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

[Al68]

It is not OK to have two different values for the simultaneity of O' ie, two different t'.

You said it was.

Do you have the proof?

When did I say that?

 

[atyy]

I do not need to include "You omitted the +- in your derivation below" +-, because the light postulate says the points are simultaneous.

This is axiomatic and I do not need to prove it.

So, given the simultaneity by the light postulate, I need only consider one point.

If what you say is true, then you should be able to obtain the same result with x'=-r.

 

[Dale]

Looks like I missed a lot last night. I don't know if this was already resolved, but cfrogue's proof is incorrect. Specifically:

We have by the SR spherical light sphere,
ct' = x', t' = x'/c
Also, by selection x' = r.

These two lines are wrong. It should be:
"We have by the SR spherical light sphere,
ct' = ±x', t' = ±x'/c
Also, by selection ±x' = r."

The fact that there are two times in the unprimed frame follows from that.

 

[cfrogue]

If what you say is true, then you should be able to obtain the same result with x'=-r.

This is curious.

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'.

I'm thinking about all this.

What is your view at this point?

 

[cfrogue]

Looks like I missed a lot last night. I don't know if this was already resolved, but cfrogue's proof is incorrect. Specifically:These two lines are wrong. It should be:
"We have by the SR spherical light sphere,
ct' = ±x', t' = ±x'/c
Also, by selection ±x' = r."

The fact that there are two times in the unprimed frame follows from that.

Yes, I agree and just caught that also at the same time you posted.

 

[cfrogue]

Looks like I missed a lot last night. I don't know if this was already resolved, but cfrogue's proof is incorrect. Specifically:These two lines are wrong. It should be:
"We have by the SR spherical light sphere,
ct' = ±x', t' = ±x'/c
Also, by selection ±x' = r."

The fact that there are two times in the unprimed frame follows from that.

This seems strange.

As t starts at zero in O and advances to t1 = r/(λ(c+v)), ct' < |x'|.
At that point in the time of O, ct' = -x'.

Then the negative x' continues to go more negative as time increases in O.

Then when the time in O reaches, t2 = r/(λ(c-v)), ct' = +x'.

Is this correct?

 

[Dale]

Yes, I agree and just caught that also at the same time you posted.

No problem. I have updated my spacetime diagram to highlight the events where the light cone strikes the ends of the rods. The green dots are the events where the light cone strikes the ends of the unprimed rod. The red dots are the events where the light cone strikes the ends of the primed rod.

 

[cfrogue]

No problem. I have updated my spacetime diagram to highlight the events where the light cone strikes the ends of the rods. The green dots are the events where the light cone strikes the ends of the unprimed rod. The red dots are the events where the light cone strikes the ends of the primed rod.

You do produce some excellent graphics.

 

[cfrogue]

No problem. I have updated my spacetime diagram to highlight the events where the light cone strikes the ends of the rods. The green dots are the events where the light cone strikes the ends of the unprimed rod. The red dots are the events where the light cone strikes the ends of the primed rod.

This implies there are two times in O t1, t2, such that the light stikes of the endpoints of the rod of O' are simultaneous in O' and t1 < t2.

Is this correct?

 

[atyy]

This is curious.

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'.

I'm thinking about all this.

What is your view at this point?

My view is that you are calculating the same thing as you did here, which was correct, and you had no problem interpreting:

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]

My view is that you are calculating the same thing as you did here, which was correct, and you had no problem interpreting:

Perhaps.

Using the expanding light sphere in O, and allowing time to increase in O based on the expanding light sphere, I have found ct' = |r| at two different times in O', ie t1', t2'. I am not sure what this means though.

 

[atyy]

Perhaps.

Using the expanding light sphere in O, and allowing time to increase in O based on the expanding light sphere, I have found ct' = |r| at two different times in O', ie t1', t2'. I am not sure what this means though.

