E(c2)=M additional formula for hyperspeed

[alexander]

Well, here are some issues with your monkey reasoning, Illiad, nobody has spoken Kaanaic in over 600 years, thus nobody knows how to speak it. On top of which there may not be enough vocabulary in Kaanaic to translate the US anthem. On top of that we have yet to see a healthy monkey that can actually speak (other then this monkey), never mind a traumatized monkey with a wooden leg... and thats before you start considering the fallen tree that nobody hears falling...

 

 

Back to Ryan:

with regards to mass, our current laws of physics assume mass, irrelevant mass, or proper mass, to be

a characteristic of the total energy and momentum of an object or a system of objects

 

Ok, ok, if you want proof that your reasoning is full of holes consider the following problem with your model, if i suppose that E=mc^2 and m=Ec^2 are both true, then we should find a place where both of these would have to be used to describe something... like lets say to describe invariant mass of a particle collision...

 

standard model suggest that invariant mass is equal to total energy of a system divided by c^2, or [math](Wc^2)^2=\left(\sum{E}\right)^2-\left|\left|\sum{pc}\right|\right|^2[/math], however your model suggests a bit of the opposite, that invariant mass is that system multiplied by c^2 or [math]\left(\frac{W}{c^2}\right)^2=\left(\sum{E}\right)^2-\left|\left|\sum{pc}\right|\right|^2[/math] so since we have both energy and mass in this equation, and assuming that somehow both energy mass equivalence and your nonsensical mass formuli are true, lets figure out a simple fully elastic collision where 2 particles collide and form one particle. For simplicity sake suppose the following:

both partciles have equal mass i.e. m1=m2

particles momentum vectors have equal magnitude, but opposite directions, or same direction but opposite magnitude, i.e. p1=-p2 either case, momentum vectors cancel out, regardless of what they are.

 

[math]

\begin{split}

M^2&=\left(\sum{E}\right)^2-\left|\left|\sum{pc}\right|\right|^2\\

(Wc^2)^2&=(E_1+E_2)^2-\left|\left|p_1c+p_2c\right|\right|^2\\\text{Given: } E=mc^2\\

&=(m_1c^2+m_2c^2)^2-\left|\left|p_1c+p_2c\right|\right|^2\\

&=(c^2(m_1+m_2))^2-\left|\left|c(p_1+p_2)\right|\right|^2\\

&=c^4((m_1+m_2))^2-c^2\left|\left|(p_1+p_2)\right|\right|^2\\\text{dividing both sides by } c^2\\

(Wc)^2&=c^2((m_1+m_2))^2-\left|\left|(p_1+p_2)\right|\right|^2\\\text{suppose } m_1=m_2=1 \text{ and } p_2=-p_1\\

(Wc)^2&=c^24\\

W^2c^2&=c^24\\

W^2&=4\\

W&=2

\end{split}[/math]

 

so as sort of easily predicted even without knowing physics, you take 2 objects, thow them at each other, and get a third object that weighs the sum of the two... one gram plus one gram is equal to two gram

 

[math]

\begin{split}

M^2&=\left(\sum{E}\right)^2-\left|\left|\sum{pc}\right|\right|^2\\

\left(\frac{W}{c^2}\right)^2&=(E_1+E_2)^2-\left|\left|p_1c+p_2c\right|\right|^2\\\text{Given: } E=mc^2\\

&=(m_1c^2+m_2c^2)^2-\left|\left|p_1c+p_2c\right|\right|^2\\

&=(c^2(m_1+m_2))^2-\left|\left|c(p_1+p_2)\right|\right|^2\\

&=c^4((m_1+m_2))^2-c^2\left|\left|(p_1+p_2)\right|\right|^2\\\text{multiplying both sides by } c^2\\

W^2&=c^8(m_1+m_2)^2-c^6\left|\left|(p_1+p_2)\right|\right|^2\\

W&=c^4(m_1+m_2)^2-c^3\left|\left|(p_1+p_2)\right|\right|\\\text{suppose } m_1=m_2=1 \text{ and } p_2=-p_1\\

W=2c^4\\\text{where }c=299792458 m/s\\

W=16155217426124980458527601492303392

\end{split}[/math]

 

so according to your model, an elastic collision of two electrons:

[math]2*0.000000000000000000000000000000910938188*299792458^4=14716.404488900313560426542451384947854733696kg[/math]

creates a particle weighting in at 14.716 metric tons (this is even ignoring the serious issue with units in this example)

or one gram plus one gram = 8.12 times the mass of the sun?

