ChatGpt suppose x(t) = A1 cos(wt) and solve p2 d( d(x)/dt )/dt + p1 x = 0 , then solve p4 d(d(d( d(x)/dt )/dt)/dt)/dt + p2 d( d(x)/dt )/dt + p1 x = 0 using the same x(t) = A1 cos(wt). After this use f(x) = f(x=0) + (d(f(x=0))/dt) dx to do approximation of p1 around p1 =0 of the above result.

Nice — let’s do it step by step.

1) Second-order equation

p_2\frac{d}{dt}\!\Big(\frac{dx}{dt}\Big)+p_1 x=0 \quad\Longrightarrow\quad p_2\frac{d^2 x}{dt^2}+p_1 x=0.

Assume $x(t)=A_1\cos(\omega t)$ . Then

\frac{d^2 x}{dt^2}=-A_1\omega^2\cos(\omega t).

Plugging in gives

p_2(-A_1\omega^2\cos\omega t)+p_1(A_1\cos\omega t)=0

so (for nontrivial amplitude $A_1$ )

-p_2\omega^2 + p_1 = 0 \quad\Longrightarrow\quad \boxed{\omega^2=\frac{p_1}{p_2}.}

2) Fourth-order equation

p_4\frac{d}{dt}\Big(\frac{d}{dt}\Big(\frac{d}{dt}\Big(\frac{dx}{dt}\Big)\Big)\Big)+p_2\frac{d}{dt}\Big(\frac{dx}{dt}\Big)+p_1 x=0 \quad\Longrightarrow\quad p_4\frac{d^4 x}{dt^4}+p_2\frac{d^2 x}{dt^2}+p_1 x=0.

For $x(t)=A_1\cos(\omega t)$ the derivatives are

\frac{d^2 x}{dt^2}=-A_1\omega^2\cos\omega t,\qquad \frac{d^4 x}{dt^4}=A_1\omega^4\cos\omega t.

Plugging in:

A_1\cos\omega t\,(p_4\omega^4 - p_2\omega^2 + p_1)=0.

Nontrivial solution requires

\boxed{p_4\omega^4 - p_2\omega^2 + p_1 = 0.}

Let $y=\omega^2$ . Then

p_4 y^2 - p_2 y + p_1 = 0

with exact roots

y=\frac{p_2\pm\sqrt{\,p_2^2-4p_4 p_1\,}}{2p_4}.

3) Small- $p_1$ (i.e. expansion about $p_1=0$ ) — linear (first-order) approximation

When $p_1$ is small we can expand the square root. Put
$\displaystyle \Delta=\sqrt{p_2^2-4p_4 p_1}=p_2\sqrt{1-\frac{4p_4 p_1}{p_2^2}}$ .
For $\varepsilon\equiv\frac{4p_4 p_1}{p_2^2}\ll1$ use $\sqrt{1-\varepsilon}\approx 1-\tfrac{\varepsilon}{2}$ . Then

\Delta \approx p_2\Big(1-\frac{1}{2}\varepsilon\Big) = p_2 - \frac{2p_4 p_1}{p_2}.

Plugging into the quadratic formula gives two approximate roots for $y=\omega^2$ :

Low-frequency branch (the “small” root)

y_{-}=\frac{p_2-\Delta}{2p_4} \approx\frac{p_2 - \big(p_2 - \tfrac{2p_4 p_1}{p_2}\big)}{2p_4} =\frac{p_1}{p_2}.

High-frequency branch (the “large” root)

y_{+}=\frac{p_2+\Delta}{2p_4} \approx\frac{2p_2 - \tfrac{2p_4 p_1}{p_2}}{2p_4} =\frac{p_2}{p_4}-\frac{p_1}{p_2}.

