Alternative Twins Paradox

Jack_ · Apr 4, 2010

AlphaNumeric said: ↑

IT follows by applying a LT to the sphere. Ignore the centre because its obviously confusing you.
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I want you to show me the math of a sphere with an infinite number of origins.

Make any model you want.

I am not the one confused.

This is a requirement for the truth of SR and now you know that.

You cannot map one to many in any universe or metric space.

Otherwise do it. If you cannot, you submit to my clear logic that SR is a false theory. That is now the challenge.

Can you map one to many?

[/QUOTE]

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Jack_ · Apr 4, 2010

AlphaNumeric said: ↑

I get that impression too. His misuse of the word 'proof' is at odds with his claim that he's familiar with mathematical areas like set theory or logic. He tries to use the word 'decidable' yet he's unable to correctly use the word 'proof'. It's not possible for someone to learn about decidability and not learn what 'proof' means. He demands people provide mathematics to justify their claims and then doesn't respond to the maths when people provide it. He claims he's found a Lorentz transformation which maps a point on the light cone to a point not on the light cone but can't respond when I prove (and I'm using the word correctly here) that such a transformation isn't a Lorentz transformation then, as it violates the definition of the transformations.

Never the less for the purposes of being complete and to bring things like the results of this post all into one thread (else it'll scatter into many different threads and its silly to say the same stuff 4 times in different threads to the same person) I'll go through the maths in more detail here.

Definition : A Lorentz transformation is one which leaves the metric $\eta$ of Minkowski space-time, hence for referred to as M, invariant.

Definition : Minkowski space-time is isotropic, homogeneous, flat and has 1 time-like direction and N space-like direction.

Given the tangent bundle TM the metric acts as $\eta \,:\, TM \times TM \to \mathbb{R}$. If $X,Y \in TM$ then $\eta(X,Y) \in \mathbb{R}$. The dimension of TM is N+1.

Definition : The Lorentz group G is therefore defined as $G = \{ \Lambda \in \textrm{GL}(N+1,\mathbb{R}) \,s.t.\, \eta(X,Y) = \eta(\Lambda \cdot X,\Lambda \cdot Y) \, \forall X,Y \in TM \} \equiv SO(1,N)$

Given coordinates of M $(t,x_{n}) \equiv x^{\mu}$ the induced coordinates of TM are $(\partial_{t},\partial_{x_{n}}) \equiv e_{\mu}$. A general element of TM has components in these coordinates $X = X^{\mu}e_{\mu}$. The components of $\eta$ are then $\eta_{\mu\nu} \equiv \eta(e_{\mu},e_{\nu})$.

The Lorentz transformations act on the components of vector as $\Lambda : X^{\mu} \to \tilde{X}^{\mu} = \Lambda^{\mu}_{\nu}X^{\nu}$. Inserting this into the definition of the Lorentz transformation we have $\eta(X,Y) = \eta(\Lambda \cdot X,\Lambda \cdot Y)$ becoming $\tilde{X}^{\mu}\tilde{Y}^{\nu}\eta_{\mu\nu} = X^{\mu}Y^{\nu} \Lambda^{\rho}_{\mu} \Lambda^{\sigma}_{\nu} \eta_{\rho\sigma}$

This is true for all X and Y and therefore we have the defining action of a Lorentz transformation in terms of coordinate induced components : $\eta_{\mu\nu} = \Lambda^{\rho}_{\mu} \Lambda^{\sigma}_{\nu} \eta_{\rho\sigma}$.

Definition : All indices are raised and lowered by $\eta$. $X_{\mu} = \eta_{\mu\nu}X^{\nu}$ and $\chi^{\mu} = \eta^{\mu\nu}\chi_{\nu}$ where $\eta^{\mu\nu}\eta_{\nu\rho} = \delta^{\mu}_{\rho}$.

Definition : A vector $p = p^{\mu}e_{\mu}$ is time-like if $p^{\mu}p_{\mu} < 0$, space-like if $p^{\mu}p_{\mu} > 0$ and null if $p^{\mu}p_{\mu} = 0$.

