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Thread: Cosmic Speed Limit

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    Cosmic Speed Limit

    Carrying over from the Carlson thread. Is C a speed limit?

    Responding to skepticalbip


    I am not thinking in Newtonian terms, that is an easy dismisal.

    E = mc^2/sqrt( 1 - v^2/C^)

    It is a simple question. Two ships are at rest. One accelerates away towards C. How is v in then equation determined, keeping in mind there is no possible absolute velocity, only change from a staring point.

    The math does blow up at v = C, properly called a singularity, or a zero in the denominator.

    In control systems when a zero or singularity occurs something can and often does 'blow up'. Math is math regardless.

    If v = C is not a limit, then theoretically C can be exceeded?

    https://en.wikipedia.org/wiki/Mass_i..._relativityNot

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    Contributor skepticalbip's Avatar
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    You are forgetting about time dilation and Lorentz contraction.

    Assume a ship leaves Earth on a trip a 1000 LY away. For the entire trip the ship accelerates at one g.

    An observer on Earth will see the ship accelerating away but the further it gets the slower the rate of acceleration will be measured to be as it approaches relativistic velocities with respect to Earth. The ship's speed will be seen as asymptotically approaching 3 x 10^8 M/s.

    An observer on the ship will experience one g acceleration throughout the trip. But remember acceleration is unit length divided by time squared. With respect to an observer on Earth the duration of second as measured on the ship will continually increase and the distance in direction of travel will be longer than that measured by those on the ship. The observer on the ship will still measure one g but with respect to Earth his second will be longer and the distance traveled shorter making the ship's acceleration much less than one g as measured from Earth.

    Because of the time dilation and Lorentz contraction the ship can make the 1000 LY trip easily possible during the lifetime of the original crew (assuming energy available for such long term acceleration) even though more than a thousand years elapsed on Earth. The Earth observer will measure that the trip took more than a thousand years at less than 3 x 10^8 M/s. Those on the ship will have seen the distance to their destination constantly shrinking because of Lorentz contraction faster than due to the distance they cover.

    Since we are talking about relativity, the observer making the measurement needs to be specified. Different observers will make different observations unlike in Newtonian mechanics where time and distance are universals.
    Last edited by skepticalbip; 06-17-2019 at 11:13 PM.

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    c as a cosmic speed limit ultimately derives from Maxwell's equations, where every vacuum wave solution has speed c.

    I will derive that speed from Maxwell-equation observables.

    Gauss's law: electric field and electric charge density
     \mathbf{\nabla} \cdot \mathbf{E} = \frac{\rho}{\epsilon_0}
    Gauss's law for magnetism: no magnetic monopoles are known to exist
     \mathbf{\nabla} \cdot \mathbf{B} = 0
    Faraday's law of induction, relating electric and magnetic fields
     \mathbf{\nabla} \times \mathbf{E} = - \frac{\partial \mathbf{B}}{\partial t}
    Ampere's circuital law and Maxwell's displacement current, relating electric and magnetic fields, and electric current
     \mathbf{\nabla} \times \mathbf{B} = \mu_0 \left( \mathbf{J} + \epsilon_0 \frac{\partial \mathbf{E}}{\partial t} \right)
    Conservation of electric charge and current
     \frac{\partial \rho} {\partial t} + \mathbf{\nabla} \cdot \mathbf{J} = 0
    The quantities ε0 and μ0 can be measured using Coulomb's law for the electric field and the Biot-Savart law for magnetic field.

    Let us specialize to variation along only one direction: x. The vacuum equations become
     E_{x,x} = 0 , B_{x,x} = 0  \\ - E_{z,x} = - B_{y,t} , E_{y,x} = - B_{z,t} \\ - B_{z,x} = \mu_0 \epsilon_0 E_{y,t} , B_{y,x} = \mu_0 \epsilon_0 E_{z,t}
    Taking the x derivative of the second equations and the time derivative of the third equations, one can eliminate the terms that depend on the magnetic field, and one finds
     E_{(y,z),xx} = \epsilon_0 \mu_0 E_{(y,z),tt}
    This is a wave equation, and one can solve it very easily by making E dependent on x - c*t for some speed c. This gives us
     \epsilon_0 \mu_0 c^2 = 1

    Thus, the speed of light in a vacuum is a fixed speed.

