# A New Light In Physics

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### #1 martillo

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Posted 26 June 2007 - 01:08 PM

A great challenge to open minds:

A New Light In Physics

"Classical Physics is coming back, RELOADED!"

### #2 Boerseun

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Posted 26 June 2007 - 01:49 PM

Did you write that yourself, or did you find it on the 'net?

I just read the "New Evidence" section, and there are some glaring mistakes in it.

### #3 martillo

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Posted 26 June 2007 - 03:39 PM

Boerseun,

Did you write that yourself, or did you find it on the 'net?

I wrote it.

I just read the "New Evidence" section, and there are some glaring mistakes in it.

### #4 Erasmus00

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Posted 26 June 2007 - 09:44 PM

Consider your claim that the "real" equation of motion is F=ma instead of $F= \frac{dp}{dt}$ This is completely false. Lets look at your rocket equation.

Consider a rocket of mass M moving at speed v, expelling exhaust of mass dm at a velocity u. Note, u is a negative quantity if v is positive.

$P(t) = (M+dm)v$
$P(t+dt)=M(v+dv)+dm(v+dv+u)$
$dP=P(t+dt)-P(t)=Mdv+udm$

Now, there is no external force on this system (nothing external). Hence,
$\frac{dP}{dt}=0=M\frac{dv}{dt}+u\frac{dm}{dt}$

This leads to the traditional (and experimentally verified) rocket equation. Using F=ma does NOT result in the correct result.
-Will

### #5 martillo

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Posted 27 June 2007 - 05:36 AM

Erasmus00,

This leads to the traditional (and experimentally verified) rocket equation. Using F=ma does NOT result in the correct result.

Yes it does and as is described at the end of the page.

By definition p = Mv and dp/dt = M(dv/dt) + v(dM/dt)
Now if F = Ma we have dp/dt = F + v(dM/dt)
Now if you apply this to the composed system of the rocket and the total fuel (the contained plus the expelled one you have:
dM/dt = 0 since the total fuel is constant and so:
dp/dt = F = 0 since there are no external forces applied.
Then we have the same initial equation.

The momentum must be conserved so the momentum of the rocket with its contained fuel must be equal and opposite to the momentum of the expelled fuel which is exactly the same conditon from which the equation of the rocket is currently derived! This leaves to the same equations in terms of the mass of the rocket and its contained fuel:
m(dv/dt) = -u(dm/dt)
the known equation of the rocket.

You must also note that in the pages cited in the text it is stated that the right term of the equation represents the force on the rocket. Then as the left part is m(dv/dt) = ma this represents ma = F and not dp/dt = F, no way.

### #6 Erasmus00

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Posted 27 June 2007 - 09:30 AM

You must also note that in the pages cited in the text it is stated that the right term of the equation represents the force on the rocket. Then as the left part is m(dv/dt) = ma this represents ma = F and not dp/dt = F, no way.

Then that particular page is wrong. The force on the rocket is a nebulous thing- considering that at any given instant much of the fuel is still part of the rocket. Thus, separating force on the rocket from force on the fuel is an impossible task.

Much better is considering the entire system, rocket+fuel and noting there are no external forces, so momentum is conserved. Further, without $F=\frac{dp}{dt}$ Newtonian mechanics would not always have conserved momentum in the absence of external forces.

Finally, its important to realize that the rocket equations you cite (while tested experimentally) where derived FROM Newtonian mechanics. They are derived in nearly every mechanics book, straight from F = \frac{dp}{dt}
-Will

### #7 martillo

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Posted 27 June 2007 - 12:09 PM

Erasmus00,

Then that particular page is wrong.

No they are right.

The force on the rocket is a nebulous thing- considering that at any given instant much of the fuel is still part of the rocket. Thus, separating force on the rocket from force on the fuel is an impossible task.

Not at all.
The mass m in the rocket equation mdv/dt = u(dm/dt) represents the mass of the rocket plus the contained fuel and u(dm/dt) is the force on the rocket with its contained fuel.
You should take more care in the analisis of the problem.

