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

[JesseM]

Do you? What are they? Anyway, the phrase "two events exclusively in one frame" doesn't seem to make sense, events are physical things that are assigned coordinates by all frames, they don't "belong" to one frame or another.

Yea, I should have said from the POV of O.

You didn't answer my question of what two events you're thinking of, though.

We now imagine space to be measured from the stationary system K by means of the stationary measuring-rod, and also from the moving system k by means of the measuring-rod moving with it; and that we thus obtain the co-ordinates x, y, z, and , , respectively. Further, let the time t of the stationary system be determined for all points thereof at which there are clocks by means of light signals in the manner indicated in § 1;
http://www.fourmilab.ch/etexts/einstein/specrel/www/

It appears that Einstein had a "clock" located at each point of the stationary measuring-rod synchronized by SR's simultaneity convention.

Do you agree?

Yes, this is the physical basis for the time coordinate assigned to any event in an inertial frame--you assume that each frame has a row of clocks attached to different positions on a measuring-rod at rest in that frame, with the clocks synchronized according to the Einstein synchronization convention, and then any event that occurs alongside the rod can be assigned a time-coordinate by looking at the reading on the clock that was right next to the event at the moment it happened.

Yea, I am simply talking about elaspsed time in O only, no other frame.

The elapsed time between two events on a worldline of a clock at rest in O?

 

[cfrogue]

You didn't answer my question of what two events you're thinking of, though.

I am thinking about the endpoint of the O' rod.

Please look for this in section 4
Between the quantities x, t, and , which refer to the position of the clock, we have, evidently, x=vt
http://www.fourmilab.ch/etexts/einstein/specrel/www/

It appears that Einstein derives time dilation based on the fact that x=vt or based on the motion of the rod.

Yes, this is the physical basis for the time coordinate assigned to any event in an inertial frame--you assume that each frame has a row of clocks attached to different positions on a measuring-rod at rest in that frame, with the clocks synchronized according to the Einstein synchronization convention, and then any event that occurs alongside the rod can be assigned a time-coordinate by looking at the reading on the clock that was right next to the event at the moment it happened.

OK, that is what I thought.

The elapsed time between two events on a worldline of a clock at rest in O?

Yes, I am only talking about rest clocks

 

[JesseM]

OK, with the twins solution on Wiki and many other places, when a clock in O elapses by t, then in a moving frame, its clock in that frame will elapse λt.

http://en.wikipedia.org/wiki/Twin_paradox

No, it's not a single clock C' at rest in the moving frame O' that measures gamma*t for the clock C in O to tick forward by t, it's the coordinate time in O' that's gamma*t, or alternately the difference in readings between two clocks at rest in O', the first measuring the time t1 in O' of the event of clock C showing its first reading, the second measuring the time t2 in O' of the event of clock C showing its second reading. If the separation between the two readings of C was t, then (t2 - t1) will equal gamma*t.

Take a look at my thread an illustration of relativity with rulers and clocks. In this thread I show how each frame assigns coordinates using local readings on a ruler with synchronized clocks attached to different markings, exactly as discussed in the quote from Einstein you provided earlier. Suppose we treat frame O in your example as synonymous with frame A in my diagrams, and frame O' as synonymous with frame B. And suppose the clock C at rest in frame A is the one at position x=692.3 meters on A's ruler, just to the right of the green clock. Finally, suppose the two events on C's worldline are the event of it showing a time of 2 microseconds, and the event of it showing a time of 3 microseconds, so t = 3 - 2 = 1 microsecond.

Now if you look at the top part of the diagram, you can see that the first event of clock C reading 2 microseconds happens right next to the clock at the 346.2 meter mark on ruler B, when that clock reads 0 microseconds. And if you look at the bottom part of the diagram, you can see that the second event of clock C reading 3 microseconds happens right next to a different clock on ruler B, namely the one at the -173.1 meter mark on ruler B, when that clock reads 2 microseconds.

So, the time between the readings of these two different clocks on ruler B is 2 microseconds, and that is t', the time in B's frame between the event of clock C reading 2 microseconds and the event of clock C reading 3 microseconds. Since gamma=2 in my diagram, this does satisfy the time dilation equation t' = gamma*t.

 

[Hans de Vries]

The OP's subject resembles the spherical mirror clock which
is a very nice example to thread all three aspects of SR:
Lorentz contraction, Time dilation and Non-simultaneity.