Under a Lorentz transformation the components of $p = p^{\mu}e_{\mu}$ are mapped to the new basis and the vector components transform as $p^{\mu} \to p^{\mu}\Lambda^{\nu}_{\mu} \equiv P^{\mu}$. Using the raising and lowering properties we have $P^{\mu}P_{\mu} = P^{\mu}P^{\nu} \eta_{\mu\nu}$. We insert the definition of $P^{\mu}$, $P^{\mu}P^{\nu} \eta_{\mu\nu} = (p^{\sigma}\Lambda^{\mu}_{\sigma})(p^{\rho}\Lambda^{\nu}_{\rho}) \eta_{\mu\nu} = (p^{\sigma}p^{\rho})(\Lambda^{\mu}_{\sigma}\Lambda^{\nu}_{\rho} \eta_{\mu\nu} )$. By definition we have that $\eta_{\sigma\rho} = (\Lambda^{\mu}_{\sigma}\Lambda^{\nu}_{\rho} \eta_{\mu\nu} )$ and therefore $(p^{\sigma}p^{\rho})(\Lambda^{\mu}_{\sigma}\Lambda^{\nu}_{\rho} \eta_{\mu\nu} ) = (p^{\sigma}p^{\rho})\eta_{\sigma\rho} = p^{\sigma}p_{\sigma}$. Therefore $P^{\mu}P_{\mu} = p^{\nu}p_{\nu}$. Lorentz transformations lead the norm of a vector unchanged.

The light cone is defined by points which are on null trajectories from the apex of the cone. The light cone whose apex is (t,x) = (0,0) is defined by points which satisfy $\sum_{i}x_{i}^{2} = t^{2}$. This is the equation of a sphere centred on the origin whose radius is proportional to time. It can be rearranged to $-t^{2}+\sum_{i}x_{i}^{2} = 0$. In terms of vector components we have $x^{\mu}x_{\mu} = 0$ and in terms of the metric $x^{\mu}x^{\nu}\eta_{\mu\nu} = 0$. By the above result this is Lorentz invariant, $y^{\mu}y_{\mu} = 0$ and therefore the light cone maps to the light cone.

We consider this explicitly for a Lorentz boost in the $x_{1}$ direction by speed v. The explicit form of the Lorentz transformation is determined by the coordinate relationships.

$\tau = \gamma (t - vx_{1})$, $y_{1} = \gamma ( x_{1} - vt)$, $y_{i} = x_{i}$ for $i \in \{2,\ldots,N\}$
Inverting these transformations
$t = \gamma (\tau + vy_{1})$, $x_{1} = \gamma ( y_{1} + v\tau)$, $x_{i} = y_{i}$ for $i \in \{2,\ldots,N\}$

The definition of the light sphere is $\sum_{m=1}^{N} x_{m}^{2} = (ct)^{2}$. We rearrange this to collect the non-trivially transformed coordinates, $\sum_{m=2}^{N}x_{m}^{2} + \Big( -t^{2} + x_{1}^{2} \Big) = 0$. The summation trivially changes to $\sum_{m=2}^{N}y_{m}^{2}$. The bracketed second term is of the form seen above and thus Lorentz invariant in structure, $-t^{2} + x_{1}^{2} = -\tau^{2} + y_{1}^{2}$. We shall demonstrate this explicitly; $-t^{2} + x_{1}^{2} = -(\gamma (\tau + vy_{1}))^{2} + (\gamma ( y_{1} + v\tau))^{2} = -\gamma^{2}(1-v^{2})\tau^{2} + \gamma^{2}(1-v^{2})y_{1}^{2} = -\tau^{2}+y_{1}^{2}.$

Therefore the light sphere $\sum_{m=1}^{N} x_{m}^{2} = (ct)^{2}$ Lorentz transforms into the light sphere $\sum_{m=1}^{N} y_{m}^{2} = (c\tau)^{2}$. We have also demonstrated that a single Lorentz transformation on an N-sphere reduces to the case of a 1-sphere as all other directions are transverse to the boost and thus unchanged.

Summary : We have constructed the component action of Lorentz transformation in a matrix representation from the definition of Lorentz transformations. We have demonstrated that such a transformation does not alter the norm of vectors in TM. We have demonstrated that the light cone in a given set of coordinates (aka frame 1) is mapped to the light cone in a second set of coordinates (aka frame 2) if the change of basis transformation is a Lorentz transformation.

