Education and Crank Claims: Special Relativity

Discussion in 'Physics & Math' started by rpenner, Oct 5, 2011.

  1. rpenner Fully Wired Staff Member

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    Hello -- the purpose of this thread is limited to just Special Relativity, the theory, the history of the theory, the math of the theory, the experiment evidence for the theory and to address any crank claims about the same.

    Please keep the posts as self-contained as possible. Helpful comments appreciated. Thank you.

    Education: What is Newton's Law of Inertia?

    Also known as Newton's first law of motion, it states that the motion of a free particle (a body without external force applied to it) has a velocity that doesn't change with time.
    \(\vec{F} = 0 \; \Rightarrow \; \frac{d \vec{v}}{d t} = 0\)
    Thus the motion may be described by a linear equation: \(\vec{r}(t) = \vec{v} t + \vec{r_0}\).

    Education: What is Newton's concept of space and time?

    Newton believed that time was absolute and space was Euclidean and absolute. Newton believed that in principle there could be a localizable center to the universe and beginning of time, although his physics did not require that there be a practical manner of actually locating such a center and beginning. If this concept of absolute time and space is explored, it becomes curious that the laws of physics seem so insensitive to the absolute position in the universe and the absolute time.

    Education: What is Galileo's concept of space and time?

    Galileo, perhaps because he focused a bit more on practical matters and leaned less on the concepts of Euclid, believed that the laws of physics were the same everywhere and written such that only relative measurements mattered. Galileo knew he could not point to the center of the universe, but discovered important things about the motion of objects nevertheless. Therefore, only relative distances, \(\Delta \vec{r}\), and elapsed time \(\Delta t\) and relative velocity, \(\vec{v} = \frac{\Delta \vec{r}}{\Delta t}\) mattered. There are two ways to spin a pinwheel: hold it in the wind, or run through still air and these two methods are similar when the relative speed of the air to the pinwheel are the same.

    A Galilean transform of coordinates in the \(\hat{x}\) direction may be written in absolute terms:
    \(\begin{pmatrix}x' \\ y' \\ z' \\ t' \end{pmatrix} = \begin{pmatrix}1 & 0 & 0 & -v \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix}x \\ y \\ z \\ t \end{pmatrix} + \begin{pmatrix}x_{\textrm{translation}} \\ y_{\textrm{translation}} \\ z_{\textrm{translation}} \\ t_{\textrm{translation}} \end{pmatrix} \) and the inverse transform has the same form, with different parameters.

    Since this is a linear transform, it can be written in a shorter form if we only talk about coordinate differences:
    \(\begin{pmatrix}\Delta x' \\ \Delta y' \\ \Delta z' \\ \Delta t' \end{pmatrix} = \begin{pmatrix}1 & 0 & 0 & -v \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \Delta x \\ \Delta y \\ \Delta z \\ \Delta t \end{pmatrix} \). The inverse is easier to compute.

    Further if we are talking about the world-line of a particle, we can write \(\Delta x = x(\Delta t)\) (and similarly for y and z). And if we are talking about inertial motion, we are talking about constant velocity and \(\Delta x = u_x \Delta t + \Delta x_0\) (and similarly for y and z). So we may compute:
    \(\begin{pmatrix}\Delta x' \\ \Delta y' \\ \Delta z' \\ \Delta t' \end{pmatrix} = \begin{pmatrix}1 & 0 & 0 & -v \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} u_x \Delta t \\ u_y \Delta t \\ u_z \Delta t \\ \Delta t \end{pmatrix} = \begin{pmatrix} ( u_x - v ) \Delta t \\ u_y \Delta t \\ u_z \Delta t \\ \Delta t \end{pmatrix} = \begin{pmatrix} ( u_x - v ) \Delta t' \\ u_y \Delta t' \\ u_z \Delta t' \\ \Delta t' \end{pmatrix}\).
    So \(u'_x = u_x - v\) is how inertial world lines transform under the Galilean transform.

