (a * t) / sqrt(1 + (a * t)^2) and call it "Final velocity".

Forgive me, but I cannot proceed further with this derivation until you clarify this issue. What is "Final velocity"? It appears very much like you are starting with a result, doing a few manipulations and arriving at the result you started with. This is no proof!

Sorry I cannot disprove what I intended to as I cannot even get your proof off the ground.

The mere fact that he said nothing does not mean he would neccessarily agree with you.

You’re right. It’s just a theory.

v = (a * t) / sqrt(1 + (a * t)^2)

This is the equation you start with, where we define:

v Final velocity, as a fraction of c; no units (Oh! but it does have units, whatever units you define c in, note that if you define c=1, then the units are unit-length/unit-time, where light travels 1 unit-length over the duration of 1 unit-time)

When c = 1, as it does here, it is dimensionless; no units. See also here (http://quantumrelativity.calsci.com/Relativity/Chapter3.html) and here (http://home.earthlink.net/~djmp/ScientificResearch.html) (search for "no units"). When v is a fraction of c, it is also dimensionless.

a Acceleration as described above, in units where c = 1 (vague? maybe my explaination above is what you meant)
t Time elapsed in the gantry’s frame, in units where c = 1 (same)

OK, so what do you mean by Final velocity? because everyone here will know velocity to be v = a * t, yet you have v = (a * t) / sqrt(1 + (a * t)^2) and call it "Final velocity".

v = a * t is Galileo’s equation, eq. 2.3. The equations you quote are from section 9, which are not mine. They source from The Relativistic Rocket (http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html), in which v is the final velocity. (Note the comment therein about improving "the effective rate of travel", the same thing I elaborate upon in section 2 and call "effective velocity".) I changed those equations only to set c = 1, making v a fraction of c. The final velocity v in section 9 has the same meaning as it does for Galileo’s equation, eq. 2.3. The equations in section 9 are valid (assuming SR—special relativity—is valid) whereas Galileo’s equation is invalid.

It appears very much like you are starting with a result, doing a few manipulations and arriving at the result you started with. This is no proof!

The equations in section 9 are published equations, which is why they are presented in an appendix. I use them in my derivations.

Sorry I cannot disprove what I intended to as I cannot even get your proof off the ground.

You might read more carefully. I make it clear in section 9 that those are not my equations.

When c = 1, as it does here, it is dimensionless; no units. See also here (http://quantumrelativity.calsci.com/Relativity/Chapter3.html) and here (http://home.earthlink.net/~djmp/ScientificResearch.html) (search for "no units"). When v is a fraction of c, it is also dimensionless. It would be more proper to state that v is non-dimensionalized by dividing it by c=1.



v = a * t is Galileo’s equation, eq. 2.3. The equations you quote are from section 9, which are not mine. They source from The Relativistic Rocket (http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html), in which v is the final velocity. (Note the comment therein about improving "the effective rate of travel", the same thing I elaborate upon in section 2 and call "effective velocity".) I changed those equations only to set c = 1, making v a fraction of c. The final velocity v in section 9 has the same meaning as it does for Galileo’s equation, eq. 2.3. The equations in section 9 are valid (assuming SR—special relativity—is valid) whereas Galileo’s equation is invalid.



The equations in section 9 are published equations, which is why they are presented in an appendix. I use them in my derivations.



You might read more carefully. I make it clear in section 9 that those are not my equations.

I must conceed that I was skimming through your appendices and missed precisely what equations you were using.

Anyway, we start with:

v = (a * t) / sqrt(1 + (a * t)^2) [9.10]

Substitute γ in eq. 9.10 for sqrt(1 + (a * t)^2)

However, from eq. 9.12: γ = 1 / sqrt(1 - v^2)

from γ = 1 / sqrt(1 - (a * t)^2) & γ = 1 / sqrt(1 - v^2), v = a * t

Now which is it? Is v = a * t or v = (a * t) / sqrt(1 + (a * t)^2). It certainly cannot be both.

Look Zanket, Aer is right. Effective velocity is meaningless. If you just want to define this term and use it somehow in derivations, that's okay. But you can't assume that in order to reach another galaxy, it is enough for the effective velocity to exceed the velocity of light. It makes some sense in talking about using time dilation to reach a far-away place during a short lifetime, as in the muon experiment. Take a look at the relevant post in this page, and also at the website.
http://www.sciforums.com/showthread.php?t=47257&page=5&pp=20
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/muon.html

It would be more proper to state that v is non-dimensionalized by dividing it by c=1.

I use the same terminology for v and c that Taylor and Wheeler use in their book Exploring Black Holes. If it’s good enough for them, it works for me.

Now which is it? Is v = a * t or v = (a * t) / sqrt(1 + (a * t)^2). It certainly cannot be both.

In my previous post to you I said, “v = a * t is Galileo’s equation, eq. 2.3. ... Galileo’s equation is invalid.” And in section 2 I say, “The equations in section 9 supersede Galileo’s formulas for acceleration. ... Eq. 2.3, which invalidly allows a velocity of c or greater, is replaced by eqs. 9.5 or 9.6.”

Look Zanket, Aer is right. Effective velocity is meaningless. If you just want to define this term and use it somehow in derivations, that's okay. But you can't assume that in order to reach another galaxy, it is enough for the effective velocity to exceed the velocity of light.

It is enough and hence not meaningless, because a given effective velocity corresponds one-to-one to an actual velocity. Eq. 2.2 converts effective velocity to actual velocity. Suppose you want to get from the Milky Way to the Andromeda galaxy, 2 million light years away as we measure it, in 2 years on your clock while moving at constant velocity between the galaxies. That requires an effective velocity of (2 million light years / 2 years) = 1 million c. Plugging 1 million into eq. 2.2 returns v = 99.99999999995% of c. That’s the actual velocity at which you must move relative to the galaxies. Effective velocity is just another way of thinking of an actual velocity.

It makes some sense in talking about using time dilation to reach a far-away place during a short lifetime, as in the muon experiment.

