One disproof of EP?

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OK, makis, I'll bite (against my better judgment)

The solution is x = -x which, absurd as that is,
still makes more sense than the rest of this entire thread,
which remains gibberish. Sorry.

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Of course that is not a solution. Your solution is an example of how people missinterpret mathematical symbols and reach wrong conclusions. 99 out of 100 students will insist that is the solution (or even 2 = -2) and then go ahead, in a totally arogant way, to insist that mathmetics is wrong. Most people lack understanding of the rules that come with those symbols and feel as they also do with Physics principles and this leads to the sort of conclusions made by Cindy and Daisy. Just remember that making extraordinary claims takes extraordinary effort and ideas, not just some simple scribings.

Without going into detalis now, the equation has complex roots. To get to your conclusion you performed an illegal operation in math.

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I agree with you. Makis's X-equation has no solution and has no relation to your analysis. Your analysis is right, although it is not sufficient to claim EP wrong. One reason for that is SR has not been proved right yet. However, you have established a new relation between SR and EP, which can be stated as the following:
1/ If SR is right, then EP is not correct.
2/ If EP is right, then SR is not correct.

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ok people, let's get the ball rolling.

Let's start first with Newton. It is a mere coincidence that mi=mg in Newtonian mechanics and has been proved only experimentally in a few parts in 10^13 to 10^15. Newton had no idea whether mi should be equal to mg, actually some think he swa it as a miracle of nature.

Newton's law is a good approximation for betta <<1. It just happens as a miracle. Try to write Newton's law in a better form to undertand what Cindy is doing wrong:

F = (GMm)/[r1(t) - r2(t)]^2

So the force depends on separation of the position of two masses in time, t. This violates the principle of no simultaneaity in relativity theory. Then, Newton's law cannot be used (even a kid knows that) when relativistic effects are present. Therefore, using it in an analysis in combination with relativistic principles generates non-sense results. It is that simple, but we can go further.

Einstein took the position that gravity and acceleration is the same thing (simplified version of EP) and made his famous "thought experiment" known as "Gedankenexperiment". Then, if we assume that EP is right ( and that is ALL we can do unless we prove otherwise experimentally), gravity has an effect on time. Einstein extended his thought experiment calculations to a non-uniform gravity potential produced by a mass M and ended-up with GR.

Therefore:

a) Newton law applies only for betta<<1
b) Newron's law is an approximation of reality
c) You cannot use an approximation of reality in conjuction with a theory where that approximation is not even good and claim you disproved the principle assumption of another theory, when it just happens that your approximating law remarkably checks that principle.
d) SR + EP

---

> GR
e) Equating symbols in Physics assumes an understanding of the conditions and reference frames. Because gravity is the symbol F in Newton and also elsewhere, it does not mean that you can start equating the right hand sides also. To give you an example, you live in a society that everybody pays with apples for anything and I live in a society doing that with oranges. We both buy the same shirt. That does not mean apples=oranges.
f) I hope I did not waist my time.

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[Samizdat :]
OK, makis, I'll bite (against my better judgment)

The solution is x = -x which, absurd as that is,
still makes more sense than the rest of this entire thread,
which remains gibberish. Sorry.

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[makis :]
Of course that is not a solution. Your solution is an example of how people missinterpret mathematical symbols and reach wrong conclusions. 99 out of 100 students will insist that is the solution (or even 2 = -2) and then go ahead, in a totally arogant way, to insist that mathmetics is wrong.

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You haven't wasted your time, yet, makis. But there comes a point where Suspense takes the next train and Boredom is the only one left
waiting at the station. So is there or is there not a solution to your equation? I at least followed the instructions intrinsic to your problem, and solved for x. Your example of "2 = -2" is therefore unfair. If you have the mathematics, shed the light of day on it, rather than keep it shrouded in the mists. You say my solution is wrong. I say your equation is nonsensical. If you have the solution, let's see it.
I have nothing against mathematics, nor magic, but I do take umbrage
at my being accused of insisting that mathematics is wrong, when what I do insist is that magic posing as mathematics is wrong. When Andrew Wiles found the proof of Fermat's Last Theorem, did he simply write down the final lines of the proof and say "Here, now go and find the intermediate steps." No, he gave the reader the benefit of the doubt, and the option of working it out for her- or himself. He didn't taunt and tease with the methods and means. It is all laid out for anyone who wishes to spend a lifetime, if that is sufficient
time for any given individual, to solve it for themselves. It would be useful to resolve this ugly business sooner rather than later,
because it is unproductive to leave the thread and therefore the board in a state of stagnancy due to a breakdown in agreement on first principles. In your defense, I will state, however, that it is very nearly miraculous that *anyone* among us can wade through the material of physics and mathematics, constrained as we are by the clunkiness of limited character sets. Let's all try to be less judgmental, considering our dependence upon the crude and easily misunderstood substitution of carets for superscripts and underlines for subscripts, just to name a couple of the more egregrious conventions we're forced to use.


I'll drop just one more name, then bugger off: Einstein once said "Don't make things simple; make them simpler."

