A tweek to Einsteins theory

[morgsboi]

S=Speed

C=3 x 10⁸ ms^-1 (speed of light)

If S>C then E=M(C+(S-C)^2

Edited November 6, 2011 by morgsboi

[timo]

Obviously, in this form the number of opening and closing parentheses in your relation between the letters E, M, C, and S does not match, which makes it very dubious. But the more fundamental issue here is: what is your point?

[swansont]

Can you derive this equation from some underlying principles?

[morgsboi]

Can you derive this equation from some underlying principles?

 

It's only something I've just though of while thinking about the Opera (Oscillation Project with Emulsion-tRacking Apparatus) experiment, which broke the speed of light and how the amount of energy would be measured but would breaking the speed of light change the mass of the particles?

[swansont]

It's only something I've just though of while thinking about the Opera (Oscillation Project with Emulsion-tRacking Apparatus) experiment, which broke the speed of light

 

Well, that's not yet been confirmed. But science abhors ad-hoc-ery. You need to justify the equation.

[morgsboi]

Well, that's not yet been confirmed. But science abhors ad-hoc-ery. You need to justify the equation.

 

Yes, I know it's not confirmed but if light speed was broken (doesn't matter what experiment) then would this work? And I don't understand how it could be justified as it's in it's simplest form.

 

Yes, I know it's not confirmed but if light speed was broken (doesn't matter what experiment) then would this work? And I don't understand how it could be justified as it's in it's simplest form.

 

Actually I've noticed a flaw, sorry. The equation should be if S=>C then E=M(C+(S-C)^2 which I will change in the forum. And as a word equation it would simply be: If the speed is greater than the speed of light then Energy is equal to the speed of light plus the sum of the speed minus the speed of light.

Make sense??

[insane_alien]

by justify i'm fairly sure swansont meant: why should that be the equation used? how did you come to that equation?

 

equations don't just pop out of nowhere.

 

also, as this is most definitely considering motion why have you used E=mc^2 (the simplification for zero momentum) rather than the full equation?

[swansont]

Yes, I know it's not confirmed but if light speed was broken (doesn't matter what experiment) then would this work? And I don't understand how it could be justified as it's in it's simplest form.

 

How was it derived, or otherwise arrived at?

[morgsboi]

How was it derived, or otherwise arrived at?

 

As I said, I was thinking about if something went faster than the speed of light then what would happen to Einstein's theory, so I developed a simple and quick tweak.

 

by justify i'm fairly sure swansont meant: why should that be the equation used? how did you come to that equation?

 

equations don't just pop out of nowhere.

 

also, as this is most definitely considering motion why have you used E=mc^2 (the simplification for zero momentum) rather than the full equation?

 

I've explained it a couple of posts below and also I am only 14 and I come on here to learn more about my thoughts and theories that I have so I'm not a big expert in science but I am hoping to be.

[Cap'n Refsmmat]

The full form of the equation for particles in motion is [math]E^2 = m^2 c^4 + p^2 c^2[/math], where p is momentum. Your equation will need to reduce to this when [math]v<c[/math].

[DrRocket]

The full form of the equation for particles in motion is [math]E^2 = m^2 c^4 + p^2 c^2[/math], where p is momentum. Your equation will need to reduce to this when [math]v<c[/math].

 

 

That is somewhat a matter of taste and definitions (in particular what is meant by "mass").

 

[math]E^2 = m^2 c^4 + p^2 c^2[/math],

 

applies with [math]m[/math] being rest mass ([math]m_0[/math]), which is currently in fashion.

 

 

[math]E = m c^2 [/math],

 

applies if m is taken to be relativistic mass ([math] \gamma m_0[/math]) and if the rest mass is non-zero. It is equivalent to [math]E^2 = m_0^2 c^4 + p^2 c^2[/math] in that case.

 

So for massive neutrinos [math]E = m c^2[/math] is fully accurate if properly interpreted

[Cap'n Refsmmat]

I suppose so; I must've been taught by fashionable physicists.

 

There's still the issue of units. If the equation is [math]E=m(c+(s-c)^2)[/math], then there are two parts:

 

[math]E=mc + m(s-c)^2[/math]

 

The left term does not have units of energy, while the right term does, so the equation doesn't work out.

 

One could interpret the missing parenthesis differently, as [math]E=m(c+(s-c))^2[/math], but that's just algebraically equivalent to [math]E=m(s)^2[/math].

 

morgsboi, which version of the equation did you mean?

[morgsboi]

The full form of the equation for particles in motion is [math]E^2 = m^2 c^4 + p^2 c^2[/math], where p is momentum. Your equation will need to reduce to this when [math]v<c[/math].