[A23]

Energy additional formula :

 

I don't know exactly what you're seeking for, but an easy method is :

 

from the most general type of equation we seek : [math]E^n=c^\alpha p^\beta m^\gamma[/math]

 

in units we get :

 

[math]\frac{m^{\alpha+\beta}kg^{\beta+\gamma}}{s^{\alpha+\beta}}=\frac{m^{2n}kg^n}{s^{2n}}[/math]

 

for [math]n=1[/math], we can with this "strange/silly", or what we call

 

"turning in round methods/mathematics" get the following relations, with for example :

 

the triples {[math]\alpha,\beta,\gamma[/math]}={1,1,0},{0,2,-1},{-1,3,-2},{-2,4,-3},... and hence a possible equation with coefficient [math]A_i[/math]

 

[math]E=A_1pc+A_2\frac{p^2}{m}+A_3\frac{cp^3}{m^2}+...[/math]

 

and if we canonically quantize, [math]E\rightarrow i\hbar\partial_t, p\rightarrow -i\hbar\partial_x[/math] :

 

[math]-i\hbar\frac{\partial\Psi}{\partial t}=A_1\frac{-i\hbar\partial\Psi}{\partial x}+A_2\frac{-\hbar^2\partial^2\Psi}{\partial x^2}+A_3\frac{-ic\hbar^3\partial^3\Psi}{m^2\partial x^3}+...[/math]

 

if but you seek variation on c, you get :

 

[math]E=pc+c^2m+c^3m^2/p+c^4m^3/p^2+c^5m^4/p^3+c^6m^5/p^4+c^7m^6/p^5+...[/math]

 

"The method of people that know nothing to do"

[A23]

Nobody knows (*wink*), since multiplication is not defined for vectors

 

Hi Ben,

 

I thought about your post,

 

speed multiplied by speed is anyhow in units : (speed squared), anyhow we have sort of "multiplication" (?) for vectors :

 

let suppose [math]\left(\begin{array}{c}a\\b\end{array}\right),\left(\begin{array}{c}d\\e\end{array}\right)[/math] were 2 vectors of norm c, describing velocities of 2 photons in 2D :

 

dot-product : [math]\left(\begin{array}{c}a\\b\end{array}\right)\cdot\left(\begin{array}{c}d\\e\end{array}\right)=ad+be[/math] (scalar[math]\leq c^2[/math])

 

Hadamard product : [math]\left(\begin{array}{c}a\\b\end{array}\right)\odot\left(\begin{array}{c}d\\e\end{array}\right)=\left(\begin{array}{c}ad\\be\end{array}\right)[/math] (vector but norm smaller equals [math]c^2[/math])

 

Tensor product :[math]\left(\begin{array}{c}a\\b\end{array}\right)\otimes\left(\begin{array}{c}d\\e\end{array}\right)=\left(\begin{array}{c}ad\\ae\\bd\\be \end{array}\right)[/math] (vector of too high dimension, norm is [math]c^2[/math]) (or in 2x2 matrix form, the energy 'tensor')

 

the only internal product would be the Hadamard (not sure about the name) product.

 

Anyhow, I think no one of these makes sense in our context.

[Illiad]

haha...alright, I made the monkey up :D. Looks like ryan theory already fell apart. I thought I was going to see some strange maths by ryan, but I guess not.

[Qfwfq]

Einstien said E=M(c2), I say additionally to that that E(C2)=M

This works providing, of course, that natural units are chosen so that c is an adimensional quantity equal to 1.

 

So, what's the point of having such an absurd discussion about it?

[alexander]

Q, it still doesnt work, if he were to mention natural units and drop c entirely, i could understand, but regardless of what c is, mass is not (or at least he has failed to show any proof) ec^2, but yes in natural units it will work the same as e/c^2, except for, ofcourse as craig pointed out earlier, the units wont make any sense what-so-ever...

[Qfwfq]

If klagrooooopf is adimensional and equals 1 it doesn't matter whether you multiply or divide by it or not, nor how many times. The simultaneous equations:

 

A = B klagrooooopf

 

B = A klagrooooopf

 

simply constrain klagrooooopf to be such that its square is 1 and in the case with klagrooooopf defined as [imath]c^2[/imath] natural units satisfy this perfectly (along with three phyisically meaningless options).

