B Relativity of Measures: A vs B Frame & Light Source S

[PeterDonis]

Thank you.
3. Then by virtue of 2, the motion of S is relative to the rest frame of E and the motion of E is relative to the rest frame of S, in that neither hold the equations 'More" valid in terms of the laws.

Yes, no?

Not as you state it. There is no such thing as "the" motion of anything. The correct statement is that S and E are moving relative to each other, so that in the rest frame of E, S is moving, and in the rest frame of S, E is moving. Both frames are equally valid.

 

[PeroK]

Thank you.
@ PeroK, yes, I will likely say more than necessary. It's a symptom of doubt.

2. The principle of relativity requires the laws do not discern one "inertial frame" from another when the Lorentz transformations hold the laws valid in all.Yes, no?

Yes. It means that the laws of physics must be Lorentz invariant. Newton's laws of motion, Hooke's law, Coulomb's law are all special case approximations of the "real" Lorentz invariant laws.

Note that Maxwell's equations are Lorentz invariant. That makes them incompatible with Galilean relativity, which was Einstein's starting point for his discovery of SR.

 

[Dale]

3. Then by virtue of 2, the motion of S is relative to the rest frame of E and the motion of E is relative to the rest frame of S, in that neither hold the equations 'More" valid in terms of the laws.

Yes, no?

That is oddly written, but I think you are saying that S is moving relative to E and E is moving relative to S, and that both are valid inertial frames so the laws of physics will be the same in both. If that is what you intended then, yes.

 

[AlMetis]

That is oddly written, but I think you are saying that S is moving relative to E and E is moving relative to S, and that both are valid inertial frames so the laws of physics will be the same in both. If that is what you intended then, yes.

Thank you, that is what I mean.

4. When the light pulse is emitted from S (t=0) the position of S and E coincide on x.
Does this corresponding time of emission with position of S and E change the path of the light pulse observed from the rest frame of either S, or E from the path agreed in #1?
Yes, no?

 

[Dale]

4. When the light pulse is emitted from S (t=0) the position of S and E coincide on x.
Does this corresponding time of emission with position of S and E change the path of the light pulse observed from the rest frame of either S, or E from the path agreed in #1?
Yes, no?

This is problematic. The event $(t,x)=(0,0)$ is the same event as $(t',x')=(0,0)$ but "coincide on x" seems to mean that you think that all events $(t,x)=(0,\chi)$ are the same events as $(t',x')=(0,\chi)$, which is not the case for any $\chi \ne 0$.

I also do not understand what you mean by "Does this corresponding time of emission with position of S and E change the path of the light pulse observed from the rest frame of either S, or E from the path agreed in #1?" The light pulse does not exist prior to emission, so I am not clear how it would make sense to speak of its path changing at emission.

 

[PeroK]

4. When the light pulse is emitted from S (t=0) the position of S and E coincide on x.
Does this corresponding time of emission with position of S and E change the path of the light pulse observed from the rest frame of either S, or E from the path agreed in #1?
Yes, no?

This may be the root of your confusion. The emission event has a single coordinate (in any frame). For convenience, we may take that event as the origin of the two inertial reference frames. There is no common path at that point. There is only the single emission event.

As coordinate time passes in each frame, the path of the light becomes a trajectory, or locus or worldline (depending on your terminology). This worldline is the same physical path, but is described by different coordinates in each frame. This idea that the same physical path can be described by two different sets of coordinates seems to be the problem.

This is tied up with your misunderstanding of the invariance of the speed of light, and your desire to make the direction of light invariant. You want the path of light in all frames to be determined by the orientation of the source at the time of emission. In this case, the source may be a laser pointing in the positive y-axis in both frames. But, if the laser is moving in one frame, then the velocity of the light it emits will not be in the positive y direction. This seems to be your stumbling block.

This goes back to an early post where I said that light inherits velocity and momentum from the motion of the source and you contradicted this. This is where you are going wrong.

Just to make this absolutely clear. The second postulate is:

The speed of light is independent of the motion of the source.

The following are not true:

[Wrong] The (vector) velocity of light is independent of the velocity of the source. (This would be physically absurd.)

