Deriving Equations for Light Sphere in Collinear Motion - O and O' Observers

[cfrogue]

Assume there is a stationary observer O and a moving observer O' at v in collinear relative motion.

When origined with O, O' emits a spherical light pulse.

Now, the light pulse is described by O as x^2 + y^2 + z^2 = (ct)^2 and by O' as x'^2 + y^2 + z^2= (ct')^2.

By considering only the x-axis points of the light sphere, it is the case that x'^2= (ct')^2. Thus, ct' = ± x'.

Here is my question. What are the equations strictly from the coordinates and proper time of O to describe ct' = ± x'. This means, what are the x points in O and what are the times in O for the light sphere of O'.

 

[Dale]

What are the equations strictly from the coordinates and proper time of O to describe ct' = ± x'. This means, what are the x points in O and what are the times in O for the light sphere of O'.

ct = ± x

 

[cfrogue]

ct = ± x

This violates R of S.

 

[Dale]

No it doesn't

 

[cfrogue]

No it doesn't

How descriptive. Perhaps you are right.

The light sphere in O' is located at ct' = ± x'.

t' = ( t - vx/c^2 )λ

x' = ( x - vt )λ

The only way your equation would work for the light sphere of O' is if v = 0.

 

[atyy]

How descriptive. Perhaps you are right.

The light sphere in O' is located at ct' = ± x'.

t' = ( t - vx/c^2 )λ

x' = ( x - vt )λ

The only way your equation would work for the light sphere of O' is if v = 0.

Substitute x=ct into your equation for t'

t' = (t - vct/c²)λ
t' = λ(1-v/c)t

Solve for t

t = t'/(λ(1-v/c))

Substitute x=ct into your equation for x', followed by substituting above equation for t

x' = (ct - vt)λ
x' = λ(c-v)t
x' = λ(c-v)t'/(λ(1-v/c))
x' = (c-v)t'/(1-v/c)
x' = c(1-v/c)t'/(1-v/c)
x' = ct'

 

[cfrogue]

Substitute x=ct into your equation for t'

t' = (t - vct/c²)λ
t' = λ(1-v/c)t

Solve for t

t = t'/(λ(1-v/c))

Substitute x=ct into your equation for x', followed by substituting above equation for t

x' = (ct - vt)λ
x' = λ(c-v)t
x' = λ(c-v)t'/(λ(1-v/c))
x' = (c-v)t'/(1-v/c)
x' = c(1-v/c)t'/(1-v/c)
x' = ct'

Solve for t

t = t'/(λ(1-v/c))

How can you do this?

Events simultaneous for O' will not be simultaneous for O.

Thus, there wil be a t1 and t2 in O, not a common t.

 

[atyy]

Simultaneous means there are two events at two different spatial coordinates at the same coordinate time in one frame.

But there's only one event here - described as the photon being at (x,t) in one frame, oe equivalently as the photon being at (x',t') in another frame.

That's what your equation t' = (t - vx/c²)λ means. I just plugged and chugged.

 

[cfrogue]

Simultaneous means there are two events at two different spatial coordinates at the same coordinate time in one frame.

But there's only one event here - described as the photon being at (x,t) in one frame, oe equivalently as the photon being at (x',t') in another frame.

That's what your equation t' = (t - vx/c²)λ means. I just plugged and chugged.

I just wrote the transformation equations down.

You cannot use the same t without violating the R of S.

Keep in mind, O' is moving.

Also, there are two events, ± x'.

Also, if the two events are simultaneous in O and O', I can run this into a contradiction of R of S with collinear relative motion.

Something is wrong here.

Since O' emitted the light, there exists points that are equidistant in O' and occur at the same time in O' but are not simultaneous on O.

These are the points I am looking for.

 

[atyy]

Event A is (xa=ct, ta=t)
Event B is (xb=-ct, tb=t)

ta=tb=t, so event A and B are simultaneous in O.

Use your equations to find ta' and tb'.

 

[Dale]

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 = ct2) First principles approach:

We know that the second postulate is that c is the same in all reference frames, so we can immediately write that the speed of the light pulse is c. This implies
2a) x = ct + B

Since we know that the origins coincided with the flash we know that x=0 and t=0 is a point on the light pulse, so we can use that to solve for B
2b) 0 = c0 + B
2c) B = 0

Substituting 2c into 2a gives
2d) x = ct

Note that approach 1 is a general approach that will work for any equation that you care to write. Approach 2 is specific to this problem since we are dealing with light pulses and will not work in general. I would typically recommend approach 1.

