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

[cfrogue]

Yes, that is correct.

Also correct.

Do you also agree, O' sees its points hit at the same time when
<br /> t = \frac{d}{2c} {\sqrt{\frac{c+v}{c-v}}<br />

in the time coordinates of O?

 

[atyy]

Do you also agree, O' sees its points hit at the same time when
<br /> t = \frac{d}{2c} {\sqrt{\frac{c+v}{c-v}}<br />

in the time coordinates of O?

"O' sees its points hit" means that there are two events, one photon hitting one endpoint of O', another photon hitting the other endpoint of O'

"at the same time" means that O' assigns both events the same t' coordinate

"when t" means you are asking for both events to be assigned the same t coordinate, which is simply impossible by the relativity of simultaneity.

 

[Dale]

Do you also agree, O' sees its points hit at the same time when
<br /> t = \frac{d}{2c} {\sqrt{\frac{c+v}{c-v}}<br />

in the time coordinates of O?

No.

<br /> \frac{d}{2c} {\sqrt{\frac{c+v}{c-v}} = \frac{d}{2 \gamma (c-v)} = t_{R&#039;}<br /><br /> \frac{d}{2c} {\sqrt{\frac{c+v}{c-v}} \neq \frac{d}{2c} {\sqrt{\frac{c-v}{c+v}} = \frac{d}{2 \gamma (c+v)} = t_{L&#039;}<br />

Your equation is only an equation for t_R', and is not correct for t_L'

 

[cfrogue]

"O' sees its points hit" means that there are two events, one photon hitting one endpoint of O', another photon hitting the other endpoint of O'

"at the same time" means that O' assigns both events the same t' coordinate

"when t" means you are asking for both events to be assigned the same t coordinate, which is simply impossible by the relativity of simultaneity.

OK, R of S says light will hit right endpoint of O' in O at

<br /> t_R = \frac{d}{2\gamma(c-v)} = \frac{d}{2c} {\sqrt{\frac{c+v}{c-v}}<br />


Is this correct?

 

[cfrogue]

No.

<br /> \frac{d}{2c} {\sqrt{\frac{c+v}{c-v}} = \frac{d}{2 \gamma (c-v)} = t_{R&#039;}<br />


<br /> \frac{d}{2c} {\sqrt{\frac{c+v}{c-v}} \neq \frac{d}{2c} {\sqrt{\frac{c-v}{c+v}} = \frac{d}{2 \gamma (c+v)} = t_{L&#039;}<br />

Your equation is only an equation for t_R', and is not correct for t_L'

Yes, from LT I think we need only do the following.

t' = d/(2c), x = -(d/2), (the left endpoint is at -(d/2) relative to the origin of O')

t = ( t' + vx/c^2 )λ

t = ( t' - vd/(2c^2) )λ

t = ( d/(2c)- vd/(2c^2) )λ


<br /> t = \frac{d}{2c} {\sqrt{\frac{c-v}{c+v}} <br />

 

[atyy]

I don't know if this will help, but even more fundamental than the relativity of simultaneity is the conventionality of simultaneity. That is, there is nothing real to O' saying that a photon hit the left end of his rod and another photon the other end of his rod "at the same time". There is no real "at the same time" for events that happen at different places. There is only a real "at the same time" for events that happen at the same place, which is really just a way of saying that only events, only the intersections of worldlines are real.

This seems especially clear to any musician who's played in an ensemble. We want different notes by different musicians in different locations to be played "at the same time". If the musicians are far apart, and musician A plays a note and musician B plays a note, and they sound "at the same time" for musician A, then the notes cannot be "at the same time" for musician B, because B must have struck his note earlier than musican A, so that although B's note travels a greater distance to A's ear than A's note travels to A's ear, both notes hit A's ear at the same time - but certainly the notes are not "at the same time" for B who struck his note earlier. Thus music ensembles show that "at the same time" is nonsense (at best an approximation) if things occur in different places, and it's only exact for things in the same place.

