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

[JesseM]

There are 5 events in question:
1) flash: t = t' = 0, x = x' = 0
2) light hits right end of unprimed rod: t = d/(2c), x = d/2
3) light hits left end of unprimed rod: t = d/(2c), x = -d/2
4) light hits right end of primed rod: t' = d/(2c), x' = d/2
5) light hits left end of primed rod: t' = d/(2c), x' = -d/2

But cfrogue didn't say the flash coincided with the origin...there's no reason it needs to, the flash can occur anywhere in spacetime we want to set it, and the origins of the two coordinate systems can occur somewhere else. On the other hand, sometimes cfrogue says things which suggest maybe he does want the flash to occur at the origin...if true that would just mean his thinking is confused, though, since having the flash occur at the origin would be incompatible with the stated assumption that the light strikes the right end of the rod in O' at t'=0 and x'=d/2.

 

[atyy]

There are 5 events in question:
1) flash: t = t' = 0, x = x' = 0
2) light hits right end of unprimed rod: t = d/(2c), x = d/2
3) light hits left end of unprimed rod: t = d/(2c), x = -d/2
4) light hits right end of primed rod: t' = d/(2c), x' = d/2
5) light hits left end of primed rod: t' = d/(2c), x' = -d/2

My interpretation too.

But cfrogue didn't say the flash coincided with the origin...there's no reason it needs to, the flash can occur anywhere in spacetime we want to set it, and the origins of the two coordinate systems can occur somewhere else. On the other hand, sometimes cfrogue says things which suggest maybe he does want the flash to occur at the origin...if true that would just mean his thinking is confused, though, since having the flash occur at the origin would be incompatible with the stated assumption that the light strikes the right end of the rod in O' at t'=0 and x'=d/2.

Yes, I think cfrogue is confused in that way, not totally sure though, I am confused as to how he is confused!

 

[Dale]

But cfrogue didn't say the flash coincided with the origin...there's no reason it needs to, the flash can occur anywhere in spacetime we want to set it, and the origins of the two coordinate systems can occur somewhere else. On the other hand, sometimes cfrogue says things which suggest maybe he does want the flash to occur at the origin...if true that would just mean his thinking is confused, though, since having the flash occur at the origin would be incompatible with the stated assumption that the light strikes the right end of the rod in O' at t'=0 and x'=d/2.

I think that is what he intended, it certainly is what he intended until the last few pages where the discussion between you and he became confusing to me. Let's wait for him to confirm or clarify.

 

[cfrogue]

Well, it is axiomatic that t=d/(2c) in O.

Um, no it isn't. The rod is at rest in O', not O, remember?

Yes, I was talking about what O sees about the flash here. I preobably need to be very specific of the frames.

Anyway, this is what O sees for its own version of the light sphere.

In O' the strikes occurred simultaneously (as you said in post 319) and at t'=0 (as you said in post 322), that was part of the original conditions of the problem! It was when and where they occurred in O that you were wondering about.

There are three ideas.
1) When does O see the strikes of the endpoints of its own rod
2) When does O see the strikes of the endpoints of the rod of O'. This has not been part of the discussions, but O sees the strikes at d/(2λ(c-v)), d/(2λ(c+v)).
3) We then explored the question, when does O calculate that O' will see the simultaneous strikes in its own frame. We came up with this equation.

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

We know that ct' = ±x' at this time t, in the frame in O. So, we then set v = c/√2.
At this v, the light sphere strikes the endpoints of the rod of O and it also strikes the endpoints of the rod of O'. But, the center of the rod/light sphere of O' is located at vt = (c/√2)(d/(2c))= d/(2 √2 ) in the coordinates of O.

 

[cfrogue]

No. The formula for t(R') is only correct if (x'(R')=d/2, t'(R')=0), which cannot be the case if the light is emitted at (x=x'=0, t=t'=0). You made the same error back in #334, and JesseM corrected your error in #335:

Thanks for the music link!

So, when does O calculate that O' sees the right and left end points struck at the same time?


This is not asking when does O see the left and right endpoint of O' struck.

 

[cfrogue]

What do you mean by "because the rod length is d and the light is centered". Is that supposed to explain why t'=0?

OK, d/2 is because the roid length is d while at rest in O' and the light source is centered, therefore, light must travel in O' from the center to the endpoints which is a length d/2.

t'=0 means that is the initial condition at which the light pulse is emitted.