I would suggest you study DaleSpam's diagram in #497 and JesseM's diagram in #182 very carefully. In DaleSpam's diagram, there are two light spheres, marked by a pair of green points and a pair of red points respectively. There is only one light cone, which are the yellow lines, showing the left going photon and the right going photon.

 

[cfrogue]

I would suggest you study DaleSpam's diagram in #497 and JesseM's diagram in #182 very carefully. In DaleSpam's diagram, there are two light spheres, marked by a pair of green points and a pair of red points respectively. There is only one light cone, which are the yellow lines, showing the left going photon and the right going photon.

I did look at them.

There is a light sphere in O and one in O'.

LT says, for two times in O, the condition ct' = ±r is true.

Also, LT says there are two different times in O' in which ct' = ±r is true based on the emerging light sphere in O.

That seems to mean there are two in O'.

I am not sure though.

Do the diagrams show this?

 

[Dale]

This implies there are two times in O t1, t2, such that the light stikes of the endpoints of the rod of O' are simultaneous in O' and t1 < t2.

Is this correct?

Yes, where t1 is the t coordinate of the event indicated by the red dot on the left and t2 is the t coordinate of the event indicated by the red dot on the right. In this drawing (c=1, v=0.6, d=2) we have t1=0.5 and t2=2.0 so t1<t2 is correct.

 

[Dale]

I have found ct' = |r| at two different times in O', ie t1', t2'. I am not sure what this means though.

It means simultaneity is relative.

 

[Dale]

There is a light sphere in O and one in O'.

Yes, the light sphere at t=1 in O is indicated by the green dots and the light sphere at t'=1 in O' is indicated by the red dots.

LT says, for two times in O, the condition ct' = ±r is true.

Yes. Note that the left and right red dots are at t=0.5 and t=2.0 respectively.

Also, LT says there are two different times in O' in which ct' = ±r is true

No. Note that the left and right red dots are both at t'=1.

 

[cfrogue]

Yes, the light sphere at t=1 in O is indicated by the green dots and the light sphere at t'=1 in O' is indicated by the red dots.Yes. Note that the left and right red dots are at t=0.5 and t=2.0 respectively.No. Note that the left and right red dots are both at t'=1.

Are you using a particular relative v?

 

[Dale]

Yes, as I mentioned in post 504 I used v=0.6, c=1, and d=2 (or r=1) for this drawing. The exact times will be different for different v, c, or d, but it is always given by the Lorentz transform.

 

[cfrogue]

Yes, as I mentioned in post 504 I used v=0.6, c=1, and d=2 (or r=1) for this drawing. The exact times will be different for different v, c, or d, but it is always given by the Lorentz transform.

OK, thanks .6c is a good one to use.

Please evaluate the below and tell me what you think. Perhaps, I made a math error or
something.


Assume v = 3/5(c). Let r be one half to rest distance of the rod in O'
λ = 5/4, (c-v) = 2/5, (c + v) = 8/5

When t reaches r/(2c), the radius of the light sphere is ct = r/2 in O.
For the left light beam, its x location is -r/2 or x = -ct = -r/2.
t' = ( t - xv/c^2)λ
t' = ( t + tv/c )λ = t ( 1 + v/c )λ = t ( 1 + 3/5 )5/4 = 2t = r/c.

Thus, (ct')^2 = r^2 and therefore, the light sphere in O' strikes both points at +r and -r.


Then when t reaches 2r/c, x = 2r, so
t' = ( 2r/c - 2rv/c^2)λ = 2r/c(1 - v/c)λ = 2r/c(2/5)5/4 = r/c
Thus, (ct')^2 = r^2 and therefore, the light sphere in O' strikes both points at +r and -r.

Hence, there are two different times in O when O' sees the simultaneous strikes.
This is different from R of S.

Thus, time proceeds from 0 to t = r/(2c) then the light sphere in O' strikes the endpoints at the same time. Then time proceeds forward as the light sphere expands in O and then again, another simultaneous strikes occurs at t = 2r/c.