[Wrong] The momentum of the light is independent of the source. (In fact, even the magnitude of the momentum is dependent on the motion of the source: this is the (relativistic) Doppler shift.

[Wrong] The direction of the light relative to three coordniate axes that coincide at emission is independent of the motion of source. (This again would be physically absurd.)

It seems to me that you've spent a lot of time ploughing your own furrow on this one. From my experience, it will now take considerable intellectual courage on your part to admit this is wrong and abandon these ideas and start with a clean slate using the correct second postulate.

 

[AlMetis]

This is problematic.

I am not thinking two events that occur in the same place and time are the same event.

In my #1 question we had not discussed the relative position of E and S.
This question is confirming that the aberration will remain as observed in #1 when the position of S and E coincide on x at the time of emission.
It’s purpose is to make clear the position of emission relative to E and S. We will call it 0x in E’s frame and for convenience we will call it 0x in S’s frame.
Unless this is the point of my mistake, we now have a symmetry of relative motion on x across y for both frames. E will observe exactly the same motion of S as S does of E with opposite sign of x.
(We will assume observers in E and S are NOT facing each other so the x direction will be common to both.)

Yes, no?

 

[Dale]

I am not thinking two events that occur in the same place and time are the same event.

Two events that occur at the same time and space coordinates with respect to a single frame are the same event. Also, any event and its Lorentz transform are the same event just with that one event’s coordinates expressed in two different frames.

In my #1 question we had not discussed the relative position of E and S.

E and S are reference frames, so I don’t know what you mean by their relative position. A reference frame extends throughout spacetime. Its position is everywhere.

Perhaps you are asking about the relative position of the spatial origin of the frames?

This question is confirming that the aberration will remain as observed in #1 when the position of S and E coincide on x at the time of emission.

Yes, the equations I wrote apply for all $0\le t$ and $0\le t’$. That does include the time of emission, $t=t’=0$.

But I still don’t understand what you mean by the path changing at the time of emission. The path didn’t exist before the time of emission and after emission the path went in a straight line. There is no change in direction. There was no direction before emission, and after emission it goes in a straight line.

Aberration is not a change in direction, it is a disagreement about direction.

It’s purpose is to make clear the position of emission relative to E and S.

Ok, I thought that was already clear $\left. r^\mu \right|_{t=0}=(0,0,0,0)$ and $\left. r^{\mu’} \right|_{t’=0}=(0,0,0,0)$

Unless this is the point of my mistake, we now have a symmetry of relative motion on x across y for both frames. E will observe exactly the same motion of S as S does of E with opposite sign of x.

Yes, although that was part of your problem setup from the beginning and is not something that we just “now have” as a result of some analysis.

 

[AlMetis]

But I still don’t understand what you mean by the path changing at the time of emission.

When I make reference to an inertial frame by its label alone (E, S, A, B etc), I’m refering to its spatial origin x0, y0, z0.
I didn’t think I ask if the path changes at time of emission. I asked if the observed path, or aberration would change from #1 when E and S coincide at the time of emission.
You answered it as I meant it.
I should have set the relative positions of E and S from the beginning. I didn’t think it would matter, but then I didn’t think I was mistaken, so I am being more careful to avoid assumptions.

5. When the light pulse is emitted from E not S, does this change the aberration observed in the rest frame of S or E from that agreed in #1?

 

[AlMetis]

The speed of light is independent of the motion of the source.

The following are not true:

[Wrong] The (vector) velocity of light is independent of the velocity of the source. (This would be physically absurd.)

[Wrong] The momentum of the light is independent of the source. (In fact, even the magnitude of the momentum is dependent on the motion of the source: this is the (relativistic) Doppler shift.

[Wrong] The direction of the light relative to three coordniate axes that coincide at emission is independent of the motion of source. (This again would be physically absurd.)

I am not suggesting any of these.
Just to make it absolutely clear, my earlier reference to the momentum of light was in reference to the example in your post I responded to which was an analogy of a ball tossed from a moving car. I thought the analogy was understood in my response. I did not expect the pedantic attack that followed.
All the physical characteristics of light emitted from a source in motion would not happen if light did not have momentum.
My point was unlike balls and cars, light does not move with speed c+v, or c-v its speed is always c.

