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

[Dale]

A more abstract approach is to start with the principle of relativity and show that there are only two coordinate transforms consistent with this - Galileo and Lorentz. There's no need to impose the invariance of light speed. Then you simply check by experiment which kind of universe we're in

Well, you do still need more than the first postulate. You can either use the second postulate or you can use experiment. But you need something more.

I do like postulating the invariance of the spacetime interval. That is one single postulate that actually does get you all the way. Of course, it isn’t so intuitive, but then again this is relativity we are talking about.

 

[Dale]

But we don’t regain the symmetry.
The symmetry is not exclusive to Galilean relativity, it is also predicted by Einsteinian relativity on the left and lost on the right.

Why did it disappear?

What symmetry? You must be misunderstanding something because all of the predicted symmetries of SR have been experimentally confirmed. So either you think some symmetry is predicted and it isn’t, or you are not actually looking at the symmetry that you think you are.

 

[AlMetis]

They give you the proper way to transform your view of reality in one inertial frame, to the correct view of reality in another inertial frame.

Yes, they do this by translating numerical values of a measure in one frame, to the numerical values of the same measure in another frame. Why do our measures of dimension change with our motion relative to another frame? Einstein explained “what” happens, time slows, length contracts, mass increases and when all of these are calculated via the Lorentz transformation they accurately predict the “amounts” that each change. The Lorentz transformation explains how to get there, it does not explain why we have to go there. What aspect of reality is defined in the need for the Lorentz equations? If they didn’t work and light was still a constant, reality would be different a very different thing.

 

[AlMetis]

What symmetry are you talking about?

I've marked the symmetry break in this copy.

 

[PeterDonis]

they do this by translating numerical values of a measure in one frame, to the numerical values of the same measure in another frame.

Lorentz transformations don't transform "measures", they transform coordinates.

What aspect of reality is defined in the need for the Lorentz equations?

The geometry of spacetime.

If they didn’t work and light was still a constant, reality would be different a very different thing.

You have no basis for this claim unless you have some alternate theory in which the transformations are different from the Lorentz transformations but the speed of light is still the same in all frames. And we already now such a theory is impossible if we accept the principle of relativity, because the POR only allows two possibilities, Galilean transformations (infinite invariant speed) and Lorentz transformations (finite invariant speed). In the former case, the speed of light is not the same in all frames. So your claim is assuming an impossibility.

I've marked the symmetry break in this copy.

Your diagram on the right is not correct for the SR case, so all you are showing is that Galilean relativity makes a wrong prediction.

 

[Dale]

I've marked the symmetry break in this copy.

I don’t get it. I don’t know what you are calling asymmetry there

 

[strangerep]

What aspect of reality is defined in the need for the Lorentz equations?

The principle that all inertial (i.e., non-accelerating) observers perceive the same laws of physics.

That, plus a couple of technical points about Lie groups implies a group of transformations among inertial reference frames that, in general, involve 2 universal constants: one has dimensions of inverse velocity squared (usually identified with $1/c^2$) and another with either dimensions of inverse time or else inverse length squared (the latter corresponding to the de Sitter group). The 2nd constant is usually ignored, leaving us with the Poincare group, of which the Lorentz transformations are a subgroup.

The value of the $1/c^2$ must then be determined from experiment.

 

[Ibix]

Why do our measures of dimension change with our motion relative to another frame?

On a Euclidean plane, why do the x and y separations between things change when you rotate coordinate axes? Because you've changed which directions in space you call the x axis and the y axis.

On a Minkowski plane, why do the x and t separations between things change when you boost coordinate axes? Because you've changed which directions in spacetime you call the x axis and the t axis.

That's really all that's going on here. Relativity allows you some freedom to choose which direction in spacetime to call "only moving in time" and "only moving in space". That choice affects how you choose to describe things in much the same way a Euclidean rotation might change a long narrow thing into a short wide one.

 

[AlMetis]

Lorentz transformations don't transform "measures", they transform coordinates.

How far apart are these coordinates?

 

[AlMetis]

The geometry of spacetime.

That’s true, I should have said the geometry of spacetime represented by the need to employ the Lorentz transformations between inertial frames expresses a continuum of fundamental dimension we don’t, but I would like to, understand.

 

[AlMetis]

You have no basis for this claim

The basis I have is they “do” work , light “is” a constant and reality “is” what it is.

 

[AlMetis]

Your diagram on the right is not correct for the SR case, so all you are showing is that Galilean relativity makes a wrong prediction.

Both sides are Galilean representations. The red line labeled ‘c_3v” on right side at t=1 represents the time dilated position of A relative to E (in principle, not to scale). If this is not correct, please explain why.

 

[jbriggs444]

How far apart are these coordinates?

What are you talking about?