4.13 Simultaneity from the Spherical Mirror clock
http://physics-quest.org/Book_Chapter_Non_Simultaneity.pdf

Regards, Hans

 

[JesseM]

I am thinking about the endpoint of the O' rod.

That's not an event though. Do you want to pick two events which both happen on the worldline of the clock at the endpoint of the O' rod?

Please look for this in section 4
Between the quantities x, t, and , which refer to the position of the clock, we have, evidently, x=vt
http://www.fourmilab.ch/etexts/einstein/specrel/www/

It appears that Einstein derives time dilation based on the fact that x=vt or based on the motion of the rod.

Yes, but he was looking at the time in the stationary frame between two events on the worldline of the moving clock--the event of it reading 0, and the event of it reading t. The origins coincide so the event of this clock reading 0 happens at a time coordinate of tau=0 in the stationary frame, while the event of it reading t happens at a time coordinate of tau=t*sqrt(1 - v^2/c^2) in the stationary frame. So, the time interval in the stationary frame is just the second time coordinate minus the first, which of course is tau=t*sqrt(1 - v^2/c^2).

The elapsed time between two events on a worldline of a clock at rest in O?

Yes, I am only talking about rest clocks

OK, if we pick two events on the worldline of a clock C at rest in O that have a time separation of t in frame O, then the time separation t' between these events in O is gamma*t.

 

[cfrogue]

No, it's not a single clock C at rest in the moving frame O' that elapses gamma*t, it's the coordinate time in O' that's gamma*t, or alternately the difference in readings between two clocks at rest in O', the first measuring the time t1 in O' of the event of clock C showing its first reading, the second measuring the time t2 in O' of the event of clock C showing its second reading. If the separation between the two readings of C was t, then (t2 - t1) will equal gamma*t.

Take a look at my thread an illustration of relativity with rulers and clocks. In this thread I show how each frame assigns coordinates using local readings on a ruler with synchronized clocks attached to different markings, exactly as discussed in the quote from Einstein you provided earlier. Suppose we treat frame O in your example as synonymous with frame A in my diagrams, and frame O' as synonymous with frame B. And suppose the clock C at rest in frame A is the one at position x=692.3 meters on A's ruler, just to the right of the green diagram. Finally, suppose the two events on C's worldline are the event of it showing a time of 2 microseconds, and the event of it showing a time of 3 microseconds, so t = 3 - 2 = 1 microsecond.

Now if you look at the top part of the diagram, you can see that the first event of clock C reading 2 microseconds happens right next to the clock at the 346.2 meter mark on ruler B, when that clock reads 0 microseconds. And if you look at the bottom part of the diagram, you can see that the second event of clock C reading 3 microseconds happens right next to a different clock on ruler B, namely the one at the -173.1 meter mark on ruler B, when that clock reads 2 microseconds.

So, the time between the readings of these two different clocks on ruler B is 2 microseconds, and that is t', the time in B's frame between the event of clock C reading 2 microseconds and the event of clock C reading 3 microseconds. Since gamma=2 in my diagram, this does satisfy the time dilation equation t' = gamma*t.

This is excellent and I completely agree with all of this.

The coordinate time interval, yes, that is will elapse slower than an interval in the stationary frame from the POV of the stationary frame.

I guess now, we agree I can apply time dilation to describe the motion of the rod to illustrate the elapsed coord time in O'.

Do you agree with the following, when the center of the rod is located at the origin of O, then the left endpoint of the rod is located at xstart = d/(2λ)?

After any time t in O, the endpoint is located at xt = d/(2λ)+vt.

Thus, ∆x = vt.

Also, I asked DR Greg, but, when will O think O' sees the endpoints of the rods struck simultaneously?

I come up with t = d/(2*λ*(c-v)).

 

[JesseM]

This is excellent and I completely agree with all of this.

The coordinate time interval, yes, that is will elapse slower than an interval in the stationary frame from the POV of the stationary frame.

I guess now, we agree I can apply time dilation to describe the motion of the rod to illustrate the elapsed coord time in O'.

Do you agree with the following, when the center of the rod is located at the origin of O, then the left endpoint of the rod is located at xstart = d/(2λ)?

Normally I'd think of the coordinates increasing from left to right so that xstart = -d/(2*gamma), but this is just a matter of convention so sure, I agree.

After any time t in O, the endpoint is located at xt = d/(2λ)+vt.

Assuming the rod is traveling to the left, yes.

Thus, ∆x = vt.