If I am mistaken, which everyone posting says I am, I will be as gracious and grateful as possible in acknowledging and thanking everyone for their help.
But as a scientist you will understand I cannot believe what I don’t understand. That is superstition, not science.

 

[jbriggs444]

When I make reference to an inertial frame by its label alone (E, S, A, B etc), I’m refering to its spatial origin x0, y0, z0.

Three numbers is enough to identify the origin of a three dimensional coordinate system. It is not enough to identify the origin of a four dimensional coordinate system.

The position of the origin is not enough to specify a coordinate system.

What about the state of motion of the system? What about the angles of its reference axes?

Note that being able to specify $x_0, y_0$ and $z_0$ assumes that you already have another coordinate system already established. If you want a new coordinate system what you want is a transformation rule. Possibly expressed with the use of a coordinate transformation matrix.

 

[PeroK]

My point was unlike balls and cars, light does not move with speed c+v, or c-v its speed is always c.

I'm sure it's already been pointed out that this is wrong. Also, critically, we are talking about two dimensional motion. In which case we have the transformation rule for velocity components.

Note that the same rule of velocity transformation applies to a ball as to light. The difference is that the speed of light is an invariant of the transformation.

But as a scientist you will understand I cannot believe what I don’t understand.

My point is simply that we all go down a rabbit-hole now and again. It's important to recognise that, stop digging and get yourself out.

 

[Dale]

When I make reference to an inertial frame by its label alone (E, S, A, B etc), I’m refering to its spatial origin x0, y0, z0.

Please don’t. You will be consistently misunderstood. The label alone should refer to the reference frame itself. If you mean “the spatial origin of $E$” then say “the spatial origin of $E$”. If writing that is too cumbersome then use a new symbol. For example you could define $e^{\mu’}=(t’,0,0,0)$ for the spatial origin of $E$.

5. When the light pulse is emitted from E not S, does this change the aberration observed in the rest frame of S or E from that agreed in #1?

I still don’t understand the use of the word change here. What does “change the aberration” mean?

If a device at rest wrt $S$ emits light along the worldline $r^\mu=(t,0,t,0)$ then the same worldline has coordinates $r^{\mu'}=(t',-t' v, t' \sqrt{1-v^2},0)$ in $E$.

Now “when the light pulse is emitted from $E$ not $S$” is ambiguous. You could mean that a device at rest wrt $E$ emits light along the same worldline as above $r^{\mu'}=(t',-t' v, t' \sqrt{1-v^2},0)$ which is still the same worldline as $r^{\mu}=(t,0,t,0)$. Alternatively you could mean that a device at rest wrt $E$ emits light along a different worldline $\rho^{\mu’}=(t’,0,t’,0)$ which is the same worldline as $\rho^{\mu}=(t,t v, t \sqrt{1-v^2},0)$

So I cannot link my statements to your question about “change the abberation”, but hopefully you can.

 

[Ibix]

I've made an animation of what I think the OP is talking about. The $x$ and $y$ directions are horizontal and vertical as usual. There is a green-painted laser lying parallel to the $y$ axis and at rest in the frame depicted (this is what the OP calls frame $S$) and a blue-painted laser also lying parallel to the $y$ axis but moving left-to-right at 0.5c (this is at rest in the frame the OP calls $E$). As the two lasers pass they emit a short pulse of green or blue light (respectively) , and each says that the pulse is emitted perpendicular to the $x$ axis.

(The above is an animated gif - depending on your browser I think you may need to click on it and/or open the image in its own tab to see the animation.) I've additionally marked concentric circles centered on the emission point in this frame, and vertical lines upwards from the lasers. You can see that the laser pulses always remain directly above their emitting lasers, and that both laser pulses are doing the same speed because they always lie on the same circles. It may be helpful to see all of the frames of the animation above superimposed:

Here you can see that both pulses move in straight lines, but at an angle to one another.

The question, I think, is what does this look like in the rest frame of the blue laser, frame $E$. The answer is that it looks exactly the same, except left-right reversed and with the colours swapped.