    If someone is running at speed u, and we run at speed v in the same direction, Galileo would predict that we see that someone moving at speed \(u' = u - v\).

    Education: What is the Special Theory of Relativity?

    The Special Theory of Relativity is the 1905 proposal, based on the empirical success of Maxwell's equations in laboratories on Earth even though the Earth is not in a uniform state of rest, that the laws of physics are everywhere the same, always written in the same form if the coordinate system is one where Newton's Law of Inertia is valid, and that the speed of light as measured in vacuum is always measured to have the same value, c. It is this last proposal which is incompatible with Newtonian and Galilean intuition.

    The consequences of this proposal is that space and time are not independent of each other, but conspire so that inertial observers (not rotating or subject to external forces) in relative motion see light (in vacuum) as always traveling with fixed speed and that the laws of physics conspire to preserve certain quantities which all inertial observers agree upon. These quantities are called Lorentz invariants because they are unchanged when a Lorentz transform is used to transform the observer physics of one inertial observer to another inertial observer's coordinate system.

    Education: What are the invariants of the Lorentz transform?

    \(c^2 (\Delta t)^2 - ( \Delta \vec{r} )^2\) which in the Cartesian coordinate system we can write as \(c^2 (\Delta t)^2 - ( \Delta x)^2- ( \Delta y)^2- ( \Delta z)^2\) is the important coordinate invariant. Holding it constant describes a hyperbolic surface, but since the Lorentz transform is a continuous transformation we only use one sheet of the hyperbolic surface.

    Education: How does one use the Lorentz transform?

    You have some choices depending what properties you are trying to exploit since all of the following are mathematically equivalent for \(-c < v < c\). All of them give coordinate differences for a new inertial observer who is moving relative to the current coordinates at speed v along the \(\hat{x}\) direction.

    \(\begin{pmatrix}\Delta x' \\ \Delta y' \\ \Delta z' \\ c \Delta t' \end{pmatrix} = \begin{pmatrix} \cosh \, \tanh^{-1} \, \frac{v}{c} & 0 & 0 & - \sinh \, \tanh^{-1} \, \frac{v}{c} \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ - \sinh \, \tanh^{-1} \, \frac{v}{c} & 0 & 0 & \cosh \, \tanh^{-1} \, \frac{v}{c} \end{pmatrix} \begin{pmatrix} \Delta x \\ \Delta y \\ \Delta z \\ c \Delta t \end{pmatrix} \).

    \(\begin{pmatrix}\Delta x' \\ \Delta y' \\ \Delta z' \\ c \Delta t' \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & 0 & 0 & \frac{-v}{c \sqrt{1 - \frac{v^2}{c^2}}} \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \frac{-v}{c \sqrt{1 - \frac{v^2}{c^2}}} & 0 & 0 & \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \end{pmatrix} \begin{pmatrix} \Delta x \\ \Delta y \\ \Delta z \\ c \Delta t \end{pmatrix} \).

    \(\begin{pmatrix}\Delta x' \\ \Delta y' \\ \Delta z' \\ \Delta t' \end{pmatrix} = \begin{pmatrix} \cosh \, \tanh^{-1} \, \frac{v}{c} & 0 & 0 & - c \, \sinh \, \tanh^{-1} \, \frac{v}{c} \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ - \frac{1}{c} \, \sinh \, \tanh^{-1} \, \frac{v}{c} & 0 & 0 & \cosh \, \tanh^{-1} \, \frac{v}{c} \end{pmatrix} \begin{pmatrix} \Delta x \\ \Delta y \\ \Delta z \\ \Delta t \end{pmatrix} \).

    \(\begin{pmatrix}\Delta x' \\ \Delta y' \\ \Delta z' \\ \Delta t' \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & 0 & 0 & \frac{-v}{\sqrt{1 - \frac{v^2}{c^2}}} \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \frac{-v}{c^2 \sqrt{1 - \frac{v^2}{c^2}}} & 0 & 0 & \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \end{pmatrix} \begin{pmatrix} \Delta x \\ \Delta y \\ \Delta z \\ \Delta t \end{pmatrix} \).

