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

[cfrogue]

The time interval for a single event is 0 so, 0 = γ 0

Obviously, there must be some specific end for two time points in a frame, but the equation is a general one.

 

[Dale]

Obviously, there must be some specific end for two time points in a frame, but the equation is a general one.

No, it is not general, it is specific to the time between two events which are co-located in one frame. I showed that explicitly previously in this thread and Einstein assumed a stronger condition in his derivation.

 

[cfrogue]

The time interval for a single event is 0 so, 0 = γ 0

I gave a silly answer.

I will call the end of the event in O' when O' sees the simultaneous strike.

 

[Dale]

I gave a silly answer.

I will call the end of the event in O' when O' sees the simultaneous strike.

Then the two events are not co-located and the time dilation formula does not apply.

Whenever you encounter a new formula in physics, the number one most important thing to learn about that formula is not the details of the equation itself, nor even the details of the derivation of the formula. Rather the single most important thing to learn is the circumstances to which the formula applies and those to which it doesn't apply.

 

[cfrogue]

Then the two events are not co-located and the time dilation formula does not apply.

Whenever you encounter a new formula in physics, the number one most important thing to learn about that formula is not the details of the equation itself. Rather the single most important thing to learn is the circumstances to which the formula applies and those to which it doesn't apply.

OK, I have a start to the time in O', and I have an end.

This implies I can never use time dilation.

Dont forget, I am not calculating anything in O. x = 0.

So, can you show some end times that are legal.

 

[cfrogue]

Then the two events are not co-located and the time dilation formula does not apply.

Whenever you encounter a new formula in physics, the number one most important thing to learn about that formula is not the details of the equation itself, nor even the details of the derivation of the formula. Rather the single most important thing to learn is the circumstances to which the formula applies and those to which it doesn't apply.

How about this.

Can you tell me in the time of O when O' sees the simultaneous strikes?

 

[Dale]

So, can you show some end times that are legal.

Sure. Pick one single x' line (e.g. x'=1) and choose any two events on that line.

 

[cfrogue]

Sure. Pick one single x' line (e.g. x'=1) and choose any two events on that line.

Oh, so distance is the key to picking this?

OK, so I see the distance, what is the time dilation?

 

[Dale]

Can you tell me in the time of O when O' sees the simultaneous strikes?

We have gone over, and over, and over this again and again already. The left one is at t=0.5 and the right one is at t=2 (for v=0.6, c=1, and r=1).

 

[cfrogue]

We have gone over, and over, and over this again and again already. The left one is at t=0.5 and the right one is at t=2 (for v=0.6, c=1, and r=1).

I am not saying when O sees the strikes. I am not saying that. O is not watching.

O calls O' on a light phone, what is the answer?

 

[atyy]

I am not saying when O sees the strikes. I am not saying that. O is not watching.

O calls O' on a light phone, what is the answer?

You've calculated this yourself, and even called it relativity of simultaneity. The following two calculations of yours match up if you set r=d/2.

When x'=-r in O', that is at the time r/(λ(c+v)) in O.
When x'=+r in O', that is at the time r/(λ(c-v)) in O.

Thus, two different times in O are producing simultaneity in O'.

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))]

 

[cfrogue]

You've calculated this yourself, and even called it relativity of simultaneity. The following two calculations of yours match up if you set r=d/2.

O calls O' on a light phone and asks the time of simultaneity.

so, what is the answer?

 

[cfrogue]

I will answer.

O calls O' on the light phone and O' says, says t' = r/c.

O has an endpoint of time.

It is time to apply time dilation.

 

[Dale]

Oh, so distance is the key to picking this?

OK, so I see the distance, what is the time dilation?

Yes, again, the position must be constant (Δx=0) in one of the frames, then the time dilation formula applies.

 

[atyy]

I will answer.

O calls O' on the light phone and O' says, says t' = r/c.

O has an endpoint of time.

It is time to apply time dilation.

Go ahead, let's see what you get. I am generally incompetent with length contraction and time dilation, but I can check your formula using the LT.

For example, back when you derived ct(R')=d/(2γ)+vt, you used length contraction, and I had to check it by Lorentz transformation of (x'(R')=d/2,t'(R')=d/2c).

Blake - Walking in the air

 

[Al68]

I will answer.

O calls O' on the light phone and O' says, says t' = r/c.

O has an endpoint of time.

It is time to apply time dilation.

Then let's apply it: t = t' * gamma, like you said. Fine, but since in O, the only point on the rod at which t = t' = 0 simultaneously with the light emission is the origin (rod midpoint), then the resulting t will be the time in O simultaneous with t' at the midpoint of the rod only (x' = 0).