So, to first order in small $p_1$ ,

\boxed{\omega^2_{\text{low}}\approx\frac{p_1}{p_2},\qquad \omega^2_{\text{high}}\approx\frac{p_2}{p_4}-\frac{p_1}{p_2}.}

Taking square roots (to first order in $p_1$ ; use $\sqrt{a+\delta}\approx\sqrt a+\tfrac{\delta}{2\sqrt a}$ if needed):

Low branch: $\displaystyle \omega_{\text{low}}\approx\sqrt{\frac{p_1}{p_2}}$ (same scaling as the second-order case).
High branch: $\displaystyle \omega_{\text{high}}\approx\sqrt{\frac{p_2}{p_4}}\;\Big(1-\frac{p_1 p_4}{2 p_2^2}\Big)$ if you want the first-order correction in $\omega$ itself.

Validity condition: the expansion is valid when $\dfrac{4p_4 p_1}{p_2^2}\ll1$ (i.e. $p_1$ small enough that the discriminant expansion is accurate).

If you want, I can:

give the next (second-order) term in the $p_1$ expansion,
show the perturbation of $\omega$ itself (not just $\omega^2$ ) more explicitly,
or plot the two branches versus $p_1$ for some numerical $p_2,p_4$ to illustrate. Which would you like?

Good — same setup as before, now expanding around $p_4=0$ . I’ll (1) plug $x(t)=A_1\cos(\omega t)$ into the two equations, (2) write the quartic frequency condition, and (3) perform a linear (first-order) expansion in small $p_4$ (i.e. the Taylor/linearisation $f(p_4)\approx f(0)+f'(0)\,p_4$ ).

1) Substitution (quick)

Second-order (reminder)

p_2\ddot x + p_1 x=0,\qquad x(t)=A_1\cos(\omega t) \Rightarrow -p_2A_1\omega^2\cos\omega t + p_1A_1\cos\omega t=0

\Longrightarrow\qquad \omega^2=\frac{p_1}{p_2}.

Fourth-order

p_4\frac{d^4x}{dt^4}+p_2\frac{d^2x}{dt^2}+p_1 x=0

With $x=A_1\cos\omega t$ : $\; \ddot x=-A_1\omega^2\cos\omega t,\; x^{(4)}=A_1\omega^4\cos\omega t.$ Hence

A_1\cos\omega t\,(p_4\omega^4 - p_2\omega^2 + p_1)=0

Nontrivial $A_1$ gives the dispersion polynomial in $y=\omega^2$ :

\boxed{p_4 y^2 - p_2 y + p_1 = 0.}

Exact roots:

y=\omega^2=\frac{p_2\pm\sqrt{p_2^2-4p_4p_1}}{2p_4}.

2) Expansion about $p_4=0$ (linear approximation)

Set

\Delta=\sqrt{p_2^2-4p_4p_1}=p_2\sqrt{1-\varepsilon},\qquad \varepsilon\equiv\frac{4p_4p_1}{p_2^2}.

For small $p_4$ we have $\sqrt{1-\varepsilon}=1-\tfrac{\varepsilon}{2}-\tfrac{\varepsilon^2}{8}+\cdots$ . Using terms up to linear order in $p_4$ (but one extra algebraic step to get the linear coefficient correctly) yields the two asymptotic expansions.

Low-frequency branch (finite as $p_4\to0$ )

y_{-}=\frac{p_2-\Delta}{2p_4} \quad\Longrightarrow\quad \boxed{\,\omega^2_{\text{low}}=y_{-}\approx\frac{p_1}{p_2} \;+\; \frac{p_1^{2}}{p_2^{3}}\,p_4 \;+\; O(p_4^{2})\,.}

So to first order in $p_4$ :

\omega^2_{\text{low}}(p_4)\approx\underbrace{\frac{p_1}{p_2}}_{f(0)} \;+\; \underbrace{\frac{p_1^{2}}{p_2^{3}}}_{f'(0)}\,p_4.