 

[Dale]

For the left light beam, its x location is -r/2 or x = -ct = -r/2.
t' = ( t - xv/c^2)?
t' = ( t + tv/c )? = t ( 1 + v/c )? = t ( 1 + 3/5 )5/4 = 2t = r/c

Don't forget to do the same for the right:

For the right light beam, its x location is r/2 or x = ct = r/2.
t' = ( t - xv/c^2)?
t' = ( t - tv/c )? = t ( 1 - v/c )? = t ( 1 - 3/5 )5/4 = t/2 = r/(4c)

 

[cfrogue]

Don't forget to do the same for the right:

For the right light beam, its x location is r/2 or x = ct = r/2.
t' = ( t - xv/c^2)?
t' = ( t - tv/c )? = t ( 1 - v/c )? = t ( 1 - 3/5 )5/4 = t/2 = r/(4c)

Of course, I did that.

But, that does not change the fact that (ct')^2 = r^2 as required by the light sphere in O'.

This would seem to indicate two light sphere in O'.

One is generated by the back beam and one is generated by the front beam.


Now, if I use the front beam and the back beam and translate into O', I find light travels to the left in O' further than it travels in the right after any time t in O. Recall time may have dilation between the frames, but time moves linearly for all frames.

On the other hand, if I use the back beam of O as radius in O', then that is one light sphere and if I use the radius of the front beam in O, I get a different light sphere in O'.

 

[Dale]

This would seem to indicate two light sphere in O'.

There are an infinite number of light spheres in O', one corresponding to each t'.

One is generated by the back beam and one is generated by the front beam.

How would you reconcile this idea with a true 4D situation (which I can't draw for obvious reasons) where there is not a front beam and a back beam, but one continuous cone?

Now, if I use the front beam and the back beam and translate into O', I find light travels to the left in O' further than it travels in the right after any time t in O.

Yes, but although it travels to the left further in O' it also takes longer to get there in O', so the speed of light is still c in O'.

On the other hand, if I use the back beam of O as radius in O', then that is one light sphere and if I use the radius of the front beam in O, I get a different light sphere in O'.

I'm not sure what you mean by this. Can you point it out on the diagram?

 

[cfrogue]

There are an infinite number of light spheres in O', one corresponding to each t'.

Agreed.

How would you reconcile this idea with a true 4D situation (which I can't draw for obvious reasons) where there is not a front beam and a back beam, but one continuous cone?

I gave two specific points in the time coords of O where the light sphere in O' is simultaneous and showed the math.

So if time is linear in O, then it should be linear in O'.

Yes, but although it travels to the left further in O' it also takes longer to get there in O', so the speed of light is still c in O'.

That is not quite true in terms of the description of time. The time of the left beam in O' is beating faster than the time of the right beam as the time in O progresses. It is true that the speed of light calculates to c in both directions though.

I'm not sure what you mean by this. Can you point it out on the diagram?

I will work on this.

 

[cfrogue]

I'm not sure what you mean by this. Can you point it out on the diagram?

I put a pciture in this post but it is incorrect.

 

[cfrogue]

I'm not sure what you mean by this. Can you point it out on the diagram?

OK, it seems this is the diagram I get when using the front and back radii of the light sphere in O as it appears in O' after any time t in O.

 

[cfrogue]

I'm not sure what you mean by this. Can you point it out on the diagram?

This is what I get when I use the front radius of O as the radius in O' and the back radius of O as the radius of O' after any time t in O.

The back radius is the larger sphere.

 

[Dale]

So if time is linear in O, then it should be linear in O'..

Yes. The Lorentz transform is linear.

That is not quite true in terms of the description of time. The time of the left beam in O' is beating faster than the time of the right beam as the time in O progresses. It is true that the speed of light calculates to c in both directions though.

What you are talking about is called the Doppler effect. You get a blueshift (higher frequencies) in one direction and a redshift (lower frequencies) in the other direction.