 

[cfrogue]

There are 5 events in question:
1) flash: t = t' = 0, x = x' = 0
2) light hits right end of unprimed rod: t = d/(2c), x = d/2
3) light hits left end of unprimed rod: t = d/(2c), x = -d/2
4) light hits right end of primed rod: t' = d/(2c), x' = d/2
5) light hits left end of primed rod: t' = d/(2c), x' = -d/2

Agreed.

Then, the question is, in the time of O, when does the light sphere satisfy 4 and 5 as simultaneous in O'.

 

[cfrogue]

But cfrogue didn't say the flash coincided with the origin...there's no reason it needs to, the flash can occur anywhere in spacetime we want to set it, and the origins of the two coordinate systems can occur somewhere else. On the other hand, sometimes cfrogue says things which suggest maybe he does want the flash to occur at the origin...if true that would just mean his thinking is confused, though, since having the flash occur at the origin would be incompatible with the stated assumption that the light strikes the right end of the rod in O' at t'=0 and x'=d/2.

You and I went through a lengthy discusstion that
x=0, t=0, x'=0, t'=0.

No, it does not have to be this way, but it is easier.

 

[JesseM]

Yes, I was talking about what O sees about the flash here. I preobably need to be very specific of the frames.

But what do you mean for "what O sees about the flash" specifically? The coordinates in O that the flash occurs? The coordinates in O that the light from the flash reaches the endpoint of a rod at rest in O? Or the coordinates in O that the light from the flash reaches the endpoint of a rod at rest in O'? It was the third one that we calculated when we found t = d/(2c)--first you had assumed the light reached the endpoint of the rod at rest in O' at t'=0 and x'=d/2 (post 322), then you had performed the Lorentz transformation on these coordinates and we found that the corresponding t coordinate in O was t = (d/2c)*1/sqrt(c^2/v^2 - 1) (posts 322, 325, and 327), then you plugged in v=c/sqrt(2) into this and we got t=d/2c (posts 330 and 331). So, the whole thing was based on assuming we were talking about the time the light reached the end of the rod at rest in O', and finding the time coordinate of this event in O assuming that the coordinates of this event in O' were t'=0 and x'=d/2. Do you disagree with this synopsis? (if you do, please review the posts I linked to and see if any of my description is inaccurate)

There are three ideas.
1) When does O see the strikes of the endpoints of its own rod
2) When does O see the strikes of the endpoints of the rod of O'. This has not been part of the discussions, but O sees the strikes at d/(2λ(c-v)), d/(2λ(c+v)).

You really are confused, 2) was the whole basis for our discussions.

3) We then explored the question, when does O calculate that O' will see the simultaneous strikes in its own frame. We came up with this equation.

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

No, that equation was only for the time coordinate in O of the light reaching the right endpoint of the rod at rest in O', again assuming the light hit the right endpoint of the rod at rest in O' at t'=0 and x'=d/2 in O'. Again, please review the previous posts.

We know that ct' = ±x' at this time t, in the frame in O. So, we then set v = c/√2.
At this v, the light sphere strikes the endpoints of the rod of O and it also strikes the endpoints of the rod of O'.

No, you specifically assumed that the light hit the right end of the rod at rest in O' at time t'=0 in O', not at time t'=d/(2c) in O'.

But cfrogue didn't say the flash coincided with the origin...there's no reason it needs to, the flash can occur anywhere in spacetime we want to set it, and the origins of the two coordinate systems can occur somewhere else. On the other hand, sometimes cfrogue says things which suggest maybe he does want the flash to occur at the origin...if true that would just mean his thinking is confused, though, since having the flash occur at the origin would be incompatible with the stated assumption that the light strikes the right end of the rod in O' at t'=0 and x'=d/2.

You and I went through a lengthy discusstion that
x=0, t=0, x'=0, t'=0.

No, it does not have to be this way, but it is easier.

More confusion on your part, we were only talking about whether the clocks and positions of the two frames were synched such that the origin of one lined up with the origin of the other, i.e. x=0 and t=0 lined up with x'=0 and t'=0 (this is always assumed in any situation where you make use of the Lorentz transformation, if it wasn't true you'd have to use a more general 'Poincaré transformation'). This has nothing to do with the question of whether the light flash happened at the origin of both frames! The light flash can happen at any coordinates we want it to. If you want to have the light flash happen at the spacetime origin of both frames that will certainly make the problem a lot simpler, but then you will have to change your original assumption from post 322 that the light reached the endpoint of the rod at rest in O' at time t'=0 and position x'=d/2, because it's incompatible with the idea that the original flash happened at x'=0 and t'=0. Instead we should say, as DaleSpam did, that the light reached the endpoint of the rod at rest in O' at time t'=d/(2c) and position x'=d/2, and then if we apply the Lorentz transformation to that in order to find the coordinates of this event in O, we will not get a time coordinate of t = (d/2c)*1/sqrt(c^2/v^2 - 1) as in the previous derivation. Instead the time coordinate of this event in O will work out to:

t = gamma*(t' + vx'/c^2) = gamma*(d/(2c) + (d/2c)*(v/c)) = (d/2c)*gamma + (d/2c)*gamma*(v/c) = (d/2c)*[1/sqrt(1 - v^2/c^2)] + (d/2c)*1/sqrt(c^2/v^2 - 1).

If you then plug in v=c/sqrt(2) and gamma=sqrt(2) you get:

t = (d/2c)*(sqrt(2) + 1)

Not t=d/2c as was the case previously when you assumed the event happened at t'=0 and x'=d/2.

 

[cfrogue]

But what do you mean for "what O sees about the flash" specifically? The coordinates in O that the flash occurs? The coordinates in O that the light from the flash reaches the endpoint of a rod at rest in O? Or the coordinates in O that the light from the flash reaches the endpoint of a rod at rest in O'? It was the third one that we calculated when we found t = d/(2c)--first you had assumed the light reached the endpoint of the rod at rest in O' at t'=0 and x'=d/2 (post 322), then you had performed the Lorentz transformation on these coordinates and we found that the corresponding t coordinate in O was t = (d/2c)*1/sqrt(c^2/v^2 - 1) (posts 322, 325, and 327), then you plugged in v=c/sqrt(2) into this and we got t=d/2c (posts 330 and 331). So, the whole thing was based on assuming we were talking about the time the light reached the end of the rod at rest in O', and finding the time coordinate of this event in O assuming that the coordinates of this event in O' were t'=0 and x'=d/2. Do you disagree with this synopsis? (if you do, please review the posts I linked to and see if any of my description is inaccurate)

So, the whole thing was based on assuming we were talking about the time the light reached the end of the rod at rest in O', and finding the time coordinate of this event in O

This is the goal.

In O', light must travel a distance d/2 in both directions so x' is OK.

The question I then asked was, when will this happen in the time of O, that the light travels d/2 in O'?

I think that t'=0 is wrong. I think I made a mistake there. Thanks for going through that.

Should it be this?

x'=d/2.

x'=ct'.

t = ( t' - vx'/c² )λ

t = ( x'/c - vx'/c² )λ

t = ( d/(2c) - vd/2c²)λ

t = d/(2c) ( 1 - (v/c) )λ

Does this look better?

 

[JesseM]

So, the whole thing was based on assuming we were talking about the time the light reached the end of the rod at rest in O', and finding the time coordinate of this event in O

This is the goal.

In O', light must travel a distance d/2 in both directions so x' is OK.

The question I then asked was, when will this happen in the time of O, that the light travels d/2 in O'?

I think that t'=0 is wrong. I think I made a mistake there. Thanks for going through that.

Should it be this?

x'=d/2.

x'=ct'.

Right, so t' = x'/c = d/2c.

t = ( t' - vx'/c² )λ

Minor mistake: it should be a + sign there since we're going from O' to O (and O' is the one moving at +v relative to O), i.e. t = gamma*(t' + vx'/c^2)

t = ( x'/c - vx'/c² )λ

t = ( d/(2c) - vd/2c²)λ

t = d/(2c) ( 1 - (v/c) )λ

Does this look better?

should be t = (d/2c)*(1 + (v/c))*gamma, but otherwise yes.

I'm taking a little trip for the weekend so I'll have to continue this on Monday, see you later...

 

[Dale]

There are 5 events in question:
1) flash: t = t' = 0, x = x' = 0
2) light hits right end of unprimed rod: t = d/(2c), x = d/2
3) light hits left end of unprimed rod: t = d/(2c), x = -d/2
4) light hits right end of primed rod: t' = d/(2c), x' = d/2
5) light hits left end of primed rod: t' = d/(2c), x' = -d/2

Agreed.

Then, the question is, in the time of O, when does the light sphere satisfy 4 and 5 as simultaneous in O'.