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    The next step is to consider Newtonian mechanics. In particular, we must consider its space-time symmetries. What does a symmetry mean mathematically? It means that certain transformations on something will keep it looking the same. So we must consider what transformations of space and time will keep Newtonian mechanics looking the same. Here is a list:
    1. Translations or shifts in space and time
    2. Space rotations
    3. Space-time boosts
    4. Combinations of the first three

    For going from space and time (x,y,z,t) to (x',y',z',t') a rotation with angle a around the x-axis is
     x' = x \\ y' = y \cos a - z \sin a \\ z' = y \sin a + z \cos a \\ t' = t
    A boost by velocity v in the x direction is
     x' = x + v t \\ y' = y \\ z' = z \\ t' = t
    This kind of boost is called a Galilean boost, after Galileo Galilei.

    Notice that time is unchanged by rotations and boosts, and notice also what happens to velocities in Galilean boosts: for u = x/t, u' = u + v. It can vary in both direction and magnitude.

    This has a very obvious inconsistency with Maxwell's equations, since electromagnetic waves have a constant speed no matter what their direction is: c.

    Since Newtonian mechanics and Maxwellian electrodynamics were both enormously successful, this was a BIG headache. Experimental tests of possible solutions, like the Michelson-Morley experiment, only made the situation worse, by agreeing with Maxwell rather than Newton.

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    I had electromagnetics and I understand Maxwell’s Equations.

    Given C is not relative across all frames and motion is..

    A jet has a Mach meter, %speed of sound. I want to build a %light speed meter for a space ship.

    A spaceship in orbit around Earth and deaprts out into space. On board I can measure acceeration and calculate change in velocity. I turn off the engine and the ship is traveling at a constant speed.

    I have a number for velocity but it is relative to the starting point. I get the same change in velocity regradless of the velocity of starting point.

    From relativity there is no absolute frame, so it is impossible to know an absolute velocity of a starting point. Given that, on the ship how do I know what percent of C the ship is traveling at?

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    Of the various symmetries of Newtonian mechanics, all the basic ones but boosts are also symmetries of Maxwellian electrodynamics. So we look for Maxwell-friendly boosts. The solution was obtained by Hendrik Antoon Lorentz over 1892 - 1899, and the resulting boosts are thus called Lorentz boosts.

    Let us now try to find a Lorentz boost:
     x' = \gamma (x + v t) \\ y' = \gamma'' y \\ z' = \gamma'' z \\ t' = \gamma' (t + w x)
    where the γ's are all functions of (v/c)2.

    Since combining two boosts must yield another boost, we have a constraint on possible boosts. This seems like a very difficult job, but one can simplify it by making one of the boosts small, and then finding the result boost's velocity parameter. That can be done by differentiating by the small boost's velocity. The first results are that w is proportional to v, and that γ'' = 1:
     x' = \gamma (x + v t) \\ y' = y \\ z' = z \\ t' = \gamma' (t + k v x)
    where k is a constant.

    Continuing the solution, we find that γ' = γ:
     x' = \gamma (x + v t) \\ y' = y \\ z' = z \\ t' = \gamma (t + k v x)
    and that
     \gamma = \frac{1}{\sqrt{1 - k v^2}
    This gives a velocity-addition law:
     v_{12} = \frac{v_1 + v_2}{1 + k v_1 v_2}

    Let us now find k. For c constant, we find k = 1/c2. Thus, the Lorentz boost is
     x' = \gamma (x + v t) \\ y' = y \\ z' = z \\ t' = \gamma (t + v x / c^2) \\ \gamma =  \frac{1}{\sqrt{1 - v^2/c^2}

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    This result was derived using the constancy of c, but it can be shown to be true for Maxwell's equations by using appropriate boosts of the electric and magnetic fields.

    One can define a space-time interval that is symmetric under Lorentz boosts by using a modification of Pythagoras's theorem:
    S = - (c t)^2 + x^2 + y^2 + z^2
    There are five types of intervals:
    • S > 0: spacelike
    • S = 0, t > 0: forward null
    • S = 0, t < 0: backward null
    • S < 0, t > 0: forward timelike
    • S < 0, t < 0: backward timelike


    I now turn to energy and momentum. Consider an object moving at a small speed vy in the y direction. Its momentum in that direction is m*vy and zero in the other coordinate directions. Now do a boost in the x direction with velocity v. The new vy will be:
     v_y = \frac{dy}{dt} = \frac{1}{\gamma} v_{y0} = \sqrt{1 - v^2/c^2} v_{y0}
    To compensate, the effective mass must be multiplied by that gamma factor, and we find that the relativistic momentum is
     p = \frac{m v}{ \sqrt{1 - v^2/c^2}  }
    To get energy, we integrate force over distance, where force is change of momentum:
     E = \int F dx = \int F \frac{dp}{dt} dx = \int m \frac{1}{(1 - v^2/c^2)^{3/2}} v dv = m c^2 \left( \frac{1}{ \sqrt{1 - v^2/c^2} } - 1 \right)

    This implies that every object has an intrinsic energy: E = m*c2

    Yes, Einstein's famous equation.