 

[cfrogue]

Yes. The Lorentz transform is linear. What you are talking about is called the Doppler effect. You get a blueshift (higher frequencies) in one direction and a redshift (lower frequencies) in the other direction.

Let me see.

If I conducted an MMX experiment in the moving frame, O', I would get a constant frequency.

Nope, that does not do it.

Also, while I am thinking about it, if I conducted one in the stationary frame, I would also get a constant frequency.

 

[Dale]

Let me see.

If I conducted an MMX experiment in the moving frame, O', I would get a constant frequency.

Nope, that does not do it.

Also, while I am thinking about it, if I conducted one in the stationary frame, I would also get a constant frequency.

True if you restrict the analysis to a single frame, but you are not restricting your "light sphere" analysis to a single frame. You are trying to look at light spheres in one frame from the coordinates of another. If you did a MMX experiment in one frame and measured the frequency of the light of the various arms in another frame then you would indeed get different frequencies forward and backwards.

Regarding the diagrams, I was actually hoping you could identify your concern on my spacetime diagram where all of the coordinates are well identified.

 

[cfrogue]

True if you restrict the analysis to a single frame, but you are not restricting your "light sphere" analysis to a single frame. You are trying to look at light spheres in one frame from the coordinates of another. If you did a MMX experiment in one frame and measured the frequency of the light of the various arms in another frame then you would indeed get different frequencies forward and backwards.

Regarding the diagrams, I was actually hoping you could identify your concern on my spacetime diagram where all of the coordinates are well identified.

First, before we continue, does MMX decide a constant speed of light?

 

[cfrogue]

Regarding the diagrams, I was actually hoping you could identify your concern on my spacetime diagram where all of the coordinates are well identified.

I am not sure.

LT proves two different time coordinates in O such that O' sees simultaneity.

Does your spacetime diagram stuff show this fact?

If not, I would abandon it.

 

[Dale]

First, before we continue, does MMX decide a constant speed of light?

The MMX shows that the speed of light is isotropic.

I am not sure.

LT proves two different time coordinates in O such that O' sees simultaneity.

Does your spacetime diagram stuff show this fact?

If not, I would abandon it.

Yes, the spacetime diagram shows everything about the LT in one space and one time dimension.

 

[cfrogue]

The MMX shows that the speed of light is isotropic.

BTW, it is well documented MMX does not prove a constant speed of light.

Here is a sample.

This rules out any conceptually coherent ballistic theory of light propagation, according to which the speed of light is the vector sum of the velocity of the source plus a vector of magnitude c. Ironically, the original Michelson-Morley experiment was consistent with the ballistic theory, but inconsistent with the naïve ether theory, whereas the Sagnac effect is consistent with the naïve ether theory but inconsistent with the ballistic theory. Of course, both results are consistent with fully relativistic theories of Lorentz and Einstein, since according to both theories light is propagated at a speed independent of the state of motion of the source.
http://www.mathpages.com/rr/s2-07/2-07.htm

Yes, the spacetime diagram shows everything about the LT in one space and one time dimension.

Show me the diagram where in the time of O, O' sees two different simultaneity strikes.

We have proven that together.

 

[Dale]

BTW, it is well documented MMX does not prove a constant speed of light...Ironically, the original Michelson-Morley experiment was consistent with the ballistic theory...

That's why I said "isotropic" rather than "constant".

Show me the diagram where in the time of O, O' sees two different simultaneity strikes.

We have proven that together.

That's exactly what the red and green dots show. We have been over this many times already.

 

[cfrogue]

That's why I said "isotropic".

If you believe in the isotropic argument, why did you mention red/blue frequency then?
Therefore, you are separating light speed and frequency with this logic.

I agree BTW.

MMX does not decide a constant speed of light and therefore, when talking about the speed of light, frequency based experiments should never be brought up.

That's what the red and green dots show. We have been over this many times already.

Can you please show me on the diagram the 2 different times in O where O' sees simultaneity?

We proved this together with LT.