Problem Of Length Contraction At Cern Due To Special Theory Of Relativity.

[Vmedvil]

Physics is all math silly. Everything described in physics is defined by math including the very word energy.

 

Energy: the ability to perform work. is a math expression.

 

Well, in physics there is experimental proof which math does not have "Observation" which is how I know it belongs there.

 

http://iopscience.iop.org/article/10.3367/UFNe.0185.201511f.1225/meta

 

http://www.physlink.com/education/askexperts/ae406.cfm

Edited December 21, 2017 by Vmedvil

[Shustaire]

You really do like making up pointless excuses don't you.

Do you honestly believe people will accept those excuses for not following the correct procedures in equation building?

 

The apple didn't hit Newton on the head and he instantly derived his Newtons laws.

 

He invented calculus to derive those equations.

Edited December 21, 2017 by Shustaire

[Vmedvil]

You really do like making up pointless excuses don't you.

Do you honestly believe people will accept those excuses for not following the correct procedures in equation building?

 

No seriously.

[Shustaire]

Come on the full energy momentum equation can easily ne treated under central potential systems just as easily as it can be undera vacuum.

[Shustaire]

Seriously under GR there is 3 primary class of solutions. Vaccuum, Newton(central porential) and Schwartzchild. They all use the energy momentum relation as one of the fundamental equations.

 

Don't try to learn a physics topic by the internet googling. Buy the textbooks and study them.

Edited December 21, 2017 by Shustaire

[Vmedvil]

Ya, right  lol................... your joking right.

[Shustaire]

Lets do an example.

 

Lets find a simplified expression for Lorentz shall we lets start with its coordinates.

[math]x^\mu=(x^0,x^1,x^2,x^3)=(ct,x,y,z)[/math] with the subscript [math]\mu=0,1,2,3[/math]

 

so [math]x^\mu[/math] is now my space-time coordinates. Wow ain't that just ducky one variable for all four....

 

So consider a Lorentz frame S where two events use [math]x^\mu[/math] and [math]\acute{x}^\mu[/math] Both observers will only agree on the Worldline invariant interval. sign convention for Lorentz (-+++).

[math]-\Delta S^2=-(\Delta x^0)^2)+(\Delta x^1)^2)+(\Delta x^2)^2)+(\Delta x^3)^2)[/math]

[math]-\delta S^2=-(\Delta x^0)^2)+(\Delta x^1)^2)+(\Delta x^2)^2)+(\Delta x^3)^2)=-\delta S^2=-(\Delta \acute{x}^0)^2)+(\Delta \acute{x}^1)^2)+(\Delta \acute{x}^2)^2)+(\Delta \acute{x}^3)^2)[/math]

in short

[math]\Delta S^2=\acute{S}^2[/math]

the minus sign on the left denotes time-like separated

[math]\Delta S^2>0[/math]

 

However you want the infinitesimal coordinate variations for curve fitting under GR, QM uses quantized units but that's easy enough to add in later on. An invariant interval [math]ds^2[/math] recall I mentioned line element?

 

so

[math]ds^2=-(dx^0)^2+(dx^1)^2+(dx^2)^2+(dx^3)^2[/math]

so

[math]ds^2=\acute{ds}^2

 

so lets do the subscripts vs superscripts. use x for the zeroth (a coordinate is scalar)

[math]dx_\mu=dx^\mu[/math] easy enough to apply that to each coordinate. I shouldn't have to show you that.

 

this gives us the change in the zeroth component (zeroth is a scalar value) the sign of

[math]dx_\mu=(dx_0,dx_1,dx_2,dx_3)=(-dx^0,dx^1,dx^2,dx^3)[/math]

so

[math]ds^2=dx_0dx^0+dx_1dx^1,dx_2dx^2,dx_3dx^3[/math]

 

we have now removed the need of the minus sign by using convariant and contravariant indices.

 

so now we can simplify the last as

[math]ds^2=\sum^3_{\mu=0}dx_\mu^\mu[/math]

 

so in Minkowskii metric by Einstein summation above [math]\eta_{\mu\nu}[/math]

[math]-ds^2=\eta_{\mu\nu}dx^\mu dx^\nu\mu[/math] notice the change in order of the basis.

Now so far we are symmetric

 

what about antisymmetric?

 

we need to decompose a two index object [math]M_{\mu\nu}[/math] into its symmetric and antisymmetric parts.

[math]M_{\mu\nu}=\frac{1}{2}(M_{\mu\nu})+M_{\nu\mu})+\frac{1}{2}(M_{\mu\nu})+M_{\nu\mu})[/math]

the antisymmetric part on the right changes sign under exchange of indices

so we denote this by

[math]\xi_{\mu\nu}=-\xi_{\mu\nu}[/math]

[math]\xi_{\mu\nu}dx^\mu dx^\nu=(-\eta_{\mu\nu}dx^\mu dx^\nu=-\xi-{\mu\nu}dx^\nu dx^\mu=-\eta_{\mu\nu}dx^\mu dx^\nu[/math]

 

so then

[math]-ds^2=\eta_{00}xx^0dx^0+\eta{01}dx^0dx^1+\eta_{11}dx^1dx^1+...[/math]

Edited December 21, 2017 by Shustaire

[Vmedvil]

Lets do an example.

 

Lets find a simplified expression for Lorentz shall we lets start with its coordinates.

[math]x^\mu=(x^0,x^1,x^2,x^3)=(ct,x,y,z)[/math] with the subscript [math]\mu=0,1,2,3[/math]

 

so [math]x^\mu[/math] is now my space-time coordinates. Wow ain't that just ducky one variable for all four....