At any other location in O', the light emission was not simultaneous with t = t' = 0, so the elapsed time will not equal coordinate time (in O).

The t obtained (t = t' * gamma) will be the elapsed time in O of any clock stationary in O', but will only be the coordinate time in O simultaneous with t' at the origin of O'.

Bottom line is that the time dilation formula is for elapsed time, which only represents coordinate time for a clock that initially read zero.

In O, if a clock at the origin of O' reads zero when the light is emitted, then clocks at the ends of the rod didn't read zero when the light is emitted, so they won't read the same t' when the light reaches them (in O).

So basically, if an observer at the origin of O' calls an observer in O "on the light phone" and says "Hey dude, my clock read zero when the light was emitted and read t' when the light reached the rod ends, dude", then the observer in O will answer "Whoa dude, when your clock read t', mine read t = t' * gamma, but unlike you, dude, my clock reading t wasn't simultaneous (in O) with the light reaching either end of the rod, dude. Bummer, dude, I might have to learn some SR, dude."

The simplified time dilation formula t = t' * gamma is only valid when t and t' are both initially zero (which means they must have been local). Otherwise the correct time dilation formula is: (final t - initial t) = (final t' - initial t') * gamma.

 

[JesseM]

OK, I have a start to the time in O', and I have an end.

This implies I can never use time dilation.

Of course you can use time dilation, you can use it when both events occur at the same position in one of the frames. But if the first event is the origin where O and O' are coincident, and the second event is where the light reached one end of the rod at rest in O', then in neither frame is the distance between the events 0, in both frames the distance is c times the time interval between the events in that frame.

Dont forget, I am not calculating anything in O. x = 0.

What are you talking about? delta-x is not 0 between these events. You can't just declare "I am not calculating anything in O", since you picked the events there is a definite distance between them in O, whether you choose to calculate it or not. Again, the time dilation formula can only be used when you have two events where the distance between them is 0 in one of the two frames.

So, can you show some end times that are legal.

If your starting event is when O and O' are coincident at x=0, t=0 and x'=0, t'=0, then you can pick any later event that occurs at the origin of one of the two frames, either at x=0 or at x'=0, and at a later time.

 

[Dale]

I am generally incompetent with length contraction and time dilation, but I can check your formula using the LT.

IMO, it is a bad idea (especially for beginners) to use the length contraction or time dilation formulas at all, they are too easy to mess up as you have seen. Instead it is best to always use the Lorentz transform, and the time dilation and length contraction formulas will automatically pop out whenever they are appropriate.

cfrogue, I am going to reiterate this advice which comes from both atyy and myself now. Don't use the length contraction and time dilation formulas, they are not worth it. They are too easy to misapply (as you have repeatedly demonstrated) and they automatically drop out of the Lorentz transform whenever they do apply. They are a minor simplification to the Lorentz transform, but a major source of error. They are just not worth the headache.

 

[cfrogue]

cfrogue, I am going to reiterate this advice which comes from both atyy and myself now. Don't use the length contraction and time dilation formulas, they are not worth it. They are too easy to misapply (as you have repeatedly demonstrated) and they automatically drop out of the Lorentz transform whenever they do apply. They are a minor simplification to the Lorentz transform, but a major source of error. They are just not worth the headache.

Trust me, I will dump junk that does not work immediately.

But, is that not a valid question?

When does O' in its own frame see the simultaneity.

I clearly see a difference of when an event will occur concerning light and time dilation.

For example, when light strikes a point, each frame has a different t for that event, but also, time dilation is built into the calculation.

Thus, there are two calculations to transform a point and t,
1) There exists a difference in simultaneity.
2) There exists time dilation.

I can be more specific.

t' = ( t - vx/c² )λ.

t'/λ = ( t - vx/c² )

So, t'/λ, thus, time dilation is already handled for LT calculations.
-vx/c² handles the simultaneity differential between the two frames.

This is why one cannot apply time dilation to an LT calculation on a spacetime point because it is already done.


But, when I am not transforming a light event, it seems to me time dilation applies.

Either way, we know t' = r/c for simultaneity in O'.

We know O disagrees with the time the points are struck.

But, I believe time dilation applies to t' = r/c because we are not transforming a space time coordinate or applying it to an already transformed coordinate.

 

[cfrogue]

Of course you can use time dilation, you can use it when both events occur at the same position in one of the frames. But if the first event is the origin where O and O' are coincident, and the second event is where the light reached one end of the rod at rest in O', then in neither frame is the distance between the events 0, in both frames the distance is c times the time interval between the events in that frame.

