I've tried several variations of this scenario and I seem to be getting contradictory results. Please explain how this and other similar scenarios work with Einstein's concepts.

There are two long rods, R1 and R2 moving relative to each other parallel to the x-axis with relative velocity |V| = c*sqrt(3)/2. Each rod as measured in their own rest frame has length L = sqrt(3). Rod R1 is just above the x-axis and rod R2 is just below the x-axis. Every point along each rod has a laser that can send out photons in the y-direction perpendicular to the x-axis as measured in their respective frames. The lasers on R1 point in the negative y-direction and the lasers on R2 point in the positive y-direction. So if the two rods are passing each other, photons from rod R1 will hit R2 and vice versa if the laser is turned on. If photons from one rod hit the other rod, they will leave markings in the rod at the point where the emitted photon hits the other rod. Rod R1 has more powerful lasers than R2. If a point on R2 is hit for one nanosecond or more by photons from R1, the laser at that point on R2 stops working.
When the rods meet at the same x coordinate, clocks on R1 all read zero as measured in the rest frame of R1, and clocks on R2 all read zero as measured in the rest frame of R2. R1 is setup so when the clocks on R1 all read 1 second minus 1 nanosecond, all the lasers on R1 are simultaneously turned on for 2 nanoseconds. When this occurs, per Einstein's concepts, due to length contraction, the entire length of rod R2 extends half the length of rod R1 (plus a few meters due to the lasers being on for 2 nanoseconds). Therefore, all the lasers on rod R2 are disabled when rod R2 has traveled just over half the length of R1, so no markings from the lasers on R2 show up on half the length of rod R1.
The lasers on R2 are setup differently. The lasers along every point of R2 are always turned on. So when rods R1 and R2 meet at the same x coordinate the lasers on R2 begin marking each point along rod R1.
Please explain why in this scenario, the lasers on R2 can only mark half the length of rod R1.
Thanks,
David Seppala
Bastrop TX

On 30-Jul-21 2:23 pm, sepp623@yahoo.com wrote:
> I've tried several variations of this scenario and I seem to be getting contradictory results. Please explain how this and other similar scenarios work with Einstein's concepts.
>
> There are two long rods, R1 and R2 moving relative to each other parallel to the x-axis with relative velocity |V| = c*sqrt(3)/2. Each rod as measured in their own rest frame has length L = sqrt(3). Rod R1 is just above the x-axis and rod R2 is just below the x-axis. Every point along each rod has a laser that can send out photons in the y-direction perpendicular to the x-axis as measured in their respective frames. The lasers on R1 point in the negative y-direction and the lasers on R2 point in the positive y-direction. So if the two rods are passing each other, photons from rod R1 will hit R2 and vice versa if the laser is turned on. If photons from one rod hit the other rod, they will leave markings in the rod at the point where the emitted photon hits the other rod. Rod R1 has more powerful lasers than R2. If a point on R2 is hit for one nanosecond or more by photons from R1, the laser at that point on R2 stops working.
> When the rods meet at the same x coordinate, clocks on R1 all read zero as measured in the rest frame of R1, and clocks on R2 all read zero as measured in the rest frame of R2. R1 is setup so when the clocks on R1 all read 1 second minus 1 nanosecond, all the lasers on R1 are simultaneously turned on for 2 nanoseconds. When this occurs, per Einstein's concepts, due to length contraction, the entire length of rod R2 extends half the length of rod R1 (plus a few meters due to the lasers being on for 2 nanoseconds). Therefore, all the lasers on rod R2 are disabled when rod R2 has traveled just over half the length of R1, so no markings from the lasers on R2 show up on half the length of rod R1.
> The lasers on R2 are setup differently. The lasers along every point of R2 are always turned on. So when rods R1 and R2 meet at the same x coordinate the lasers on R2 begin marking each point along rod R1.
> Please explain why in this scenario, the lasers on R2 can only mark half the length of rod R1.
> Thanks,
> David Seppala
> Bastrop TX
>

Express it in coordinate terms, then apply the Lorentz transformation.
Show your working.

Sylvia.