    Or with the short-hand notation: \(\gamma \equiv \gamma(v) \equiv \cosh \, \tanh^{-1} \, \frac{v}{c} = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\) we can also write:

    \(\begin{pmatrix}\Delta x' \\ \Delta y' \\ \Delta z' \\ c \Delta t' \end{pmatrix} = \begin{pmatrix} \gamma & 0 & 0 & \frac{-v \gamma}{c} \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \frac{-v \gamma}{c} & 0 & 0 & \gamma \end{pmatrix} \begin{pmatrix} \Delta x \\ \Delta y \\ \Delta z \\ c \Delta t \end{pmatrix} \).

    \(\begin{pmatrix}\Delta x' \\ \Delta y' \\ \Delta z' \\ \Delta t' \end{pmatrix} = \begin{pmatrix} \gamma & 0 & 0 & - v \gamma \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \frac{-v \gamma}{c^2} & 0 & 0 & \gamma \end{pmatrix} \begin{pmatrix} \Delta x \\ \Delta y \\ \Delta z \\ \Delta t \end{pmatrix} \).

    Naturally, these expressions simplify when physical units are used such that c = 1. Also note that when |v| << c, this transform is close to the Galilean transform, so one would suspect only sensitive tests could distinguish the prediction results of Galileo and Special Relativity.

    Education: How does one demonstrate that the Lorentz transform preserves the invariant?
    \(\begin{pmatrix}\Delta x' \\ \Delta y' \\ \Delta z' \\ c \Delta t' \end{pmatrix} = \begin{pmatrix} \cosh \, \tanh^{-1} \, \frac{v}{c} & 0 & 0 & - \sinh \, \tanh^{-1} \, \frac{v}{c} \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ - \sinh \, \tanh^{-1} \, \frac{v}{c} & 0 & 0 & \cosh \, \tanh^{-1} \, \frac{v}{c} \end{pmatrix} \begin{pmatrix} \Delta x \\ \Delta y \\ \Delta z \\ c \Delta t \end{pmatrix} = \begin{pmatrix} \Delta x \, \cosh \, \tanh^{-1} \, \frac{v}{c} \; - \; c \Delta t \, \sinh \, \tanh^{-1} \, \frac{v}{c} \\ \Delta y \\ \Delta z \\ c \Delta t \, \cosh \, \tanh^{-1} \, \frac{v}{c} \; - \; \Delta x \, \sinh \, \tanh^{-1} \, \frac{v}{c}\end{pmatrix}\)
    Therefore:
    \(\begin{eqnarray} c^2 (\Delta t')^2 - ( \Delta x')^2- ( \Delta y')^2- ( \Delta z')^2 & = & c^2 (\Delta t)^2 \cosh^2 \, \tanh^{-1} \, \frac{v}{c} - 2 c (\Delta t) (\Delta x) \left(\cosh \, \tanh^{-1} \, \frac{v}{c} \right) \left( \sinh \, \tanh^{-1} \, \frac{v}{c} \right) + (\Delta x)^2 \sinh^2 \, \tanh^{-1} \, \frac{v}{c} \\ & & - \left( (\Delta x)^2 \cosh^2 \, \tanh^{-1} \, \frac{v}{c} - 2 c (\Delta t) (\Delta x) \left(\cosh \, \tanh^{-1} \, \frac{v}{c} \right) \left( \sinh \, \tanh^{-1} \, \frac{v}{c} \right) + c^2 (\Delta t)^2 \sinh^2 \, \tanh^{-1} \, \frac{v}{c} \right) \\ & & - ( \Delta y)^2- ( \Delta z)^2 \\ & = & c^2 (\Delta t)^2 \left( \cosh^2 \, \tanh^{-1} \, \frac{v}{c} - \sinh^2 \, \tanh^{-1} \, \frac{v}{c} \right) - ( \Delta x)^2 \left( \cosh^2 \, \tanh^{-1} \, \frac{v}{c} - \sinh^2 \, \tanh^{-1} \, \frac{v}{c} \right) - ( \Delta y)^2- ( \Delta z)^2 \\ & = & c^2 (\Delta t)^2 - ( \Delta x)^2- ( \Delta y)^2- ( \Delta z)^2 \end{eqnarray}\)

    Education: How does the Lorentz transformation transform inertial motion?