 

[PeroK]

I've made an animation of what I think the OP is talking about. The $x$ and $y$ directions are horizontal and vertical as usual. There is a green-painted laser lying parallel to the $y$ axis and at rest in the frame depicted (this is what the OP calls frame $S$) and a blue-painted laser also lying parallel to the $y$ axis but moving left-to-right at 0.5c (this is at rest in the frame the OP calls $E$). As the two lasers pass they emit a short pulse of green or blue light (respectively) , and each says that the pulse is emitted perpendicular to the $x$ axis.
View attachment 322484
(The above is an animated gif - depending on your browser I think you may need to click on it and/or open the image in its own tab to see the animation.) I've additionally marked concentric circles centered on the emission point in this frame, and vertical lines upwards from the lasers. You can see that the laser pulses always remain directly above their emitting lasers, and that both laser pulses are doing the same speed because they always lie on the same circles. It may be helpful to see all of the frames of the animation above superimposed:
View attachment 322485
Here you can see that both pulses move in straight lines, but at an angle to one another.

The question, I think, is what does this look like in the rest frame of the blue laser, frame $E$. The answer is that it looks exactly the same, except left-right reversed and with the colours swapped.

I believe that @AlMetis thinks your animation contradicts the second postulate.

 

[Ibix]

I think we all see that, but @AlMetis thinks your animation contradicts the second postulate.

I'm completely lost about what @AlMetis thinks because he (I guess) seems to agree with everything that we say but still thinks something's wrong somewhere. I'm curious to see if I get any commentary on the animation.

 

[jbriggs444]

I'm completely lost about what @AlMetis thinks because he (I guess) seems to agree with everything that we say but still thinks something's wrong somewhere. I'm curious to see if I get any commentary on the animation.

Not sure, but perhaps the disconnect is with the idea of a launcher which is unambiguously vertical and a resulting trajectory which is at an angle that depends on the state of motion of the launching mechanism.

One can explore different techniques for collimation to see how these translate to a trajectory. The fun one is the direction of the wave form being emitted by the flat top surface of a laser. The easy one is a light rifle.

 

[AlMetis]

I'm completely lost about what @AlMetis thinks because he (I guess) seems to agree with everything that we say but still thinks something's wrong somewhere. I'm curious to see if I get any commentary on the animation.

First I want to thank you Dale for your patients and help. My language, and lack of knowledge has made it far more difficult for you than it should be, so for that I am very grateful.
I don’t want to derail our conversation, and I will come back to if I still don’t get it, but I think Ibix’s animation will help clear up my problem quicker than my ability to explain it.@Ibix, thank you. I think your animation should solve my problem. (I was in the middle of making an animation myself.)
You have set the viewer at rest with the green source. The emission point is illustrated by its constant position relative to the concentric circles and its constant position relative to the viewer, i.e. the rest frame of the green emitter.
As you mentioned, this “event” observed at rest with the blue frame, is identical except mirrored across y with colors reversed.

How does the center of the concentric circles remain at rest with both green and blue when the pulse is only emitted by green?
I understand (at least I think I do) the reciprocal nature of all the observations of this “one” event resulting from the relativity of motion between the two observers. I don’t understand how the center of the concentric rings can follow/remain at rest with, both frames when it represents a single event.

 

[Ibix]

How does the center of the concentric circles remain at rest with both green and blue when the pulse is only emitted by green?

The concentric circles are drawn at rest with respect to green. Blue would see them moving and length contracted into ellipses. A similar set of circles could be drawn for the blue emitter; they would remain centered on the blue laser and would appear elliptical in the animation.

I don’t understand how the center of the concentric rings can follow/remain at rest with, both frames when it represents a single event.

The center of the circles is not a single event. It is a place in space, extended in time. The two frames agree that the pulses were emitted when the two lasers passed one another, but their ideas of "where the emission was in space" differ after that.

 

[AlMetis]

The concentric circles are drawn at rest with respect to green. Blue would see them moving and length contracted into ellipses. A similar set of circles could be drawn for the blue emitter; they would remain centered on the blue laser and would appear elliptical in the animation.