On Thursday, July 29, 2021 at 11:23:28 PM UTC-5, sep...@yahoo.com wrote:
> I've tried several variations of this scenario and I seem to be getting contradictory results. Please explain how this and other similar scenarios work with Einstein's concepts.
>
> There are two long rods, R1 and R2 moving relative to each other parallel to the x-axis with relative velocity |V| = c*sqrt(3)/2. Each rod as measured in their own rest frame has length L = sqrt(3). Rod R1 is just above the x-axis and rod R2 is just below the x-axis. Every point along each rod has a laser that can send out photons in the y-direction perpendicular to the x-axis as measured in their respective frames. The lasers on R1 point in the negative y-direction and the lasers on R2 point in the positive y-direction. So if the two rods are passing each other, photons from rod R1 will hit R2 and vice versa if the laser is turned on. If photons from one rod hit the other rod, they will leave markings in the rod at the point where the emitted photon hits the other rod. Rod R1 has more powerful lasers than R2. If a point on R2 is hit for one nanosecond or more by photons from R1, the laser at that point on R2 stops working.
> When the rods meet at the same x coordinate, clocks on R1 all read zero as measured in the rest frame of R1, and clocks on R2 all read zero as measured in the rest frame of R2. R1 is setup so when the clocks on R1 all read 1 second minus 1 nanosecond, all the lasers on R1 are simultaneously turned on for 2 nanoseconds. When this occurs, per Einstein's concepts, due to length contraction, the entire length of rod R2 extends half the length of rod R1 (plus a few meters due to the lasers being on for 2 nanoseconds). Therefore, all the lasers on rod R2 are disabled when rod R2 has traveled just over half the length of R1, so no markings from the lasers on R2 show up on half the length of rod R1.
> The lasers on R2 are setup differently. The lasers along every point of R2 are always turned on. So when rods R1 and R2 meet at the same x coordinate the lasers on R2 begin marking each point along rod R1.
> Please explain why in this scenario, the lasers on R2 can only mark half the length of rod R1.
> Thanks,
> David Seppala
> Bastrop TX
Let me make things slightly different, that doesn't change things for rod R1. When the clocks on R1 all read one second, the lasers along the first half length of R1 plus delta are turned on for 2 nanoseconds. The others remain off throughout the scenario.
David Seppala
Bastrop TX

On Thursday, July 29, 2021 at 10:43:24 PM UTC-7, sep...@yahoo.com wrote:
> > I've tried several variations of this scenario and I seem to be getting contradictory results.

What you've described (in overly convoluted terms) is nothing but the trivial barn pole scenario, and as always you are failing to correctly account for the relativity of simultaneity. You should follow Sylvia's advice.

> > The lasers on R2 are setup differently. The lasers along every point of R2 are always
> > turned on. So when rods R1 and R2 meet at the same x coordinate the lasers on R2
> > begin marking each point along rod R1. Please explain why in this scenario, the lasers
> > on R2 can only mark half the length of rod R1.

The whole "2 nanosecond" thing is needlessly elaborate. Just say that momentary illumination from R1 disables the adjacent light on R2. Let S1 and S2 denote inertial coordinate systems in which R1 and R2 respectively are at rest, and let the rods first begin to overlap at the common origin of S1 and S2. In terms of S1 the lights on R2 are all disabled along the locus of simultaneous events extending linearly from x=0,t=1 to x=sqrt(3)/2, t=1. This disables all of the lights on R2. In terms of S2, the lights of R2 are disabled along the locus of non-simultaneous events extending linearly between x=-sqrt(3), t=2 and x=0, t=1/2. This extinguishes all the lights of R2, and it proceeds sequentially as R1 sweeps past R2, always extinguishing the lights of R2 before the trailing half of R1 reaches them. Understand?

On Friday, July 30, 2021 at 3:02:14 AM UTC-5, Al Coe wrote:
> On Thursday, July 29, 2021 at 10:43:24 PM UTC-7, sep...@yahoo.com wrote:
> > > I've tried several variations of this scenario and I seem to be getting contradictory results.
> What you've described (in overly convoluted terms) is nothing but the trivial barn pole scenario, and as always you are failing to correctly account for the relativity of simultaneity. You should follow Sylvia's advice.
> > > The lasers on R2 are setup differently. The lasers along every point of R2 are always
> > > turned on. So when rods R1 and R2 meet at the same x coordinate the lasers on R2
> > > begin marking each point along rod R1. Please explain why in this scenario, the lasers
> > > on R2 can only mark half the length of rod R1.
> The whole "2 nanosecond" thing is needlessly elaborate. Just say that momentary illumination from R1 disables the adjacent light on R2. Let S1 and S2 denote inertial coordinate systems in which R1 and R2 respectively are at rest, and let the rods first begin to overlap at the common origin of S1 and S2. In terms of S1 the lights on R2 are all disabled along the locus of simultaneous events extending linearly from x=0,t=1 to x=sqrt(3)/2, t=1. This disables all of the lights on R2. In terms of S2, the lights of R2 are disabled along the locus of non-simultaneous events extending linearly between x=-sqrt(3), t=2 and x=0, t=1/2. This extinguishes all the lights of R2, and it proceeds sequentially as R1 sweeps past R2, always extinguishing the lights of R2 before the trailing half of R1 reaches them. Understand?