    If we take the coordinate difference between two events on the same inertial world line, we have :
    \(\begin{pmatrix}\Delta x' \\ \Delta y' \\ \Delta z' \\ \Delta t' \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & 0 & 0 & \frac{-v}{\sqrt{1 - \frac{v^2}{c^2}}} \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \frac{-v}{c^2 \sqrt{1 - \frac{v^2}{c^2}}} & 0 & 0 & \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \end{pmatrix} \begin{pmatrix} u_x \Delta t \\ u_y \Delta t \\ u_z \Delta t \\ \Delta t \end{pmatrix} = \begin{pmatrix} \frac{u_x - v}{\sqrt{1 - \frac{v^2}{c^2}}} \Delta t \\ u_y \Delta t \\ u_z \Delta t \\ \frac{1 - \frac{u_x v}{c^2}}{\sqrt{1 - \frac{v^2}{c^2}}} \Delta t \end{pmatrix} = \begin{pmatrix} \frac{u_x - v}{1 - \frac{u_x v}{c^2}} \Delta t' \\ \frac{u_y \sqrt{1 - \frac{v^2}{c^2}}}{1 - \frac{u_x v}{c^2}} \Delta t' \\ \frac{u_z \sqrt{1 - \frac{v^2}{c^2}}}{1 - \frac{u_x v}{c^2}} \Delta t' \\ \Delta t' \end{pmatrix}\)
    So if \(\vec{u} = \left( u_x , \, u_y , \, u_z \right)\), then \(\vec{u'} = \left( \frac{u_x - v}{1 - \frac{u_x v}{c^2}} , \, \frac{u_y \sqrt{1 - \frac{v^2}{c^2}}}{1 - \frac{u_x v}{c^2}} , \, \frac{u_z \sqrt{1 - \frac{v^2}{c^2}}}{1 - \frac{u_x v}{c^2}} \right) = \left( c \, \tanh \left( \tanh^{-1} \, \frac{u_x}{c} \; - \; \tanh^{-1} \, \frac{v}{c} \right) , \, u_y \, \frac{\cosh \tanh^{-1} \, \frac{u_x}{c} }{ \cosh \left( \tanh^{-1} \, \frac{u_x}{c} \; - \; \tanh^{-1} \, \frac{v}{c} \right) } \, u_z \, \frac{\cosh \tanh^{-1} \, \frac{u_x}{c} }{ \cosh \left( \tanh^{-1} \, \frac{u_x}{c} \; - \; \tanh^{-1} \, \frac{v}{c} \right) } \right) \).

    So although the velocity changes, the Lorentz transform preserves inertial motion.
     
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  3. DonQuixote Registered Senior Member

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    Would it be appropriate to discuss other, non-mathematical aspects of SR in this thread? I am not going to make "crank claims". I will admit to being sceptical to SR, but to tell you the truth, I find it really difficult to even read the "On the Electrodynamics of Moving Bodies". Would it be OK to use this thread to clear up what it says? Example: It states that the speed of light is independent of the movement of the source. It never states that it is independent of the movement of the observer. Yet everybody takes it to mean exactly that. Is it just implied, or does it follow from the postulate, or what? If you don't think it is appropriate, I will respect that.

    Unfortunately, I am unable to comment on the math. All those equations just scare me.
     
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  5. prometheus viva voce! Moderator

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    Suppose you have a source S and two observers O and P. P is in motion relative to O and S, and from P's point of view he sees whatever he sees. The point is, the fact that P can see O means some light is being reflected from somewhere and is falling on P's light sensor. That means effectively O is a source, even though he doesn't carry a "true source," like a torch.
     