 

Okay, thanks. So what is V?

 

I suppose so; I must've been taught by fashionable physicists.

 

There's still the issue of units. If the equation is [math]E=m(c+(s-c)^2)[/math], then there are two parts:

 

[math]E=mc + m(s-c)^2[/math]

 

The left term does not have units of energy, while the right term does, so the equation doesn't work out.

 

One could interpret the missing parenthesis differently, as [math]E=m(c+(s-c))^2[/math], but that's just algebraically equivalent to [math]E=m(s)^2[/math].

 

morgsboi, which version of the equation did you mean?

Yes, the right hand version is a lot better.

[Fuzzwood]

velocity.

[DrRocket]

Okay, thanks. So what is V?

 

 

Yes, the right hand version is a lot better.

 

But [math]E=ms^2[/math] would imply that [math]E \rightarrow 0[/math] as [math]s \rightarrow 0[/math] which we know is grossly wrong (or nuclear weapons and nuclear power plants would not work). This also flies in the face of Einstein's recognition of the equivalence of mass and energy, so it is a long way from a mere tweak -- it is just plain wrong.

 

And THAT is why swansont's challenge to you to derive your equation from some set of basic principles is important. You don't revise a theory that is supported by a mountain of empirical data and developed logically from simple principles by simply pulling an equation out of ................ the air perhaps.

[morgsboi]

But [math]E=ms^2[/math] would imply that [math]E \rightarrow 0[/math] as [math]s \rightarrow 0[/math] which we know is grossly wrong (or nuclear weapons and nuclear power plants would not work). This also flies in the face of Einstein's recognition of the equivalence of mass and energy, so it is a long way from a mere tweak -- it is just plain wrong.

 

And THAT is why swansont's challenge to you to derive your equation from some set of basic principles is important. You don't revise a theory that is supported by a mountain of empirical data and developed logically from simple principles by simply pulling an equation out of ................ the air perhaps.

 

Yes, it was "out of the air" and I put it on here to get help with it.

[Mystery111]

That is somewhat a matter of taste and definitions (in particular what is meant by "mass").

 

[math]E^2 = m^2 c^4 + p^2 c^2[/math],

 

applies with [math]m[/math] being rest mass ([math]m_0[/math]), which is currently in fashion.

 

 

[math]E = m c^2 [/math],

 

applies if m is taken to be relativistic mass ([math] \gamma m_0[/math]) and if the rest mass is non-zero. It is equivalent to [math]E^2 = m_0^2 c^4 + p^2 c^2[/math] in that case.

 

So for massive neutrinos [math]E = m c^2[/math] is fully accurate if properly interpreted

 

For a nuetrino at rest, and even at that it is missing some important details in describing neutrino's. In fact, some theories of neutrino's would not have a mass term in it under a Weyl Limit.

 

The mass does the same thing to coupling particles and antiparticles together as you would find from a majorana equation. Nuetrino's are spin 1/2 so they are fermions:

 

[math]-i(\alpha \hat{p})c\psi + \beta M c^2 \psi = i\hbar \partial_t \psi[/math]

 

This can be re-written as:

 

[math]i\hbar \frac{\partial \phi}{\partial t} = -ic\hbar \vec{\sigma} \cdot \nabla_{\phi} + M c^2 \phi'[/math]

 

[math]i\hbar \frac{\partial \chi}{\partial t} = -ic\hbar \vec{\sigma} \cdot \nabla_{\chi} + M c^2 \chi'[/math]

 

as two component equations, when this equation is under a Weyl representation

 

[math]i\hbar \frac{\partial \eta}{\partial t} = -ic\hbar \vec{\xi} \cdot \nabla \eta + M c^2 \xi[/math]

 

and

 

[math]i\hbar \frac{\partial \xi}{\partial t} = -ic\hbar \vec{\eta} \cdot \nabla \xi - M c^2 \eta[/math]

 

[math]\eta[/math] and [math]\xi[/math] are in fact coupled to a limit where [math]M[/math] is nonzero. Under mathematcial strutiny, the fact that the Nuetrino has such a ridiculously small mass incorporates the similar contention that the nuetrino could act more or less like a particle with no mass. So in this case, the non-relativistic case [math]Mc^2[/math] will not suffice.

 

With the limit where [math]M=0[/math] reduces to the Weyl equation

 

[math]\frac{\partial \xi v'}{\partial} = -c\vec{\sigma} \cdot \nabla \xi v'[/math]

 

This only suits right solutions for antiparticles for nuetrinos [math]v'[/math].