Definition : Given a pseudo-Riemannian space with n space-like directions and N-n time-like directions normal coordinates $x^{\mu}$ are such that $\eta_{\mu\nu} = \eta(e_{\mu},e_{\nu})$ is diagonal with N-n entries of -1 and n entries of +1.

If $\delta(k)$ is the k by k Kronecker delta then $\eta = (-\delta(N-n)) \oplus (\delta(n))$. We take $x^{\mu} \to (x^{a},x^{A})$ under this decomposition.

Definition : The Weyl rotation trick is the Euclideanisation of M by the coordinate rotation $x^{a} \to -ix^{a} \equiv y^{a}$ and $y^{A} = x^{A}$ and thus $e_{a} \to -if_{a}$ and $e_{A} \to f_{A}$ such that $\eta(f_{\mu},f_{\nu}) \sim \delta(N)$.

Definition : Given a fibre bundle E over base M, $E \to^{\pi} M$ whose fibres $\pi^{-1}(p)$ for $p\in M$ are morphic to some vector space V then there is a principle G-bundle such that a representation of G is the symmetry group of V. Or more clearly given a vector space you can examine it either by examining its vector structure or its symmetry structure.

Definition : The symmetry group associated to a space with n space-like directions and N-n time-like directions is SO(N-n,n). Euclidean space $\mathbb{R}^{n}$ has symmetry group SO(n). Minkowski space-time previously defined has symmetry group SO(1,N).

Under the Weyl rotation trick Minkowski space-time is Euclideanised and has symmetry group SO(N+1), which is that of N+1 dimensional space in Newtonian physics. Hence the mathematical structure of Minkowski space-time, and by corollary special relativity and electromagnetism, is morphic to that of Euclidean space. Hence an inconsistency in one feeds through into the other. Furthermore by the equivalence of vector spaces and their symmetry groups if the mathematical structure of geometry in Minkowski space-time is flawed then it feeds through into all group theory and all Lie theory.

Hence a flaw in special relativity wouldn't be tucked away in some convoluted physical set up, it would be endemic to pretty much all maths and physics for the last 400+ years. Given various areas have been proven to be mathematical consistent this would contradict the notion the geometry of Minkowski space-time is inconsistent.

Feel free to ask any questions on any algebra or logic I use Jack, I have nothing to hide. I'll provide references to specific results if you don't believe me or you want to find out more. Most of it is stuff I'd expect any competent first year to know and be able to work through themselves. Other than the fibre bundles any one whose done more than 2nd year vector calc would be able to grasp it. Its covered in all relativity textbooks too and is assumed knowledge in any graduate course. The fact null vectors map to null vectors proves your claim about LTs mapping points on a light cone to points off a light cone false. If you claim otherwise explain where my algebra is wrong and provide your alternative working.

Failure to respond will mean I simply post a link to this post every time you whine about wanting someone to 'show you the math' or provide mathematical justification or to 'put up or shut up'. If you can't convince a 'primitive' theoretical physics PhD then you've got no chance with professors who'd run rings around me. Try not to fall at the first hurdle.
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You are wasting your time.

The objective is to map the stationary light sphere such that at any one time in the stationary frame, the light sphere will be spherical in the moving frame.

Do not forget, if SR is true, it must be spherical in the moving frame when mapped.

As you mentioned and I agree, it is distorted by time after mapping. It must be spherical for the theory to be consistent without distortion.

That requires one time in the stationary frame for the mapping and also requires the mapping of the origin of the light sphere which is not the origin in the stationary sphere.

Good luck.

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Jack_ · Apr 4, 2010

RJBeery said: ↑

To be fair I think it's pretty easy to say that he's got people thinking. If you guys thought he was an absolute crank, rather than just stubborn, arrogant, and wrong, then you would've left the thread long ago.

His paper has plenty of math, btw. Maybe he has already submitted to a journal and was rejected..?
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I can already tell you know the train embankment experiment is wrong because the train has its valid emission points also.

So, is this your way of getting along?

As you can see, I am a crasher.

Cept, they are only dimly aware I am crashing SR.

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Jack_ · Apr 4, 2010

AN, are you goiing to at least attack my first post?

