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

[AlMetis]

Why not?

Because you just said the electrons are fired at 0.8c.
If I can measure where the electron left the gun and time of its travel over a fixed distance, I will find it to be 0.8c.

 

[AlMetis]

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

 

[PeterDonis]

@AlMetis in your post #122, there is no asymmetry. The motion of Red and Blue is equal and opposite relative to Green in every frame. You have just chosen to draw your diagram 2 in a frame other than Green's rest frame, which makes things look "asymmetrical" even though they are not. The physical fact is that the motion is symmetrical relative to Green, and so the best frame to view the motion is Green's rest frame, as in diagram 1. But that doesn't mean the symmetry goes away in some other frame; it just means you've picked a bad frame to view it in.

 

[PeroK]

Because you just said the electrons are fired at 0.8c.
If I can measure where the electron left the gun and time of its travel over a fixed distance, I will find it to be 0.8c.

I was hoping for a serious answer.

 

[Ibix]

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.

This still doesn't make sense. The light cone doesn't form a standard of rest. It's always the same cone in all frames. And it isn't a worldline either.

I still don't understand what your problem is. You seem to be able to draw accurate diagrams, you just don't seem to believe them, or you see some contradiction between them.

 

[Dale]

View attachment 322741View attachment 322742

What is the SPECIFIC symmetry and the asymmetry you are asking about here?

I have asked a long time ago for you to identify the specific symmetry, and you still have not done so. You seemed to be making progress in that direction with your points 1-5. But after 5 you jumped to these diagrams. That is fine, but I still don’t know what is the symmetry in question:

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.

From my perspective we have not made it past my post 72, now more than 50 posts ago.

(For the future, please don’t post images of text. The figures are great, but the images of text are unquotable)

By the way, in your second figure the velocities are wrong. You need to use relativistic velocity addition.

 

[AlMetis]

But that doesn't mean the symmetry goes away in some other frame; it just means you've picked a bad frame to view it in.

Thank you PeterDonis

I think you have just made it clear to me why everyone here keeps asking me the same question. I thought we were on the same page and looking for where I’ve misunderstood SR. But it now appears we are reading different books.
All of which makes me that much more grateful for everyone’s help and patients.

I agree, the relative motion of Red, Green and Blue remains constant, that is crucial to my question.
If I am free to choose any rest frame (inertial frame) from which to define the laws, and I choose the rest frame of Green which sets the motion of Red and Blue identical relative to Green, how do I explain the change in the time of light between Red and Green, and Blue and Green which is clearly shown in the light cone diagrams?

 

[Dale]

how do I explain the change in the time of light between Red and Green, and Blue and Green which is clearly shown in the light cone diagrams?

It is not clearly shown to me. What do you mean by "the change in the time of light"?

 

[AlMetis]

This still doesn't make sense. The light cone doesn't form a standard of rest. It's always the same cone in all frames. And it isn't a worldline either.

As I understand it, a light cone is a causal (light speed) limit representing the spatial separation on the horizontal axis x and time on the vertical axis y. If I diagram the spatial separation as constant on x, over increasing y, I have set the viewer at rest with the event with respect to the causal horizon of the event.

 

[Ibix]

how do I explain the change in the time of light between Red and Green, and Blue and Green which is clearly shown in the light cone diagrams?

You mean, in the rest frame of green, why does it take longer for the reflections to reach red and blue? Because red and blue are moving and it takes more time for the light to reach them because it has further to go.

In the rest frame of red, it takes longer for the light to reach green and even longer to reach blue, and for the same reason.

 

[Ibix]

As I understand it, a light cone is a causal (light speed) limit representing the spatial separation on the horizontal axis x and time on the vertical axis y. If I diagram the spatial separation as constant on x, over increasing y, I have set the viewer at rest with the event with respect to the causal horizon of the event.

None of that makes sense. A light cone is a causal boundary - events in its interior or on it can be affected by the event at its tip, events outside cannot. You can't be at rest with respect to one.

 

[PeroK]

By the way, in your second figure the velocities are wrong. You need to use relativistic velocity addition.

Look at the above posts:

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?

Because you just said the electrons are fired at 0.8c.
If I can measure where the electron left the gun and time of its travel over a fixed distance, I will find it to be 0.8c.

This suggests to me that the OP needs to learn more about the basics of SR. We are now 130+ posts down a rabbit hole, with little or no prospect of ever emerging into daylight.

 

[AlMetis]

It is not clearly shown to me. What do you mean by "the change in the time of light"?

The diagrams show the time of light from Green to Red is longer (higher on the time axis) than the time of light from Green to Blue.