    Collecting these results,
     p = \frac{mv}{\sqrt{1 - v^2/c^2}} \\ E = \frac{m c^2}{\sqrt{1 - v^2/c^2}} \\ E^2 - (p c)^2 = (m c^2)^2
    Last edited by lpetrich; 06-18-2019 at 06:19 PM. Reason: v dt -> v dv

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    Quote Originally Posted by steve_bank View Post
    I had electromagnetics and I understand Maxwell’s Equations.

    Given C is not relative across all frames and motion is..

    A jet has a Mach meter, %speed of sound. I want to build a %light speed meter for a space ship.

    A spaceship in orbit around Earth and deaprts out into space. On board I can measure acceeration and calculate change in velocity. I turn off the engine and the ship is traveling at a constant speed.

    I have a number for velocity but it is relative to the starting point. I get the same change in velocity regradless of the velocity of starting point.

    From relativity there is no absolute frame, so it is impossible to know an absolute velocity of a starting point. Given that, on the ship how do I know what percent of C the ship is traveling at?
    I'm usually amazed by posts like this because you just know it is going to get humbled quickly.

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    If one applies rotations and boosts to momentum and energy, one finds that they behave like space and time in a space-time interval.

    Hermann Minkowski's paper: Space and Time - Wikisource, the free online library


    Now a detailed derivation of Lorentz boosts. First start with a general expression for a boost:
     B(v) = \begin{pmatrix} \gamma(v) & \gamma(v) v \\ \zeta(v) w(v) & \zeta(v) \end{pmatrix}
    Two boosts must combine into a third boost:
     B(v_{12}) = B(v_1) \cdot B(v_2)

    Make one of the boosts small:
     v_1 = v \\ v_2 = dv \\ v_{12} = v + q(v) dv
    One should not assume that q(v) = 1.

    This gives us
     \begin{pmatrix} \gamma + \gamma' q (dv) & \gamma v + \gamma' v q (dv) + \gamma q (dv) \\ \zeta w + \zeta' w q (dv) + \zeta w' q (dv) & \zeta + \zeta' q (dv)  \end{pmatrix} = \begin{pmatrix} \gamma & \gamma v  \\ \zeta w & \zeta \end{pmatrix} \cdot \begin{pmatrix} 1 & (dv) \\ k (dv) & 1 \end{pmatrix}
    where k = w'(0) giving us equations
     \gamma' q = \gamma v k \\ \gamma' v q + \gamma q = \gamma \\ \zeta' w q + \zeta w' q = \zeta k \\ \zeta' q = \zeta w
    The first and second equations give us
     q = 1 - k v^2 \\ \gamma = (1 - k v^2)^{-1/2}
    The third and fourth equations give us
     w^2 + (1 - k v^2) w' = k
    giving us
     w = k v \\ \zeta = \gamma
    Thus,
     B(v) = (1 - k v^2)^{-1/2} \begin{pmatrix} 1 & v \\ k v & 1 \end{pmatrix}

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    Quote Originally Posted by steve_bank View Post
    I had electromagnetics and I understand Maxwell’s Equations.

    Given C is not relative across all frames and motion is..

    A jet has a Mach meter, %speed of sound. I want to build a %light speed meter for a space ship.
    A mach meter measures the speed of the aircraft with respect to the air. If you are imagining your %light speed meter to be measuring the ship's speed with respect to light then it will always register zero regardless of how much or how long the ship has accelerated.

    A spaceship in orbit around Earth and deaprts out into space. On board I can measure acceeration and calculate change in velocity. I turn off the engine and the ship is traveling at a constant speed.

    I have a number for velocity but it is relative to the starting point. I get the same change in velocity regradless of the velocity of starting point.

    From relativity there is no absolute frame, so it is impossible to know an absolute velocity of a starting point. Given that, on the ship how do I know what percent of C the ship is traveling at?
    There is no such thing as absolute velocity in relativity. That is a Newtonian concept. Newtonian imagined a fixed universal reference frame. Relativity assumes that measurements in all reference frames are equally valid regardless of their relative motions with respect to each other. In relativity, the second, meter, kilogram, etc. measured in a reference frame in motion with respect to the observer are all dependent on the relative velocity between those two reference frames.

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