If you pick two events on the worldline of the endpoint which have a time separation of t in frame O, then yes, their spatial separation ∆x in frame O will be ∆x = vt.

Also, I asked DR Greg, but, when will O think O' sees the endpoints of the rods struck simultaneously?

You mean, if the ends of the rod are struck simultaneously in O', what will be the time interval between the strikes in frame O?

I come up with t = d/(2*λ*(c-v)).

Let's say in frame O' the left end is struck at x'=d,t'=0 and the right end is struck at x'=-d,t'=0. Then the time of the left end being struck in frame O is:

gamma*(t' + vx'/c^2) = gamma*v*d/c^2

And the time of the right end being struck is:

gamma*(t' + vx'/c^2) = -gamma*v*d/c^2

So, the time interval between strikes in O would be 2*gamma*v*d/c^2. What was the reasoning behind your answer?

 

[cfrogue]

Normally I'd think of the coordinates increasing from left to right so that xstart = -d/(2*gamma), but this is just a matter of convention so sure, I agree.

Assuming the rod is traveling to the left, yes.

If you pick two events on the worldline of the endpoint which have a time separation of t in frame O, then yes, their spatial separation ∆x in frame O will be ∆x = vt.

You mean, if the ends of the rod are struck simultaneously in O', what will be the time interval between the strikes in frame O?

Let's say in frame O' the left end is struck at x'=d,t'=0 and the right end is struck at x'=-d,t'=0. Then the time of the left end being struck in frame O is:

gamma*(t' + vx'/c^2) = gamma*v*d/c^2

And the time of the right end being struck is:

gamma*(t' + vx'/c^2) = -gamma*v*d/c^2

So, the time interval between strikes in O would be 2*gamma*v*d/c^2. What was the reasoning behind your answer?

I calculated x' = ( x - vt )λ, I then put x = ct, but I could not come up with a good reason so that is why I asked.

I see, you applied the translation of t' back to t. You concluded that x'=d/c, I think it should be d/(2c) but that is no deal.

Why do you have a negative sign?

 

[JesseM]

I calculated x' = ( x - vt )λ, I then put x = ct, but I could not come up with a good reason so that is why I asked.

I see, you applied the translation of t' back to t. You concluded that x'=d/c, I think it should be d/(2c) but that is no deal.

Oh yeah, I guess I accidentally gave the rod a length of 2d rather than d.

Why do you have a negative sign?

Well, if the left end was at x'=d at t'=0 in O' (should have been x'=d/2, but never mind), then the right end would have to be at x'=-d, at t'=0, no? Both ends are equidistant from the origin of the x' axis since that's where the center of the rod is located.

 

[cfrogue]

Oh yeah, I guess I accidentally gave the rod a length of 2d rather than d.

Well, if the left end was at x'=d at t'=0 in O' (should have been x'=d/2, but never mind), then the right end would have to be at x'=-d, at t'=0, no? Both ends are equidistant from the origin of the x' axis since that's where the center of the rod is located.

I think we only need to calculate one anyway since they are both the same time.

It looks good to me.

 

[cfrogue]

Oh yeah, I guess I accidentally gave the rod a length of 2d rather than d.

Well, if the left end was at x'=d at t'=0 in O' (should have been x'=d/2, but never mind), then the right end would have to be at x'=-d, at t'=0, no? Both ends are equidistant from the origin of the x' axis since that's where the center of the rod is located.

OK, I wrote it in this format.

t = d/(2c) ( λ/c )

Is this OK?

 

[Dale]

I know how to derive that equation.

I know it is when Δx = 0.

Can you then use LT and tell me the time dilation for the scenerio we are talking about?

Sure, as always
\Delta t' = \gamma (\Delta t - v \Delta x/c^2).

In this case Δx = ±1, c=1, v=0.6, Δt = 1, so
Δt' = 1.25 (1 - 0.6 (+1)/1²) = 0.5
and
Δt' = 1.25 (1 - 0.6 (-1)/1²) = 2

 

[JesseM]

OK, I wrote it in this format.

t = d/(2c) ( vλ/c )

Is this OK?

If the length is d then it should just be (d/c)*(v*gamma/c). The strike at the left end happens at t' = +gamma*v*d/(2*c^2) and the strike at the right end occurs at t' = -gamma*v*d/(2*c^2), so the time between them is gamma*v*d/c^2.

 

[cfrogue]

Sure, as always
\Delta t' = \gamma (\Delta t - v \Delta x/c^2).