How about any of you?

You all will attack me personally but will not respond to several proofs I have given.

If you all are intelligent, I assume you would focus on my presentations.

If it is the case that the mods here felt this is just off science, it would seem you all could do something to prove something to stop for example the twins contradiction.

Here is the bottom line.

That post is sufficient to take out SR.

Personally I refuse to believe a failed theory.

Looks like I am different froms others.

So, who here can refute the first post in this thread?

Jack_ · Apr 4, 2010

RJBeery
Armchair Physicist (532 posts)
Today, 10:55 AM #27

Jack, I've read your proposed solution to the SR problems we've been discussing (i.e. Absolute Motion). Wouldn't absolute motion imply that we could detect redshift/blueshift from an inertial light source?
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No it does not.

Frequency is a function of wavelength and light speed.

Therefore, since MMX detects a constant frequency, then the absolute motion is built into the speed of light, ie Ritz's theory, which has been refuted by experiments from moving lights sources, or the absolute motion is built into the wavelength.

Since light has been shown to be constant in the vacuum of space, and frequency is constant in a frame that in reality is moving somehow, then the wavelength must be compressed in the absolute direction of travel causing a constant frequency in a frame that has some kind of motion.

So, no, we would not see doppler under these required conditions.

James R · Apr 4, 2010

Jack_:

Jack_ said: ↑

Post one is still sitting in this thread without any refutation.
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I can't comment on that because I haven't bothered reading the thread. Probably, given that people like Pete and Alphanumeric have been posting here, post #1 was refuted long ago.

Since, you clearly are more intelligent than I am, ad I am stupid from your view, I will expect you will refute it quickly.

How long will this take?
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I gave you a chance when you first arrived here, Jack_. I quickly established from my interactions with you and from observing your interactions with Pete that your aim is to troll here and you post in bad faith.

My time is too valuable to waste on you. Goodbye.

Jack_ · Apr 4, 2010

James R said: ↑

Jack_:

I can't comment on that because I haven't bothered reading the thread. Probably, given that people like Pete and Alphanumeric have been posting here, post #1 was refuted long ago.

I gave you a chance when you first arrived here, Jack_. I quickly established from my interactions with you and from observing your interactions with Pete that your aim is to troll here and you post in bad faith.

My time is too valuable to waste on you. Goodbye.
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James, you are superior as you say.

Quickly dispel the first post according to your truthfullness.

Rpenner and AN would do this if they could and have done it on several occasions to others.

If I an so inferior, why can't you folks take out even the first post which I consider primitive.

So, if I am so stupid, narrow minded, ignorant and all the other names I have been called, what does that make everyone that cannot refute my simple little first post?

RJBeery · Apr 4, 2010

Jack said:

I can already tell you know the train embankment experiment is wrong because the train has its valid emission points also.

So, is this your way of getting along?

As you can see, I am a crasher.

Cept, they are only dimly aware I am crashing SR.
Click to expand...

Dude I don't try getting along around here. I do what you are doing - I pick fights to test and expand my knowledge. I've butt heads with AN, BenTheMan and a few other less capable guys...

Here's a thread on Einstein Locality.

Here's one where I declare that black holes don't exist.

Here's one where I point out that, without BLOCK TIME, SR and QM are incompatible!

Jack_ · Apr 4, 2010

RJBeery said: ↑

Dude I don't try getting along around here. I do what you are doing - I pick fights to test and expand my knowledge. I've butt heads with AN, BenTheMan and a few other less capable guys...

Here's a thread on Einstein Locality.

Here's one where I declare that black holes don't exist.

Here's one where I point out that, without BLOCK TIME, SR and QM are incompatible!
Click to expand...

OK, I stand corrected.

RJBeery · Apr 4, 2010

Now, despite my last post, when I'm wrong I admit it and move on. I also respect the knowledge and math skills of a few folks here, even if they lack vision...anyway I am reserving judgment on your posts until I have enough time to understand them.

Jack said:

No it does not.

Frequency is a function of wavelength and light speed.

Therefore, since MMX detects a constant frequency, then the absolute motion is built into the speed of light, ie Ritz's theory, which has been refuted by experiments from moving lights sources, or the absolute motion is built into the wavelength.

