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

[cfrogue]

Do the Lorentz transform of x=vt into the primed coordinates:
1) x = γ (x' + v t')
2) t = γ (t' + v x'/c²)

Then by substitution into x = v t

3) γ (x' + v t') = v γ (t' + v x'/c²)

Which simplifies to

4) x' = 0

Not only does that require that all events share the same x' coordinate (co-local in the primed frame with Δx'=0), but it requires that those events lie exclusively on the x'=0 line. So the condition x=vt that Einstein uses here is actually more restrictive than the co-local condition. It is not necessary to use Einstein's more restrictive condition, although he does not show it in his seminal paper it is shown elsewhere.

However, again, the time dilation formula does not apply for this measurement. The events are not co-local in any frame, and they are certainly not co-local with the origin in any frame.

Regarding the worldline diagram, I think I understand the two different paths we have been on.

You were talking about R of S, which I agree exists and I was thinking about the light sphere.

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/

Now, I assumed I could follow this argument above and look at the light sphere in O'.

Well, I could not as we showed. It depends on whether we use the left ray in O or the right ray in O.

 

[JesseM]

Let me ask this.

Assume I have twin1 at rest and twin2 moving in collinear relative motion.

Does time dilation exist?

If you pick two events on the worldline of twin1, the time dilation equation tells you how the time between these events in twin2's frame is greater than the time between them in twin1's frame; likewise, if you pick two events on the worldline of twin2, the time dilation equation tells you how the time between these events in twin1's frame is greater than the time between them in twin2's frame. In each twin's own rest frame it is the other twin's clock that's running slow by the amount predicted by the time dilation equation, but of course since all inertial motion is relative, there is no objective truth about whose clock is "really" running slow.

 

[cfrogue]

If you pick two events on the worldline of twin1, the time dilation equation tells you how the time between these events in twin2's frame is greater than the time between them in twin1's frame; likewise, if you pick two events on the worldline of twin2, the time dilation equation tells you how the time between these events in twin1's frame is greater than the time between them in twin2's frame. In each twin's own rest frame it is the other twin's clock that's running slow by the amount predicted by the time dilation equation, but of course since all inertial motion is relative, there is no objective truth about whose clock is "really" running slow.

I am saying this.

Twin1 is stationary.

We are viewing this from twin1.

Is the clock for twin2 beating slower?

 

[JesseM]

I am saying this.

Twin1 is stationary.

We are viewing this from twin1.

Is the clock for twin2 beating slower?

In twin1's frame, twin2's clock is beating slower, yes.

 

[cfrogue]

In twin1's frame, twin2's clock is beating slower, yes.

Now, say twin2 is sitting in a rigid body sphere and everything else is the same.

Is there time dilation for twin2 as calculated by twin1.

 

[JesseM]

Now, say twin2 is sitting in a rigid body sphere and everything else is the same.

Is there time dilation for twin2 as calculated by twin1.

Yes, time dilation is just a feature of the two events used, the surroundings don't make a difference.

 

[cfrogue]

Yes, time dilation is just a feature of the two events used, the surroundings don't make a difference.

Now say twin2 is dancing with a flashlight.

Twin2 is flashing that light every which away.

If the elapsed time in twin2 is t', will it be t'λ in the frame of twin1 from the POV of twin1?

 

[Dale]

If the elapsed time in twin2 is t', will it be t'λ in the frame of twin1 from the POV of twin1?

The elapsed time of what specifically in twin2's frame?

 

[cfrogue]

The elapsed time of what specifically in twin2's frame?

Oh, when O and O' are coincident, t = t' = 0.

Why does an event matter?

They start at the same time and for any time t', t' = t/λ.

I looked at Einstein's chapter 4 and I am not seeing that a specific end is needed.

It just writes about A and B in general.

I am not doing acceleration and integration.

 

[Dale]

Oh, when O and O' are coincident, t = t' = 0.

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

I looked at Einstein's chapter 4 and I am not seeing that a specific end is needed.

I showed this earlier today. Einstein also required that the time be measured at a single location in the primed frame. I am asking you to specify what elapsed time you are referring to because it makes a difference. If you are referring to the elapsed time between two events on twin2's worldline (like turning on and turning off the flashlight) then those events are co-located (Δx=0) and the time dilation formula applies. If you are referring to the elapsed time between two events that are not both on twin2's worldline (like turning on the flashlight and the light hitting the end of a rod) then the events are not co-located and the time dilation formula does not apply.

 

[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.