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

[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.

 

[cfrogue]

The MMX shows that the speed of light is isotropic.
Yes, the spacetime diagram shows everything about the LT in one space and one time dimension.

Here are some past discussions you had on this subject.

The equation of the sphere of light in the stationary frame is:
c²t² = x² + y² + z²

Transforming to the moving frame is:
(ct'γ-vx'γ/c)² = γ²(x'-vt')² + y'² + z'²

Which simplifies to:
c²t'² = x'² + y'² + z'²

https://www.physicsforums.com/showpost.php?p=2142933&postcount=7
Re: moving light bulb sphere of photons

Originally Posted by Dreads
Rotate this malformed wagon wheel thru 360 degrees around the x axs and it is not a sphere

Yes it is a sphere; I derived it above.

Since your conclusion is demonstrably wrong, why don't you go back and see if you can spot the mistake.

https://www.physicsforums.com/showpost.php?p=2144836&postcount=17

 

[atyy]

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

Yes. For t > 0, and x=±r, where r > 0, there is one light sphere in O at t=r/c. For t' > 0, and x'=±r, where r > 0, there is one light sphere in O' at t'=r/c. In DaleSpam's diagram, c=1 and r=1, so the light sphere of O are the green dots at (x=-1, t=1) and (x=1,t=1), and the light sphere of O' are the red dots at (x'=-1, t'=1) and (x'=1,t'=1)

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'.

If r > 0, since c > 0 then some of the times t or t' you are talking about are negative. These are not from the LT. I believe you are referring to the past light sphere of O', which is not shown on the diagram. I haven't been very careful about t>0 and t'>0 in my previous discussion, so maybe this was what was confusing you. Yes, there are two light cones - one future and one past. So far I thought we've been only discussing the future light cone, but I think you are inferring a past light cone from the negative t in the condition x²=t².

Rachmaninov - Prelude in D major, opus 23 no. 4

 

[Dale]

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 don't understand this comment at all. Do you believe that the Doppler effect is somehow inconsistent with the isotropy of the speed of light? It is not; in fact, the Doppler effect can be derived from the isotropy of the speed of light.

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.

I would say that the MMX is a wavelength based experiment. If the speed of light were not isotropic then the number of wavelengths in the different arms would be different and an interference pattern would emerge. However, I don't see how any of this discussion of the MMX is relevant to your ongoing light cone questions.

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

We proved this together with LT.

Certainly, as we have discussed several times already that is represented by the red dots. The red dots are simultaneous in O' (both occur at t'=1 as you can see by the white lines), but they occur at different times in O (t=0.5 and t=2.0 as you can see by the black lines).

 

[cfrogue]

I don't understand this comment at all. Do you believe that the Doppler effect is somehow inconsistent with the isotropy of the speed of light? It is not; in fact, the Doppler effect can be derived from the isotropy of the speed of light.I would say that the MMX is a wavelength based experiment. If the speed of light were not isotropic then the number of wavelengths in the different arms would be different and an interference pattern would emerge. However, I don't see how any of this discussion of the MMX is relevant to your ongoing light cone questions.

I believe the doppler effect is consistent with a constant speed of light.

I only mentioned MMX because you brought the red blue argument.

I am OK with dropping this part. Thanks.

Certainly, as we have discussed several times already that is represented by the red dots. The red dots are simultaneous in O' (both occur at t'=1 as you can see by the white lines), but they occur at different times in O (t=0.5 and t=2.0 as you can see by the black lines).

I agree that there does not exist a time in O such that the points are struck simultaneously in O' while using LT.

However, x' is also the x-radius of the expanding light sphere as calculated by O and (ct')^2 = x'^2.

It is true at two different times in O where (ct')^2 = r^2. Is this correct?

Finally, if O called O' on a light phone and asked the time it took for the simultaneity, then O' would answer t' = r/c because of the constant speed of light. Then O would use time dilation and convert this time as (r/c)λ.

Now, the reason I wrote this is because Einstein said in chapter 4, Physical Meaning of the Equations Obtained in Respect to Moving Rigid Bodies and Moving Clocks

Between the quantities x, t, and τ, which refer to the position of the clock, we have, evidently, x=vt and

τ = ( t - vx/c^2 )λ.

τ = ( t - (tv^2)/c^2 )λ.

τ = t( 1 - (v^2)/c^2 )λ.

τ = t/λ.

It is the case that x=vt at any time t for the moving frame.

I am not sure how all this works together.
1) There is LT which shows there is no time t for which O can declare simultaneity in O'.
2) There are two times in t where the light sphere satisfies (ct')^2 = x'^2.
3) Since O' is moving relative to O, it would seem time dilation also applies such that simultaneity occurs in O' at (r/c)λ in the time of O.

 

[atyy]

It is true at two different times in O where (ct')^2 = r^2. Is this correct?