In this case Δx = ±1, c=1, v=0.6, Δt = 1, so
Δt' = 1.25 (1 - 0.6 (+1)/1²) = 0.5
and
Δt' = 1.25 (1 - 0.6 (-1)/1²) = 2

What do you get for the time dilation for this?

x = λvt/ (1 + λ)

Note, y > 0 for this case and z = 0.

 

[cfrogue]

If the length is d then it should just be (d/c)*(v*gamma/c). The strike at the left end happens at t' = +gamma*v*d/(2*c^2) and the strike at the right end occurs at t' = -gamma*v*d/(2*c^2), so the time between them is gamma*v*d/c^2.

I cannot see that.

They are struck at the same time in O', no?

 

[JesseM]

I cannot see that.

They are struck at the same time in O', no?

Sorry I got my t' and t mixed up, I should have written that in frame O, the strike at the left end happens at t = +gamma*v*d/(2*c^2) and the strike at the right end occurs at t = -gamma*v*d/(2*c^2), so the time between them in frame O is gamma*v*d/c^2.

 

[cfrogue]

Sorry I got my t' and t mixed up, I should have written that in frame O, the strike at the left end happens at t = +gamma*v*d/(2*c^2) and the strike at the right end occurs at t = -gamma*v*d/(2*c^2), so the time between them in frame O is gamma*v*d/c^2.

OK, this difference can be written as

(d/(2c)) ( 2*gamma*v/c )

Thus, this occurs before the stike in O at slow speeds since a rod in O is struck at t = d/(2c)

Something is wrong with this

 

[JesseM]

OK, this difference can be written as

(d/(2c)) ( 2*gamma*v/c )

Thus, this occurs before the stike in O at slow speeds since a rod in O is struck at t = d/(2c)

Something is wrong with this

Huh? What "strike in O" are you talking about? I thought there were only two strikes at either end of the rod at rest in O', are you imagining a third strike at one end of a rod at rest in O?

 

[cfrogue]

Sorry I got my t' and t mixed up, I should have written that in frame O, the strike at the left end happens at t = +gamma*v*d/(2*c^2) and the strike at the right end occurs at t = -gamma*v*d/(2*c^2), so the time between them in frame O is gamma*v*d/c^2.

Let me tell you what I have been thinking about.

The right end point of the rod in O' crosses the right end point of the stationary frame at time

vt + r/λ = r
vt = r ( 1 - 1/λ )
t = (r/v) ( 1 - 1/λ )

I also know, the center point of the light sphere is located at vt at any time t in O for the light sphere in O'

So, what I am missing is the time in O when the simultaneous strikes occur in O'.

This will give me a detailed analysis of the light sphere in O' from the coords of O.

I am having a mental block on the time in O when the simultaneous strikes occur in O'.


[Edited to add: There are two rods. One in O and one in O'. They have the same rest length in the respective frames of d. When they are centered, the light source centered on the O' rod, emits a light.]

 

[cfrogue]

Huh? What "strike in O" are you talking about? I thought there were only two strikes at either end of the rod at rest in O', are you imagining a third strike at one end of a rod at rest in O?

I gave you a detailed objective in the next post.

 

[JesseM]

Let me tell you what I have been thinking about.

The right end point of the rod in O' crosses the right end point of the stationary frame at time

vt + r/λ = r
vt = r ( 1 - 1/λ )
t = (r/v) ( 1 - 1/λ )

OK

I also know, the center point of the light sphere is located at vt at any time t in O.

The point that marks the center of the light sphere according O' is located at vt in O, but the center of the light sphere according to O is always at x=0 at any time t in O. Do you disagree?

So, what I am missing is the time in O when the simultaneous strikes occur in O'.

I already gave you that, assuming the strikes happened at t'=0 in O'. Or do you want the strikes to happen at the moment the light sphere reaches the ends of the rod at rest in O', assuming the original flash that created the sphere happened at t'=0 and x'=0 in O'? Or do you want the strikes to happen at the moment the light sphere reaches the ends of the rod in O', but you want to time the original flash so that the light sphere reaches the right end of the rod in O' at the same moment it reaches the right end of the rod in O? Or something else? You really need to explain what you're thinking in clearer terms.

 

[cfrogue]

The point that marks the center of the light sphere according O' is located at vt in O, but the center of the light sphere according to O is always at x=0 at any time t in O. Do you disagree?

Yes, I agree what O sees is a light sphere origined at x = 0.