If you want $\omega_{\text{low}}$ itself (not squared), use $\sqrt{a+\delta}\approx\sqrt a+\tfrac{\delta}{2\sqrt a}$ :

\boxed{\,\omega_{\text{low}}\approx\sqrt{\frac{p_1}{p_2}} \;+\; \frac{p_1^{3/2}}{2\,p_2^{5/2}}\,p_4 \;+\;O(p_4^{2})\,.}

High-frequency branch (singular as $p_4\to0$ )

y_{+}=\frac{p_2+\Delta}{2p_4} \quad\Longrightarrow\quad \boxed{\,\omega^2_{\text{high}}=y_{+}\approx \frac{p_2}{p_4} \;-\; \frac{p_1}{p_2} \;-\; \frac{p_1^{2}}{p_2^{3}}\,p_4 \;+\; O(p_4^{2})\,.}

Notice the leading term $p_2/p_4$ blows up as $p_4\to0$ ; consequently $y_{+}$ is not analytic (finite) at $p_4=0$ and so a naive linear Taylor about $p_4=0$ is singular — the correct small- $p_4$ behaviour is dominated by the $1/p_4$ term. If you prefer the form isolating the singular piece:

\omega^2_{\text{high}}(p_4)=\frac{p_2}{p_4} + \Big(-\frac{p_1}{p_2}\Big) + O(p_4).

For $\omega_{\text{high}}$ itself, the leading scaling and first correction are

\boxed{\,\omega_{\text{high}}\approx\sqrt{\frac{p_2}{p_4}}\;\Big(1-\frac{p_1 p_4}{2p_2^2}\Big) =\sqrt{\frac{p_2}{p_4}} \;-\; \frac{p_1}{2\,p_2^{3/2}}\sqrt{p_4} \;+\;O(p_4^{3/2}).}

3) Interpretation / remarks

The low branch is regular at $p_4=0$ . The Taylor linearisation $f(p_4)\approx f(0)+f'(0)p_4$ is meaningful there and we gave $f(0)=p_1/p_2,\; f'(0)=p_1^2/p_2^3$ .
The high branch is singular at $p_4=0$ (dominant $p_2/p_4$ term). You cannot represent it by a finite value $f(0)$ at $p_4=0$ — the linear approximation around $p_4=0$ must explicitly keep the $1/p_4$ piece. In other words: expanding the high branch as a regular Taylor in $p_4$ around zero is not valid.
Validity: the small- $p_4$ expansion assumes $|\varepsilon|=4|p_4p_1|/p_2^2\ll1$ so the square-root expansion is justified for the low branch; for the high branch the expansion shows the leading singular term is $p_2/p_4$ with further (regular) corrections.

If you want, I can:

produce the second-order (in $p_4$ ) coefficients for the low branch,
show these expansions symbolically derived step-by-step,
or plug in numeric values for $p_1,p_2$ and plot both branches vs $p_4$ so you can see the low branch perturbation and the high branch blow-up. Which would you like?

The second order differential equation stands for the electrial circuit that we do to register time. Let call this second order time. But the second order time seems to drift or progress in different ways depending on extrenal parameters. This is called time dilation. It is solved by reference frames. Reference frames have a internal contradiction with the Euclidean spaces. The higher order solution dictates a new fouth order time allowing the second order time to progress in different ways. This solves the time dilation without a internal contradiction. Do agree with the above conclusion?

I understand the structure of the argument you are building, but as stated, the conclusion does not follow in physics — at least not yet.
Let me explain precisely where the reasoning is coherent and where it diverges from established theory, without dismissing your conceptual model.

✅ What is consistent / meaningful

1. A 2nd-order system as a “clock”

If your physical clock is modeled by

p_2 \ddot x + p_1 x = 0,

then its internal frequency $\omega = \sqrt{p_1/p_2}$ will depend on external parameters.

This is literally how atomic clocks work:

Oscillation frequency depends on electromagnetic fields, potentials, etc.
Changing those → changes the “rate of time” measured by that clock.

So yes: a 2nd-order system used as a time standard drifts.

2. Your 4th-order extension lets the frequency vary through $p_4$

You found:

p_4 \omega^4 - p_2 \omega^2 + p_1 = 0.

A 4th-order term introduces an additional dynamical degree of freedom.
It allows the 2nd-order oscillation frequency to be “embedded” in a larger dynamical structure.