I posted a simple variation of the scenario, called "Who wins - relativity contradiction" Please explain that scenario.
David Seppala
Bastrop TX

On Friday, July 30, 2021 at 6:26:37 AM UTC-7, sep...@yahoo.com wrote:
> > > > I've tried several variations of this scenario and I seem to be getting contradictory results.
> > What you've described (in overly convoluted terms) is nothing but the trivial barn pole scenario, and as always you are failing to correctly account for the relativity of simultaneity. You should follow Sylvia's advice.
> > > > The lasers on R2 are setup differently. The lasers along every point of R2 are always
> > > > turned on. So when rods R1 and R2 meet at the same x coordinate the lasers on R2
> > > > begin marking each point along rod R1. Please explain why in this scenario, the lasers
> > > > on R2 can only mark half the length of rod R1.
> > The whole "2 nanosecond" thing is needlessly elaborate. Just say that momentary illumination from R1 disables the adjacent light on R2. Let S1 and S2 denote inertial coordinate systems in which R1 and R2 respectively are at rest, and let the rods first begin to overlap at the common origin of S1 and S2. In terms of S1 the lights on R2 are all disabled along the locus of simultaneous events extending linearly from x=0,t=1 to x=sqrt(3)/2, t=1. This disables all of the lights on R2. In terms of S2, the lights of R2 are disabled along the locus of non-simultaneous events extending linearly between x=-sqrt(3), t=2 and x=0, t=1/2. This extinguishes all the lights of R2, and it proceeds sequentially as R1 sweeps past R2, always extinguishing the lights of R2 before the trailing half of R1 reaches them.. Understand?
>
> I posted a simple variation of the scenario, called "Who wins - relativity contradiction" Please explain that scenario.

Didn't you forget something? You posed a question, and I answered it. The normal decent human being would say "Okay, I understand, thanks."

By the way, for those who are keeping score, this is the 784th challenge to the consistency of special relativity that Barnpole Dave has posed, and so far the score is Special Relativity 784, Dave "Barnpole" Seppala 0. In 784 of those cases, Dave's mistake was failing to correctly account for the relativity of simultaneity. Well, you know what they say: The 785th time is the charm!

On Thursday, July 29, 2021 at 9:23:28 PM UTC-7, sep...@yahoo.com wrote:
> I've tried several variations of this scenario and I seem to be getting contradictory results. Please explain how this and other similar scenarios work with Einstein's concepts.
>
> There are two long rods, R1 and R2 moving relative to each other parallel to the x-axis with relative velocity |V| = c*sqrt(3)/2. Each rod as measured in their own rest frame has length L = sqrt(3). Rod R1 is just above the x-axis and rod R2 is just below the x-axis. Every point along each rod has a laser that can send out photons in the y-direction perpendicular to the x-axis as measured in their respective frames. The lasers on R1 point in the negative y-direction and the lasers on R2 point in the positive y-direction. So if the two rods are passing each other, photons from rod R1 will hit R2 and vice versa if the laser is turned on. If photons from one rod hit the other rod, they will leave markings in the rod at the point where the emitted photon hits the other rod. Rod R1 has more powerful lasers than R2. If a point on R2 is hit for one nanosecond or more by photons from R1, the laser at that point on R2 stops working.
> When the rods meet at the same x coordinate, clocks on R1 all read zero as measured in the rest frame of R1, and clocks on R2 all read zero as measured in the rest frame of R2. R1 is setup so when the clocks on R1 all read 1 second minus 1 nanosecond, all the lasers on R1 are simultaneously turned on for 2 nanoseconds. When this occurs, per Einstein's concepts, due to length contraction, the entire length of rod R2 extends half the length of rod R1 (plus a few meters due to the lasers being on for 2 nanoseconds). Therefore, all the lasers on rod R2 are disabled when rod R2 has traveled just over half the length of R1, so no markings from the lasers on R2 show up on half the length of rod R1.
> The lasers on R2 are setup differently. The lasers along every point of R2 are always turned on. So when rods R1 and R2 meet at the same x coordinate the lasers on R2 begin marking each point along rod R1.
> Please explain why in this scenario, the lasers on R2 can only mark half the length of rod R1.
> Thanks,
> David Seppala
> Bastrop TX

Have we ever measured distance to go away?
Then what is the evidence for that being real?
It doesn't belong instead.

Either atoms contract or they do not.
and if they do their orbital distortion would cause the
atoms to fail in chemistry.