I am trying to completely map how it proceeds in O' in terms of the time and measurements of O

I already gave you that, assuming the strikes happened at t'=0 in O'. Or do you want the strikes to happen at the moment the light sphere reaches the ends of the rod at rest in O', assuming the original flash that created the sphere happened at t'=0 and x'=0 in O'? Or do you want the strikes to happen at the moment the light sphere reaches the ends of the rod in O', but you want to time the original flash so that the light sphere reaches the right end of the rod in O' at the same moment it reaches the right end of the rod in O? Or something else? You really need to explain what you're thinking in clearer terms.

Please look at what you gave me.

I have changed it at the end.

It will strike the right endpoint at t = ( t' + vx'/(c^2))λ in O'.

t'=0 and x'=d/2 because the rod length is d and the light is centered.

So, t = vdλ/(2*c^2)

t = (d/(2c)) (v/c)λ
t = (d/(2c)) sqrt(1/((c/v)^2 - 1))

How does the last equation look?

 

[Dale]

What do you get for the time dilation for this?

x = λvt/ (1 + λ)

Note, y > 0 for this case and z = 0.

I'm not sure what you mean. Is x = λvt/ (1 + λ) the equation of the worldline of some object? If so, what time coordinates do you wish to evaluate it at?

 

[cfrogue]

I'm not sure what you mean. Is x = λvt/ (1 + λ) the description of the worldline of some object? If so, what time coordinates do you wish to evaluate it at?

It is an event, but y >> 0.

What is t'?

I come up with t' = t with v > 0.

 

[JesseM]

Please look at what you gave me.

I have changed it at the end.

It will strike the right endpoint at t = ( t' + vx'/(c^2))λ in O'.

t'=0 and x'=d/2 because the rod length is d and the light is centered.

You want the light to strike the right endpoint at t'=0? That would imply that the original flash was set off at t'=-d/2c, correct? This will make the problem more complicated, since the center of the light sphere will not be at x=0 in frame O...it would be easier if you had the light flash set off at the spacetime origin of both frames, so the light would reach the right end at t'=d/(2c) in frame O'.

So, t = vdλ/(2*c^2)

t = (d/(2c)) (v/c)λ
t = (d/(2c)) sqrt(1/(c/v)^2 - 1))

How does the last equation look?

(v/c)*gamma = sqrt(v^2/c^2)/sqrt(1 - v^2/c^2) = 1/[sqrt(c^2/v^2)*sqrt(1 - v^2/c^2)] = 1/sqrt(c^2/v^2 - 1). That's not the same as the factor after (d/(2c)) in your last equation.

 

[Dale]

It is an event, but y >> 0.

What is t'?

I come up with t' = t with v > 0.

x = λvt/ (1 + λ) is the equation of a worldline in spacetime (has the form t = mx + b). It can't be an event. An event is the equivalent of a point in spacetime, e.g. the intersection of two worldlines.

 

[cfrogue]

You want the light to strike the right endpoint at t'=0? That would imply that the original flash was set off at t'=-d/2c, correct?

(v/c)*gamma = sqrt(v^2/c^2)/sqrt(1 - v^2/c^2) = 1/[sqrt(c^2/v^2)*sqrt(1 - v^2/c^2)] = 1/sqrt(c^2/v^2 - 1). That's not the same as the factor after (d/(2c)) in your last equation.

Well, that is what I meant, I gess I left something off.

To make sure, this is what I intended and I want us to agree.

t = ( d/(2c) ) ( √(1 / ((c/v)² - 1) ) )

 

[JesseM]

Well, that is what I meant, I gess I left something off.

To make sure, this is what I intended and I want us to agree.

t = ( d/(2c) ) ( √(1 / ((c/v)² - 1) ) )

Yes, that's the same as what I got.

 

[cfrogue]

Yes, that's the same as what I got.

<br /> t = (\frac{d}{2c}) \frac{1}{\sqrt{c^2/v^2 - 1}}<br />

OK, I did it in this format.

Is this correct?

How do you enlarge these things?

 

[cfrogue]

Yes, that's the same as what I got.

Set v = c/√ 2

 

[JesseM]

Set v = c/√ 2

OK, so t=d/2c.

 

[cfrogue]

OK, so t=d/2c.

the frame has moved vt, where is the center of the light sphere in O' in the coords of O?

 

[JesseM]

the frame has moved vt, where is the center of the light sphere in O' in the coords of O?

