Light sphere question

DaleSpam

OK, so what do you think is wrong about 1d)? It is simply the substitution of 1a) and 1b) into 1c). You gave 1c) so I assume that either you think I wrote the Lorentz transform wrong in 1a) and 1b) or you think that for some reason substitution is no longer a valid algebraic operation in relativity. If you think that I wrote it wrong, then what do you think is the right formula for the Lorentz transform, and if you think that substitution is no longer a valid algebraic operation can you express why you believe that?

DrGreg

There is no need to worry about the simultaneity, or not, of A and B. Just consider A on its own and do the transformation.

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.

cfrogue

Let me think about this a while.

I see some sense here.

cfrogue

How do you connect this to ta given that there exists a center of the light sphere located at vt?

JesseM

Different frames disagree about where the center of the sphere is. Suppose there are two objects A and B moving inertially at different velocities, and at the moment they cross paths, A emits a flash of light in all directions. In this case, in A's rest frame the sphere's center is at the position of A at later times, while in B's rest frame the sphere's center is at the position of B at later times. Also, because of the relativity of simultaneity, if you pick two events on opposite sides of the sphere at a single moment in A's frame, then in B's frame these events will happen at two different times, when the sphere had two different radii.

cfrogue

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.

JesseM

Yes, of course--this is implied by the fact that the light moves at c in all directions.
This is already implied by the fact that x = ±ct in the first frame and x' = ±ct' in the other. Obviously this means that the center of the 1D sphere will be x=0 in the first frame and x'=0 in the second, and by the Lorentz transformation, an object at rest at x=0 in the first frame is moving at velocity -v in the second frame, while an object at rest at x'=0 in the second frame is moving at velocity +v in the first.

cfrogue

Yes, of course--this is implied by the fact that the light moves at c in all directions.

Think about this for a light sphere.

One light sphere will have two origins in two different places in space.

How?

JesseM

What do you mean "one light sphere"? If you are talking about a light sphere at a single instant in one frame (i.e. the set of all points that lie a distance ct from the origin of that frame at time t), then the points that make up that one sphere are part of many different light spheres at different moments in the second frame, thanks to the relativity of simultaneity. Each light sphere is really the intersection between the light cone and a surface of simultaneity, and each frame has different surfaces of simultaneity.

DaleSpam

This is most certainly not true. You have been given multiple correct math derivations, all leading to the same conclusion all of which you have summarily rejected with no substantiation.

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?

jtbell

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.

atyy

OK, I think I understand your scenario. So just do it the other way round.

Event A is (xa'=ct', ta'=t')
Event B is (xb'=-ct', tb'=t')

Use your formulas to calculate ta and tb.

A.T.

What do you mean by events in one frame being simultaneous with the events in a different frame? Two events can be simultaneous (or not simultaneous) within one frame, not across frames.

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' = ( cT-vT)*gamma, ta' = (T-vT/c)*gamma)
B : (xb' = (-cT-vT)*gamma, tb' = (T+vT/c)*gamma)

they are not simultaneous, because

ta' <> tb'

You can then swap the apostrophe annotation my formulas, so the events are simultaneous in O' but not O.

cfrogue

cfrogue, atyy already presented some pretty convincing math, but let me try two other ways:

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.

matheinste

What does it mean for two points to appear synchronous?

Matheinste.

cfrogue

I am OK with everything above.

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'.

cfrogue

But, ta' = tb' since x' and -x' are synchronous in O'.

So, this does not work.