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So is there or is there not a solution to your equation? I at least followed the instructions intrinsic to your problem, and solved for x. Your example of "2 = -2" is therefore unfair. If you have the mathematics, shed the light of day on it, rather than keep it shrouded in the mists. You say my solution is wrong. I say your equation is nonsensical. If you have the solution, let's see it.

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If x = -x then 2 = -2 and 3 = -3 and k = -k and A = -A. So why did you complain my friend? You said x = -x. Why is it unfair then for someone to conclude that 2 = -2 based on your statement?

Let Z = (x+1)/(x-1). Then Z is a valid definition if x is not equal to 1. Then the equation (x+1)/(x-1) + (x-1)/(x+1)=2 becomes:

Z+1/Z =2. We can multiply through by Z, iF Z not equal to 0. Then x must not be -1. Then we get Z^2-2Z+1 = 0. Solving for Z we get:

Z1 = Z2 = 1

Then, we can recall our definition for Z in terms of x:

Z1=Z2 =(x+1)/(x-1) = 1. This step was necessary to break the two ratios and determine that each must be equal to 1.

Your invalid operation was that you multiplied through by x-1 or x+1, without marking that x must not be equal to 1 and/or -1. That's where your wrong conclusion came from. The solution:

Replace x by a sequence Xn that is increasing an unbound in n. Then Z becomes:

Z(Xn) = (Xn+1)/(Xn-1) Divide through by Xn:

Z(Xn) = [1+1/Xn]/[1-1/Xn] Then take the limit as Xn --> inf

Z(x) = limZ(Xn) = [1+lim(1/Xn)]/[1-lim(1/Xn] = 1 (because lim(1/Xn)=0). Therefore, the equation Z(x)=1 is satisfied in a limiting sense by x = inf.

Of course, there is quite a number of theorems that go along to satisfy my solution, which I cannot possibly state here.

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In your defense, I will state, however, that it is very nearly miraculous that *anyone* among us can wade through the material of physics and mathematics, constrained as we are by the clunkiness of limited character sets. Let's all try to be less judgmental, considering our dependence upon the crude and easily misunderstood substitution of carets for superscripts and underlines for subscripts, just to name a couple of the more egregrious conventions we're forced to use.

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Thank you for your kindness.

Makis,
You misunderstood about Cindy's issues. Let me make them clearer.

1/ For m_g.

She said by definition, m_g = F/g
where F is gravitational force, and g = GM/R^2

It does not matter what you think about this kind of "force", and what theory you choose, Newtonian or relativitis.
She claim that,

At a time point t, at a distance R, for a speed v of m, you always get <u>only one</u> value of m_g.

Do you agree?

2/ For m_i, Cindy do require you use SR.
With definition m_i = F/a
Where F is a mechanical true force
a is acceleration
then she claim that

At a time point t, at a distance R, for a speed v of m, you get <u> more than one </u> value of m_i. (Values of m_i vary form gamma.m to (gamma^3)m depend on direction of velocity of m).

Do you agree?

I do agree to her both issues and her conclusion m_i differ from m_g (if SR used to consider m_i). Therefore, I made a statement for her

If SR is right, then EP is wrong and vice versa.

For performed experiments, Cindy's analysis implies an explanation. Reason for these experiments failed to distinguish m_i and m_g is factor gamma. If they have an experiment, in which velocity v of m has a proper direction and fast enough compared to c, then they can distinguish m_i and m_g.

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2/ For m_i, Cindy do require you use SR, then she claim that

At a time point t, at a distance R, for a speed v of m, you get more than one value of m_i. (Values of m_i vary form gamma.m to (gamma^3)m depend on direction of velocity of m).

Do you agree?


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I disagree with the way language is used. At time t and given a direction for v you can only have one value for mi. You cannot have multiple values at the same time, although the value changes for v directions. That value you get will be equal to mg if the correct detrmination for mg is made. Your statement below is also wrong:

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1/ For m_g.

She said by definition, m_g = F/g
where F is gravitational force, and g = GM/R^2

It does not matter what you think about this kind of "force", and what theory you choose, Newtonian or relativitis.
She claim that,

At a time point t, at a distance R, for a speed v of m, you always get only one value of m_g.

Do you agree?

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It makes a difference whether you use the correct theory. For relativistic frames, the equation you state for mg is not valid, but only for betta <<1. But if betta<<1 then gamma =1 and your previous statement is irrelevant since mi will not depend on the direction of v.(gamma = gamma^3=1). Again, g = GM/R^2 is an approximation valid only for beta<<1 and the fact that it approximates reality is a coincidence and miracle.

Something else. Physics and language games have no relation to each other. If they do, let's propose Cindy for a Nobel price.

Makis,

1/ For m_i

She does not mean that you get multiple values of m_i at the same time.
She do mean that there exist at least one direction, which makes m_i differs from m_g.

2/ For m_g

m_g = gravitational force / gravitation g
This is <u>definition</u> of m_g

and also g = GM/R^2, is <u>definition</u> of gravitation g.

Don't you feel good about these definition? If not, you are free to suggest other sensible definitions. And again, with your definitions, you still have only one value of m_g, my friend.