If it makes sense for that scenario, then how is it meaningless? And how is reaching a galaxy in any time on your clock not reaching a far-away place during a short lifetime?

The Usenet Physics FAQ mentions effective velocity here:

From The Relativistic Rocket (http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html):
If a rocket accelerates at 1g (9.81 m/s²) the crew will experience the equivalent of a gravitational field with the same strength as that on Earth. If this could be maintained for long enough they would eventually receive the benefits of the relativistic effects which improve the effective rate of travel.

Effective velocity applies to you traversing between galaxies the same as it does to the muon traversing the Earth’s atmosphere. There is no pertinent difference in those scenarios. When the muon’s velocity relative to the Earth is 0.999c, its effective velocity, given by eq. 2.1, is about 22c.

In my previous post to you I said, “v = a * t is Galileo’s equation, eq. 2.3. ... Galileo’s equation is invalid.” And in section 2 I say, “The equations in section 9 supersede Galileo’s formulas for acceleration. ... Eq. 2.3, which invalidly allows a velocity of c or greater, is replaced by eqs. 9.5 or 9.6.” !!! Not only do you have total disregard for what I just said, you seem to not be able to follow your own derivation!

First of all, these derivations should not be thrown into an appendix as you have done. Appendicies are used for reference of other peoples work or side explainations - not new derivations that are at the core of the abstract of your paper. The mere fact that I have to produce the derivation here in an ordered manner that would be expected of a scientific paper because the way the derivation is presented in your paper is inadequate is very telling.

Here is your derivation:

Starting with the equation for a relativistic rocket as shown in eq. 1 (reference to paper on relativistic rocket would be included here), we substitute the quantity of γ in eq. 2 (reference) into eq. 1 to get eq. 3.

v = (a * t) / sqrt(1 + (a * t)^2) [1]

γ = sqrt(1 + (a * t)^2) [2]

v = (a * t) / γ [3]

Eq 3. can be rewritten as:

v * γ = a * t [4]

The quantity of γ as shown in eq. 5 (reference) is substituted into eq. 4 to obtain eq. 6.

γ = 1 / sqrt(1 - v^2) [5]

v / sqrt(1 - v^2) = a * t [6]

From the defintion of "effective velocity", we have eq. 7.

veff = a * t [7].



This is your derivation as it sort of appears on your page (your version isn't very good). Now back to what I said before, you use

γ = sqrt(1 + (a * t)^2) [2]

and

γ = 1 / sqrt(1 - v^2) [5]

which means v = a * t.

but this contradicts the fact that you have

v = (a * t) / sqrt(1 + (a * t)^2) [1]

So as you can see, your derivation is flawed!

!!! Not only do you have total disregard for what I just said, you seem to not be able to follow your own derivation!

I didn’t disgregard it; see below.

First of all, these derivations should not be thrown into an appendix as you have done. Appendicies are used for reference of other peoples work or side explainations - not new derivations that are at the core of the abstract of your paper.

I googled for “derivation appendix” and see lots of derivations in appendices; they can’t all be wrong. It’s a formatting issue, and I like it better that way, so it is easier to tell the details from the larger points. I also like to refer to equations that are not mine, rather than copy them into my derivations, as you did for yours. Not everyone will like it—I can’t please everyone.

Now back to what I said before, you use

γ = sqrt(1 + (a * t)^2) [2]

and

γ = 1 / sqrt(1 - v^2) [5]

which means v = a * t.

but this contradicts the fact that you have

v = (a * t) / sqrt(1 + (a * t)^2) [1]

So as you can see, your derivation is flawed!

Now I see why you think I disregarded what you said. In your previous post, I didn’t realize that “v = a * t” was a conclusion that you were reaching on the line that begins with “from”. On that line, where or how did you get that γ = 1 / sqrt(1 - (a * t)^2)?

Above, your [2] is not in your previous post on the line that begins with “from”. How you are concluding that v = a * t from [2] and [5]? The only way I can get the right-hand side of [2] to equal the right-hand side of [5] is when I substitue the right-hand side of [1] for v in [5].

I googled for “derivation appendix” and see lots of derivations in appendices; they can’t all be wrong. It’s a formatting issue, and I like it better that way, so it is easier to tell the details from the larger points. I also like to refer to equations that are not mine, rather than copy them into my derivations, as you did for yours. Not everyone will like it—I can’t please everyone. Have you written any scientific papers/reports at an institutional level (university or maybe even a company)? If so, what institution?



Now I see why you think I disregarded what you said. In your previous post, I didn’t realize that “v = a * t” was a conclusion that you were reaching on the line that begins with “from”. On that line, where or how did you get that γ = 1 / sqrt(1 - (a * t)^2)? None of that is my work, I merely copied the equations from your paper. Honest - I didn't type out a single one of those equations except for the last one, veff = a * t ( I forgot to copy it originally and typed it instead of going back to your page and doing the ol' copy&paste). Also, I copied the equations in the order that you presented or mentioned them, so no "trickeration" was done on my part. It still shocks me that you cannot follow your own derivation.



Above, your [2] is not in your previous post on the line that begins with “from”. How you are concluding that v = a * t from [2] and [5]? The only way I can get the right-hand side of [2] to equal the right-hand side of [5] is when I substitue the right-hand side of [1] for v in [5]. Why are you bringing up my previous post, that post was completely separate from the post before. The post before I was referring to the derivation in your paper and outlining a few flaws. The next post I reproduced your derivation since you seemed to not be able to follow your own derivation in your own paper. That derivation above is not my work, it is yours.

Edit: Ok, the wording is my work, the equations are your work. I overlooked the fact that I had put in my own wording.

Have you written any scientific papers/reports at an institutional level (university or maybe even a company)? If so, what institution?

Off topic.

None of that is my work, I merely copied the equations from your paper. Honest - I didn't type out a single one of those equations except for the last one, veff = a * t ( I forgot to copy it originally and typed it instead of going back to your page and doing the ol' copy&paste). Also, I copied the equations in the order that you presented or mentioned them, so no "trickeration" was done on my part. It still shocks me that you cannot follow your own derivation.