So when we say two different events happened at the same time for O', that is nothing real - that happens because of a particular clock synchronization *convention* use to define an inertial frame. If two photons hitting the left and right ends of O's rod "at the same time" for O' is not even real for O', then it really is even less sensible to ask that they be real for O.

 

[Dale]

Yes, from LT I think we need only do the following.

t' = d/(2c), x = -(d/2), (the left endpoint is at -(d/2) relative to the origin of O')

t = ( t' + vx/c^2 )λ

t = ( t' - vd/(2c^2) )λ

t = ( d/(2c)- vd/(2c^2) )λ


<br /> t = \frac{d}{2c} {\sqrt{\frac{c-v}{c+v}} <br />

Yes, for the left endpoint only.

 

[cfrogue]

Yes, for the left endpoint only.

I should have said t_L, but agreed.

 

[cfrogue]

OK, so, O calls O' on a light phone and asks the time in which the points were struck at the same time in its frame.

What does O' say?

I bet O' says it reads t' = d/(2c).

Then O calculates and concludes, O' saw its points hit at the same time at

<br /> t = \frac{d}{2c} {\sqrt{\frac{c+v}{c-v}}<br />

or should O use time dilation and conclude

t = d/(2λc).

 

[atyy]

OK, so, O calls O' on a light phone and asks the time in which the points were struck at the same time in its frame.

What does O' say?

I bet O' says it reads t' = d/(2c).

Then O calculates and concludes, O' saw its points hit at the same time at

<br /> t = \frac{d}{2c} {\sqrt{\frac{c+v}{c-v}}<br />

or should O use time dilation and conclude

t = d/(2λc).

O must also ask O' where the light struck the rods at the same t' coordinate. The different x' coordinates for the two events lead O to conclude they occurred at different t coordinates.

 

[cfrogue]

Also, using LT, we are able to conclude the points of simultaneity are located at

vt - d/(2λ) for the left points and the right points is at vt + d/(2λ)

These are therefore, the endpoints of the light sphere in the coords of O.

Also, the center of the lights sphere of O' is located at the emission point in O' which for any time t is located at vt in the coords of O.

Is this correct?

 

[cfrogue]

O must also ask O' where the light struck the rods at the same t' coordinate. The different x' coordinates for the two events lead O to conclude they occurred at different t coordinates.

Well, O' only has one answer. What should I tell him? He is waiting on the line? Just a joke.

O' says the points were struck at d/(2c) and at -d/(2c) at time t' = d/(2c).

Do you agree O' is correct about his frame?

 

[atyy]

Well, O' only has one answer. What should I tell him? He is waiting on the line? Just a joke.

O' says the points were struck at d/(2c) and at -d/(2c) at time t' = d/(2c).

Do you agree O' is correct about his frame?

Yes.

Edit. No! At d/2 and -d/2 at t'=d/2c

 

[cfrogue]

Yes.

Edit. No! At d/2 and -d/2 at t'=d/2c

Agreed.

So, where are these points d/2, -d/2 in O?

x' = (x - vt)λ

Right point
d/2 = (x1 - vt)λ
x1 = d/(2λ) + vt

Left Point
-d/2 = (x2 - vt)λ
x2 = vt - d/(2λ)

Now, since these are different distances from the origin in O, they will not be simultaneous in O but are in O'.

These, given d in O', these are the x points of simultaneity for O' in the x coords of O.

Also, the origin of O' is located at vt.

Is this correct?

 

[atyy]

Agreed.

So, where are these points d/2, -d/2 in O?

x' = (x - vt)λ

Right point
d/2 = (x1 - vt)λ
x1 = d/(2λ) + vt

Left Point
-d/2 = (x2 - vt)λ
x2 = vt - d/(2λ)

Points or events? The Lorentz transformations only apply to events.

 

[cfrogue]

Points or events? The Lorentz transformations only apply to events.

I guess one could say the events occur at those points on the xaxis.

 

[atyy]

I guess one could say the events occur at those points on the xaxis.

If the events are the left going photon hitting the left end, and the right going photon hitting the right end of the primed rods, then t must be t(L') and t(R').