Mitchell Raemsch

sepp623@yahoo.com wrote:

> I posted a simple variation of the scenario, called "Who wins -
> relativity contradiction" Please explain that scenario.
> David Seppala Bastrop TX

here we go, then

Nurse Warns - Stay away from Vaxxed Zombies!
https://www.bitchute.com/video/VXFU03p9Imw8/

On Friday, July 30, 2021 at 1:48:44 PM UTC-7, mitchr...@gmail.com wrote:
> On Thursday, July 29, 2021 at 9:23:28 PM UTC-7, sep...@yahoo.com wrote:
> > I've tried several variations of this scenario and I seem to be getting contradictory results. Please explain how this and other similar scenarios work with Einstein's concepts.
> >
> > There are two long rods, R1 and R2 moving relative to each other parallel to the x-axis with relative velocity |V| = c*sqrt(3)/2. Each rod as measured in their own rest frame has length L = sqrt(3). Rod R1 is just above the x-axis and rod R2 is just below the x-axis. Every point along each rod has a laser that can send out photons in the y-direction perpendicular to the x-axis as measured in their respective frames. The lasers on R1 point in the negative y-direction and the lasers on R2 point in the positive y-direction. So if the two rods are passing each other, photons from rod R1 will hit R2 and vice versa if the laser is turned on. If photons from one rod hit the other rod, they will leave markings in the rod at the point where the emitted photon hits the other rod. Rod R1 has more powerful lasers than R2. If a point on R2 is hit for one nanosecond or more by photons from R1, the laser at that point on R2 stops working.
> > When the rods meet at the same x coordinate, clocks on R1 all read zero as measured in the rest frame of R1, and clocks on R2 all read zero as measured in the rest frame of R2. R1 is setup so when the clocks on R1 all read 1 second minus 1 nanosecond, all the lasers on R1 are simultaneously turned on for 2 nanoseconds. When this occurs, per Einstein's concepts, due to length contraction, the entire length of rod R2 extends half the length of rod R1 (plus a few meters due to the lasers being on for 2 nanoseconds). Therefore, all the lasers on rod R2 are disabled when rod R2 has traveled just over half the length of R1, so no markings from the lasers on R2 show up on half the length of rod R1.
> > The lasers on R2 are setup differently. The lasers along every point of R2 are always turned on. So when rods R1 and R2 meet at the same x coordinate the lasers on R2 begin marking each point along rod R1.
> > Please explain why in this scenario, the lasers on R2 can only mark half the length of rod R1.
> > Thanks,
> > David Seppala
> > Bastrop TX
> Have we ever measured distance to go away?
> Then what is the evidence for that being real?
> It doesn't belong instead.
>
> Either atoms contract or they do not.
> and if they do their orbital distortion would cause the
> atoms to fail in chemistry.
>
>
> Mitchell Raemsch

Where has science measured size contraction?
Where is that evidence?

On 30-Jul-21 2:23 pm, sepp623@yahoo.com wrote:
> I've tried several variations of this scenario and I seem to be getting contradictory results. Please explain how this and other similar scenarios work with Einstein's concepts.
>
> There are two long rods, R1 and R2 moving relative to each other parallel to the x-axis with relative velocity |V| = c*sqrt(3)/2. Each rod as measured in their own rest frame has length L = sqrt(3). Rod R1 is just above the x-axis and rod R2 is just below the x-axis. Every point along each rod has a laser that can send out photons in the y-direction perpendicular to the x-axis as measured in their respective frames. The lasers on R1 point in the negative y-direction and the lasers on R2 point in the positive y-direction. So if the two rods are passing each other, photons from rod R1 will hit R2 and vice versa if the laser is turned on. If photons from one rod hit the other rod, they will leave markings in the rod at the point where the emitted photon hits the other rod. Rod R1 has more powerful lasers than R2. If a point on R2 is hit for one nanosecond or more by photons from R1, the laser at that point on R2 stops working.
> When the rods meet at the same x coordinate, clocks on R1 all read zero as measured in the rest frame of R1, and clocks on R2 all read zero as measured in the rest frame of R2. R1 is setup so when the clocks on R1 all read 1 second minus 1 nanosecond, all the lasers on R1 are simultaneously turned on for 2 nanoseconds. When this occurs, per Einstein's concepts, due to length contraction, the entire length of rod R2 extends half the length of rod R1 (plus a few meters due to the lasers being on for 2 nanoseconds). Therefore, all the lasers on rod R2 are disabled when rod R2 has traveled just over half the length of R1, so no markings from the lasers on R2 show up on half the length of rod R1.
> The lasers on R2 are setup differently. The lasers along every point of R2 are always turned on. So when rods R1 and R2 meet at the same x coordinate the lasers on R2 begin marking each point along rod R1.
> Please explain why in this scenario, the lasers on R2 can only mark half the length of rod R1.
> Thanks,
> David Seppala
> Bastrop TX
>

Why do you keep doing this? Your supposed contradictions are always due
to errors in your analysis.

Sylvia.