
#19
Nov2709, 08:13 PM

Sci Advisor
PF Gold
P: 1,806

Ignore B and consider A. Show that when x_{a} = ct_{a} then x'_{a} = ct'_{a}. (The details of this have already been given in several posts.) Do you disagree with this? Then, as an entirely separate argument, ignore A and consider B. Show that when x_{b} = −ct_{b} then x'_{b} = −ct'_{b}. 



#20
Nov2709, 08:25 PM

P: 687

Let me think about this a while. I see some sense here. 



#21
Nov2709, 08:27 PM

P: 687





#22
Nov2709, 08:36 PM

Sci Advisor
P: 8,470





#23
Nov2709, 08:40 PM

P: 687

Two different frames will disagree where the center of the one light sphere is located? These are the equations I am looking for. I am not looking for talk, just the math. The light postulate is clear. Light proceeds spherically in the frame from the emission point regardless of the motion of the light source. 



#24
Nov2709, 08:43 PM

Sci Advisor
P: 8,470





#25
Nov2709, 08:47 PM

P: 687

Think about this for a light sphere. One light sphere will have two origins in two different places in space. How? 



#26
Nov2709, 09:20 PM

Sci Advisor
P: 8,470





#27
Nov2709, 10:57 PM

Mentor
P: 16,469

I reiterate my question: what specifically is wrong with my eq 1d)? I have gone back and verified that the Lorentz transform equations were copied down correctly in 1a) and 1b). So the only possible objection is that you think substitution is no longer a valid algebraic operation in relativity. Why do you think that? 



#28
Nov2809, 01:13 AM

Mentor
P: 11,220

To keep things simple, suppose that the origins of frames A and B coincide at [itex]x_A = x_B = 0[/itex] and [itex]t_A = t_B = 0[/itex], and that in frame A, frame B moves in the +x direction with speed v. Then coordinates in the two frames are related by the Lorentz transformation
[tex]x_B = \gamma (x_A  vt_A)[/tex] [tex]y_B = y_A[/tex] [tex]z_B = z_A[/tex] [tex]t_B = \gamma \left( t_A  \frac{v x_A}{c^2} \right)[/tex] where as usual [tex]\gamma = \frac {1} {\sqrt {1  v^2 / c^2}}[/tex] Suppose that when the origins of the two frames coincide, a light flash goes off at the momentarily mutual origin. In frame A, the light expands as a sphere centered at the point [itex]x_A = y_A = z_A = 0[/itex], described by the equation [tex]x_A^2 + y_A^2 + z_A^2 = c^2 t_A^2[/tex] In frame B, the the light expands as a sphere centered at the point [itex]x_B = y_B = z_B = 0[/itex], described by the equation [tex]x_B^2 + y_B^2 + z_B^2 = c^2 t_B^2[/tex] (Recall that the equation of a sphere centered at the origin is [itex]x^2 + y^2 + z^2 = R^2[/itex]. In frame A, [itex]R_A = ct_A[/itex] and in frame B, [itex]R_B = ct_B[/itex].) You can verify that these two equations are consistent by substituting the Lorentz transformation equations into the second one and showing that it reduces to the first one. This is the same thing that DaleSpam did in post #11, except that I'm using three spatial dimensions and he used only one. The points [itex]x_A = y_A = z_A = 0[/itex] and [itex]x_B = y_B = z_B = 0[/itex] are the same point only when [itex]t_A = t_B = 0[/itex]. At all other times, in either frame, they are different points. Therefore, at those other times the light sphere is centered at different points in the two frames. 



#29
Nov2809, 03:10 AM

Sci Advisor
P: 8,004

Event A is (xa'=ct', ta'=t') Event B is (xb'=ct', tb'=t') Use your formulas to calculate ta and tb. 



#30
Nov2809, 03:50 AM

P: 3,536

Two events are simultaneous in a frame if their t coordinates in that frame are equal. For example the two light sphere events A : (xa = cT, ta = T) B : (xb = cT, tb = T) are simultaneous in the frame O where they have the above coordinates, because ta = tb But in the frame O' where they have the coordinates A : (xa' = ( cTvT)*gamma, ta' = (TvT/c)*gamma) B : (xb' = (cTvT)*gamma, tb' = (T+vT/c)*gamma) they are not simultaneous, because ta' <> tb' 



#31
Nov2809, 03:47 PM

P: 687

1) Lorentz transform approach: We have the standard form of the Lorentz transform 1a) t' = ( t  vx/c^2 )γ 1b) x' = ( x  vt )γ And we have any arbitrary equation in the primed frame 1c) x' = ct' To obtain the corresponding equation in the unprimed frame we simply substitute 1a) and 1b) into 1c) 1d) ( x  vt )γ = c(( t  vx/c^2 )γ) Which simplifies to 1e) x = ct My problem with this method is that this causes the x locations in O, x and x to be simultaneous in both O and O'. Thus, ct = +x and ct' = + x'. They are in relative motion and therefore, the two points cannot appear synchronous to both frames. 



#32
Nov2809, 03:55 PM

P: 1,060

Matheinste. 



#33
Nov2809, 03:55 PM

P: 687

However, these conditions must be met. 1) When the light sphere strikes two equidistant x points in O', say x' and x', this cannot be synchronous to O. 2) When two equidistant x points in O are struck, this cannot be synchronous in O' Finally, one light sphere must have two different origins in space. I di not see how this is possible. Since we are able to translate O' to O, then O will conclude once light sphere will have two different origins in its own space and worse, the light sphere origin of O' moves with the frame of O'. This is the very definition of light speed anisotropy. Anyway, what I was hoping to see is x' and x' translated into the coordinates of O such that it is clear, these corresponding x1 and x2 and clearly not synchronous in O but are synchronous in O'. 



#34
Nov2809, 03:57 PM

P: 687

So, this does not work. 



#35
Nov2809, 04:03 PM

P: 687

Then, I would like to figure out how a light sphere will have a fixed origin in O while at the same time also have a moving origin in O located at vt based on the necessary conditions of the light postulate for O'. 



#36
Nov2809, 04:19 PM

P: 3,536




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