How can I follow your derivation (yes, it becomes yours when you copy and paste it—you could just refer to mine instead, then it would be mine) when your story changes? Yesterday, 9:55pm you post:

from γ = 1 / sqrt(1 - (a * t)^2) & γ = 1 / sqrt(1 - v^2), v = a * t

Today, 10:07am you post:

Now back to what I said before, you use

γ = sqrt(1 + (a * t)^2) [2]

and

γ = 1 / sqrt(1 - v^2) [5]

which means v = a * t.

The same conclusion based on different equations given by you, which I boldfaced. The first one boldfaced is not in my paper, and never was. You can say I lie, but what are the odds I could have corrected something at the beginning of a progression of derivations of my paper and yet come to all the same conclusions? Nil.

Off topic. It would help me to know :-)



How can I follow your derivation (yes, it becomes yours when you copy and paste it—you could just refer to mine instead, then it would be mine) when your story changes? Yesterday, 9:55pm you post: It is plagarism if I claimed it to be mine. But anyway, that has no importance in analyzing your paper.



Today, 10:07am you post:



The same conclusion based on different equations given by you, which I boldfaced. The first one boldfaced is not in my paper, and never was. You can say I lie, but what are the odds I could have corrected something at the beginning of a progression of derivations of my paper and yet come to all the same conclusions? Nil. Did you point this out earlier? That is indeed a mistake on my part, I do agree that &gamma; = 1 / sqrt(1-(at)^2) is not your equation. I'll reexamine that derivation and examine the others in a more careful manner. I suppose I at least owe you that much now.

It would help me to know :-)

I’ll answer it later, perhaps privately. For now, I wish to not digress.

It is plagarism if I claimed it to be mine.

By my comment I mean that, if you copy & paste a derivation of mine into your post, then I’m following what you pasted, and not necessarily what is in my paper.

Did you point this out earlier?

Not initially; I didn’t see it. I pointed it out in today’s 4:40pm post.

I'll reexamine that derivation...

All your input is greatly appreciated, especially when you think I’m wrong.

I've not really examined any of your derivations yet (sorry). However, I've so far put together the following equations that all come directly from your paper in each of your derivations. The equations are ordered as you mentioned them in your derivations.

10.1

v = (a * t) / sqrt(1 + (a * t)^2)

γ = sqrt(1 + (a * t)^2)

v = (a * t) / γ

v * γ = a * t

γ = 1 / sqrt(1 - v^2)

v / sqrt(1 - v^2) = a * t

veff ≡ v / sqrt(1 - v^2)

veff = a * t



10.2

v = (a * t) / sqrt(1 + (a * t)^2)

veff = a * t

v = veff / sqrt(1 + veff^2)



10.3

Newton: v = sqrt(R / r), since Newton: v = a * t,

veff = sqrt(R / r)

v = veff / sqrt(1 + veff^2)

v = sqrt(R / r) / sqrt(1 + sqrt(R / r)^2)

v = sqrt((R / r) / (1 + (R / r)))

v = sqrt((R / r) / ((r + R) / r))

v = sqrt(R / (r + R))



10.4

v = sqrt(R / r)

1 / γ = sqrt(1 - v^2)

1 / γ = sqrt(1 - sqrt(R / r)^2)

1 / γ = sqrt(1 - (R / r))



10.5

v = sqrt(R / (r + R))

1 / γ = sqrt(1 - v^2)

1 / γ = sqrt(1 - sqrt(R / (r + R))^2)

1 / γ = sqrt(1 - (R / (r + R)))

1 / γ = sqrt((r + R - R) / (r + R))

1 / γ = sqrt(r / (r + R))






Before I went any further, I noticed one glaring problem with the following equations you derived from 10.4 and 10.5:

1 / γ = sqrt(1 - (R / r))

1 / γ = sqrt(r / (r + R))

If we equate the right hand side of the two equations above, we have:

sqrt(1 - (R / r)) = sqrt(r / (r + R))

1- R/r = r / (r + R)

(r - R) / r = r / (r + R)

r^2 = (r - R)(r + R)

r^2 = r^2 - R^2

0 = R

Can you explain?

Before I went any further, I noticed one glaring problem with the following equations you derived from 10.4 and 10.5:
...
Can you explain?

See section 4. The problem you notice is because one of the equations derived is invalid—intentionally. Section 10.4 shows the derivation of Einstein’s gravitational distortion equation (eq. 4.2), which I claim to be invalid (the equation that is, not the derivation). Section 10.5 shows the derivation of the new equation for gravitational distortion (eq. 4.3).

Section 10.4 is my derivation, to show that I can derive Einstein’s equation using the same logic in section 4, my logic, that I use to derive the new equation. This helps to validate my logic and hence the new equation. (And, to my knowledge, this simple derivation of Einstein’s equation—the sole equation that makes the Schwarzschild metric differ from the metric for flat spacetime—has not been done before.) The derivations differ because they incorporate different escape velocity equations, as explained in section 4 and leading up to that point.

See section 4. The problem you notice is because one of the equations derived is invalid—intentionally. This is not good practice. Why include an erroroneous derivation?

Anyway, that is not the heart of the problem here. As I pointed out before, the manner in which you have presented your derivations is not commonly acceptable as a scientific paper. Your version is not a matter of preference as you believe, rather your paper is in the format of a followup paper to a paper that would have been devoted to the derivations that you use. Only derivations that have been presented elsewhere go into an appendix if you so choose - but it is just as equally ok to merely reference the equations from the paper in which they were derived.

For a person analyzing your derivations before they go onto your followup sections, they have to flip endlessly through your appendices and back to the followup sections to find out in what context these derivations are used. This is not very efficient as well as quite annoying which is why the conventions I mentioned above have been adopted by scientists.