DaleSpam and cfrogue, please ignore me if this is irrelevant, since I think you guys are getting somewhere. But is cfrogue's concern here with the formal solution t'=±r/c, for r > 0, ie. the past light cone?

The light sphere loci are given by x'²=ct'². For fixed r > 0 and t' > 0 then there are two x' values sharing one t' value on the light sphere (x'=±r, t'=r/c). For fixed r > 0 and t' < 0, there are also two x' values sharing one t' value (x'=±r, t'=-r/c). So considering past and future light cones there are two t' values for fixed r. (Not sure if I got the equations all right, but am wondering something along those lines.)

 

[Dale]

I agree that there does not exist a time in O such that the points are struck simultaneously in O' while using LT.

Excellent.

However, x' is also the x-radius of the expanding light sphere as calculated by O and (ct')^2 = x'^2.

Remember, the light sphere in O', when transformed into the coordinates of O is a non-simultaneous ellipsoid. I don't know how it would even be possible to define its radius. So without a crystal clear definition of the radius of a non-simultaneous ellipsoid I can't make any conclusions about this.

It is true at two different times in O where (ct')^2 = r^2. Is this correct?

Yes, again see the red dots.

Finally, if O called O' on a light phone and asked the time it took for the simultaneity, then O' would answer t' = r/c because of the constant speed of light. Then O would use time dilation and convert this time as (r/c)λ.

Now, the reason I wrote this is because Einstein said in chapter 4, Physical Meaning of the Equations Obtained in Respect to Moving Rigid Bodies and Moving Clocks

Between the quantities x, t, and τ, which refer to the position of the clock, we have, evidently, x=vt and

τ = ( t - vx/c^2 )λ.

τ = ( t - (tv^2)/c^2 )λ.

τ = t( 1 - (v^2)/c^2 )λ.

τ = t/λ.

It is the case that x=vt at any time t for the moving frame.

We have already discussed this extensively. The time dilation formula only applies when the two events are co-local in one of the frames (Δx=0). You are using it where it doesn't apply. Suppose I were to try and use the compound interest formula A = P (1+r/n)^(nt). It is also a correct formula for t, but it simply doesn't apply.

I am not sure how all this works together.
1) There is LT which shows there is no time t for which O can declare simultaneity in O'.
2) There are two times in t where the light sphere satisfies (ct')^2 = x'^2.
3) Since O' is moving relative to O, it would seem time dilation also applies such that simultaneity occurs in O' at (r/c)λ in the time of O.

Again, the time dilation formula only applies if Δx=0 (co-local events), which is not the case here, the ends of the rods are at different positions from the flash in all coordinate systems.

 

[cfrogue]

Remember, the light sphere in O', when transformed into the coordinates of O is a non-simultaneous ellipsoid. I don't know how it would even be possible to define its radius. So without a crystal clear definition of the radius of a non-simultaneous ellipsoid I can't make any conclusions about this.

Yes, I have decided it is meaningless also from the position of O since the each side of the sphere has a different event timing associated with it due to R of S.

The time dilation formula only applies when the two events are co-local in one of the frames (Δx=0). You are using it where it doesn't apply.

I do not agree with this since Einstein derived it not by using x = 0 but by using x = vt.

However, the light sphere with LT already has it built into so it does not apply a second time.

However, I am not sure how you can justify not using it. And I am not using it to look at the end points of the rods.

I used to to determine when in O' it sees the simultaneity. Since its time is ct = r when the strikes occur, its seems natural to apply time dilation.

 

[cfrogue]

Excellent.

BTW, you do good work, thanks.

 

[cfrogue]

Remember, the light sphere in O', when transformed into the coordinates of O is a non-simultaneous ellipsoid.

Here is what I was trying to achieve.

Assume a rigid body sphere is moving in relative motion.

I can map this into the corrds of O from O' and it looks like an ellipsoid.

It is centered at vt with a x-radius of r/λ and y, z radii of r in O.

Yet, I cannot map the expanding light sphere using just LT. I am probably going to have to adjust the light sphere someone to make it work.

Do you have any ideas as to why I can map this O' rigid body sphere into O but cannot map the light sphere expanding in it?

I suppose I need to back out the simultaneity right and left issues somehow.

As we have seen, (c't)^ = x'^2 is not working because we get two of them.

 

[Dale]

I do not agree with this since Einstein derived it not by using x = 0 but by using x = vt.

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.

 

[JesseM]

I do not agree with this since Einstein derived it not by using x = 0 but by using x = vt.

If you look at section 4 of the 1905 paper he specifically assumes a clock which is at rest in the inertial frame k, and which elapses a time of tau in that frame. So, we are considering two events on the clock's worldline which have a coordinate separation of 0 in the k frame (and earlier he referred to the first position coordinate of this frame with the greek letter xi, not the roman letter x), and a time separation of tau in that frame, and then figuring how this relates to the time separation t between the same pair of events in a different frame. x=vt is an equation of motion for the clock that's supposed to apply in the separate "stationary" frame K; you can tell it's a different frame because it uses t instead of tau for the time coordinate.