As I mentioned before, you made things more complicated by requiring that the light sphere reached x'=d/2 at t'=0, this means that the original flash (the tip of the light cone) must have happened at x'=0, t'=-d/(2c). Translating this into frame O:

x = gamma*(x' + vt) = -gamma*d*v/(2c). With v = c/sqrt(2) this becomes:
-(1/sqrt(1/2))*d/(2*sqrt(2)) = -d/2

t = gamma*(t' + vx'/c^2) = -gamma*d/(2c). With v=c/sqrt(2) this becomes:
-(1/sqrt(1/2))*d/(2c) = -d/(c*sqrt(2))

So in frame O, the flash happened at x=-d/2, t=-d/(c*sqrt(2)). If you want to figure out the coordinates in O of an object which remains at the center of the light sphere in O', you'd need an object with velocity v which passes through those coordinates, so its position as a function of time would be:

x(t) = vt - v*(-d/(c*sqrt(2))) - d/2

So at time t = d/(2c), this object would be at:

x = vd/(2c) + vd/(c*sqrt(2)) - d/2 = vd/(2c) + (vd*sqrt(2))/(2c) - dc/2c
= d*[v*(1 + sqrt(2)) - c]/2c

 

[cfrogue]

As I mentioned before, you made things more complicated by requiring that the light sphere reached x'=d/2 at t'=0, this means that the original flash (the tip of the light cone) must have happened at x'=0, t'=-d/(2c). Translating this into frame O:

x = gamma*(x' + vt) = -gamma*d*v/(2c). With v = c/sqrt(2) this becomes:
-(1/sqrt(1/2))*d/(2*sqrt(2)) = -d/2

t = gamma*(t' + vx'/c^2) = -gamma*d/(2c). With v=c/sqrt(2) this becomes:
-(1/sqrt(1/2))*d/(2c) = -d/(c*sqrt(2))

So in frame O, the flash happened at x=-d/2, t=-d/(c*sqrt(2)). If you want to figure out the coordinates in O of an object which remains at the center of the light sphere in O', you'd need an object with velocity v which passes through those coordinates, so its position as a function of time would be:

x(t) = vt - v*(-d/(c*sqrt(2))) - d/2

So at time t = d/(2c), this object would be at:

x = vd/(2c) + vd/(c*sqrt(2)) - d/2 = vd/(2c) + (vd*sqrt(2))/(2c) - dc/2c
= d*[v*(1 + sqrt(2)) - c]/2c

How do you figure x=-d/2, t=-d/(c*sqrt(2)).

The flash happened at the center of O when x = 0.


Let's keep this simple.

Only look at O for the time being.

The flash occurred at the origin and proceeds spherically and stikes the points r, -r at time r/c.

That is a fact.

Now, we looked at a general equation when O calculated O' saw its points struck at the same time.

We did that. Here is the equation.
<br /> t = \frac{d}{2c} \frac{1}{\sqrt{c^2/v^2 - 1}}<br />
OK, that is when O believes O' sees the simultaneous strike.

I said set v = c/√2.

Guess what, they occur are at the same time in O.

 

[JesseM]

How do you figure x=-d/2, t=-d/(c*sqrt(2)).

The flash happened at the center of O when x = 0.

Not if you wanted to have it so the flash happened at the center of O', and reached the right endpoint at x'=d/2 at time t'=0 in O'. In that case the flash cannot have happened at x=0 in O.

Let's keep this simple.

Only look at O for the time being.

The flash occurred at the origin and proceeds spherically and stikes the points r, -r at d/r.

That is a fact.

OK, then when you translate into O', it won't be true that the light reached x'=d/2 at time t'=0. Everything will be simpler if you assume the flash happened at the spacetime origin, in which case it reaches the right endpoint of the rod at rest in O' at time t'=d/(2c). That would mean you'd have to revise your calculation of the time t in frame O that the light reached the right endpoint of the rod at rest in O'.

Now, we looked at a general equation when O calculated O' saw its points struck at the same time.

We did that. Here is the equation.
<br /> t = \frac{d}{2c} \frac{1}{\sqrt{c^2/v^2 - 1}}<br />
OK, that is when O believes O' sees the simultaneous strike.

That equation was based on the assumption that the light reached the right endpoint of the rod at rest in O' at time t'=0 in O', which means if you want the flash to have happened at the center of this rod at x'=0 in O', it must have happened at t'=-d/(2c), which means (by applying the Lorentz transformation to x'=0, t'=-d/(2c)) it did not happen at x=0 in O.