With regard to presentation, it is, of course, a matter of preference. Certainly when I prepare an article for publication, I start by assuming that people will not read it if they can possibly avoid it. So I would do my best to make it easy for them, even if they only wanted to skim.

I would most certainly avoid using terms like "A flaw in...." or "Death of...." in the title. I never, ever read such stuff.

Except that I did read the "Flaw in...." stuff. And there is, it seems to me, a major "flaw" in the premise. Let's see.

The assertion is that, for an accelerating body, provided the time interval under consideration is infinitely small, a = v, hence the special theory applies, and time dilation, length contraction follows.

The new "theory" suggests that it is valid to regard each successive time increment in which a = v as inertial frames in relative motion. This cannot be.

For, in the special theory, Einstein quite explicitly says (in his 1905 classic) that the time of an event is nothing more or less than a judgement about the simultaneity of that event and the hands on my watch. In other words, time dilation and what a senior member here calls "realtivity of simultaneity" are two sides of the same coin.

But obviously, it makes no sense to talk about the the simultaneity of events that occur in different time frames, so by the same token it makes no sense to talk about the same material object at infinitely small but successive time intervals as being capable of exhibiting time time dilation.

This is not good practice. Why include an erroroneous derivation?

I amply explained why in my previous post, and said that the equation is invalid, not the derivation. I’ve got a book, The Riddle of Gravitation. Four of the sections in its appendix are derivations. One of them is an intentionally invalid derivation, at the end of which the author says, “The calculation just made certainly has its never-never aspects.” Should Barnes & Noble yank this book from its shelves? I was in no way confused by the author’s practice. And I appreciated that the author stuffed the derivations in the appendix, so they did not cloud the larger points.

As I pointed out before, the manner in which you have presented your derivations is not commonly acceptable as a scientific paper. Your version is not a matter of preference as you believe, rather your paper is in the format of a followup paper to a paper that would have been devoted to the derivations that you use. Only derivations that have been presented elsewhere go into an appendix if you so choose - but it is just as equally ok to merely reference the equations from the paper in which they were derived.

Then why do I get 67,000 hits when I google for “derivation appendix arxiv”? I see lots of original derivations in appendices of scientific papers in that search. From where are you getting these commonly accepted practices?

For a person analyzing your derivations before they go onto your followup sections, they have to flip endlessly through your appendices and back to the followup sections to find out in what context these derivations are used.

You can open the doc in two windows both displayed, so you don’t have to flip back and forth. Or you can print out sections 9 and 10. Given the ease of that, I think you’re making a mountain out of molehill. If I did have the derivations embedded in the sections from which they are referenced, the reader would still have to jump down to the equations in section 9, to which the derivations often refer.

This is not very efficient as well as quite annoying which is why the conventions I mentioned above have been adopted by scientists.

It is less annoying for those who wish to see the larger points unclouded by the details. And if this convention has been adopted by scientists, then how do you explain my Google search?

I would most certainly avoid using terms like "A flaw in...." or "Death of...." in the title. I never, ever read such stuff.

“Death,” I agree. But I’m correctly using the word “flaw”. How would you reference a flaw of GR?

The new "theory" suggests that it is valid to regard each successive time increment in which a = v as inertial frames in relative motion. This cannot be.

It does not suggest that. It gets annoying here when claims are made about the paper without a reference to the paper. Is such a reference too much to ask?

I amply explained why in my previous post, and said that the equation is invalid, not the derivation. If there is no error in a derivation but you prove the equation to be in error, then one or more of the assumptions in the derivation are incorrect. For your case, you would be arguing that one or more of the postulates of relativity are incorrect - yet I've not seen you make this statement.

I’ve got a book, The Riddle of Gravitation. Four of the sections in its appendix are derivations. A book is not a scientific paper even if it covers a scientific topic. You cannot compare the two.

One of them is an intentionally invalid derivation, at the end of which the author says, “The calculation just made certainly has its never-never aspects.” Should Barnes & Noble yank this book from its shelves? I was in no way confused by the author’s practice. And I appreciated that the author stuffed the derivations in the appendix, so they did not cloud the larger points. You are still using a book as an example, not a scientific paper - do not try to tell me they are they same!



Then why do I get 67,000 hits when I google for “derivation appendix arxiv”? Your google search is flawed. Just because a paper on arxiv.org has the words "derivation" and "appendix" in it does not mean that the new derivations contained within the paper are in the appendix.

I see lots of original derivations in appendices of scientific papers in that search. Are you sure they are original to that paper? Or are the derivations not related directly to the abstract of the paper?

From where are you getting these commonly accepted practices? Experience in reading and understanding other people's derivations. Any derivation I've ever wanted to analyze has always been in the body of the paper, not an attachment.



You can open the doc in two windows both displayed, so you don’t have to flip back and forth. Or you can print out sections 9 and 10. Given the ease of that, I think you’re making a mountain out of molehill. If I did have the derivations embedded in the sections from which they are referenced, the reader would still have to jump down to the equations in section 9, to which the derivations often refer. As I said before, the presentation you seem to want to show is that of a followup paper that expresses results and implications of a theory. You don't have any papers that are devoted to the derivation which is what scientists would want to review first in a peer-review. From your response to James R, I assumed you had intended this paper for such a peer-review.



It is less annoying for those who wish to see the larger points unclouded by the details. And if this convention has been adopted by scientists, then how do you explain my Google search?[/QUOTE]

If there is no error in a derivation but you prove the equation to be in error, then one or more of the assumptions in the derivation are incorrect. For your case, you would be arguing that one or more of the postulates of relativity are incorrect - yet I've not seen you make this statement.

It does not have to be a postulate of relativity that is incorrect. GR is not derived solely from the postulates of relativity. GR shares Newton’s equation for escape velocity. I show that this equation is invalid, in section 3.

Are you sure they are original to that paper? Or are the derivations not related directly to the abstract of the paper?

Lots of them sure seem original to me. Take a look for yourself.

From your response to James R, I assumed you had intended this paper for such a peer-review.