The point is, you can only use the time dilation equation when you are considering to events that have a spatial separation of zero in one of the two frames you're considering; if the first spatial coordinate of the two frames are referred to as xi and x, then it can be either x=0 or xi=0, it doesn't matter (likewise if the coordinates are x and x', then it can be either x'=0 or x=0). Whichever frame is the one where the two events are co-local, the time dilation equation always takes the form:

t_noncolocal = t_colocal/sqrt(1 - v^2/c^2)

or equivalently:

t_colocal = t_noncolocal*sqrt(1 - v^2/c^2)

The second form of the equation is the one that appears in section 4 of Einstein's 1905 paper, with tau as the time separation in the frame where the events are colocal and t as the time separation in the frame where they aren't.

 

[Dale]

BTW, you do good work, thanks.

Thanks!

Here is what I was trying to achieve.

Assume a rigid body sphere is moving in relative motion.

I can map this into the corrds of O from O' and it looks like an ellipsoid.

It is centered at vt with a x-radius of r/λ and y, z radii of r in O.

Yet, I cannot map the expanding light sphere using just LT. I am probably going to have to adjust the light sphere someone to make it work.

Do you have any ideas as to why I can map this O' rigid body sphere into O but cannot map the light sphere expanding in it?

I suppose I need to back out the simultaneity right and left issues somehow.

As we have seen, (c't)^ = x'^2 is not working because we get two of them.

When we say "looks like an ellipsoid" that usually means that it has the equation of an ellipsoid at a given (simultaneous) instant in time, so it would be difficult to relate that to a non-simultaneous ellipsoid. I am sure that it could be done, but it just wouldn't be very natural.

Another way to see the difficulty is to think in terms of 4D geometry (or at least 3D geometry, one time and two space). The light cone is, obviously, a cone in 4D, but a rigid spherical body would be a cylinder in 4D. What you are essentially trying to do is to take the cone, slice it diagonally to get an ellipsoidal cross section, then take a cylinder, tilt it in 4D so that it intersects the cone along that ellipsoid, then somehow find the radius of that cylinder, not in the tilted direction, but in some original non-diagonal plane.

It could probably be done, but it would be a lot of work and a rather weird result.

 

[JesseM]

Thanks!When we say "looks like an ellipsoid" that usually means that it has the equation of an ellipsoid at a given (simultaneous) instant in time, so it would be difficult to relate that to a non-simultaneous ellipsoid. I am sure that it could be done, but it just wouldn't be very natural.

It wouldn't be physically very natural, but you could just take the equation of a light sphere in O', then figure out the position coordinates in O of all the events that make up this sphere using the Lorentz transformation, ignoring the time coordinate...in terms of the position coordinates of O the result should be an ellipsoid. In a spacetime diagram this is would just look like the "shadow" of of a light sphere in O' when cast down vertically onto a plane of simultaneity in O.

 

[cfrogue]

If you look at section 4 of the 1905 paper he specifically assumes a clock which is at rest in the inertial frame k, and which elapses a time of tau in that frame. So, we are considering two events on the clock's worldline which have a coordinate separation of 0 in the k frame (and earlier he referred to the first position coordinate of this frame with the greek letter xi, not the roman letter x), and a time separation of tau in that frame, and then figuring how this relates to the time separation t between the same pair of events in a different frame. x=vt is an equation of motion for the clock that's supposed to apply in the separate "stationary" frame K; you can tell it's a different frame because it uses t instead of tau for the time coordinate.

The point is, you can only use the time dilation equation when you are considering to events that have a spatial separation of zero in one of the two frames you're considering; if the first spatial coordinate of the two frames are referred to as xi and x, then it can be either x=0 or xi=0, it doesn't matter (likewise if the coordinates are x and x', then it can be either x'=0 or x=0). Whichever frame is the one where the two events are co-local, the time dilation equation always takes the form:

t_noncolocal = t_colocal/sqrt(1 - v^2/c^2)

or equivalently:

t_colocal = t_noncolocal*sqrt(1 - v^2/c^2)

The second form of the equation is the one that appears in section 4 of Einstein's 1905 paper, with tau as the time separation in the frame where the events are colocal and t as the time separation in the frame where they aren't.

Let me ask this.

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

Does time dilation exist?

 

[cfrogue]

It wouldn't be physically very natural, but you could just take the equation of a light sphere in O', then figure out the position coordinates in O of all the events that make up this sphere using the Lorentz transformation, ignoring the time coordinate...in terms of the position coordinates of O the result should be an ellipsoid. In a spacetime diagram this is would just look like the "shadow" of of a light sphere in O' when cast down vertically onto a plane of simultaneity in O.

This I want to try to do.

I am thinking just put a rigid body sphere in relative motion at radius r.

Then, match the light sphere to that one for all r.

 

[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.