 

[cfrogue]

If the events are the left going photon hitting the left end, and the right going photon hitting the right end of the primed rods, then t must be t(L') and t(R').

Yea, I am more thinking about how the light sphere proceeds in O' from the coords of O.

I used these rods to look at it in a fixed way.

I do know the center of the light sphere for O' is located at vt in the coords of O.

I am going to think about it more.

 

[atyy]

I do know the center of the light sphere for O' is located at vt in the coords of O.

Yes. Since the primed rod is centred at O' , this is exactly the statement that the leftward photon hits the left end of the primed rod at the same t' as the rightward photon hits the right end of the primed rod.

 

[cfrogue]

Yes. Since the primed rod is centred at O' , this is exactly the statement that the leftward photon hits the left end of the primed rod at the same t' as the rightward photon hits the right end of the primed rod.

I am putting a clock on the right end of the rod of O'.

From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B
http://www.fourmilab.ch/etexts/einstein/specrel/www/

I am beginning to think I cannot sync to zero when the origins of O and O' meet.

Oh, I could, but how can all the clocks on the rod of O' sync up?

It seems all clocks on the rod of O' must be synched in advance.

Thus, there must be a time in O and a time in O' that may or may not be equal when the centers of O and O' are coincident.

Do you agree?

yanni - the rain must fall

 

[atyy]

I am putting a clock on the right end of the rod of O'.

From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B
http://www.fourmilab.ch/etexts/einstein/specrel/www/

I am beginning to think I cannot sync to zero when the origins of O and O' meet.

Oh, I could, but how can all the clocks on the rod of O' sync up?

It seems all clocks on the rod of O' must be synched in advance.

Thus, there must be a time in O and a time in O' that may or may not be equal when the centers of O and O' are coincident.

Do you agree?

yanni - the rain must fall

There are two meanings of synch.

In relativity, the common one is synch all clocks that are stationary with respect to an inertial observer. So for example, if O' places 3 clocks, one at the left end of the primed rod, one at his location at the centre of the primed rod, and one at the right end of the primed rod, he can synch all 3 clocks - these 3 clocks will not be synched for O. Similarly O can place clocks at the left, centre and right of the unprimed rod, and synched them. these clocks will not be synched for O'.

There is a different operation, which I think you call synching, but is not called synching in relativity. When the worldlines of O and O' intersect, that is an event, and O can do a global shift of his unprimed time and unprimed space coordinates so that the event occurs at (x=0, t=0) for him, and O' can also do a global shift of his primed time and primed space coordinates so that the event occurs at (x'=0, t'=0) for him.

Love the guitar solo!

 

[cfrogue]

There are two meanings of synch.

In relativity, the common one is synch all clocks that are stationary with respect to an inertial observer. So for example, if O' places 3 clocks, one at the left end of the primed rod, one at his location at the centre of the primed rod, and one at the right end of the primed rod, he can synch all 3 clocks - these 3 clocks will not be synched for O. Similarly O can place clocks at the left, centre and right of the unprimed rod, and synched them. these clocks will not be synched for O'.

There is a different operation, which I think you call synching, but is not called synching in relativity. When the worldlines of O and O' intersect, that is an event, and O can do a global shift of his unprimed time and unprimed space coordinates so that the event occurs at (x=0, t=0) for him, and O' can also do a global shift of his primed time and primed space coordinates so that the event occurs at (x'=0, t'=0) for him.

Yes, but if I use Einstein's logic, he syncs clocks on each rod in their own frame, ie simultaneity convention. So, if I want to put a clock on the endpoints of the rod in O', then I am not going to be able to sync time to zero, but I can sync positions to zero at the intersection of O and O'.


Einstein:

We imagine further that at the two ends A and B of the rod, clocks are placed which synchronize with the clocks of the stationary system, that is to say that their indications correspond at any instant to the ``time of the stationary system'' at the places where they happen to be. These clocks are therefore ``synchronous in the stationary system.''
We imagine further that with each clock there is a moving observer, and that these observers apply to both clocks the criterion established in § 1 for the synchronization of two clocks.
http://www.fourmilab.ch/etexts/einstein/specrel/www/
Here, each point of each rod is a clock synched in its own frame.


