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|    sci.physics.relativity    |    The theory of relativity    |    225,861 messages    |
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|    Message 225,457 of 225,861    |
|    Paul B. Andersen to All    |
|    Re: Galaxies don't fly apart because the    |
|    03 Feb 26 14:26:53    |
      90794c30       From: relativity@paulba.no              Den 02.02.2026 21:16, skrev Maciej Woźniak:       > On 2/2/2026 8:28 PM, Paul B. Andersen wrote:       >>       >> | You sit in your car with an accelerometer in your hand.       >> | The accelerometer shows that your acceleration is a = 2 m/s².       >> | According to the accelerometer dv/dt = 2 m/s².       >>       >> Can you explain why this is not a normal accelerometer?              > Because, poor trash, a driver that wouldn't       > have windows in his car and was unable to       > distinguish between the acceleration of his       > car and gravity - could only survive       > in your moronic gedankenwelt.              So your answer is that the accelerometer is       not a normal accelerometer because you have no       windows in your car!              Well done, Maciej. You are obviously a very smart guy.              > So, what is your v(t)? The one that       > gives your "proper acceleration" as       > the derivative?              OK, I will explain.              But first we have to address your belief       that the derivative of a function which is zero       must be zero.              | Den 30.01.2026 15:59, skrev Maciej Woźniak:       |> Velocity "in own       |> rest frame" is always 0, no acceleration.       |> Sorry, trash.              You have to learn some elementary calculus first.              Let the speed be a function of time v(t).       At the instant, t = t₀, the speed is zero. v(t₀) = 0.       Let Δt be a time interval.              The definition of the derivative of v(t₀) is:        dv(t₀)/dt = lim (v(t₀+Δt)-v(t₀))/Δt when Δt->0              To see the result of this definition, we have to       know the function.       Let the function be: v(t) = k⋅(t-t₀) where k is a constant.       v(t₀+Δt)-v(t₀) = k⋅(t₀+Δt-t₀)-k⋅(t₀-t₀) = k⋅Δt              So the definition above becomes:        dv(t₀)/dt = lim k⋅Δt/Δt when Δt->0 = k              The acceleration dv/dt when v = 0 is k.              That the derivative of a function must be zero       when the function is zero is indeed a weird idea.              You know that sin(0) = 0 and dsin(0)/dt = cos(0) = 1       Or don't you?              The speed of a piston in a car engine is sinusoidal.       The acceleration is highest when the speed is zero,       the acceleration is zero when the speed is highest.              An accelerometer has no windows, so it cannot see       any outside reference frame.       The only reference is itself, and its speed is       always zero relative to itself (or relative to       the comoving inertial frame).              The acceleration measured in the objects rest frame       is called the objects "proper acceleration".              > Still no answer, trash? Of course.       > Invent some nice slander instead, that       > will show me The One And Only Truth Of       > The Shit you're sworn to.              Was the above the best slander you could invent? :-D              --       Paul              https://paulba.no/              --- SoupGate-Win32 v1.05        * Origin: you cannot sedate... all the things you hate (1:229/2)    |
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