I suppose you could drop the assumption that the flash happened at the center of the rod at x'=0 in O'. Then if we say it hit the right endpoint at t = \frac{d}{2c} \frac{1}{\sqrt{c^2/v^2 - 1}} in O, and we plug in v=c/sqrt(2) so t=d/(2c). And in frame O the right endpoint has x(t) = d/(2*gamma) + vt, so at t=d/(2c) it is at:
x = d/(2*gamma) + vd/(2c)

And with v=c/sqrt(2) and gamma=sqrt(2), this becomes:

x = d/(2*sqrt(2)) + d/(2*sqrt(2)) = d/sqrt(2)

So if we want the flash to have happened at x=0, it must have happened at a time d/(c*sqrt(2)) earlier than the moment the light reached the right end in frame O, meaning in frame O the flash happened at t = d/(2c) - d/(c*sqrt(2)) = d/(2c) - (d*sqrt(2))/(2c) = d*(1 - sqrt(2))/(2c). But of course if the flash happened at x=0 and t=d*(1 - sqrt(2))/(2c), that means it did not happen at x'=0 in frame O'.

I said set v = c/√2.

Guess what, they occur are at the same time in O.

What do you mean "they"? You didn't calculate the time of the light hitting the left endpoint in O.

 

[cfrogue]

Not if you wanted to have it so the flash happened at the center of O', and reached the right endpoint at x'=d/2 at time t'=0 in O'. In that case the flash cannot have happened at x=0 in O.

So, when did it happen?

We have no distance to consider.

OK, then when you translate into O', it won't be true that the light reached x'=d/2 at time t'=0. Everything will be simpler if you assume the flash happened at the spacetime origin, in which case it reaches the right endpoint of the rod at rest in O' at time t'=d/(2c). That would mean you'd have to revise your calculation of the time t in frame O that the light reached the right endpoint of the rod at rest in O'.

The light flashed in the frame of O at the origin. That is the experiment and you are saying something different.

 

[cfrogue]

That equation was based on the assumption that the light reached the right endpoint of the rod at rest in O' at time t'=0 in O', which means if you want the flash to have happened at the center of this rod at x'=0 in O', it must have happened at t'=-d/(2c), which means (by applying the Lorentz transformation to x'=0, t'=-d/(2c)) it did not happen at x=0 in O.

What??

And,
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/

Thus, the above is not correct.

 

[JesseM]

The light flashed in the frame of O at the origin. That is the experiment and you are saying something different.

Did it happen at t=0, or do you allow it to have happened at an earlier time in O? If it happened at x=0 and t=0 in O, then by the Lorentz transformation it also happened at x'=0 and t'=0 in O', which means it's impossible that the light from the flash arrived at x'=d/2 at t'=0 in O'. On the other hand, if you're willing to drop the assumption that the flash happened at t=0 in O, and also drop the assumption that it happened at x'=0 in O', then it can work...as I showed, if you start with these assumptions:

1. The flash happened at x=0 in O
2. the light from the flash reached x'=d/2 at t'=0 in O'
3. v = c/sqrt(2)

...then you get the conclusion that the flash happened at t= - d*(sqrt(2) - 1)/(2c) in O.

 

[JesseM]

What??

And,
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/

Thus, the above is not correct.

"Thus"? You're not making any sense, there's no reason the highlighted sentence contradicts my own statement that "(by applying the Lorentz transformation to x'=0, t'=-d/(2c))".

 

[cfrogue]

Did it happen at t=0, or do you allow it to have happened at an earlier time in O? If it happened at x=0 and t=0 in O, then by the Lorentz transformation it also happened at x'=0 and t'=0 in O', which means it's impossible that the light from the flash arrived at x'=d/2 at t'=0 in O'. On the other hand, if you're willing to drop the assumption that the flash happened at t=0 in O, and also drop the assumption that it happened at x'=0 in O', then it can work...as I showed, if you start with these assumptions:

1. The flash happened at x=0 in O
2. the light from the flash reached x'=d/2 at t'=0 in O'
3. v = c/sqrt(2)

...then you get the conclusion that the flash happened at t= - d*(sqrt(2) - 1)/(2c) in O.

Let's see.

What stops us from synching the clocks at zero with the light flash since we have collinear relative motion?

Sure, any angled motion would prevent this.