Now to the origin of one of the two systems (k) let a constant velocity v be imparted in the direction of the increasing x of the other stationary system (K), and let this velocity be communicated to the axes of the co-ordinates, the relevant measuring-rod, and the clocks. To any time of the stationary system K there then will correspond a definite position of the axes of the moving system
http://www.fourmilab.ch/etexts/einstein/specrel/www/
Here, I can sync the positions.

 

[Al68]

Thus, there must be a time in O and a time in O' that may or may not be equal when the centers of O and O' are coincident.

Sure, for any single event, there is a unique t for each t' at the location of the event.

Since the light reaches the ends of the rod simultaneously in O', t'(L)=t'(R), and there is a corresponding t(L) and t(R) in O, but they are not equal.

Actually, when O' tells O what time it is in O' when the light reaches the ends of the rod simultaneously in O', O' will need to know what location in O' to calculate t, and while that t' is valid for any location in O', any corresponding t will only be valid for a single location in O', and will be a different value for other locations in O'.

For that matter, O can just look at any clock at rest in O' and compare it to a clock at rest in O when the clocks pass each other. But the value of t that matches the t' observed is only valid for that single location in O'. Other locations in O' will have a different value of t for that same t'.

 

[atyy]

Yes, but if I use Einstein's logic, he syncs clocks on each rod in their own frame, ie simultaneity convention. So, if I want to put a clock on the endpoints of the rod in O', then I am not going to be able to sync time to zero, but I can sync positions to zero at the intersection of O and O'.


Einstein:

We imagine further that at the two ends A and B of the rod, clocks are placed which synchronize with the clocks of the stationary system, that is to say that their indications correspond at any instant to the ``time of the stationary system'' at the places where they happen to be. These clocks are therefore ``synchronous in the stationary system.''
We imagine further that with each clock there is a moving observer, and that these observers apply to both clocks the criterion established in § 1 for the synchronization of two clocks.
http://www.fourmilab.ch/etexts/einstein/specrel/www/
Here, each point of each rod is a clock synched in its own frame.


Now to the origin of one of the two systems (k) let a constant velocity v be imparted in the direction of the increasing x of the other stationary system (K), and let this velocity be communicated to the axes of the co-ordinates, the relevant measuring-rod, and the clocks. To any time of the stationary system K there then will correspond a definite position of the axes of the moving system
http://www.fourmilab.ch/etexts/einstein/specrel/www/
Here, I can sync the positions.

Yes, that's good too. There is a slight difference between my procedure and Einstein's, but they are essentially the same.

 

[atyy]

Yea, I am more thinking about how the light sphere proceeds in O' from the coords of O.

I used these rods to look at it in a fixed way.

I do know the center of the light sphere for O' is located at vt in the coords of O.

I am going to think about it more.

Another way of thinking that may help.

Fundamentally, there are only two photons, one going left and one going right. The left photon encounters two endpoints, the left endpoint of the primed rod and the left endpoint of the unprimed rod. The right photon similarly encounters two endpoints, the right endpoint of the primed rod and the right endpoint of the unprimed rod.

To O', the left photon reaches the left endpoint of the primed rod at the same t' as the right photon reaches the right endpoint of the primed rod. These two events that are simultaneous for O' are his light sphere. For O, the events are not simultaneous and don't form any sphere, but it doesn't matter since O is going to judge if there is a light sphere according to when the photons hit the ends of the unprimed rods, not when they hit the ends of the primed rods.

Similarly, to O, the left photon reaches the left endpoint of the unprimed rod at the same t as the right photon reaches the right endpoint of the unprimed rod. These two events that are simultaneous for O are his light sphere. For O', the events are not simultaneous and don't form any sphere, but it doesn't matter since O' is going to judge if there is a light sphere according to when the photons hit the ends of the primed rods, not when they hit the ends of the unprimed rods.