There is a formula that we can use to determine "how far apart" two coordinate four-tuples are from one another. Let us identify two events $E_1$ and $E_2$ with coordinates $(x_1, y_1, z_1, t_1)$ and $(x_2, y_2, z_2, t_2)$ respectively. The "how far apart", $s$ between the two events is given by:$$s^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2 + (z_1 - z_2)^2 - (t_1 - t_2)^2$$If we shift to an a new reference frame where the same two events $E_1$ and $E_2$ are identified with coordinates $({x'}_1, {y'}_1, {z'}_1, {t'}_1)$ and $({x'}_2, {y'}_2, {z'}_2, {t'}_2)$ then the same formula applies:$${s'}^2 = ({x'}_1 - {x'}_2)^2 + ({y'}_1 - {y'}_2)^2 + ({z'}_1 - {z'}_2)^2 - ({t'}_1 - {t'}_2)^2$$Further, it will turn out to be the case that $s^2 = s'^2$.

So "how far apart" is an invariant measure. It looks like it is a function of coordinates, but that is just because we specified our events with coordinates. It is actually a function that takes two events as input and spits out a [squared] separation as the output.

This "how far apart" formula is the "metric" for the flat space-time of special relativity. When you get to the curved space-times that general relativity can contemplate, the metric will not have this exact form. Though in the limit of increasingly nearby events, it will approximate this form ever more closely.

If you are evaluating the separation between the two ends of a physical rod, then you are asking for the separation between the world-line corresponding to the one end and the world-line corresponding to the other end. That is not the separation between two events -- it is the separation between two trajectories.

In order to apply the "how far apart" formula, you need to identify two events. We conventionally pick out two events that are "at the same time" according to some agreed-upon simultaneity convention.

Each frame has its own simultaneity convention. So each frame will be comparing different pairs of events when they measure the separation between two trajectories.

The two frames are not measuring the same thing. They are measuring different frame-specific hings. This is the same way that the measured width of a strip of paper depends on the angle at which you hold the ruler.

 

[AlMetis]

I don’t get it. I don’t know what you are calling asymmetry there

You and a number of people have said I have unknowingly switched frames of reference resulting in me mistaking the aberration observed in a “different” frame as a change in the observations in the same frame.
From everything posted, this seems to be the most consistent advice.
I need to understand at what point in time, I have switched frames of reference in context of what I think is being observed, in order to see/understand what would be observed.
So please indulge me one step at a time.

In the rest frame of the light source S and a mirror M at rest with S a fixed distance on +y, the path of the light pulse is observed perpendicular to x between emission at t=0 and reflection t=1, while S is in motion along x in the +x direction relative to the frame E.
Then
1. E will see the aberration of the light pulse from t=0 to t=1 as a path at an angle from x toward -x.
(when increasing x is in the right hand direction relative to you, the reader)Yes, no?

 

[Dale]

1. E will see the aberration of the light pulse from t=0 to t=1 as a path at an angle from x toward -x.
(when increasing x is in the right hand direction relative to you, the reader)Yes, no?

Yes. The worldline of the light in $S$ is $$r^\mu=(t,x,y,z)=(t,0,t,0)$$ and in $E$ is $$r^{\mu'}=(t',x',y',z')=(t',-t' v, t' \sqrt{1-v^2},0)$$ using units where $c=1$.

 

[PeroK]

In the rest frame of the light source S and a mirror M at rest with S a fixed distance on +y, the path of the light pulse is observed perpendicular to x between emission at t=0 and reflection t=1, while S is in motion along x in the +x direction relative to the frame E.
Then
1. E will see the aberration of the light pulse from t=0 to t=1 as a path at an angle from x toward -x.
(when increasing x is in the right hand direction relative to you, the reader)Yes, no?

Probably yes. I would have said simply:

In a frame where the source and mirror are at rest, the light path is along the positive y axis.

In a frame, E, where the source and mirror are moving to the right (along the positive x axis), then the light path is at an angle to the x axis.

In terms of aberration, this means that when the light is observed in frame E, the source is further to the right than it appears when the light is received.

 

[AlMetis]

Yes.

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?

 

[Dale]

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. Another way of stating this is that the laws of physics, both the differential equations and any associated constants, are the same in all inertial frames.

 

[PeterDonis]

How far apart are these coordinates?

The question makes no sense. You can ask what the spacetime interval is between two events (points in spacetime), but that's all.

I should have said the geometry of spacetime represented by the need to employ the Lorentz transformations between inertial frames expresses a continuum of fundamental dimension we don’t, but I would like to, understand.

Why do you think we don't understand it? You should not generalize from your own experience, because you evidently don't have a good grasp of SR, but that doesn't mean nobody does.

The basis I have is they “do” work , light “is” a constant and reality “is” what it is.

None of this justifies the claim I responded to. I gave you reasons why your claim was not justified. You haven't responded to those reasons at all.

Both sides are Galilean representations.

Which means, as I said, that they tell us nothing about SR. But you said your claim about some symmetry not being there after switching frames applies to SR. What are you basing that on?

 

[AlMetis]

Yes.

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?

 

[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.