 

[AlMetis]

You mean, in the rest frame of green, why does it take longer for the reflections to reach red and blue? Because red and blue are moving and it takes more time for the light to reach them because it has further to go.

In the rest frame of Green it takes longer for the unreflected light from Green to reach Red than it does to reach Blue.

 

[AlMetis]

You can't be at rest with respect to one.

How should I refer to a frame that remains at a constant distance from the boundary?

 

[DrGreg]

How should I refer to a frame that remains at a constant distance from the boundary?

First of all, a frame is everywhere. I presume you mean "the spatial origin of a frame".

The boundary is expanding at the speed of light, so no origin of a frame can be a constant distance from something moving at the speed of light.

 

[Dale]

The diagrams show the time of light from Green to Red is longer (higher on the time axis) than the time of light from Green to Blue.

OK, so on your diagram I have labeled four events $E_1=(t(E_1),x(E_1))$, etc.

If I understand you correctly the asymmetry that you are worried about is that $t(E_2)-t(E_1)<t(E_4)-t(E_3)$. Is that a correct understanding of the asymmetry you are looking at?

If not then please be explicit what it is. Preferably in less than 50 posts more.

 

[robphy]

In the last diagram of @AlMetis, the green event should not be collinear with the blue and red events (what @Dale labels as E2 and E4),
if that green event is the same event in the first spacetime-diagram of that series of diagrams where the three events lie on a common hyperbola... which means that those time-intervals from the meeting event are equal in magnitude).

 

[PeterDonis]

a frame that remains at a constant distance from the boundary?

If by "the boundary" you mean the light cone, there is no such frame, because there is no such thing as "constant distance from the light cone".

 

[PeterDonis]

As I understand it, a light cone is a causal (light speed) limit

No, it's the set of curves through a given point that are null, i.e., that are possible worldlines for light rays.

representing the spatial separation on the horizontal axis x and time on the vertical axis y.

That's not a light cone, that's a spacetime diagram.

If I diagram the spatial separation as constant on x

What spatial separation?

I have set the viewer at rest with the event with respect to the causal horizon of the event.

This is word salad. It makes no sense.

 

[AlMetis]

If I understand you correctly the asymmetry that you are worried about is that t(E2)−t(E1)<t(E4)−t(E3). Is that a correct understanding of the asymmetry you are looking at?

Yes.
With one correction, that does not change your question or my answer.
As robphy points out I have the green too far up the timeline, it should be below the blue.

 

[Dale]

Yes.
With one correction, that does not change your question or my answer.
As robphy points out I have the green too far up the timeline, it should be below the blue.

OK, then the answer is

... you think some symmetry is predicted and it isn’t, ...

Relativity does not predict that $t(E_2)-t(E_1)=t(E_4)-t(E_3)$ in all frames

 

[AlMetis]

Relativity does not predict that t(E2)−t(E1)=t(E4)−t(E3) in all frames

I am not saying it does.

I am saying that what relativity does predict will be observed when at rest with Green, t ( E 2 ) − t ( E 1 ) = t ( E 4 ) − t ( E 3 ) in diagram 1. changes when the information not available in diagram 1 is revealed in diagram 2. when t ( E 2 ) − t ( E 1 ) ≠ t ( E 4 ) − t ( E 3 ).

We are still at rest with Green and the symmetry of the motion between Red, Green and Blue still exists, but the time of the light pulse, i.e. the distance travelled observed at rest with Green, is not what is predicted from the information available in diagram 1.

I thought about this last night and realized there is a much more familiar way to pose my question.
I’ll post it shortly.

 

[Dale]

I am not saying it does.

Yes, you did:

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?

(emphasis added)

I don’t need you to defend your words, but that was my reason to be involved in the thread.

the information not available in diagram 1 is revealed in diagram 2

The amount of information is the same in both diagrams. The Lorentz transform is a lossless transformation in terms of information. The representation is different, but the amount of information is the same. There is no unavailable information that is revealed.

I thought about this last night and realized there is a much more familiar way to pose my question.
I’ll post it shortly.

I think not. This is useless. We are nearly 150 posts in and you are still trying to formulate your question. Nobody coming to this thread later will know to look on page 5 for the question.

I am going to close this thread. Please do not post your new question shortly. There are only a few days left in the month. Take that time to formulate the question well and do not post a new thread until the beginning of next month.

When you do, please label all important events, worldlines, measures, and frames. Do not refer to any frame variant quantity (speed, distance, time) without explicitly mentioning the frame. And use LaTeX to write symbols.