Thus O' and O judge their light spheres using different pairs of events. The underlying reality is the same, one photon going left and encountering two left endpoints, and one photon going right and encountering two right endpoints. This underlying reality is the light cone.

 

[heldervelez]

Dalespam is right since post 2

Dalespam was right since post #2.

No, it does not. The light flash moves spherically from the origin of the flash regardless of the subsequent motion of the source. That is the http://en.wikipedia.org/wiki/Postulates_of_special_relativity" : "As measured in an inertial frame of reference, light is always propagated in empty space with a definite velocity c that is independent of the state of motion of the emitting body."

Both O OR O' could emit the light, and the problem would be the same.

Take O' out of the problem, and the problem would be the same.

The postulate of independence of 'c' from the state of motion of either the emitting and receiving bodys prevails over any other considerations.

O is stationary and is also the observer, then we do not need equations at all.
x=+-ct.

 

[cfrogue]

Another way of thinking that may help.

Fundamentally, there are only two photons, one going left and one going right. The left photon encounters two endpoints, the left endpoint of the primed rod and the left endpoint of the unprimed rod. The right photon similarly encounters two endpoints, the right endpoint of the primed rod and the right endpoint of the unprimed rod.

To O', the left photon reaches the left endpoint of the primed rod at the same t' as the right photon reaches the right endpoint of the primed rod. These two events that are simultaneous for O' are his light sphere. For O, the events are not simultaneous and don't form any sphere, but it doesn't matter since O is going to judge if there is a light sphere according to when the photons hit the ends of the unprimed rods, not when they hit the ends of the primed rods.

Similarly, to O, the left photon reaches the left endpoint of the unprimed rod at the same t as the right photon reaches the right endpoint of the unprimed rod. These two events that are simultaneous for O are his light sphere. For O', the events are not simultaneous and don't form any sphere, but it doesn't matter since O' is going to judge if there is a light sphere according to when the photons hit the ends of the primed rods, not when they hit the ends of the unprimed rods.

Thus O' and O judge their light spheres using different pairs of events. The underlying reality is the same, one photon going left and encountering two left endpoints, and one photon going right and encountering two right endpoints. This underlying reality is the light cone.

First off, do you find this false?

If so, why?

This is the time in O when O' sees the points of its light sphere struck simultaneously.

<br /> t = \frac{d}{2c} {\sqrt{\frac{c+v}{c-v}}<br />

 

[cfrogue]

Dalespam was right since post #2.


Both O OR O' could emit the light, and the problem would be the same.

Take O' out of the problem, and the problem would be the same.

The postulate of independence of 'c' from the state of motion of either the emitting and receiving bodys prevails over any other considerations.

O is stationary and is also the observer, then we do not need equations at all.
x=+-ct.

How so?

Are you ignoring the light sphere in O'?

 

[atyy]

First off, do you find this false?

If so, why?

This is the time in O when O' sees the points of its light sphere struck simultaneously.

<br /> t = \frac{d}{2c} {\sqrt{\frac{c+v}{c-v}}<br />

It is incorrect because to derive that formula, you chose one event - the right photon hitting the right endpoint of the primed rod. That event has coordinates (x'=d/2, t'=d/2c). You used the Lorentz transformation to find the t coordinate of that event. Thus the t you found only refers to one event - the right photon hitting the right endpoint of the primed rod. Your statement interprets t as applying to "points of its light sphere struck simultaneously", which are two events, so it is incorrect.

 

[cfrogue]

It is incorrect because to derive that formula, you chose one event - the right photon hitting the right endpoint of the primed rod. That event has coordinates (x'=d/2, t'=d/2c). You used the Lorentz transformation to find the t coordinate of that event. Thus the t you found only refers to one event - the right photon hitting the right endpoint of the primed rod. Your statement interprets t as applying to "points of its light sphere struck simultaneously", which are two events, so it is incorrect.

Yes, that is the correct coords of that event in O'.