 

[cfrogue]

"Thus"? You're not making any sense, there's no reason the highlighted sentence contradicts my own statement that "(by applying the Lorentz transformation to x'=0, t'=-d/(2c))".

ok, we then know we can sync the positions to zero, agreed?

 

[JesseM]

Let's see.

What stops us from synching the clocks at zero with the light flash since we have collinear relative motion?

Sure, any angled motion would prevent this.

Of course you can synch the clocks so that the flash happens at t=0 in frame O and t'=0 in frame O'. But if the flash happens at t'=0 in frame O', how can the light from the flash be at position x=d/2 at t'=0 in frame O'? Light can only expand away from the flash at c in frame O', just like any other frame. Are you suggesting the flash itself happened on the endpoint of the rod in O' at t'=0?

 

[JesseM]

ok, we then know we can sync the positions to zero, agreed?

If you want to set the positions and times so the flash happens at x'=0 and t'=0 in frame O', then it's not possible for the light from the flash to have reached x'=d/2 at t'=0, the light's position as a function of time in the right direction would have to be x'(t')=ct' so it wouldn't reach x'=d/2 until t'=d/(2c).

 

[cfrogue]

Of course you can synch the clocks so that the flash happens at t=0 in frame O and t'=0 in frame O'. But if the flash happens at t'=0 in frame O', how can the light from the flash be at position x=d/2 at t'=0 in frame O'? Light can only expand away from the flash at c in frame O', just like any other frame. Are you suggesting the flash itself happened on the endpoint of the rod in O' at t'=0?

All I want to know right now is can we sync the clocks and the positions.

All evidence says we can.

 

[JesseM]

All I want to know right now is can we sync the clocks and the positions.

All evidence says we can.

I don't know what you mean by "sync the clocks and the positions", you're speaking too vaguely. There is no way to sync the clocks and positions in frame O' so that it is both true that the flash happened at x'=0 and t'=0 and that the light from the flash reached x'=d/2 at time t'=0. On the other hand, if you're just asking whether we can construct an inertial coordinate system out of clocks and rulers of the type Einstein described for frame O', of course we can, and we can construct another one for frame O such that x'=0 and t'=0 in O' lines up with x=0 and t=0 in O.

 

[cfrogue]

I don't know what you mean by "sync the clocks and the positions", you're speaking too vaguely. There is no way to sync the clocks and positions in frame O' so that it is both true that the flash happened at x'=0 and t'=0 and that the light from the flash reached x'=d/2 at time t'=0. On the other hand, if you're just asking whether we can construct an inertial coordinate system out of clocks and rulers of the type Einstein described for frame O', of course we can, and we can construct another one for frame O such that x'=0 and t'=0 in O' lines up with x=0 and t=0 in O.

Alright, do you agree we can sync the positions?

Now if O' syncs t to zero, and so does O, what is the time difference to make these two start the event at time t = 0 in both?

 

[JesseM]

Alright, do you agree we can sync the positions?

Sync the positions of what? Again, are you just asking whether we can have x=0 and t=0 in O match up with x'=0 and t'=0 in O'? If so, that's what I just said:

if you're just asking whether we can construct an inertial coordinate system out of clocks and rulers of the type Einstein described for frame O', of course we can, and we can construct another one for frame O such that x'=0 and t'=0 in O' lines up with x=0 and t=0 in O.

But if you mean something different than "sync the positions", you have to actually explain what you're talking about rather than speaking so cryptically.

Now if O' syncs t to zero, and so does O, what is the time difference to make these two start the event at time t = 0 in both?

Time difference between what and what? And when you say "start the event", start what event? The original flash of light?

 

[cfrogue]

Sync the positions of what? Again, are you just asking whether we can have x=0 and t=0 in O match up with x'=0 and t'=0 in O'? If so, that's what I just said:

So, Yes we can?

 

[JesseM]

So, Yes we can?

Yes, we can sync things up so x=0 and t=0 in O matches up with x'=0 and t'=0 in O'. I was already assuming these events matched up in all my previous calculations (since it's a requirement for them to match up in order for the Lorentz transformation to work).

 

[cfrogue]

Yes, we can sync things up so x=0 and t=0 in O matches up with x'=0 and t'=0 in O'. I was already assuming these events matched up in all my previous calculations (since it's a requirement for them to match up in order for the Lorentz transformation to work).

OK, post 334 is based purely on LT and not opinions.

I do not see you matching all the consequences.

Post 334 uses LT to give a stunning conclusion. You and I both derived this with you being the major player.

How do you square all this with your "new" assertions?