Light sphere question


by cfrogue
Tags: light, sphere
DrGreg
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Nov27-09, 08:13 PM
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Quote Quote by cfrogue View Post
Event A is (xa=ct, ta=t)
Event B is (xb=-ct, tb=t)


This assumes the events in O will be simultaneous with the events in O' given ta = t = tb.

You should already know events like this cannot be simultaneous in both frames with v ≠ 0.

So we see that we cannot attach any absolute signification to the concept of simultaneity, but that two events which, viewed from a system of co-ordinates, are simultaneous, can no longer be looked upon as simultaneous events when envisaged from a system which is in motion relatively to that system.
http://www.fourmilab.ch/etexts/einstein/specrel/www/
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 xa = cta 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 xb = −ctb then x'b = −ct'b.
cfrogue
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Nov27-09, 08:25 PM
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Quote Quote by DrGreg View Post
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 xa = cta 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 xb = −ctb then x'b = −ct'b.

Let me think about this a while.

I see some sense here.
cfrogue
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Nov27-09, 08:27 PM
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Quote Quote by DrGreg View Post
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 xa = cta 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 xb = −ctb then x'b = −ct'b.

Wait, I have a problem,

Then, as an entirely separate argument, ignore A and consider B. Show that when xb = −ctb then x'b = −ct'b.
How do you connect this to ta given that there exists a center of the light sphere located at vt?
JesseM
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Nov27-09, 08:36 PM
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Quote Quote by cfrogue View Post
How do you connect this to ta given that there exists a center of the light sphere located at vt?
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
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Nov27-09, 08:40 PM
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Quote Quote by JesseM View Post
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.

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
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Nov27-09, 08:43 PM
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Quote Quote by cfrogue View Post
Two different frames will disagree where the center of the one light sphere is located?
Yes, of course--this is implied by the fact that the light moves at c in all directions.
Quote Quote by cfrogue
These are the equations I am looking for.

I am not looking for talk, just the math.
This is already implied by the fact that x = ąct in the first frame and x' = ąct' 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
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Nov27-09, 08:47 PM
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Quote Quote by JesseM View Post
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 = ąct in the first frame and x' = ąct' 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.
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
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Nov27-09, 09:20 PM
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Quote Quote by cfrogue View Post
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?
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
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Nov27-09, 10:57 PM
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Quote Quote by cfrogue View Post
I am not looking for talk, just the math.
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
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Nov28-09, 01:13 AM
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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
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Nov28-09, 03:10 AM
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Quote Quote by cfrogue View Post
Event A is (xa=ct, ta=t)
Event B is (xb=-ct, tb=t)


This assumes the events in O will be simultaneous with the events in O' given ta = t = tb.

You should already know events like this cannot be simultaneous in both frames with v ≠ 0.

So we see that we cannot attach any absolute signification to the concept of simultaneity, but that two events which, viewed from a system of co-ordinates, are simultaneous, can no longer be looked upon as simultaneous events when envisaged from a system which is in motion relatively to that system.
http://www.fourmilab.ch/etexts/einstein/specrel/www/
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.
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Nov28-09, 03:50 AM
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Quote Quote by cfrogue View Post
This assumes the events in O will be simultaneous with the events in O' given ta = t = tb.
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'

OK, I think I understand your scenario. So just do it the other way round.
You can then swap the apostrophe annotation my formulas, so the events are simultaneous in O' but not O.
cfrogue
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Nov28-09, 03:47 PM
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Quote Quote by DaleSpam View Post
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?
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
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Nov28-09, 03:55 PM
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Quote Quote by cfrogue View Post
They are in relative motion and therefore, the two points cannot appear synchronous to both frames.
What does it mean for two points to appear synchronous?

Matheinste.
cfrogue
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#33
Nov28-09, 03:55 PM
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Quote Quote by jtbell View Post
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.
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
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#34
Nov28-09, 03:57 PM
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Quote Quote by atyy View Post
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.
But, ta' = tb' since x' and -x' are synchronous in O'.

So, this does not work.
cfrogue
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#35
Nov28-09, 04:03 PM
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Quote Quote by A.T. View Post
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.
The goal is to use x' and -x' which will be struck by the light sphere at the same time in O' and then prove the corresponding x values in O are not synchronous by using the standard LT translations.

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'.
A.T.
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#36
Nov28-09, 04:19 PM
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Quote Quote by cfrogue View Post
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'.
Events can be simultaneous, not locations.
Quote Quote by cfrogue View Post
Thus, ct = +-x and ct' = +- x'.
These equations do not describe two events in each frame, they describe all events on the light cone in each frame. The derivation by atyy only shows that all events on the light cone in one frame will also be on the light cone in the other frame, after LT. But It says nothing about whether two simultaneous events on the light cone will still be simultaneous after LT. Just because they both still are on the light cone, doesn't mean that they still have the same t-coordinate.
Quote Quote by cfrogue View Post
They are in relative motion and therefore, the two points cannot appear synchronous to both frames.
Correct. I show this in post #30

Quote Quote by cfrogue View Post
The goal is to use x' and -x' which will be struck by the light sphere at the same time in O' and then prove the [B]corresponding x values in O are not synchronous by using the standard LT translations.
That is what I did in post #30, but I just swaped O and O', and instead of x and -x I express the positions in the syncronised frame as cT and -cT, where T can be any value.

Quote Quote by cfrogue View Post
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'.
A light sphere has a fixed center in every frame.


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