As I said to him, I’ve established to my satisfaction that the odds of my paper being peer-reviewed are nil, regardless of content beyond the title, hence regardless of formatting. So no, I do not intend that. That has freed me to format as I wish. For example, the journal I submitted to demanded that I reference books by, besides the title, the publisher name and one of the names of the cities in which the publisher has an office (only the city mind you, not the state or province or country that might uniquely identify the city), rather than simply use the international-standard book number (ISBN) that can be exactly searched. A 19th-century practice from a popular journal! I have learned that commonly accepted scientific practices leave a lot to be desired.

See section 4. The problem you notice is because one of the equations derived is invalid—intentionally. Section 10.4 shows the derivation of Einstein’s gravitational distortion equation (eq. 4.2), which I claim to be invalid (the equation that is, not the derivation). Section 10.5 shows the derivation of the new equation for gravitational distortion (eq. 4.3).

Section 10.4 is my derivation, to show that I can derive Einstein’s equation using the same logic in section 4, my logic, that I use to derive the new equation. This helps to validate my logic and hence the new equation. (And, to my knowledge, this simple derivation of Einstein’s equation—the sole equation that makes the Schwarzschild metric differ from the metric for flat spacetime—has not been done before.) The derivations differ because they incorporate different escape velocity equations, as explained in section 4 and leading up to that point.

10.1

v = (a * t) / sqrt(1 + (a * t)^2)

γ = sqrt(1 + (a * t)^2)

v = (a * t) / γ

v * γ = a * t

γ = 1 / sqrt(1 - v^2)

v / sqrt(1 - v^2) = a * t

veff = v / sqrt(1 - v^2)

veff = a * t



10.2

v = (a * t) / sqrt(1 + (a * t)^2)

veff = a * t

v = veff / sqrt(1 + veff^2)



10.3

Newton: v = sqrt(R / r), since Newton: v = a * t,

veff = sqrt(R / r)

v = veff / sqrt(1 + veff^2)

v = sqrt(R / r) / sqrt(1 + sqrt(R / r)^2)

v = sqrt((R / r) / (1 + (R / r)))

v = sqrt((R / r) / ((r + R) / r))

v = sqrt(R / (r + R))




10.5

v = sqrt(R / (r + R))

1 / γ = sqrt(1 - v^2)

1 / γ = sqrt(1 - sqrt(R / (r + R))^2)

1 / γ = sqrt(1 - (R / (r + R)))

1 / γ = sqrt((r + R - R) / (r + R))

1 / γ = sqrt(r / (r + R))




First of all, let's establish some definitions for the variables you use which are v, a, t, γ, veff, r, and R.

You've non-dimensionalized "v" by dividing it by "c" as is apparent in your change to the relativistic rocket equaton. Also, you divided "a" by "c" as well giving "a" the units of frequency (e.g. hz, 1/s). "t" is apparently left alone so we can assume it has the units of time. From the equations for "γ", it is easily seen that "γ" is non-dimensional. "veff" is also non-dimensional by definition. "r" and "R" are given in units of length. So we are dealing with some non-dimensional variables (v, γ, and veff) and some variables with altered dimensions (a) and other variables with their original dimensions (t, r, and R). This is very strange and I am sure this is the origin of your error somewhere.

Now these are not physical definitions, just merely unit analysis. But this should be sufficient enough to continue.

Everything seems fine until we get to section 10.3. You say Newton's equation for velocity is v = a * t. Which is true when v is a velocity, a is acceleration, and t is time. But your definitions of v, a, and t seem to be different as only t has the units of time. In a Newtonian world, it is meaningless to divide velocity and acceleration by some arbitrary constant (in this case, c). I don't think your v and a as you've defined them can be considered exactly velocity and acceleration in a Newtonian world. Therefore, your jump to veff = sqrt(R / r) is incorrect.

That means, Einstein's derivation is correct and your derivation is not. And your veff is still meaningless (at least in a Newtonian world).

“Death,” I agree. But I’m correctly using the word “flaw”. How would you reference a flaw of GR?



Right. So there is a "flaw" in the general theory. Then you are under an obligation to show where, in the derivation of the field equations, the logical fault lies.

And your school arithmetic won't do it.

Here's a clue:

G_&mu;&nu; - 1/2g_&mu;&nu;G = -&kappa;T_&mu;&nu;.

Take us through the derivation, and show where it is "flawed". Or at the very least explain what you think it's saying.

First of all, let's establish some definitions for the variables you use which are v, a, t, γ, veff, r, and R.

There’s no need to establish the definitions here, because all the variables are defined in the “Conventions Herein” section.

This is very strange and I am sure this is the origin of your error somewhere.

There is nothing strange about how I have defined the variables. The Spacetime Wheel (http://casa.colorado.edu/~ajsh/sr/wheel.html) not only sets c in The Relativistic Rocket (http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html) equations to 1, it also sets a to 1.

In a Newtonian world, it is meaningless to divide velocity and acceleration by some arbitrary constant (in this case, c). I don't think your v and a as you've defined them can be considered exactly velocity and acceleration in a Newtonian world. Therefore, your jump to veff = sqrt(R / r) is incorrect.

It is not meaningless to do that, even in a Newtonian world. There is no problem with making v a fraction of c. There is no problem with setting c = 1. That's allowed by basic algebra! Instead of harping on this point, why not look around at all the sources that set c = 1? Google for “geometric units”. Geometric units work for Newton as well.

And your veff is still meaningless (at least in a Newtonian world).

It is not meaningless. It is a valid substitution for the right-hand side of eq. 2.1, in the Newtonian world as well, because it is basic algebraic substitution.

Right. So there is a "flaw" in the general theory. Then you are under an obligation to show where, in the derivation of the field equations, the logical fault lies.

And your school arithmetic won't do it.

There is no such obligation, and my arithmetic does suffice, because it shows that the escape velocity equation in GR is invalid. That’s all I need show, to show a flaw of GR.