 

Now of course [imath]m = Ec^2[/imath] is conceptually wrong and it's even dumb to write it, but natural units make it not matter; they make it the same as writing [imath]m = E[/imath] along with [imath]E = m[/imath]. So the OP is just an odd way of proposing the use of natural units. No more, no less, nothing to make a great fuss over.

[CraigD]

So the OP is just an odd way of proposing the use of natural units. No more, no less, nothing to make a great fuss over.

Methinks you over-interpret the original post, Q, and ignore its title, “E(c2)=M additional formula for hyperspeed”.

 

This thread’s OP, and Ryan’s later posts, appear to me to be installments in his long quest for a design approach for a vehicle that can travel faster than light. These seem to be proposing some sort of vehicle propelled by lightning.

 

Ryan admits to a “lack of an education in physics” and the resulting inability to “prove” his claims, so approaches his quest by making claims like these, and challenging people who have studied physics to “disprove” them.

 

Ryan, I've written before, and continue to believe, you’d do far better to correct you state of lacking an education in physics by studying math, physics, and a well-balanced mix of other subjects, than to post strange claims and challenge others to prove or disprove them. In short, go to school, stay in school, and actively participating in your education. If your experience is much like mine, you find it’s ultimately much more fun and rewarding to make claims you understand and can support than doing what you’ve done in this thread.

[Eclogite]

On top of that we have yet to see a healthy monkey that can actually speak (other then this monkey), never mind a traumatized monkey with a wooden leg

That's a matter of a pinion.

[Qfwfq]

Do monkeys use a wooden leg instead of the rack? :phones: :rolleyes:

 

Methinks you over-interpret the original post, Q, and ignore its title, “E(c2)=M additional formula for hyperspeed”.

My first post was just a dumber way of saying what I said in my second one. If Ryan thinks we might discover hyperspeed in this manner then yes, he needs an increased understanding of physics.

[alexander]

Do monkeys use a wooden leg instead of the rack?

 

Depends on who's rack....

[Qfwfq]

The pinion's rack, of course.

[alexander]

The pinion's rack, of course.

hope she's hot...

[Qfwfq]

hope she's hot...

Nah, maybe Ruby's Rack might be hotter...

[ryan2006]

apology

[ryan2006]

Apology accepted.

[Boerseun]

Ryan, I'm sorry to see your self-destructive postings here.

 

There's a flaw in your argument that we've been trying to point out to you since post 1.

 

I will try again, because I know even after threatening not to come back, you will watch the replies here. Here goes:

 

We know that e=mc². On top of this, you're saying that ec²=m.

 

That's simply wrong. And this is why - follow closely, it's really easy:

 

You are saying that 10=(5)(2) is equivalent to 10(2)=5.

 

It is clear that you're simply taking the one unit from the right and putting it on the left with the same operator. It was a multiply on the right - if you want that variable on the left, you have to divide it on both sides. If it was a division on the right, only then do you get to multiply on the left. This is honest-to-god second grade arithmetic.

 

And don't flatter yourself with these illusions of grandeur of being the next Einstein or expect to be the next Nobel Prize winner if you can't see that simple failure in your logic.

 

Criticism is one thing. Insulting others because you cannot take it is quite another, and quite unwelcome here. If you feel that the world of science and peer review is a total cock-up because it won't let you through to round 2, then you won't be alone. It's an evolutionary process. Only those ideas and theories that survive the harsh searchlight of logic and reason is let through. Your idea of ec²=m finds itself in the crowded space populated by the intellectual versions of the dodo, the dinosaur, the mastodon, the mammoth, the glyptodon and the vicious sable-tooth tit-mouse. They were all tried, chewed over and spat out by nature in a grimace of disgust. Some were chewed longer than others. Probably because they had less hair. Be that as it may, we should be happy when a theory we put forward is shown to be in error, because that just chips away at the obstacles standing in the way of scientific progress. It bevels the edges of the piston of science moving up the cylinder shaft of knowledge, so to speak (damn, I'm good!).

 

But be that as it may, after insulting our crew in such a fashion, I have infracted you for both your last posts. You did not leave me with a choice. You have now been suspended for a few days. Oops. If you decide to participate in Hypo again after this rather nasty faux pas on your side, please behave. If not, I fare thee well.