This is found simply through the Lorentz transform:
t = \gamma (t&#039; - v x&#039;/c^2)
x = \gamma (x&#039; - v t&#039;)

So we just plug in and evaluate:
4)
t_4 = \gamma \left( \frac{d}{2c} - v \frac{d}{2c^2} \right)
x_4 = \gamma \left( \frac{d}{2} - v \frac{d}{2c} \right)

5)
t_5 = \gamma \left( \frac{d}{2c} + v \frac{d}{2c^2} \right)
x_5 = \gamma \left( -\frac{d}{2} - v \frac{d}{2c} \right)

 

[cfrogue]

Right, so t' = x'/c = d/2c.

Minor mistake: it should be a + sign there since we're going from O' to O (and O' is the one moving at +v relative to O), i.e. t = gamma*(t' + vx'/c^2)

should be t = (d/2c)*(1 + (v/c))*gamma, but otherwise yes.

I'm taking a little trip for the weekend so I'll have to continue this on Monday, see you later...

Great thanks, later.

Looks correct. I am going to work on this a while.

 

[Dale]

Minor mistake: it should be a + sign there since we're going from O' to O (and O' is the one moving at +v relative to O), i.e. t = gamma*(t' + vx'/c^2)

My personal preference is to leave the formula the same and just let v be negative if it is going to the left. Either way is fine.

 

[cfrogue]

Here is the summary.

A light source is centered on a moving rod, moving at v, of rest length d, with the center at O'.

The stationary frame also has a rod of rest length d and the center at O.

When O and O' are coincident, light emits at O' which is a light source.

The questions is, in the time of O, when does O' see the endpoints of its rod struck simultaneously.

 |-------->v
               Light is emitted
                     |
 |-------------------O'------------------|
|--------------------O--------------------|
-d/2                                       d/2

Below, the left endpoint and right endpoint are struck simultaneously in O'.
   |-------->v
   |-------------------O'--------------------|
|--------------------O--------------------|
-d/2                                       d/2

This is the time in O when O' sees the simultaneous stikes.

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

Please correct any errors in the above.

 

[cfrogue]

My personal preference is to leave the formula the same and just let v be negative if it is going to the left. Either way is fine.

How do I see formulas in lex here?

All I see is black with some very light white.

I have to copy them elsewhere to see them.

 

[DrGreg]

How do I see formulas in lex here?

All I see is black with some very light white.

I have to copy them elsewhere to see them.

You must have an out-of-date internet browser. Update to a more recent version if you can. (e.g. Internet Explorer 7 or 8, Firefox 3, Safari 4.)

 

[cfrogue]

You must have an out-of-date internet browser. Update to a more recent version if you can. (e.g. Internet Explorer 7 or 8, Firefox 3, Safari 4.)

Thanks DrGreg.

 

[atyy]

There are 5 events in question:
1) flash: t = t' = 0, x = x' = 0
2) light hits right end of unprimed rod: t = d/(2c), x = d/2
3) light hits left end of unprimed rod: t = d/(2c), x = -d/2
4) light hits right end of primed rod: t' = d/(2c), x' = d/2
5) light hits left end of primed rod: t' = d/(2c), x' = -d/2

I'm working from the above.

Here is the summary.

A light source is centered on a moving rod, moving at v, of rest length d, with the center at O'.

The stationary frame also has a rod of rest length d and the center at O.

When O and O' are coincident, light emits at O' which is a light source.

The questions is, in the time of O, when does O' see the endpoints of its rod struck simultaneously.

 |-------->v
               Light is emitted
                     |
 |-------------------O'------------------|
|--------------------O--------------------|
-d/2                                       d/2

Below, the left endpoint and right endpoint are struck simultaneously in O'.
   |-------->v
   |-------------------O'--------------------|
|--------------------O--------------------|
-d/2                                       d/2

In your drawing for the emission, you took the unprimed point of view and drew the primed rod contracted. If your drawing for light striking an endpoint (there are 4 such events) is also from the unprimed point of view, then the primed rod should remain contracted.

This is the time in O when O' sees the simultaneous stikes.

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

I haven't checked the details, but you have only one time where you should have two different times by plugging DaleSpam's (x'4, t'₄) into the Lorentz transformations to get (x₄, t₄), and similarly with (x'₅, t'₅) to get (x₅, t₅), which should match DaleSpam's:

This is found simply through the Lorentz transform:
t = \gamma (t&#039; - v x&#039;/c^2)
x = \gamma (x&#039; - v t&#039;)

So we just plug in and evaluate:
4)
t_4 = \gamma \left( \frac{d}{2c} - v \frac{d}{2c^2} \right)
x_4 = \gamma \left( \frac{d}{2} - v \frac{d}{2c} \right)

5)
t_5 = \gamma \left( \frac{d}{2c} + v \frac{d}{2c^2} \right)
x_5 = \gamma \left( -\frac{d}{2} - v \frac{d}{2c} \right)

Edit: In the above formulas, I believe DaleSpam's "v" is cfrogue's "-v".