Here's a clue:

G_&mu;&nu; - 1/2g_&mu;&nu;G = -&kappa;T_&mu;&nu;.

Take us through the derivation, and show where it is "flawed". Or at the very least explain what you think it's saying.

Unnecessary.

There is no such obligation,....Then you obviously have no idea what the general theory is all about. If you have the temerity to describe the field equations as "unnecessary" in the context of the general theory, then you are big trouble.

Have you read the oiginal papers? I would recommend Einstein's synopsis in Ann. Phys. 49, 1916.

If you have the temerity to describe the field equations as "unnecessary" in the context of the general theory,...

I did not say or imply that.

Have you read the oiginal papers?

Off topic.

There’s no need to establish the definitions here, because all the variables are defined in the “Conventions Herein” section. Those are the most vague definitions I could possibly hope to find. As I even stated, I made no attempt to redefine your variables, establishing definitions was probably the wrong terminology on my part. All I did was establish the units of each variable.

Do you disagree with the units I established for the variables in your equations?



There is nothing strange about how I have defined the variables. The Spacetime Wheel (http://casa.colorado.edu/~ajsh/sr/wheel.html) not only sets c in The Relativistic Rocket (http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html) equations to 1, it also sets a to 1. If you think I am arguing over the fact that c=1, you are very mistaken. My argument has nothing to do with what c is equal to. My argument is only concerned with the units of the variables in your equations, I suggest you take another look at what I did.



It is not meaningless to do that, even in a Newtonian world. "It" - what are you refering to now? I think you are telling me it is not meaningless to set c=1 in a Newtonian world. I can only hope this is not what you meant because that is absurd. The concept of c has no meaning in a Newtonian world since the speed of light wouldn't be considered constant.


There is no problem with making v a fraction of c. There is no problem with setting c = 1. That's allowed by basic algebra! What is c in a Newtonian world?

Instead of harping on this point, why not look around at all the sources that set c = 1? Google for “geometric units”. Geometric units work for Newton as well. Again, I never brought up this point. Of course c can equal 1.




It is not meaningless. It is a valid substitution for the right-hand side of eq. 2.1, in the Newtonian world as well, because it is basic algebraic substitution. You are substituting apples for oranges as I suspected a long time ago. Why can you not see that?

There is no such obligation, and my arithmetic does suffice, because it shows that the escape velocity equation in GR is invalid. That’s all I need show, to show a flaw of GR. I second QuarkHead and not because he simply disagrees with you but because what he is saying is parallel to what I told you:
If there is no error in a derivation but you prove the equation to be in error, then one or more of the assumptions in the derivation are incorrect. For your case, you would be arguing that one or more of the postulates of relativity are incorrect - yet I've not seen you make this statement.

There is no such obligation, to show the field equations and their derivations are flawed: my addition and my arithmetic does suffice, because it shows that the escape velocity equation in GR is invalid. That’s all I need show, to show a flaw of GR. If you think that the general theory can be explained, derived or refuted with a bit of high school arithmetic you are wildly wrong.
Me:Here's a clue:

Gμν - 1/2gμνG = -κTμν.

Take us through the derivation, and show where it is "flawed". Or at the very least explain what you think it's saying.Zanket: Unnecessary. So..an understanding of the General Theory of Relativity can be taught in elementary math classes at school?

So here we have it. You want to pull down the general theory, and yet when I ask you to derive or explain it you say it is "unnnecessry".

And when I asked you if you had read the original literature you said it was "off topic".

Which more or less tells us where you're coming from

Hi zanket,
I've done some maths and found some very interesting results.
Firstly, I found the accelerating cliff problem to be quite difficult. It's not the same as the relativistic rocket - the rocket is essentially a point, while the cliff is obviously not.

I didn't manage to complete the problem, but I did reach some tentative conclusions that might or might not agree with your own analysis.

Here is my setup:
Imagine an incredibly high cliff face drifting in flat space.
The cliff face is marked at one-meter intervals.
An observer is at rest with respect to the cliff face, floating beside at the 0 meter mark.
At t=0, the entire cliff face begins accelerating past the inertial observer at g. What happens?


After much sweating and number crunching, I came to the conclusions that:
If the observer directly measures the velocity of each mark on the cliff face as it goes past, the velocity will approach but never reach c. The acceleration of the cliff face thus appears to slow down asymptotically to zero as its relative velocity approaches c.
If the observer uses the marks on the cliff face to calibrate distance (each mark represents one meter) and use them to measure velocity and acceleration, then the calculated velocity will increase at a constant rate of g, passing c without pause - at time t on the inertial observer's clock, they will pass meter mark 1/2 g.t²
An observer on the cliff face watching the inertial observer "fall" past will never see them fall faster than c

The faster-than-c velocity in the second point is only there because distance in the cliff frame is mixed with time in the inertial observer frame - this is much like your v_eff - it's not a real velocity, in some sense.

It's the third point that is the conclusion most relevant to your paper. It agrees with your own conclusion that escape velocity can never be larger than c.

I was intrigued.
I next looked up how the Newtonian escape velocity equation was derived, and discovered (rediscovered, really) that it came from Energy considerations. As an object falls into a gravity well, its gain in Kinetic Energy at any level must be the same as its loss in Potential energy.

That suggested an approach to find Relativistic escape velocity, since Kinetic Energy has its own equation in Special Relativity.

Crunching some more numbers, I again found agreement with your analysis. At the Swarzchild radius, the escape velocity determined by this technique is 0.745c.

More intrigue!

Now - what conclusions are to be drawn? This is where care must be taken. We must carefully consider our premises and conclusions. Take extreme care! Spotting premises is neither trivial nor easy!

In the derivation from energy conservation principles, I think my premises were:
The Newtonian equation for gravitational potential energy is correct.
The SR equation for kinetic energy is correct
Energy is conserved
No other energy sources or sinks are involved
My conclusion from these premises is:
At the Swarzchild radius (2GM/c²), escape velocity is 0.745c

Considering the further premise that GR predicts that the escape velocity at the Swarzchild radius is c, the further conclusion is that GR is incorrect.