Are you saying this is not simultaneous for ±x'. Keep in mind, we are looking inside the logic of O'.

O' says ±x' is simultaneous in its frame.

SR is clear,

(ct’)^2 = x’^2 + y^2 + z^2 for the light sphere.

Now, by looking at the x coords only in O',

ct' = ± x'

O' says these points are struck simultaneously by the light postulate.

Is this false?

 

[atyy]

Yes, that is the correct coords of that event in O'.

Are you saying this is not simultaneous for ±x'. Keep in mind, we are looking inside the logic of O'.

O' says ±x' is simultaneous in its frame.

SR is clear,

(ct’)^2 = x’^2 + y^2 + z^2 for the light sphere.

Now, by looking at the x coords only in O',

ct' = ± x'

O' says these points are struck simultaneously by the light postulate.

Is this false?

All that is good, because you refer to simultaneity and keep entirely to primed coordinates. If you use an unprimed coordinate like t, you cannot say that two simultaneous events in primed coordinates correspond to a single "when" for unprimed coordinates, because the two events belong to two "whens" for unprimed coordinates.

 

[Al68]

Yes, that is the correct coords of that event in O'.

Are you saying this is not simultaneous for ±x'. Keep in mind, we are looking inside the logic of O'.

O' says ±x' is simultaneous in its frame.

O' will easily recognize that the light reaching the ends of the rod are two different events with times of t'(L) and t'(R) and those values are equal in O', but not equal in O.

O' will easily recognize that the light reaching the ends of the rod are not simultaneous in O.

I'm assuming that O' above represents a hypothetical observer at rest in O' that is familiar with SR.

 

[heldervelez]

Mr cfrog.
put O as the emiter.
light propagate as if the emiter was O'.
reread the OP, then remove all references to O'.
solved.
O' is there only to trouble minds.
'independent from... is the key'

 

[cfrogue]

All that is good, because you refer to simultaneity and keep entirely to primed coordinates. If you use an unprimed coordinate like t, you cannot say that two simultaneous events in primed coordinates correspond to a single "when" for unprimed coordinates, because the two events belong to two "whens" for unprimed coordinates.

I proceed by reductio ad absurdum there exists only time t in O for this t'.

Assume there exists a tx < t such that the points in O' are struck at the same time.
Then tx < r/(λ(c-v))
tx = (t' + vx'/c^2)λ < r/(λ(c-v))
We have by the SR spherical light sphere,
ct' = x', t' = x'/c
Also, by selection x' = r.
( x'/c + vx'/c^2)λ < r/(λ(c-v))
(r/c + rv/c^2)λ < r/(λ(c-v))
(1/c + 1v/c^2)λ < 1/(λ(c-v))
((c + v)/c^2)λ < 1/(λ(c-v))
(c + v) < c^2/(λ^2(c-v))
(c + v) < (c^2/(c-v))((c^2 - v^2)/c^2)
c + v < (c^2 - v^2)/(c - v)
c + v < c + v
0 < 0

This is a contradiction. The same argument hold for tx > t.
Thus, the calculated t is the unique time in O when the points of O' are struck at the same time.

 

[cfrogue]

Mr cfrog.
put O as the emiter.
light propagate as if the emiter was O'.
reread the OP, then remove all references to O'.
solved.
O' is there only to trouble minds.
'independent from... is the key'

Are you aware everyone on this thread has concluded the light sphere in O' is located at vt in the coords of O at any time t?

Can you prove otherwise?

 

[atyy]

I proceed by reductio ad absurdum there exists only time t in O for this t'.

Assume there exists a tx < t such that the points in O' are struck at the same time.
Then tx < r/(λ(c-v))
tx = (t' + vx'/c^2)λ < r/(λ(c-v))
We have by the SR spherical light sphere,
ct' = x', t' = x'/c
Also, by selection x' = r.
( x'/c + vx'/c^2)λ < r/(λ(c-v))
(r/c + rv/c^2)λ < r/(λ(c-v))
(1/c + 1v/c^2)λ < 1/(λ(c-v))
((c + v)/c^2)λ < 1/(λ(c-v))
(c + v) < c^2/(λ^2(c-v))
(c + v) < (c^2/(c-v))((c^2 - v^2)/c^2)
c + v < (c^2 - v^2)/(c - v)
c + v < c + v
0 < 0

This is a contradiction. The same argument hold for tx > t.
Thus, the calculated t is the unique time in O when the points of O' are struck at the same time.