LT already handles time dilation for one event point.

In my view, you apply time dilation for elapsed time differentials for the start and stop points to a time interval in a frame. It has nothing to do with events in the general sense. Sure, events may trigger the stop of the watch or start, but teim dilation applies in general to generic time intervals.

If you are deciding light events, you must use LT and time dilation is already handled.

 

[JesseM]

LT already handles time dilation for one event point.

In my view, you apply time dilation for elapsed time differentials for the start and stop points to a time interval in a frame. It has nothing to do with events in the general sense. Sure, events may trigger the stop of the watch or start, but teim dilation applies in general to generic time intervals.

But as I pointed out in an earlier post, if you pick two times (not events) t0 and t1 in a given frame O, there is no unique answer to how long this interval lasted in another frame O', since if you pick one pair of events A and B that occurred at t0 and t1 in O, and another pair of events C and D that also occurred at t0 and t1 in O (so in O the time interval between A and B is obviously the same as the time interval between C and D ), then in frame O' the time interval between A and B can be different than the time interval between C and D. Assuming you don't disagree with this, how do you propose to define the "length of this time interval" in O' such that there is a unique answer?

 

[cfrogue]

OK, I have the light sphere in O' working correctly.

In section 3, Einstein said the following.

At the time t = τ = 0, when the origin of the co-ordinates is common to the two systems, let a spherical wave be emitted therefrom, and be propagated with the velocity c in system K. If (x, y, z) be a point just attained by this wave, then

x²+y²+z²=c²t².
Transforming this equation with the aid of our equations of transformation we obtain after a simple calculation

x'² + y'² + z'² = τ²c²
The wave under consideration is therefore no less a spherical wave with velocity of propagation c when viewed in the moving system. This shows that our two fundamental principles are compatible.5

http://www.fourmilab.ch/etexts/einstein/specrel/www/


It has to be done just like the above, point by point.

It is false that

ct' = ±x'.

Normally we think of a light sphere this way, but this cannot work under LT.

Each point is LT transformed from O to O' and then the radius for the sphere in O' comes out correctly because of the t' adjustment for each direction of the light ray.

 

[cfrogue]

But as I pointed out in an earlier post, if you pick two times (not events) t0 and t1 in a given frame O, there is no unique answer to how long this interval lasted in another frame O', since if you pick one pair of events A and B that occurred at t0 and t1 in O, and another pair of events C and D that also occurred at t0 and t1 in O (so in O the time interval between A and B is obviously the same as the time interval between C and D ), then in frame O' the time interval between A and B can be different than the time interval between C and D. Assuming you don't disagree with this, how do you propose to define the "length of this time interval" in O' such that there is a unique answer?

Simple, use the universal constant

t' = r/c.

Now bring this back to O.

 

[JesseM]

Simple, use the universal constant

t' = r/c.

Now bring this back to O.

I don't understand your response at all. If I pick a pair of times t0 and t1 in O, so the time interval's size in O is (t1 - t0), can you show me how the above comment is supposed to apply to calculating the time interval in O'? In other words, show me the actual equations used in the derivation which will end up yielding a final equation for the time interval in O'.

Also, what does r represent here? Is it the radius of the light sphere in O' at time t' in O'?

 

[JesseM]

OK, I have the light sphere in O' working correctly.

In section 3, Einstein said the following.

At the time t = τ = 0, when the origin of the co-ordinates is common to the two systems, let a spherical wave be emitted therefrom, and be propagated with the velocity c in system K. If (x, y, z) be a point just attained by this wave, then

x²+y²+z²=c²t².
Transforming this equation with the aid of our equations of transformation we obtain after a simple calculation

x'² + y'² + z'² = τ²c²
The wave under consideration is therefore no less a spherical wave with velocity of propagation c when viewed in the moving system. This shows that our two fundamental principles are compatible.5

http://www.fourmilab.ch/etexts/einstein/specrel/www/


It has to be done just like the above, point by point.

It is false that

ct' = ±x'.

Why do you say it's false? If we set y and z to zero (so we're only looking at where the light is on the x-axis), then x'² + y'² + z'² = t'²c² reduces to t'²c² = x'² which is the same as ct' = ±x'.

 

[cfrogue]

I don't understand your response at all. If I pick a pair of times t0 and t1 in O, so the time interval's size in O is (t1 - t0), can you show me how the above comment is supposed to apply to calculating the time interval in O'? In other words, show me the actual equations used in the derivation which will end up yielding a final equation for the time interval in O'.