Logically, it is clear that either:
GR is wrong, or
One of my premises is wrong, or
I've made a mistake setting out my premises and conclusions

I don't know which of these is true... but I'd place my money on one of my premises being wrong.

Thanks zanket!

In the derivation from energy conservation principles, I think my premises were:
The Newtonian equation for gravitational potential energy is correct.
The SR equation for kinetic energy is correct
Energy is conserved
No other energy sources or sinks are involved
My conclusion from these premises is:
At the Swarzchild radius (2GM/c²), escape velocity is 0.745c

Considering the further premise that GR predicts that the escape velocity at the Swarzchild radius is c, the further conclusion is that GR is incorrect.

Logically, it is clear that either:
GR is wrong, or
One of my premises is wrong, or
I've made a mistake setting out my premises and conclusions

I don't know which of these is true... but I'd place my money on one of my premises being wrong.

Thanks zanket! Post your derivation - I'll let you know what your premises are as best I can. However, I'm pretty sure your mistake is in the very first premise you listed. Zanket has already been proven in error mixing Newtonian equations with Relativity equations. See the 3rd post on this page.

Thanks Aer:

(1) Gravitational potential energy = -GMm/r
(2) Kinetic energy = gamma.mc² - mc²

When v = escape velocity:
(3) GPE + KE = 0

At the Swarzchild radius,
(4) r = 2GM/c²

Escape velocity at the Swarzchild radius was derived by substituting (4) into (1), then (1) and (2) into (3):

gamma.mc² - mc² = GMm / (2GM/c²)
gamma = 1.5
v = 0.745c

Thanks Aer:

(1) Gravitational potential energy = -GMm/r
(2) Kinetic energy = gamma.mc² - mc²

When v = escape velocity:
(3) GPE + KE = 0

At the Swarzchild radius,
(4) r = 2GM/c²

Escape velocity at the Swarzchild radius was derived by substituting (4) into (1), then (1) and (2) into (3):

gamma.mc² - mc² = GMm / (2GM/c²)
gamma = 1.5
v = 0.745c Firstly, let me say that "considering such and such equation to be correct" is not a sound premise. By assuming such a thing you are including a whole range of premises that were used to derive that equation. And since you are assuming equations for two different theories which contradict each other (that is their premises contradict each other) to derive the result you obtained, then your result is only going to be valid if all these premises are true. Two glaring premises that must hold true are (1) the speed of light is constant in all inertial reference frames, and (2) the speed of light is constant in an ether - that is only one absolute reference frame.

I hope you see why this line of thinking is flawed.

Firstly, let me say that "considering such and such equation to be correct" is not a sound premise.
I agree!
And since you are assuming equations for two different theories which contradict each other...
zanket appears to coming from the direction that if one theory is derived using an earlier theory, that some premises must be shared... which seems reasonable, as long as you know which premises are not shared.

My difficulty is that I personally have no idea exactly which premises of GR are shared with SR or Newtonian gravity and which aren't.

My hope is that zanket might realise that he has similar difficulties.

I agree!

zanket appears to coming from the direction that if one theory is derived using an earlier theory, that some premises must be shared... which seems reasonable, as long as you know which premises are not shared.

My difficulty is that I personally have no idea exactly which premises of GR are shared with SR or Newtonian gravity and which aren't.

My hope is that zanket might realise that he has similar difficulties. Perhaps a link to the derivation being challenged would be helpful.

Those are the most vague definitions I could possibly hope to find.

All are complete. Name one that you think is vague.

As I even stated, I made no attempt to redefine your variables, establishing definitions was probably the wrong terminology on my part. All I did was establish the units of each variable.

OK, but why, when units are already specified in the doc for every variable?

Do you disagree with the units I established for the variables in your equations?

Agreeing or disagreeing with your interpretation of the units is a rat hole. I won’t verify that your wording is equivalent to mine. Instead, I refer you to the units specified in the doc. If any are unclear to you, please point them out and I will clarify.

So we are dealing with ... some variables with altered dimensions (a)

The acceleration a is “in units where c = 1”. Nothing strange about that. The doc says, “In units of light years for distance and years for time, c = 1 ly / yr and g at Earth’s surface = 1.03 ly / yr^2”. This matches the value at the website The Relativistic Rocket.

"It" - what are you refering to now?

Look at what I replied to. There’s only one “meaningless” in the quote of yours to which I replied, so it is clear to what “it” refers.

I think you are telling me it is not meaningless to set c=1 in a Newtonian world. I can only hope this is not what you meant because that is absurd. The concept of c has no meaning in a Newtonian world since the speed of light wouldn't be considered constant.

Please! The constant c just means 299792458 meters per second. It is also the speed of light. “299792458 meters per second” has meaning in a Newtonian world, of course. A constant is just a value that doesn’t change. It doesn’t have to have anything to do with anything else.

What is c in a Newtonian world?

299792458 meters per second, the same as in an Einsteinian world.

You are substituting apples for oranges as I suspected a long time ago. Why can you not see that?

Because your case is invalid. The case you’ve made disagrees with elementary algebra. I'd comment as well on your frame argument, but it would be overkill so I won't.

I second QuarkHead and not because he simply disagrees with you but because what he is saying is parallel to what I told you:

You have not made your case. I replied to your point about the postulates of GR. You have not responded.

All are complete. Name one that you think is vague.



OK, but why, when units are already specified in the doc for every variable?



Agreeing or disagreeing with your interpretation of the units is a rat hole. I won’t verify that your wording is equivalent to mine. Instead, I refer you to the units specified in the doc. If any are unclear to you, please point them out and I will clarify.



The acceleration a is “in units where c = 1”. Nothing strange about that. The doc says, “In units of light years for distance and years for time, c = 1 ly / yr and g at Earth’s surface = 1.03 ly / yr^2”. This matches the value at the website The Relativistic Rocket.



Look at what I replied to. There’s only one “meaningless” in the quote of yours to which I replied, so it is clear to what “it” refers.