Further, without F=dp/dt Newtonian mechanics would not always have conserved momentum in the absence of external forces.

Wrong. It is conserved as I have already shown.
The relation dp/dt = m(dv/dt) + v(dm/dt) = F + v(dm/dt) shows that when
F=0 and dm/dt=0 the momentum is conserved. The condition dm/dt=0 is the same as the condition that the conservation of momentum be valid for "closed systems" only (constant mass) as presented in current Physics. There are cases to show that when mass varies the momentum is not conserved. With the new aproach the condition surges naturally from the rellation which requires dm/dt=0.

Finally, its important to realize that the rocket equations you cite (while tested experimentally) where derived FROM Newtonian mechanics. They are derived in nearly every mechanics book, straight from F = frac{dp}{dt}

The same can be derived with the new approach.

### #8 Erasmus00

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Posted 27 June 2007 - 04:08 PM

The condition dm/dt=0 is the same as the condition that the conservation of momentum be valid for "closed systems" only (constant mass) as presented in current Physics. There are cases to show that when mass varies the momentum is not conserved. With the new aproach the condition surges naturally from the rellation which requires dm/dt=0.

When mass varies, if there is no external force, momentum is STILL conserved. Consider the following: a raft of mass m is moving at a speed v in still water. A person (of mass M) falls into the boat with a downward velocity of V.

In this problem, the mass of the "raft" system is not conserved (it goes up by M). However, momentum IS conserved in the x direction (though not in the y).
-Will

### #9 martillo

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Posted 28 June 2007 - 01:00 PM

The system of the rocket with its contained fuel can be an example of how momentum is not conserved when mass varies.
Consider the rocket in the steady state of travelling with constant velocity. The net force acting on it is zero since the force of the expelled fuel is opposite to the resistence of the air. F=0. The mass m of the system is continuosly decreasing (and thus dm/dt is not zero) while the velocity v is mantained.
It is obvious that the momentum p = mv is not conserved for this system. So the momentum is not always conserved when the mass of the considered system varies.
I think this consideration is in basic texts of Physics!

### #10 Erasmus00

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Posted 28 June 2007 - 03:54 PM

Consider the rocket in the steady state of travelling with constant velocity. The net force acting on it is zero since the force of the expelled fuel is opposite to the resistence of the air. F=0. The mass m of the system is continuosly decreasing (and thus dm/dt is not zero) while the velocity v is mantained.

Again, treating the rocket as a system separate from its fuel can only lead to confusion, as your system is dynamically changing as the rocket expels mass.

It is far simpler to treat rocket+fuel as one system, in which case you can note that there IS a net external force, the air resistance, so obviously momentum is not conserved.
-Will

### #11 martillo

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Posted 29 June 2007 - 06:23 AM

It can lead to confusion for you only.
Momentum is not always conserved when the mass of the considered system is not conserved.
See the part "Conservation of momentum and collissions" at wikipedia: Momentum - Wikipedia, the free encyclopedia
Note the special mention about "closed system" in the principle.

### #12 CraigD

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Posted 29 June 2007 - 07:44 AM

I’ve been avoiding posting to this thread, as it seems to have diverged from its original purpose of discussing the martillo’s paper, “A New Light in Physics”. This paper appears to be an earnest attempt to, among other goals, refute the special theory of relativity by showing contradictory paradoxes that appear to result from it. Considering these paradoxes, and determining if they are due to a flaw with SR, or errors in its application, seem to me a worthwhile exercise.

I think the thread has become tangled in a semantic back-and-forth, as illustrated the following excerpts:

Consider your claim that the "real" equation of motion is F=ma instead of $F= \frac{dp}{dt}$ This is completely false.

Again, treating the rocket as a system separate from its fuel can only lead to confusion, as your system is dynamically changing as the rocket expels mass.

It can lead to confusion for you only.
Momentum is not always conserved when the mass of the considered system is not conserved.