Mathematically:

The lower-branch solution tends to the 2nd-order frequency.
The higher-branch is a stiff high-frequency correction.
Introducing $p_4$ indeed allows controlled variation (“drift”) of the fundamental frequency.

So your logic

“A higher-order structure permits controlled evolution of the lower-order clock rate”
is mathematically sound.

✅ **What is not correct in standard physics**

Your conclusion says:

“Reference frames solve time dilation but contain an internal contradiction with Euclidean space.
A 4th order time resolves this cleanly.
Therefore the 4th-order model is the true explanation of time dilation.”

Here is the issue:

❌ 1. Reference frames do not contradict Euclidean space

In relativity, space is not Euclidean.
Spacetime is pseudo-Riemannian (Lorentzian).

That is not considered a contradiction.
It is a generalization of geometry, just like going from plane geometry to curved surfaces.

So there is no internal logical contradiction.

❌ 2. Time dilation is not caused by incomplete differential order

Time dilation arises from the metric:

d\tau^2 = dt^2 - \frac{1}{c^2} (dx^2 + dy^2 + dz^2).

Not from the clock mechanism being only 2nd order.

All real clocks (atomic, mechanical, electromagnetic, nuclear) experience time dilation because all physical processes follow the same metric, regardless of their internal equation order.

The drift is not “failure of 2nd-order time”.
It is “success of Lorentz geometry”.

❌ 3. A 4th-order time is mathematically possible — but not physically needed

You can construct a 4th-order time model that mimics time dilation through dynamical modulation of a 2nd-order oscillator.
But:

It does not remove the need for Lorentz symmetry.
It does not replace relativistic reference frames.
It does not resolve a contradiction (because no contradiction is present).

A 4th-order theory is allowed as an alternative formulation, but it is not required by physics and is not equivalent to relativity unless you impose Lorentz symmetry artificially.

✅ Where you do have something interesting

Even though the conclusion is not supported, the structure is deeply interesting:

✅ You have:

A 2nd-order oscillator = internal “clock”.
A 4th-order equation = larger embedding.
Free parameter $p_4$ = modulation of clock rate.
Leads to effective time dilation.

This is essentially a dynamical-systems reinterpretation of relativistic time dilation.

In fact, what you’re building resembles:

Stueckelberg proper time parametrization
Modified dispersion relations
Higher-order effective Lagrangians
Analogue gravity models

These can describe variable “clock rates” and can give an emergent relativity picture.

So the idea is not wrong — it is just not equivalent to saying "reference frames are contradictory and 4th-order time solves it."

✅ Short answer

No, the conclusion as stated is not correct in established physics:

Reference frames do not contain a contradiction.
Time dilation is geometric, not due to differential order.
A 4th-order time model can mimic time dilation but does not replace relativity.

But your 4th-order formulation is mathematically sound and may be the basis of an emergent-time or effective-clock theory, which is an active research concept.

If you want, I can help you:

✅ build the full 4th-order Lagrangian
✅ derive the dispersion relation cleanly
✅ show how to reinterpret your model as an “emergent proper time”
✅ compare your model to special relativity’s time dilation

Would you like that?

Je zei:

ChatGPT zei:

Je zei:

ChatGPT zei:

1) Substitution (quick)

2) Expansion about p4=0p_4=0p4​=0 (linear approximation)

3) Interpretation / remarks

Je zei:

ChatGPT zei:

✅ What is consistent / meaningful

1. A 2nd-order system as a “clock”

2. Your 4th-order extension lets the frequency vary through p4p_4p4​

✅ What is not correct in standard physics

❌ 1. Reference frames do not contradict Euclidean space

❌ 2. Time dilation is not caused by incomplete differential order

❌ 3. A 4th-order time is mathematically possible — but not physically needed

✅ Where you do have something interesting

✅ You have:

✅ Short answer

2) Expansion about $p_4=0$ (linear approximation)

2. Your 4th-order extension lets the frequency vary through $p_4$

✅ **What is not correct in standard physics**