If Blue sees one ring (for simplicity) moving away, its width (on axis of motion) contracted, this is because the light was emitted by Green.
Green sees the same ring, same event of emission, remain at rest (increasing diameter at c) and it remains circular.
If Blue had emitted the light, Green would see the (Blue) ellipse moving away, and Blue would see it remain at rest and remain circular.

Green and Blue disagree after emission on were in space the emission took place, i.e. where their positions coincided, because that place is relative to their frame of reference and their motion means they are in different frames of reference.
Have I got that correct?

 

[PeroK]

If Blue sees one ring (for simplicity) moving away, its width (on axis of motion) contracted, this is because the light was emitted by Green.
Green sees the same ring, same event of emission, remain at rest (increasing diameter at c) and it remains circular.
If Blue had emitted the light, Green would see the (Blue) ellipse moving away, and Blue would see it remain at rest and remain circular.

The speed of light is independent of the motion of the source. If we are now talking about a spherical wave of light emitted in all directions, then both light spheres are spheres in both frame.

Green and Blue disagree after emission on were in space the emission took place, i.e. where their positions coincided, because that place is relative to their frame of reference and their motion means they are in different frames of reference.

I don't understand what you mean.

Have I got that correct?

Sadly not.

 

[Ibix]

If Blue sees one ring (for simplicity) moving away, its width (on axis of motion) contracted, this is because the light was emitted by Green.

No, the rings are (or could be) material objects tied to the lasers. If blue sees a ring moving away, it's because green fitted one around their laser. Blue does not see the laser pulses reaching green's ring at the same time, but does see them reaching their own ring at the same time. Green does not see the pulses reaching blue's rings at the same time but does see them reaching their own ring at the same time.

If we have an all-directions light flash instead of a laser emission, both frames see the flash as circular. Both say their own (material) ring is illuminated all at once. Both see the other's ring moving, and thus off-center except at the moment of emission, length contracted, and will say the light did not reach all parts of it at the same time.

 

[robphy]

No, the rings are (or could be) material objects tied to the lasers. If blue sees a ring moving away, it's because green fitted one around their laser. Blue does not see the laser pulses reaching green's ring at the same time, but does see them reaching their own ring at the same time. Green does not see the pulses reaching blue's rings at the same time but does see them reaching their own ring at the same time.

If we have an all-directions light flash instead of a laser emission, both frames see the flash as circular. Both say their own (material) ring is illuminated all at once. Both see the other's ring moving, and thus off-center except at the moment of emission, length contracted, and will say the light did not reach all parts of it at the same time.

Elaborating on this explanation by @Ibix ...

The rings could be a circular arrangement of mirrors facing the source.
Each observer has its own set. It's a "circular light clock".

At the event O when the worldlines meet,
a flash of light begins to trace out the future light cone of event O.
It doesn't matter which worldline (if any) is "the source" of the light.
It is the flash from event O.From my Circular Light Clock visualization: https://www.geogebra.org/m/pr63mk3j ,
(T_MOV is tunable).

The future light cone of O (cyan) is intersected
by the worldtube of the arrangement of mirrors of each inertial observer.

As @Ibix says, each sees the other arrangement as length-contracted...

• Blue observes the Blue-mirror-intersection-events are simultaneous in Blue-time and circular in Blue-space,
  but observes the Green-mirror worldtube as elliptical in Blue-space (contracted in the direction of relative-motion)
  and observes the Green-mirror-intersections-events are not-simultaneousin Blue-time and are not-circular in Blue-space.
• In fact, Blue observes the Green-mirror-intersections-events trace out points over Blue-time an ellipse in Blue-space where the meeting event is a focus of that ellipse.
• The other focus of that ellipse is the Blue-space position of where those Green-reflected signals refocus at a common event.
  (This is due to the reflective properties of an ellipse.)
• The elapsed Blue-time of the Green-refocused signal is longer than
  the elapsed Blue-time of the Blue-refocused signal. The ratio is time-dilation factor.
• The spacetime-volume enclosed by these "causal diamonds" is invariant, proportional to the square-interval between the emission and refocusing events.