Also, what does r represent here? Is it the radius of the light sphere in O' at time t' in O'?

Oh, that is the time the light sphere strikes all points.

The rest radius is r in O', c is constant, thus t' = c/r.

The emission of light is t0=0 and the termination of the interval is t1=c/r.

Note I am not picking a light event as the termination point. It is an elapsed time.

 

[cfrogue]

Why do you say it's false? If we set y and z to zero (so we're only looking at where the light is on the x-axis), then x'² + y'² + z'² = t'²c² reduces to t'²c² = x'² which is the same as ct' = ±x'.

Because we proved there exists two different t's in O such that ct' = ±x' is satisified and this violates logical consistancy.

So, I am saying that is not the way to approach the SR light sphere.

If it is approached as t increases and each point is transformed to satisfy,
x'² + y'² + z'² = τ²c²
then we have logical consistancy.

On the other hand, if we include ±x', we get different length radii for each directiion due to the simultaneity difference in each direction of the light ray in O.

It can only be approached as taking each spacetime coord in O and LT'ing it, then applying
x'² + y'² + z'² = τ²c²


That is why I say the normal logic of ct' = ±x' does not work.

The following points in O satisify that condition

t1=r/(λ(c-v)) for the ray point right and
t2=r/(λ(c+v)) for the ray point left.

If you run these through LT with x=+ct and x=-ct as appropriate, you will find two different satisfactions of the equation ct' = ±x' at two different times in O which creates two different light spheres which is a logical inconsistancy.

 

[cfrogue]

I have concluded LT will not give me the center of the light sphere after and time t in the frame of O.

I realize Einstein assumed it is at vt.

Does anyone have anything else on this that I am not seeing?

 

[cfrogue]

Also, if you run this point through LT,

x = λvt/(1+λ)

t' = t( t - vx/c² )λ

You will find t' = t with collinear relative motion.

Thus, we must be very careful not to conclude LT decides time dilation between frames.

 

[JesseM]

Because we proved there exists two different t's in O such that ct' = ±x' is satisified and this violates logical consistancy.

What violates logical consistency? I don't see anything illogical about the fact that there can be two events with values of (x,t) that satisfy ct = ±x, and with t being different for each event, such that when you apply the Lorentz transformation to the (x,t) coordinates of each event to get the (x',t') coordinates of the same event, then both events can have (x',t') coordinates that satisfy ct' = ±x' but with the new feature that the t' coordinate is the same for each event. Are you saying this is illogical or impossible?

That is why I say the normal logic of ct' = ±x' does not work.

The following points in O satisify that condition

t1=r/(λ(c-v)) for the ray point right and
t2=r/(λ(c+v)) for the ray point left.

If you run these through LT with x=+ct and x=-ct as appropriate, you will find two different satisfactions of the equation ct' = ±x' at two different times in O which creates two different light spheres which is a logical inconsistancy.

Again, what's a logical inconsistency? These events occur at different times in O, so naturally they are part of two different light spheres at different times in O. The equation ct = ±x was never supposed to define a single light sphere, rather it is an equation defining all points that lie on the light cone in 1D, it can include events at different times in O. Exactly the same is true of the equation x'² + y'² + z'² = τ²c², τ is a variable so this too is the 3D version of a light cone which includes events that lie on the path of the light at different times in O.

 

[JesseM]

I have concluded LT will not give me the center of the light sphere after and time t in the frame of O.

I realize Einstein assumed it is at vt.

Why have you concluded that? The center of the light sphere in O' is always at x'=0, and if you apply Lorentz transformation to some event which has an x' coordinate of 0 you'll always get an event whose x and t coordinates satisfy x=vt. Do you disagree? If not, what's the problem?

 

[cfrogue]

Why have you concluded that? The center of the light sphere in O' is always at x'=0, and if you apply Lorentz transformation to some event which has an x' coordinate of 0 you'll always get an event whose x and t coordinates satisfy x=vt. Do you disagree? If not, what's the problem?

I get x=(vt)λ

 

[JesseM]

I get x=(vt)λ

Suppose the event has coordinates x'=0, t'=T' in O'. Then the coordinates in O are:

x = gamma*(x' + vt') = gamma*vT'
t = gamma*(t' + vx'/c^2) = gamma*T'

So, x/t = (gamma*vT')/(gamma*T') = v, and if x/t=v then x=vt.

 

[cfrogue]

Originally Posted by cfrogue
Because we proved there exists two different t's in O such that ct' = ±x' is satisified and this violates logical consistancy.