Please! The constant c just means 299792458 meters per second. It is also the speed of light. “299792458 meters per second” has meaning in a Newtonian world, of course. A constant is just a value that doesn’t change. It doesn’t have to have anything to do with anything else.



299792458 meters per second, the same as in an Einsteinian world.



Because your case is invalid. The case you’ve made disagrees with elementary algebra. I'd comment as well on your frame argument, but it would be overkill so I won't.



You have not made your case. I replied to your point about the postulates of GR. You have not responded. Everything you just said is BS. You don't even take the time to analyze the disproof of your theory. The fact that you insist that v = a * t corresponding to (non-dim) = (1/time)*time is the Newtonian equation for velocity says it all!!

If you think that the general theory can be explained, derived or refuted with a bit of high school arithmetic you are wildly wrong.

Uh, with what part of the paper do you disagree? Your words are empty so far.

So..an understanding of the General Theory of Relativity can be taught in elementary math classes at school?

That question is unrelated to what you quoted. You said, “Take us through the derivation, and show where it is "flawed". Or at the very least explain what you think it's saying.” I said “unnecessary.” Allow me to elaborate: It is unnecessary for me to take y’all through the derivation of the equation you posted, or to explain what I think it’s saying, to show a flaw of GR. I have already shown a flaw of GR in the paper.

So here we have it. You want to pull down the general theory, and yet when I ask you to derive or explain it you say it is "unnnecessry".

As it is.

And when I asked you if you had read the original literature you said it was "off topic".

Yes, I said that. What I have read or not is irrelevant to this topic.

Which more or less tells us where you're coming from

It tells us that you post things unrelated to the thread.

For your reading pleasure, here is an explaination of why everything you just said is BS.

All are complete. Name one that you think is vague.
Every single one.
First in line:
d See definition in section 9
t See definition in section 9
T See definition in section 9
etc.

Next:
a Acceleration as described above, in units where c = 1
d Distance traversed by the rocket in the gantry’s frame, in units where c = 1
t Time elapsed in the gantry’s frame, in units where c = 1
T Time elapsed in the rocket’s frame (i.e. how much the crew ages), in units where c = 1
v Final velocity, as a fraction of c; no units

Next:
relativistic rocket A rocket accelerating under the conditions specified in section 9

Just like your derivation section, your definitions section has the reader flipping around all over the place. On top of that, take a look at your definition for "a". By that definition, I have no idea that I am suppose to divide an acceleration term by c to get "a". Now allow me to let you in on a little secret: your "a" is not acceleration. if you non-dimentionalized "a" by dividing it by an acceleration term, that would be fine. By you've divided it by a velocity term thus rendering it a frequency!






OK, but why, when units are already specified in the doc for every variable? You don't even know what the units are of your own variables! Your terminology is testimony to that.



Agreeing or disagreeing with your interpretation of the units is a rat hole. Yes it is a rat hole because your 5,900 word+ paper is disproven with just analyzing section 10.3 of your appendix. I tried to emphasize to you the importance scientist put in the derivation aspect of your paper before you go and do all that other analysis work which had to be very time consuming. And for what? Nothing, because everything in your paper starting with your Abstract is useless. Maybe now you see that there is much to desire in the commonly accepted scientifc convensions? (I already know that answer to this as you have contempt for science)



I won’t verify that your wording is equivalent to mine. Instead, I refer you to the units specified in the doc. If any are unclear to you, please point them out and I will clarify. Oh boy, that was a huge mistake on your part. READ THE ABOVE CAREFULLY.



The acceleration a is “in units where c = 1”. Nothing strange about that. WHAT DOES THAT MEAN? I know I already figured it out, but I didn't figure it out from your definitions!

The doc says, “In units of light years for distance and years for time, c = 1 ly / yr and g at Earth’s surface = 1.03 ly / yr^2”. This matches the value at the website The Relativistic Rocket. This is useless, this doesn't tell me to divide acceleration by c in order to obtain "a".



Look at what I replied to. There’s only one “meaningless” in the quote of yours to which I replied, so it is clear to what “it” refers. I was trying to ask a dumb question (did I fail that?)



Please! The constant c just means 299792458 meters per second. It is also the speed of light. “299792458 meters per second” has meaning in a Newtonian world, of course. A constant is just a value that doesn’t change. It doesn’t have to have anything to do with anything else. Well now we have a new definiton for c because a "light year" doesn't make any sense in a Newtonian world!



299792458 meters per second, the same as in an Einsteinian world. well then use that instead of a light year! Note: you cannot set c=1. (also note: once you do this and completely finish changing your paper, I have a new error waiting for you that I know you haven't thought of)





Because your case is invalid. The case you’ve made disagrees with elementary algebra. Well what can I say, you are wrong again. you are substituting a non-dimensional number for v, a frequency for a, and leaving time alone in Newtons equation v = a * t. Now THAT disagrees with elementray algebra. Lookup: substitution.

I'd comment as well on your frame argument, but it would be overkill so I won't. Brain overload?



You have not made your case. I replied to your point about the postulates of GR. You have not responded. Where? Surely I didn't miss this. This should be good, I can't wait.

My conclusion from these premises is:
At the Swarzchild radius (2GM/c²), escape velocity is 0.745c

Considering the further premise that GR predicts that the escape velocity at the Swarzchild radius is c, the further conclusion is that GR is incorrect.

That is interesting. After thinking about this for quite a while, I need more time tomorrow before I can comment.

Zanket has already been proven in error mixing Newtonian equations with Relativity equations. See the 3rd post on this page.

This is rude. You have not proven your case.

This is rude. You have not proven your case. Sorry, but the burden of proof lies with the author (that is you, you wrote that entire paper).

I've shown your paper to be riddled with stupidity and am only this blunt about it because of your crackpot attitude when someone presents a possible problem with your paper (in this case, it is not a "possible" problem as I have clearly proven my point without doubt).

That is interesting. After thinking about this for quite