It is good to see you starting to use diagrams, you are almost ready for actual spacetime diagrams.

Expanding \gamma in the expressions I gave for t_4 and t_5 above and simplifying I get
<br /> t_4 = \frac{d}{2c} {\sqrt{\frac{c+v}{c-v}}<br />
which is what you had, and
<br /> t_5 = \frac{d}{2c} {\sqrt{\frac{c-v}{c+v}}<br />
which you omitted

OK, so how did DaleSpam get two different times which, according to your argument, would both be a "time in O when the points of O' are struck at the same time"?

 

[Al68]

Thus, the calculated t is the unique time in O when the points of O' are struck at the same time.

So, I guess you have proven Einstein's entire 1905 paper and all of SR since then to be false.

Time to close this thread then. Please.

 

[cfrogue]

OK, so how did DaleSpam get two different times which, according to your argument, would both be a "time in O when the points of O' are struck at the same time"?

His point arrives at a time before the simultaneity in O' as the argument proves.

There is a unique point.

Is this not natural? Surely, the left endpoint should occur before the right endpoint according to R of S in the coords of O.

This is eactly the case. When the left point is struck in O, O' concludes the light sphere has not yet struck both points.

 

[cfrogue]

So, I guess you have proven Einstein's entire 1905 paper and all of SR since then to be false.

Time to close this thread then. Please.

How do you figure this?

We are exploring the light sphere.

Do you have a calculation that is in error?

Please tell me.

 

[atyy]

His point arrives at a time before the simultaneity in O' as the argument proves.

There is a unique point.

Is this not natural? Surely, the left endpoint should occur before the right endpoint according to R of S in the coords of O.

This is eactly ther case. When the left point is struck in O, O' concludes the light sphere has not yet struck both points.

By "His point" do you mean light hitting the left endpoint of the unprimed rod, or light hitting the left endpoint of the primed rod?

 

[atyy]

I proceed by reductio ad absurdum there exists only time t in O for this t'.

Assume there exists a tx < t such that the points in O' are struck at the same time.
Then tx < r/(λ(c-v))
tx = (t' + vx'/c^2)λ < r/(λ(c-v))
We have by the SR spherical light sphere,
ct' = x', t' = x'/c
Also, by selection x' = r.
( x'/c + vx'/c^2)λ < r/(λ(c-v))
(r/c + rv/c^2)λ < r/(λ(c-v))
(1/c + 1v/c^2)λ < 1/(λ(c-v))
((c + v)/c^2)λ < 1/(λ(c-v))
(c + v) < c^2/(λ^2(c-v))
(c + v) < (c^2/(c-v))((c^2 - v^2)/c^2)
c + v < (c^2 - v^2)/(c - v)
c + v < c + v
0 < 0

This is a contradiction. The same argument hold for tx > t.
Thus, the calculated t is the unique time in O when the points of O' are struck at the same time.

BTW, this looks correct, except for the final sentence "Thus, the calculated t is the unique time in O when the points of O' are struck at the same time." So what we are having is a problem in interpreting the mathematics.

 

[cfrogue]

By "His point" do you mean light hitting the left endpoint of the unprimed rod, or light hitting the left endpoint of the primed rod?

I am talking about both.

O sees the left hit before the right.

O' contends at the time in O when the left point is hit, neither right or left is hit yet in its own frame.

O' sees the strikes as simultaneous whereas O sees them at different times.

 

[cfrogue]

BTW, this looks correct, except for the final sentence "Thus, the calculated t is the unique time in O when the points of O' are struck at the same time." So what we are having is a problem in interpreting the mathematics.