Here, Erasmus asserts that the basic mechanical identity $\mbox{force} = \mbox{mass} \cdot \mbox{acceleration} = \mbox{mass} \cdot \frac{\Delta \mbox{velocity}}{\Delta \mbox{time}}$ is “completely false”, while the slightly less common form of the identity $\mbox{force} = \frac{d\mbox{momentum}}{d\mbox{time}} = \frac{d(\mbox{mass} \cdot \mbox{velocity})}{d\mbox{time}}$ is correct. Aside from my use of $\Delta$ to indicate ordinary real-number quantities vs. Erasmus’s use of the calculus’s “$d$” notation to indicate the equivalent infinitesimal quantities, the two equations appear to me derivable from one another, and the assertion that one is less “real” or “wrong”, not useful. The basic identity doesn’t allow mass to change, while the second one does, so, each has utility, depending on whether the mass of the body being considered changes or does not.

The argument over whether it is more or less confusing, and to whom, to calculate momentum in a closed system, in which it is conserved, or in an open one, in which it is not, likewise seems to me to be hair-splitting.

In a classical mechanical system, momentum is conserved if all particles are considered. If some particles are ignored, or removed from the system, the system’s momentum must be adjusted. Neither approach is inherently difficult to calculate, and the choice of which to use one of utility for the specific problem at hand. Even complications such as

It is far simpler to treat rocket+fuel as one system, in which case you can note that there IS a net external force, the air resistance, so obviously momentum is not conserved.

are reconcilable via an “include all or exclude some bodies” approach decision: The open rocket+fuel with external force from air resistance system can be “closed” by expanding it to be the rocket+fuel+air system, restoring conservation of momentum, restoring its simplicity, at the cost of a lot more computation, or left open, if preserving conservation of momentum is not worth the additional computation.

I think it a good idea to continue the debate over classical mechanical momentum (If the people involved continue to feel there’s really anything to debate) as a separate thread, beginning with post #2, and give the original thread a chance to return to martillo’s paper.

### #13 martillo

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Posted 29 June 2007 - 01:17 PM

CraigD,
I agree the thread may be diverging a little from the main subject F=ma without reaching any good conclusion...

May be you are misunderstanding something:

The basic identity doesn’t allow mass to change, while the second one does, so, each has utility, depending on whether the mass of the body being considered changes or does not.

It seems you haven't noted that I propose that the real equation of force actually is F=ma even when the mass m varies!

The equation of motion of the rocket is m(dv/dt) = u(dm/dt) with a varying mass m.
As stated in the pages I cite in my page, the right term represents the force on the rocket and its contained fuel (varying mass m) while the left term is just ma. Then we have that the case of the rocket motion shows us that actually the equation of the force is F = ma and not F = dp/dt!

### #14 CraigD

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Posted 29 June 2007 - 05:59 PM

May be you are misunderstanding something:

It seems you haven't noted that I propose that the real equation of force actually is F=ma even when the mass m varies!

I accept $F = m a$ as an equation of force – given that it’s the fundamental definition of the term for the past few centuries, it would be difficult not to. As all of the terms to the right of the equals sign can be considered instantaneous (no $\Delta$s in them), there’s no reason they both can’t vary, if that’s useful – as in the case of a rocket.

… Then we have that the case of the rocket motion shows us that actually the equation of the force is F = ma and not F = dp/dt!

I’m just not grasping the point you’re making, martillo. By definition, $a=\frac{dv}{dt}$, so $ma=m \frac{dv}{dt}$. By definition, $p = m v$, so $dp = m dv$ (strictly speaking, $\partial m \partial v$, but we’re already mixing notations, so a see no need to further complicate things) and $\frac{dp}{dt} = m \frac{dv}{dt} = ma$. I simply don’t see that the two equations are not equally valid.

From a pragmatic perspective, I also don’t see how the choice of notation is physically relevant. No matter how you represent it, the motion of a rocket with a particular exhaust velocity and energy is precisely determined. It seems to me, as I said before, that the debate over $F = m a$ vs. $F = \frac{dp}{dt}$ is pure hair-splitting.