 

[cfrogue]

Event A is (xa=ct, ta=t)
Event B is (xb=-ct, tb=t)

ta=tb=t, so event A and B are simultaneous in O.

Use your equations to find ta' and tb'.

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/

 

[atyy]

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.

What? OK, I don't understand your scenario. What events are simultaneous in which frame?

 

[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


2) First principles approach:

We know that the second postulate is that c is the same in all reference frames, so we can immediately write that the speed of the light pulse is c. This implies
2a) x = ct + B

Since we know that the origins coincided with the flash we know that x=0 and t=0 is a point on the light pulse, so we can use that to solve for B
2b) 0 = c0 + B
2c) B = 0

Substituting 2c into 2a gives
2d) x = ct

Note that approach 1 is a general approach that will work for any equation that you care to write. Approach 2 is specific to this problem since we are dealing with light pulses and will not work in general. I would typically recommend approach 1.

You do not have vt as the origin of the light sphere in O' with this approach. O' is moving with relative motion v. Recall, the light postulate proclaims light emits from the emission point in the frame spherically in all directions. Thus, since O' emitted the light pulse, these conditions must be met. We must have simultaneity in O' that is not simultaneous in O.

I have worked on this for days and cannot make it happen.

I will take you down the road of my failures.

This math is not convincing and is not correct.

 

[cfrogue]

What? OK, I don't understand your scenario. What events are simultaneous in which frame?

I am clearer in my last post.

I cannot do the math and have tried.

I get junk I keep throwing it out.

 

[Dale]

I have worked on this for days and cannot make it happen.

I will take you down the road of my failures.

Please post your work, I am sure I can point out the place you went wrong.

This math is not convincing and is not correct.

I accept that it is not convincing to you, but it is correct. It might help if you identified which equation you think is wrong, why you think it is wrong, and what you think it should be instead. That might be easier than posting your work if there is a lot of material.

 

[cfrogue]

Please post your work, I am sure I can point out the place you went wrong.I accept that it is not convincing to you, but it is correct. It might help if you identified which equation you think is wrong, why you think it is wrong, and what you think it should be instead. That might be easier than posting your work if there is a lot of material.

No this ain't about me, it is about your equations.

1d) ( x - vt )γ = c(( t - vx/c^2 )γ)

You cannot use t in the context of O.

There are two t's t1 and t2 since,

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/

I have the following


x = (v( t1 + t2))/2

for the x point and all this is useless.

Remember, if we fail to realize that what is simultaneous in O' is NOT simultaneous in O, then we are committing a serious act of crack pottery.

 

[Dale]

No this ain't about me, it is about your equations.

1d) ( x - vt )γ = c(( t - vx/c^2 )γ)

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]

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 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]

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.

Let me think about this a while.

I see some sense here.

 

[cfrogue]

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.

Wait, I have a problem,

Then, as an entirely separate argument, ignore A and consider B. Show that when x_b = −ct_b 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]

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]

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]

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.

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 = ±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.

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.

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]

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.

 

[Dale]

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]

To keep things simple, suppose that the origins of frames A and B coincide at x_A = x_B = 0 and t_A = t_B = 0, 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

x_B = \gamma (x_A - vt_A)

y_B = y_A

z_B = z_A

t_B = \gamma \left( t_A - \frac{v x_A}{c^2} \right)

where as usual

\gamma = \frac {1} {\sqrt {1 - v^2 / c^2}}

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 x_A = y_A = z_A = 0, described by the equation

x_A^2 + y_A^2 + z_A^2 = c^2 t_A^2

In frame B, the the light expands as a sphere centered at the point x_B = y_B = z_B = 0, described by the equation

x_B^2 + y_B^2 + z_B^2 = c^2 t_B^2

(Recall that the equation of a sphere centered at the origin is x^2 + y^2 + z^2 = R^2. In frame A, R_A = ct_A and in frame B, R_B = ct_B.)

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 x_A = y_A = z_A = 0 and x_B = y_B = z_B = 0 are the same point only when t_A = t_B = 0. 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]

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.]

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.