 

So consider a Lorentz frame S where two events use [math]x^\mu[/math] and [math]\acute{x}^\mu[/math] Both observers will only agree on the Worldline invariant interval. sign convention for Lorentz (-+++).

[math]-\Delta S^2=-(\Delta x^0)^2)+(\Delta x^1)^2)+(\Delta x^2)^2)+(\Delta x^3)^2)[/math]

[math]-\delta S^2=-(\Delta x^0)^2)+(\Delta x^1)^2)+(\Delta x^2)^2)+(\Delta x^3)^2)=-\delta S^2=-(\Delta \acute{x}^0)^2)+(\Delta \acute{x}^1)^2)+(\Delta \acute{x}^2)^2)+(\Delta \acute{x}^3)^2)[/math]

in short

[math]\Delta S^2=\acute{S}^2[/math]

the minus sign on the left denotes time-like separated

[math]\Delta S^2>0[/math]

 

However you want the infinitesimal coordinate variations for curve fitting under GR, QM uses quantized units but that's easy enough to add in later on. An invariant interval [math]ds^2[/math] recall I mentioned line element?

 

so

[math]ds^2=-(dx^0)^2+(dx^1)^2+(dx^2)^2+(dx^3)^2[/math]

so

[math]ds^2=\acute{ds}^2

 

so lets do the subscripts vs superscripts. use x for the zeroth (a coordinate is scalar)

[math]dx_\mu=dx^\mu[/math] easy enough to apply that to each coordinate. I shouldn't have to show you that.

 

this gives us the change in the zeroth component (zeroth is a scalar value) the sign of

[math]dx_\mu=(dx_0,dx_1,dx_2,dx_3)=(-dx^0,dx^1,dx^2,dx^3)[/math]

so

[math]ds^2=dx_0dx^0+dx_1dx^1,dx_2dx^2,dx_3dx^3[/math]

 

we have now removed the need of the minus sign by using convariant and contravariant indices.

 

so now we can simplify the last as

[math]ds^2=\sum^3_{\mu=0}dx_\mu^\mu[/math]

 

so in Minkowskii metric by Einstein summation above [math]\eta_{\mu\nu}[/math]

[math]-ds^2=\eta_{\mu\nu}dx^\mu dx^\nu\mu[/math] notice the change in order of the basis.

Now so far we are symmetric

 

what about antisymmetric?

 

we need to decompose a two index object [math]M_{\mu\nu}[/math] into its symmetric and antisymmetric parts.

[math]M_{\mu\nu}=\frac{1}{2}(M_{\mu\nu})+M_{\nu\mu})+\frac{1}{2}(M_{\mu\nu})+M_{\nu\mu})[/math]

the antisymmetric part on the right changes sign under exchange of indices

so we denote this by

[math]\xi_{\mu\nu}=-\xi_{\mu\nu}[/math]

[math]\xi_{\mu\nu}dx^\mudx^\nu=(-\eta_{\mu\nu}dx^\mudx^\nu=-\xi-{\mu\nu}dx^\nu dx^\mu=-\eta_{\mu\nu}dx^\mu dx^\nu[/math]

 

so then

[math]-ds^2=\eta_{00}xx^0dx^0+\eta{01}dx^0dx^1+\eta_{11}dx^1dx^1+...[/math]

That is incompatible by looking at it with QM and GR instantly failure

Edited December 21, 2017 by Vmedvil

[Shustaire]

That is incompatible by looking at it with QM instantly failure

 

that's next stupid that's just a coordinate basis for summation idiot

 

Your one to FFF>>>ing tlk with the friggin mess your equation had in mixed metric idiot.

 

Did you think its eaasy to latex all that and keep track you flipping moron.

Edited December 21, 2017 by Shustaire

[Vmedvil]

that's next stupid

Go on, I will watch she who "Knows it all" I wouldn't miss this for the world.

 

But remember you cannot use my method it is wrong remember.

Edited December 21, 2017 by Vmedvil

[Shustaire]

No FU.  I'm done wasting my time trying to teach your sorry arse. Your a complete waste of effort and skin thaat shows absolutely no flipping appreciation for the effort others put into teaching you you sorry sack of SH.T

Edited December 21, 2017 by Shustaire

[Vmedvil]

No FU.  I'm done wasting my time trying to teach your sorry arse. Your a complete waste of effort and skin thaat shows absolutely no flipping appreciation for the effort others put into teaching you you sorry sack of SH.T

 

Oh, "Teach me then" go on solve Quantum gravity prove it that you are something that knows what you are talking about. 

[Shustaire]

No you figure out how to get your position and momentum operators under Klein_ Gordon to the above

 

here is a flipping hint.

[math][x^i,p_j]=i\hbar\partial^i_j[/math]

 

I'm done with you

[Vmedvil]

No you figure out how to get your position and momentum operators under Klein_ Gordon to the above

 

here is a flipping hint.

[math][x^i,p_j]=i\hbar\partial^i_j[/math]

 

I'm done with you

 

They are.

[Shustaire]

Yeah f you they are your full of BS. You didnt even use the right flipping equation for the momentum operator dumbass

Edited December 21, 2017 by Shustaire

[Vmedvil]

Yeah f you they are your full of BS. You didnt even use the right flipping equation for the momentum operator dumbass

 

I am waiting for you at the end of this read it all the way through.

 

 

http://www.scienceforums.net/topic/111607-wormhole-metric-how-is-this-screwed-up/

[Shustaire]

So youve been told before your equation is messed up and you should start over. Why didnt you listen then?

 

Obviously they tried to point this out to you but you ignored their advice.

 

Glad I didn't waste as much time as they did. They have far greater patience than I do.

Edited December 21, 2017 by Shustaire