I also fail to see what this has to do with a refutation of the special relativity presented in the first section of the first chapter of “A New Light in Physics”, or the connection of the brief appendix in which it appears to the beginning.

It seems to me a waste of time and effort to argue over the notation of such a well-defined phenomena as rocket motion, when martillo’s paper has such provocative text as

It is proposed that a photon is a pair of a positrin and a negatrin traveling together at light speed ç and at the force equilibrium distance λ determined by the Corrected De Broglie formula.

The photon has [non-zero] mass m and verifies the Corrected De Broglie formula

, and the claim that the speed of a photon is not constant! (from A New Light In Physics, chapter 4.1 “The Photon”)

As far as I’m able to determine, “positron” and “negatrin” are terms unique to martillo’s writing.

In searching for the term “negatrin”, I came across a post in another physics forum that gives me some cause for concern, however:

I have discovered those unlucky and unhappy errors and I suggest some corrections to fix them, but I'm not a physicist and I will not be. Is up to you real physicists to take my work and develop a New Physics.

This sounds suspiciously like a position demonstrated not to endear one to the members of science sites like hypography: “I don’t have a theory, but I know current theory is wrong, and would like someone capable of developing a theory based on what I find wrong with current theory to do so”.

A better approach, I think, is to examine martillo’s central objections to the theory of relativity, and see if a mutually acceptable resolution of them can be reached.
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### #15 martillo

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Posted 30 June 2007 - 03:05 PM

CraigD,

I simply don’t see that the two equations are not equally valid.

dp/dt is not the same as mdv/dt when the mass m varies as in a rocket.
p = mv and so dp/dt = m(dv/dt) + v(dm/dt) = ma + v(dm/dt)
You see there is a complete term of difference: v(dm/dt).
So there is a big difference between dp/dt and ma when the mass varies.

Is stated in the cited pages on my page that the equation:
m(dv/dt) = u(dm/dt)
is the equation of motion of a rocket with varying mass m and that the right term is the force applied to the rocket.
It is obvious from this that the equation of motion of a rocket obeys the relation F=ma and not F=dp/dt!

I assume enough experiments have been made with rockets and that the equation is right and also the statement that the right term is the force.
Then I propose that this problem shows that the real equation of force is F=ma and not F=dp/dt.

This makes a big difference. Is not just a difference in notation as you said, they are different equations!
Particularly, in Relativity Theory, the famous equation E=mc2 is derived from the assumption that F=dp/dt and not F=ma. E=mc2 cannot be derived with F=ma.
May be you should search a little about this. Here is a link of the derivation of E=mc2: http://www.phys.ufl....lativity_8b.pdf

As far as I’m able to determine, “positron” and “negatrin” are terms unique to martillo’s writing.

Yes, I have created those names for the most elementary particles of nature proposed in my new theory.

This sounds suspiciously like a position demonstrated not to endear one to the members of science sites like hypography: “I don’t have a theory, but I know current theory is wrong, and would like someone capable of developing a theory based on what I find wrong with current theory to do so”.

The difference is that I have already developed the theory, wrote it and is available in a website for "worldwide peer review" and further development.

A better approach, I think, is to examine martillo’s central objections to the theory of relativity, and see if a mutually acceptable resolution of them can be reached.

F=ma is also a very important objection to Relativity Theory and may be more important than the frames inconsistency of Section 1.1 (more convincent)!
From my point of view, better than this would be to look for what of good the new theory could bring to Physics but it seems nobody wants to "waste time" in a non recognized theory and the common aproach is to see first if the objections to Relativity are founded... It's a matter of choice.

### #16 Qfwfq

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Posted 03 July 2007 - 08:53 AM

...the two equations appear to me derivable from one another...

...providing, of course, mass is considered time independent.

If you read Newton's actual wording of the second axiom, and the definitions preceding them, you can see that in more modern terms his actual statement was that the change in quantity of motion in a given time interval is vector equal to the impulse over the same time. His terminology was somewhat different and used the same term force ambiguously for impulse and what we now call force (and he occasionally distinguished as "instantaneous force").