 

[math]\frac{\partial \xi v'}{\partial} = -c\vec{\sigma} \cdot \nabla \xi v'[/math]

 

I think we established that it described antineutrinos.... The second equation in your coupled set up vanish because [math]\phi = \chi[/math] when [math]M=0[/math]. Now this is actually related to parity (you may understand parity under CPT-symmetry.) The Weyl equation is only permitted for right handed antineutrino's since a nuetrino is only ever left handed.

 

For that purpose you may introduce a transformation [math]\alpha \rightarrow -\alpha[/math]. Anticommutator relations are preserved in the Dirac Equation which would describe the neutrino. The transformation also effects [math]\sigma \rightarrow -\sigma[/math] so the Weyl equation becomes

 

[math]\frac{\partial \xi v}{\partial} = c\vec{\sigma} \cdot \nabla \xi v[/math]

 

For a nuetrino. Simply a neutrino must have a mass at a specific limit which allows us to make these valid transformations. The non-relativistic case of [math]Mc^2[/math] also fails to descrive fermions correctly. As you will surely know, the correct energy condition has both a positive and negative solution [math]E= \pm Mc^2[/math]. So you will also deal with

 

[math]\psi^1 = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} e^{\frac{iMc^2 t}{\hbar}}[/math]

 

[math]\psi^2 = \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix} e^{\frac{iMc^2 t}{\hbar}}[/math]

 

Are your two positive energy solutions with opposite spin states

 

[math]\psi^3 = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix} e^{\frac{+iMc^2 t}{\hbar}}[/math]

 

[math]\psi^4 = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} e^{\frac{+iMc^2 t}{\hbar}}[/math]

 

Are your two negative energy states with opposite spin

Edited November 6, 2011 by Mystery111

[swansont]

As I said, I was thinking about if something went faster than the speed of light then what would happen to Einstein's theory, so I developed a simple and quick tweak.

 

 

That's not how science is done. You can get relationships by matching the pattern of some data, but mainly you start with some physical principle and derive the equation.

[morgsboi]

That's not how science is done. You can get relationships by matching the pattern of some data, but mainly you start with some physical principle and derive the equation.

 

Yes, but as I also said, I put it on here to get information to learn about it. I am only 14 so I haven't exactly been able to get a PHD with years of studying. But I am hoping too.

[Mystery111]

You are 14, and you want to know about neutrino's, or how they are applied in relativity? All due respect but you really are jumping in the deep end my friend first.

[swansont]

Yes, but as I also said, I put it on here to get information to learn about it. I am only 14 so I haven't exactly been able to get a PHD with years of studying. But I am hoping too.

 

My point is that you are approaching this from the wrong direction. I understand the situation, because I was 14 once and didn't know how how scientific results were achieved. But you need to start with some scientific principle. Einstein started with the speed of light being the same in all reference frames and incorporated that into kinematic analysis. E=mc^2 was one result.

[between3and26characterslon]

Yes, the right hand version is a lot better.

 

I don't think you've quite understood what Cap'n was getting at here. Have a read of this http://www.efm.leeds...al_analysis.htm

 

What it is saying is that all the terms in an equation not only have units but have dimensions as well, so for instance:

 

velocity 'v' has units of ms^-1 and dimensions of LT ^-1 (where L=length and T=time)

acceleration 'a' has units of ms^-2 and dimensions of LT ^-2

 

so if you wrote an equation

 

v=a

 

you can test its validity by checking its dimensions and you can see.

 

LT ^-1 ≠ LT ^-2

 

In terms of dimensions used in an equation the lefthand side must equal the righthand side.

 

 

In regards to your equation to 'tweek' Einstein's theory I'm affraid you can not pull an equation out of thin air, presume it's correct and then reverse engineer it to find out why it's correct. What Swansont is saying is the E=mc² was the result of a lot of work not the beggining.

[michel123456]

S=Speed

C=3 x 10⁸ ms^-1 (speed of light)

If S>C then E=M(C+(S-C)^2

 

Maybe Morgsboi is 14 and missing knowledge he will acquire no doubt about that, but his mind is working.

I suppose from his OP he was basically asking whether Energy increases over mc^2 for hypothetical FTL particles.

[morgsboi]

You are 14, and you want to know about neutrino's, or how they are applied in relativity? All due respect but you really are jumping in the deep end my friend first.

 

I have done lots of studying and have a good understanding, also I'm a very quick learner. What do you expect me to do? Light magnesium with a bunsen burner? I'm much more advanced than a lot of people my age.

[IM Egdall]

You go Morgsboi! Someday soon, you will be teaching us.