I do not think so since I showed no other t fits the bill > t or < t.

 

[matheinste]

So, I guess you have proven Einstein's entire 1905 paper and all of SR since then to be false.

Time to close this thread then. Please.

It would be a shame if this thread was to be closed now as it would stop it from beating the record number of replies for a topic in the Special and General Relativity forum which currently stands at 444. However it is still a long way down the table for the number of views.

Matheinste.

 

[Al68]

Thus, the calculated t is the unique time in O when the points of O' are struck at the same time.

So, I guess you have proven Einstein's entire 1905 paper and all of SR since then to be false.

Time to close this thread then. Please.

How do you figure this?

We are exploring the light sphere.

Do you have a calculation that is in error?

Please tell me.

Einstein's 1905 paper and all of SR says that for two nonlocal events that are simultaneous in O', they occur at two different times in O.

SR also says that for any t', there is a separate t for each location in O'.

You said: "the calculated t is the unique time in O when the points of O' are struck at the same time.", which contradicts the above.

This has been pointed out dozens of times in hundreds of posts to no avail, so there is no point for this thread to continue.

 

[Al68]

Thus, the calculated t is the unique time in O when the points of O' are struck at the same time.

I am talking about both.

O sees the left hit before the right.

O' contends at the time in O when the left point is hit, neither right or left is hit yet in its own frame.

O' sees the strikes as simultaneous whereas O sees them at different times.

Well, which is it? I assume it is obvious that these posts directly contradict each other.

 

[cfrogue]

Einstein's 1905 paper and all of SR says that for two nonlocal events that are simultaneous in O', they occur at two different times in O.

SR also says that for any t', there is a separate t for each location in O'.

You said: "the calculated t is the unique time in O when the points of O' are struck at the same time.", which contradicts the above.

This has been pointed out dozens of times in hundreds of posts to no avail, so there is no point for this thread to continue.

You remind me of someone.

Noone here has refuted any of these facts.

It is agreed that R of S applies in O.

It is agreed they are simultaneous in O'.

What is the problem?

And this unique t does not contradict the above.

In fact it reinforces R of S.

It shows the left is struck before the right in O and when the left is struck, it is impossible for simultaneity in O'.

 

[cfrogue]

Well, which is it. I assume it is obvious that these posts directly contradict each other.

No they do not.

They only say the time O' sees the points struck at the same time is unique in O.

Do you think there should be two times in O when the points of O' are struck at the same time?

That would contradict relativity.

For O, L is struck before the R.

For O', when L is struck in O, the light sphere is not yet to L and R in O' for the simultaneous strike.

This is natural.

 

[atyy]

By "His point" do you mean light hitting the left endpoint of the unprimed rod, or light hitting the left endpoint of the primed rod?

I am talking about both.

O sees the left hit before the right.

O' contends at the time in O when the left point is hit, neither right or left is hit yet in its own frame.

O' sees the strikes as simultaneous whereas O sees them at different times.

Note that light striking the left endpoint of the primed rod is a different event from light striking the right endpoint of the primed rod.

For O, light hits the left endpoint of the unprimed rod at the same t as light hits the right endpoint of the unprimed rod.

For O, light hits the left endpoint of the primed rod at an earlier t than light hits the right endpoint of the primed rod.

For O', light hits the left endpoint of the primed rod at the same t' as light hits the right endpoint of the primed rod.

For O', light hits the left endpoint of the unprimed rod at a later t' than light hits the right endpoint of the unprimed rod.

 

[Al68]

They only say the time O' sees the points struck at the same time is unique in O.

Are you referring to a time at which an observer physically sees light reflected back from the rod's endpoints? Or the time the endpoints are struck? If the latter, then the times are different in O.

Do you think there should be two times in O when the points of O' are struck at the same time?

According to SR, there are. And you say so next:

For O, L is struck before the R.

Do you not realize that means that t(L) is a different value than t(R), meaning more than one t in O for the single t' in O' [t'=t'(L)=t'(R)]?

There is obviously a major communication problem here.

BTW, who do I remind you of?