What violates logical consistency? I don't see anything illogical about the fact that there can be two events with values of (x,t) that satisfy ct = ±x, and with t being different for each event, such that when you apply the Lorentz transformation to the (x,t) coordinates of each event to get the (x',t') coordinates of the same event, then both events can have (x',t') coordinates that satisfy ct' = ±x' but with the new feature that the t' coordinate is the same for each event. Are you saying this is illogical or impossible?

No, I am saying uisng that method creates two different light spheres. I gave you the points. Use them and apply the radius in both directions.

[
Again, what's a logical inconsistency? These events occur at different times in O, so naturally they are part of two different light spheres at different times in O. The equation ct = ±x was never supposed to define a single light sphere, rather it is an equation defining all points that lie on the light cone in 1D, it can include events at different times in O. Exactly the same is true of the equation x'² + y'² + z'² = τ²c², τ is a variable so this too is the 3D version of a light cone which includes events that lie on the path of the light at different times in O.

I have been down this road.

It does not achieve anything. I created two different light spheres in O' because of the simultaneity differential of the direction of the light rays in O.

Yet, if I do strictly point by point transforms as suggested by Einstein, I have a consistant resolution.

 

[cfrogue]

Suppose the event has coordinates x'=0, t'=T' in O'. Then the coordinates in O are:

x = gamma*(x' + vt') = gamma*vT'
t = gamma*(t' + vx'/c^2) = gamma*T'

So, x/t = (gamma*vT')/(gamma*T') = v, and if x/t=v then x=vt.

Good one.

Thanks. That works.

 

[Dale]

I clearly see a difference of when an event will occur concerning light and time dilation.

Obviously. Time dilation requires Δx=0 and Δx never equals 0 for light, so there will always be a difference.

It is false that

ct' = ±x'

You already agreed it was true, and it was proved conclusively several different ways. If you want to suddenly go back and ignore the previous 570 posts then I am done with this. What is the point of continuing the conversation if you will just ignore that amount of proof?

 

[atyy]

Yet, if I do strictly point by point transforms as suggested by Einstein, I have a consistant resolution.

Yes, that is the only way to do it, and that is what I've been doing all along, although that was not apparent to you. Regardless, this is correct, and fundamental.

 

[cfrogue]

Obviously. Time dilation requires Δx=0 and Δx never equals 0 for light, so there will always be a difference.
You already agreed it was true, and it was proved conclusively several different ways. If you want to suddenly go back and ignore the previous 570 posts then I am done with this. What is the point of continuing the conversation if you will just ignore that amount of proof?

I am the one that introduced ct' = +-x'.

That was my argument.

You did not prove anything to me about this.

Then, I discovered it created two light spheres. You agreed.

The question for you is how do you create logical consistancy with ct' = +-x'?

I would like to see this.

 

[cfrogue]

Yes, that is the only way to do it, and that is what I've been doing all along, although that was not apparent to you. Regardless, this is correct, and fundamental.

Agreed.

 

[cfrogue]

Yes, that is the only way to do it, and that is what I've been doing all along, although that was not apparent to you. Regardless, this is correct, and fundamental.

Well, you could be more specific.

http://www.youtube.com/watch?v=N9QosirK-Gw"

 

[Dale]

The question for you is how do you create logical consistancy with ct' = +-x'?

I would like to see this.

We already demonstrated this more than 500 posts ago. If you would like to see it just use your browser to review the thread.

 

[atyy]

That is why I say the normal logic of ct' = ±x' does not work.

The following points in O satisify that condition

t1=r/(λ(c-v)) for the ray point right and
t2=r/(λ(c+v)) for the ray point left.

If you run these through LT with x=+ct and x=-ct as appropriate, you will find two different satisfactions of the equation ct' = ±x' at two different times in O which creates two different light spheres which is a logical inconsistancy.

Yes, these imply that there are two light spheres in O at two different times. But that is not a problem, since there are an infinite number of light spheres in O, one for each t.

Thus t1=r/(λ(c-v)), x1R=ct1 is one side of the light sphere in O at t1, whose other side must be t1=r/(λ(c-v)), x1L=-ct1, which by the LT you can check does lie on a light sphere of O'.

Smilarly, t2=r/(λ(c+v)), x2L=-ct2 is one side of the light sphere in O at t2, whose other side must be t2=r/(λ(c+v)), x2R=ct2, which by the LT you can check does lie on a light sphere of O'.

 

[jtbell]

With 592 posts plus this one, this thread has by far the most posts of any thread in the Relativity forum, ever. It's long overdue time to give it a rest. Thread closed.