Of course, the situation is symmetric.
Green observes the same things about Blue.... in accord with the Relativity Principle.Here's the overhead profile (the xy-plane):

Here's the side profile (the tx-plane),
where one can see the (1+1)-dim causal diamonds
used in my "rotated graph paper" method.

 

[Ibix]

I think it's the last graph above that encapsulates everything.

The same light cone (pale blue) interacts very differently with a ring that is stationary in this frame and a ring that is moving in this frame. The latter is length contracted, which can be seen from the narrow horizontal cross section of the brown region, but its interaction with the light is spread out in space and time (green diagonal). If you include reflections from the rings (pale red) they return to their respective centers and do so at the same proper time, which can be seen from the black dotted hyperbola.

 

[AlMetis]

The rings could be a circular arrangement of mirrors facing the source.
Each observer has its own set. It's a "circular light clock".

Thank you Rob, this is an excellent visual technique and very helpful in communicating what I have failed to do for days.
I have taken the liberty of using your design to demonstrate the mechanics I don’t understand.
The origin of the Green inertial frame is represented by the green dot.
The origin of the Blue inertial frame is represented by the blue dot.
Rob’s mirror rings are fixed around the origin of each frame, hence the reflected (top) light cones.
The light green cone that continues beyond the reflection shows the future without the mirror rings.

Diag. 1 is similar to Rob’s with the colours changed to match our Green and Blue examples.
It is assumed in these diagrams that the viewer takes the symmetry of the cone across the time axis and across the reflection axis (ring of mirrors) to indicate you, the viewer are at rest with the world line of the event 0. When/if the light cone of an inertial frame holds this same symmetry across both axis, it is at rest with the world line of the event 0.
As Rob said:
”At the event O when the world-lines meet, a flash of light begins to trace out the future light cone of event 0. It doesn't matter which world-line (if any) is "the source" of the light.
It is the flash from event O.” (my emphasis)

In diagram 2.0 Green is at rest with the world line of the event.
In diagram 3.0 Green is moving to the right of the world line of the event.
In diagram 4.0 Green and Blue are moving to the right of the world line event, with Blue moving faster than Green.
If you look at the difference between diagram 1.0 and 4.0, the symmetry of observations (the light cones) predicted after the event based on the symmetry of motion before the event, is determined by whether the motion of each frame is known/measured relative to the world line of the event, or relative to the motion each frame.Diagram 5.0 shows the predicted observations based soley on the symmetry of motion between Red, Green and Blue. But we can see from diagram 4.0 the same motion (one side only shown in 4.0) is not predicted if we know the motion of Green relative to the world line of event 0.

 

[robphy]

Thank you Rob, this is an excellent visual technique and very helpful in communicating what I have failed to do for days.
I have taken the liberty of using your design to demonstrate the mechanics I don’t understand.
[snip]

I haven't been following the conservation.
What is the specific question? (I couldn't find it in the previous post.)

 

[Ibix]

the world line of the event,

This is nonsense. Events don't have worldlines. Thinking that they do may be part of your problem.

 

[PeroK]

Thank you Rob, this is an excellent visual technique and very helpful in communicating what I have failed to do for days.

I have a question for you.

We have an electron gun that fires electrons at $0.8c$. Now, we have a reference frame where the gun is moving at $0.5c$ in the direction the gun is pointing. In this reference frame, the speed of the electrons is $0.5c + 0.8c = 1.3c$?

Why not?

 

[Ibix]

But we can see from diagram 4.0 the same motion (one side only shown in 4.0) is not predicted if we know the motion of Green relative to the world line of event 0.

Reading through this more closely, I really think your problem is "the worldline of the event". You seem to be using it to mean "the time axis of the arbitrary frame you chose to draw the diagram in". Of course you can't draw the diagram without choosing a frame, but that makes no difference to the physics and "worldline of the event" is meaningless.

The other thing you seem to do is draw different rings of mirrors in different states of motion and then observe that the results are different. That shouldn't really be surprising.

 

[AlMetis]

This is nonsense. Events don't have worldlines. Thinking that they do may be part of your problem.

I’m sorry, I am talking about the light cone of the flash form the event. Similar to world lines in Minkowski space, rest is a chosen frame, I as did Rob, chose the light cone of the event.