clj4 said:
Hmmm,
It is clear how you got the second formula but it is not clear at all how you got the first formula [tex]c(v,\theta)=c_0[/tex]. Would you care to show the intermediate steps?
This is just a statement that in SR one-way light speed is isotropic, but I will examine it anyway.
SR, RMS (Robertson-Mansouri-Sexl text theory), and Lorentz ether theory are all in agreement that the speed of light [tex]c_0[/tex] is
defined to be isotropic with respect to at least one particular inertial frame (e.g., within the ether frame [tex]\frac{X_2-X_1}{T_2-T_1}=c_0[/tex] -- actually,
all velocites are isotropic within this frame, so this equation really relates just to this example, I suppose that a more general equation like this [tex]\frac{X_2-X_1}{T_2-T_1}=v_0[/tex] is always valid). In SR however, the Lorentz transform relates this particular (arbitrary) inertial frame with every other inertial frame while explicitly preserving light speed, and therefore the speed of light is further
defined (within SR) to be isotropic with respect to every other inertial frame as well.
For example, these are (two sets of) the Lorentz transforms for motion along the x-axis:
[tex]t_1=((1-v^2/c_0^2)^{1/2}T_1-vx_1/c_0^2)[/tex],
[tex]x_1=((X_1-vT_1)/(1-v^2/c_0^2)^{1/2})[/tex],
[tex]t_2=((1-v^2/c_0^2)^{1/2}T_2-vx_2/c_0^2)[/tex],
[tex]x_2=((X_2-vT_2)/(1-v^2/c_0^2)^{1/2})[/tex].
Rather than
defining light-speed isotropy for all (inertial) frames, I will justify this equation [tex]c(v,\theta)=c_0[/tex] by
defining light-speed isotropy only in the ether frame (e.g., [tex]\frac{X_2-X_1}{T_2-T_1}=c_0[/tex]) and then combine this definition with the Lorentz transformations to show that light speed is therefore isotropic in all (inertial) frames [tex]\frac{x_2-x_1}{t_2-t_1}=c_0[/tex]; such a showing is ultimately coordinate-system dependent, and it is definitely
not something that could ever be proven by an experiment.
So, to verify that this equation [tex]\frac{x_2-x_1}{t_2-t_1}=c_0[/tex] is true (when [tex]\frac{X_2-X_1}{T_2-T_1}=c_0[/tex]) then simply transform the four coordinates from any arbitrary inertial frame (e.g., use any
v you like) into the ether frame using the Lorentz transform equations given above and verify that you always arrive back at this equation [tex]\frac{X_2-X_1}{T_2-T_1}=c_0[/tex] which we have
defined above to be true.
Start with:
[tex]\frac{x_2-x_1}{t_2-t_1}=c_0[/tex]
Transform the four space-time coordinates (by substitution) using the Lorentz transforms:
[tex]\frac{((X_2-vT_2)/(1-v^2/c_0^2)^{1/2})-((X_1-vT_1)/(1-v^2/c_0^2)^{1/2})}{((1-v^2/c_0^2)^{1/2}T_2-vx_2/c_0^2)-((1-v^2/c_0^2)^{1/2}T_1-vx_1/c_0^2)}=c_0[/tex]
Now, simply reduce this equation to [tex]\frac{X_2-X_1}{T_2-T_1}=c_0[/tex] to show that [tex]c(v,\theta)=c_0[/tex] is true (this doesn't actually
prove that it is true for all [tex]\theta[/tex] because we're only looking at motion along the x-axis).
[tex]\frac{X_2-vT_2-X_1+vT_1}{(1-v^2/c_0^2)T_2-\frac{v(X_2-vT_2)}{c_0^2}-(1-v^2/c_0^2)T_1-\frac{v(X_1-vT_1)}{c_0^2}}=c_0[/tex]
[tex]\frac{(X_2-vT_2-X_1+vT_1)c_0^2}{(1-v^2/c_0^2)T_2c_0^2-v(X_2-vT_2)-(1-v^2/c_0^2)T_1c_0^2-v(X_1-vT_1)}=c_0[/tex]
[tex]\frac{(X_2-vT_2-X_1+vT_1)c_0^2}{( c_0^2-v^2)(T_2-T_1)-v(X_2-vT_2-X_1+vT_1)}=c_0[/tex]
[tex](X_2-vT_2-X_1+vT_1)c_0^2=c_0(( c_0^2-v^2)(T_2-T_1)-v(X_2-vT_2-X_1+vT_1))[/tex]
[tex](X_2-vT_2-X_1+vT_1)c_0^2=c_0( c_0^2-v^2)(T_2-T_1)-v c_0(X_2-vT_2-X_1+vT_1)[/tex]
[tex](X_2-vT_2-X_1+vT_1)(c_0^2+v c_0)=c_0( c_0^2-v^2)(T_2-T_1)[/tex]
[tex](X_2-vT_2-X_1+vT_1)c_0(c_0+v)=c_0( c_0+v)( c_0-v)(T_2-T_1)[/tex]
[tex](X_2-vT_2-X_1+vT_1)c_0=c_0( c_0-v)(T_2-T_1)[/tex]
[tex]X_2-vT_2-X_1+vT_1=c_0(T_2-T_1)-v(T_2-T_1)[/tex]
[tex]X_2-vT_2-X_1+vT_1+vT_2-vT_1=c_0(T_2-T_1)[/tex]
[tex]X_2-X_1=c_0(T_2-T_1)[/tex]
[tex]\frac{X_2-X_1}{T_2-T_1}=c_0[/tex]