 

[Dale]

Here is the summary.

A light source is centered on a moving rod, moving at v, of rest length d, with the center at O'.

The stationary frame also has a rod of rest length d and the center at O.

When O and O' are coincident, light emits at O' which is a light source.

The questions is, in the time of O, when does O' see the endpoints of its rod struck simultaneously.

 |-------->v
               Light is emitted
                     |
 |-------------------O'------------------|
|--------------------O--------------------|
-d/2                                       d/2

Below, the left endpoint and right endpoint are struck simultaneously in O'.
   |-------->v
   |-------------------O'--------------------|
|--------------------O--------------------|
-d/2                                       d/2

This is the time in O when O' sees the simultaneous stikes.

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

Please correct any errors in the above.

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

 

[cfrogue]

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

The question that was asked is what is the time in O in which O' sees the strikes as simultaneous.

The equations you listed are the times O sees the two different strikes.

That is not the issue.

O' sees the endpoints of the rod struck at the same time at some time t'

There exists a time in O when this occurs.

That is the answer I was looking for.

So, again, the question is not when does O see the endpoints of O' struck.

The question is when in the time of O does O' see the strikes of its endpoints.

 

[atyy]

There are 4 strikes of endpoints - 2 endpoints per rod.

For O , 2 of them are simultaneous, 2 are not simultaneous.

For O' , 2 of them are simultaneous, 2 are not simultaneous.

The strikes simultaneous for O are not simultaneous for O'

The strikes simultaneous for O' are not simultaneous for O.

 

[cfrogue]

There are 4 strikes of endpoints - 2 endpoints per rod.

For O , 2 of them are simultaneous, 2 are not simultaneous.

For O' , 2 of them are simultaneous, 2 are not simultaneous.

The strikes simultaneous for O are not simultaneous for O'

The strikes simultaneous for O' are not simultaneous for O.

This is all true.

For O' , 2 of them are simultaneous
This occurs for the endpoints of O'.

Now, the question that was asked is when does this happen in the time of t.

 

[Al68]

This is all true.

For O' , 2 of them are simultaneous
This occurs for the endpoints of O'.

Now, the question that was asked is when does this happen in the time of t.

DaleSpam gave that answer as:

<br /> t_4 = \frac{d}{2c} {\sqrt{\frac{c+v}{c-v}}<br />
and
<br /> t_5 = \frac{d}{2c} {\sqrt{\frac{c-v}{c+v}}<br />

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.

 

[Dale]

The question that was asked is what is the time in O in which O' sees the strikes as simultaneous.

The equations you listed are the times O sees the two different strikes.

That is not the issue.

O' sees the endpoints of the rod struck at the same time at some time t'

There exists a time in O when this occurs.

That is the answer I was looking for.

So, again, the question is not when does O see the endpoints of O' struck.

The question is when in the time of O does O' see the strikes of its endpoints.

It seems like you don't yet fully grasp the relativity of simultaneity. That is OK, it is one of the most difficult concepts to master.

Because the strikes are simultaneous in O' they are not simultaneous in O. That means that the strikes have two different times in O, as shown by the Lorentz transform above. So your question, as posed, has no answer. There is no single time in O when both endpoints of O' are struck by the light.

The key error is here:

There exists a time in O when this occurs.

There does not exist a single such time. There exist two, one for each end of the rod.

 

[cfrogue]

DaleSpam gave that answer as:

<br /> t_4 = \frac{d}{2c} {\sqrt{\frac{c+v}{c-v}}<br />
and
<br /> t_5 = \frac{d}{2c} {\sqrt{\frac{c-v}{c+v}}<br />

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.

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

 

[cfrogue]

It seems like you don't yet fully grasp the relativity of simultaneity. That is OK, it is one of the most difficult concepts to master.

Because the strikes are simultaneous in O' they are not simultaneous in O. That means that the strikes have two different times in O, as shown by the Lorentz transform above. So your question, as posed, has no answer. There is no single time in O when both endpoints of O' are struck by the light.

The key error is here:

There does not exist such a time.

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))]

 

[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.

 

[atyy]

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?

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"

 

[Al68]

t_L < t_R
Is this R of S?

Yes, and it's also the only logical way that the light is "spherical" in O when the rod is in motion.