Undergrad Does this make a cubic out of a quadratic system?

[jk22]

Suppose the system of equations (coming from invariance of the wave equation) :

$$B=-vE\\A^2-B^2/c^2=1\\E^2-c^2D^2=1\\AD=EB/c^2\\B=vA\\AE-BD=1$$

If one adds a lightspeed movement like

$$A=a+f\\B=b-cf\\D=d+h\\E=e-ch$$

Then solving equ 1 for f gives

$$f=(b+ve-vch)/c$$

Equ 4 for h implies

$$h=(ve^2+ac^2d+bcd+vecd)/(vec-ac^2-bc+vc^2d)$$

And finally equ 3 for e yields

$$vce^3\\+e^2(ac^2-vc^2d+bc+2vc^2d)\\+e(vc+vc^3d^2+2c(ac^2d+bcd+c^3vd^2))\\+(-ac^4d^2-bc^3d^2+vc^4d^3-ac^2-bc+vc^2d+2ac^4d^2+2bc^3d^2)=0$$

Hence this should make cubic roots appear in the solving of the system of equation.

Is it a mistake or is it possible to make a cubic come out of a quadratic system of equations ?

 

[fresh_42]

It is hard to tell what you consider a variable and what a parameter. Anyway.

We have a system of $n$ quadratic polynomials in, say $\mathbb{R}[X_k]$.
If you now consider them all simultaneously, which you did via the coordinate transformation, which combines them, then we get polynomials from $\mathbb{R}[X_1,X_2,\ldots,X_n]$. Their degrees are thus up to $n^2$.

 

[jk22]

So normally these cubic root shall simplify to the usual square root of Lorentz transform ? But I don't see how this were possible.

 

[fresh_42]

So normally these cubic root shall simplify to the usual square root of Lorentz transform ? But I don't see how this were possible.

I only see an alphabet soup. I'll try to analyse it more specifically.We have polynomials $p_1,\ldots,p_6 \in \mathbb{R}[A,B,D,E]$ with $\deg p_i \leq 2$. Then you change variables by setting $A=a+ f\, , \, B=b-c f\, , \, D=d+ h\, , \,E=e-c h$ with two additional auxiliary variables $f,h$.

The resulting equations in $a,b,d,e,f,h$ are still of degree at most two even though in more variables, since we introduced only linear modifications. Sorry, I didn't look close enough on the degree of your transformations before.

Now you should write down your new equations:
\begin{align}
B=-vE &\longrightarrow b-c f = - ve + v c h \\
A^2-\frac{1}{c^2}B^2=1&\longrightarrow (a+f)^2-\frac{1}{c^2}(b-c f)^2=1\\
&\ldots
\end{align}

If you will have written all six equations in the new variables, out multiplied or not, then you can proceed.
Obey the following rules:

• do not divide
• trace the numbers of the equations you used to avoid doubles and reversals
• for the calculation you could rename $c=\gamma, v=\sigma$ in order to distinguish them from variables. Rename them back afterwards.

You should not get polynomials of degree higher than two. The resulting formulas shouldn't be very different to the original ones.

 

[mfb]

A minimal example: $a=(b+1)^2$ and $b=(c-1)^2$. Plugging the second equation into the first we get $a=c^4 - 3 c^3 + 8 c^2 - 8 c + 4$. The two simple quadratic equations produced a messy-looking quartic equation.

 

[jk22]

So it can go up to $2^n$ ?

 

[fresh_42]

So it can go up to $2^n$ ?

ät
Only if the transformation formulas have an according degree. The degree doesn't change in your case, since the transformations are linear: e.g. $p_1(B,E)=B+vE$ is the first polynomial, $p_1(B,E)=0$ the first equation, and $B=b-cf\, , \,E=e-ch$ the transformations. We therefore get $0=B+vE=(b-cf)+v(e-ch)=b+ve-c(f+h)$ which is still linear in the new variables $b,e,f,h$. The second polynomial is quadratic: $p_2(A,B)=A^2-\frac{1}{c^2}B^2-1$. Now we change variables again by linear transformations: $A=a+f\, , \,B=b-cf$, and we expect the degree to be $2$ again:
$$
p_2(A,B)=A^2-\frac{1}{c^2}B^2-1=(a+f)^2-\frac{1}{c^2}(b-cf)^2-1=a^2+\frac{1}{c^2}b^2+2af-\frac{2}{c}bf-1
$$
Now $\deg p_2= max\{\deg (a^2),\deg(b^2),\deg(af),\deg(bf)\} =2$ as expected for $p_2\in \mathbb{R}[a,b,f,]$.

The degree only raises if the transformations have a degree strictly greater than one, which is not the case in our example.

 

[jk22]

But AE is of degree 2 ?

 

[fresh_42]

But AE is of degree 2 ?

Yes, assuming $A,B,D,E$ are the indeterminates.

 

[jk22]

What if another solution involving cubics could exist as keeping the wave-equation invariant but allowing for any speed of the observer, obtained by superposing to a general uniform motion a light speed movement like here above ?

This could be the theory EPR sought and the one Bell pointed out with his theorem ?

On the other hand it is not the direction wanted since infinite speeds were said to be non physical by Newton, so it is coming back or going back and forth between different trends in history of science masters...