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  7. DonQuixote Registered Senior Member

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    While I don't disagree with anything of this, I see no connection to the question.
     
  8. Motor Daddy ☼☼☼☼☼☼☼☼☼☼☼ Valued Senior Member

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    So the light that is reflected off O hits P at t=0. At t=1 how far did light travel since t=0? How far did O travel? How far did S travel? What is the distance between O and S at t=1? What is the distance between O and P at t=1?

    According to my universe, SR is the crank theory, and Einstein was the biggest crank of all. Anyone who leads people to BS whilst knowing it's BS is a crank of the worst kind.
     
    Last edited: Oct 5, 2011
  9. Tach Banned Banned

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    There are (at least) two answers to your question:

    1. There is no difference between the light source moving wrt the observer and the observer moving wrt the light source. In fact, you can't tell one from the other. So, Einstein did not have to state the obvious.

    2. There is ample experimental confirmation that the light speed in vacuum is independent of the speed of the light source.
     
  10. Tach Banned Banned

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    Can't be, you are the biggest of them all.
     
  11. prometheus viva voce! Moderator

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    None of this stuff matters for the purposes of answering Don's question

    No comment.
     
  12. DonQuixote Registered Senior Member

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    So this obviousness is contained in the "Principle of Relativity", then?

    Quote Einstein:

    If so, I have the answer I was looking for. You may say that Einstein don't have to state the obvious, but it wasn't obvious to me that it followed from his postulates. If you say it does, I'm prepared to accept it. I thought it did, but it wasn't obvious to me.

    I am aware of that, although Einstein doesn't mention it in the original paper. At this point I am not concerned with whether E's postulates are correct or not. I am trying to sort out exactly what he is saying, true or not.

    Thanks for the link. Will study.
     
  13. Tach Banned Banned

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    Yes.




    Good.


    The experiments came AFTER his paper. For obvious reasons.
     
  14. DonQuixote Registered Senior Member

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    The Michelson-Morley experiment was in 1887, i believe. E's paoer is from 1905? I Don't know if Eisnstein was aware of it. But if this counts as an experiment on the independence of speed of light of the source or the observer, I'm not sure. Point is that they expected light speed to vary, but could not find it did.

    Or have I got all this wrong?
     
  15. rpenner Fully Wired Staff Member

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    http://www.fourmilab.ch/etexts/einstein/specrel/www/
    Paragraph 1 discusses that there is no known electromagnetic phenomenon which depends on absolute motion -- the math of Maxwell's equations takes a different form if the source is moving or the target is moving, but the physical result depends only on relative motion.

    The first part of paragraph 2 defines a class of inertial observers where all the laws of physics are in the same form.
    Since for whatever instantaneous state of motion a source may have there is always a potential inertial observer sharing that state of motion, there is always a potential observer where any source is stationary and for any observer in motion relative to that potential observer the source is moving. So there is no absolute meaning to "stationary" or "moving", since these terms should always be used in relation to a specific inertial observer and another might well disagree and be correct, relative to himself.

    Also, inertial observers cannot be in relative motion to themselves, so a single inertial observer cannot meaningfully talk about his own motion.

    So observer 1 sees a myriad of sources with varying velocities and measures the speed of light from those sources always to be c. Observer 2, in relative motion to 1, sees the same sources with different velocities and equally validly measures the speed of light to always be c. Thus the speed of light doesn't depend on the state of motion of the source or the inertial observer or the relative motion between the source and the inertial observer.
     
  16. Motor Daddy ☼☼☼☼☼☼☼☼☼☼☼ Valued Senior Member

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    True, the speed of light is independent of sources and observers, which proves that the light sphere expands from the point in space the light was emitted, regardless of what the source does after it emitted the light.

    Do you understand that?
     
  17. Tach Banned Banned

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    No, it doesn't count. The experiments I listed for you, DO.
     
