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

[cfrogue]

For consistency with your earlier notation it should be

O sees the strikes of O' at
t_L' = d/(2cλ(c+v))
t_R' = d/(2cλ(c-v))

t_L' < t_R'

Yes, this is relativity of simultaneity.

Thanks, that is better notation.

 

[cfrogue]

It implies neither. Neither are logical implications of my post.

You said,

Those are the times in O that the 2 strikes occur simultaneously in O'. A single t' in O' for two nonlocal events correspond to two different t values in O.

 

[atyy]

For consistency with your earlier notation it should be

O sees the strikes of O' at
t_L' = d/(2cλ(c+v))
t_R' = d/(2cλ(c-v))

t_L' < t_R'

Yes, this is relativity of simultaneity.

http://www.youtube.com/watch?v=kJB0...4E9CDF2E&playnext=1&playnext_from=PL&index=4"

I made an algebraic error, there's a an extra factor of c in the denominator that shouldn't be there.

 

[Al68]

This implies one light sphere hits the points of O' simultaneously twice and at two different radius lengths of the light sphere.

It implies neither. Neither are logical implications of my post.

You said,

Those are the times in O that the 2 strikes occur simultaneously in O'. A single t' in O' for two nonlocal events correspond to two different t values in O.

That's right. That statement doesn't even remotely imply what you said.

 

[cfrogue]

I made an algebraic error, there's a an extra factor of c in the denominator that shouldn't be there.

LOL, I see it now, I really should read more closely.

Eddie Vedder, I like his music.

 

[cfrogue]

That's right. That statement doesn't even remotely imply what you said.

Those are the times in O that the 2 strikes occur simultaneously in O'. A single t' in O' for two nonlocal events correspond to two different t values in O.

t1 makes t' true and
t2 make t' true

t1 != t2.

 

[Al68]

t1 makes t' true and
t2 make t' true

t1 != t2.

OMG?! Do you seriously mean that two events that are simultaneous in O' (occur at the same t') actually occur at two different times in O (t1 and t2)?!

How could that possibly be? :!):!)

 

[cfrogue]

OMG?! Do you seriously mean that two events that are simultaneous in O' (occur at the same t') actually occur at two different times in O (t1 and t2)?!

How could that possibly be? :!):!)

LOLOLOLOL
I did not say there were two events in O'. You said one t' and I said what you did.

But, it is the case that the left are right endpoints of the O' rod are not hit at the same time in O, agreed?

 

[Al68]

I did not say there were two events in O'.

You were referring to the time t' that the light hit the endpoints of the rod. That t' represents the time for both events in O'.

But, it is the case that the left are right endpoints of the O' rod are not hit at the same time in O, agreed?

Agreed, if they are hit at the same time (t') in O', then they are not hit at the same time in O (unless v = 0).

 

[Dale]

Let's see if I understand R of S.

O sees the strikes of O' at
t_L = d/(2cλ(c+v))
t_R = d/(2cλ(c-v))

t_L < t_R
Is this R of S?

[Edited for correction]

[t_L = d/(2λ(c+v))
t_R = d/(2λ(c-v))]

Yes, that is correct.

But, it is the case that the left are right endpoints of the O' rod are not hit at the same time in O, agreed?

Also correct.

 

[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