    Last edited: Oct 5, 2011
  18. OnlyMe Valued Senior Member

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    Einstein does not reference the Michelson-Morley experiment directly but he does make reference, the following reference, in the second paragraph of that paper..,
    There has been some debate as to whether he was referring to Michelson-Morley or one of the other experiments taking place prior to 1905. In any event he was aware that no motion of the Earth relative to the aether of Maxwell or Lorentz was detectable. The main issue here was not the existence of the aether, it was whether the Earth moved through the aether.

    Additionally Einstein only raised the speed of light in vacuum to the level of a universal constant, in the following quote from the same paper. This was most likely drawn from the experimental work of Fizeau, who had done a great deal of work in the laboratory measuring the speed of light in a variety of conditions and mediums.

    This was likely drawn from the work of Fizeau rather than the Michelson-Morley experiments. Fizeau had conducted a number of experiments in the laboratory measuring the speed of light under various conditions and through different mediums.
     
    Last edited: Oct 5, 2011
  19. rpenner Fully Wired Staff Member

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    Education: How does the Lorentz transformation transform inertial motion of something moving at the speed of light?

    Start with \(\vec{u} = \left( u_x , \, u_y , \, u_z \right) = \left( c \cos \theta , \, c \sin \theta \cos \phi, \, c \sin \theta \sin \phi \right) \) so that \((\vec{u})^2 = c^2\).

    \(\vec{u'} = \left( \frac{c \cos \theta - v}{1 - \frac{v \cos \theta}{c}} , \, \frac{c \sin \theta \cos \phi \sqrt{1 - \frac{v^2}{c^2}}}{1 - \frac{v \cos \theta}{c}} , \, \frac{ c \sin \theta \sin \phi \sqrt{1 - \frac{v^2}{c^2}}}{1 - \frac{v \cos \theta}{c}} \right) \)
    so \((\vec{u'})^2 = c^2 \frac{ \left( \cos \theta - \frac{v}{c} \right)^2 + \sin^2 \theta \left( \cos^2 \phi + \sin^2 \phi \right) \left( 1 - \frac{v^2}{c^2} \right) }{\left(1 - \frac{v}{c} \cos \theta \right)^2} = c^2 \frac{ \cos^2 \theta - 2 \frac{v}{c} \cos \theta + \frac{v^2}{c^2} + \sin^2 \theta - \frac{v^2}{c^2} \sin^2 \theta }{1 - 2 \frac{v}{c} \cos \theta + \frac{v^2}{c^2} \cos^2 \theta} = c^2\)

    So the math of the Lorentz transform says if something is measured by one inertial observer to be moving at the speed of light, any other inertial observer will also see that moving at the speed of light.

    This follows quickly from the invariant since for any two events along the world-line of something moving at the speed of light, \(c^2 (\Delta t)^2 = (\Delta \vec{r})^2\) so that \(c^2 (\Delta t)^2 - (\Delta \vec{r})^2 = 0 = c^2 (\Delta t')^2 - (\Delta \vec{r'})^2\).
     
  20. DonQuixote Registered Senior Member

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    It's a bit odd, though. You say the independence of the SOL to the speed of the observer is contained in postulate 1.

    Why does he have to postulate the independence of SOL of the source?

    In the day, it wasn't controversial, was it? SOL was thought of as analoguous (spelling?) to the speed of sound, which is not dependent of the speed of the source. The aether was supposed to be the medium, as air is for sound.

    Does he need this postulate (for the maths, maybe)? Does he replace the need for an aether with a posulate?

    I am aware that this post isn't exactly rigid logic. Just thinking out loud.
     
  21. DonQuixote Registered Senior Member

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    It's not hard to understand. But if this was the case, the movement of an observer relative to the point of origin should have an effect on the *meassured* speed of light, should it not?
     
  22. billvon Valued Senior Member

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    Thanks for the post. That's a good concise summary.
     
  23. billvon Valued Senior Member

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    Sort of. All observers see the speed of light as identical.

    Correct! And the key here is that is _always_ true no matter which frame you choose. _All_ observers see it expand as a sphere, even the observer in a frame moving relative to the source.
     

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