Since light has been shown to be constant in the vacuum of space, and frequency is constant in a frame that in reality is moving somehow, then the wavelength must be compressed in the absolute direction of travel causing a constant frequency in a frame that has some kind of motion.

So, no, we would not see doppler under these required conditions.
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I need to think about this more...I not used to thinking in terms of c+v. Speaking of red-shift, it would clearly be detectable by either X or Y observers in Einstein's train embankment (whichever observer was in relative motion to the sources). This red-shift could be measured to adjust the calculated emission times and come to full agreement on simultaneity of the flashes, I believe. Not sure it adds much to the discussion but it's something to think about.

James R · Apr 4, 2010

Jack_ said: ↑

James, you are superior as you say.

Quickly dispel the first post according to your truthfullness.
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What will you do for me if I spend time refuting your first post? Will you agree, for example, never to post on the subject of relativity here again?

I want some concession from you if I'm going to spend my valuable time on this.

RJBeery · Apr 4, 2010

Jack said:

OK, I stand corrected.
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So my Crank Credentials check out OK? haha
Please Register or Log in to view the hidden image!

Anyway my biggest problem with this forum is that many people seem to be pretty quick to rudely dismiss things, (regardless of how wrong or right the poster is). Call me naive but I think a science forum should be a place where new ideas are floated out to the community, and rejected, modified or accepted, with all participants being polite and positive. Pete is pretty good about remaining polite and mature so the fact that you got him so riled up made me assume you were riding the woo-woo train until I spent some time reading.

I think I would shit my pants if I ever saw someone post "Wow, that's incredibly interesting!" In this thread, for example, AN wrote

AlphaNumeric said:

The only non-trivial point you raise is that each frame will see a particular point in space-time as the centre of that sphere and they are not mapped to one another by the corresponding Lorentz transformation.
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Coming from him, this is the biggest compliment you're going to get. And after 100 posts it's obvious that THIS IS THE WHOLE POINT YOU'RE TRYING TO MAKE IN THIS THREAD. Now I cannot tell you whether or not this is the "death-knell" of SR as we know it and as I said, the jury is still out until I can sit down and understand it more.

James R · Apr 4, 2010

RJBeery:

RJBeery said: ↑

Anyway my biggest problem with this forum is that many people seem to be pretty quick to rudely dismiss things, (regardless of how wrong or right the poster is). Call me naive but I think a science forum should be a place where new ideas are floated out to the community, and rejected, modified or accepted, with all participants being polite and positive. Pete is pretty good about remaining polite and mature so the fact that you got him so riled up made me assume you were riding the woo-woo train until I spent some time reading.
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He is riding the woo-woo train. Pete and I gave him a fair hearing, but he didn't want to learn anything. He just wants to troll on the subject of relativity.

Now I cannot tell you whether or not this is the "death-knell" of SR as we know it and as I said, the jury is still out until I can sit down and understand it more.
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I'd offer the following advice to those who want to "refute" relativity:

1. If your aim is to show that the universe is not as the theory of relativity describes it, then you need to come up with some observation or experimental evidence that contradicts the predictions of relativity. Mere thought experiments won't do the trick. While you might be able to imagine a universe in which Lorentz invariance doesn't hold, that in no way means that our universe is in any way similar to your imaginary universe. Also, what many anti-relativity people fail to bear in mind is that their new theory must also be consistent with (i.e. able to explain) all the experiments that have been done that appear to support relativity (unless you can point to flaws in the experiments).

2. If your aim is to show that relativity is inconsistent with our universe, then you need to disprove at least one of the two postulates of special relativity, or the principle of equivalence that underlies general relativity.

3. If your aim is to prove that relativity is "self-refuting" - i.e. internally inconsistent - then you might as well give up now. All of relativity follows from the basic postulates of the theory. If you accept those, then you are forced to accept all mathematical consequences of the postulates as well. Countless people have already tried to find a flaw in things like the relativity of simultaneity, and every single one of them has failed. Besides, the mathematics of relativity has advanced to such an extent in the past 100 years that it is relatively easy for any mathematically knowledgeable person to just see that the theory is mathematically self-consistent. It helps tremendously that it has so few axioms. So, see point 2.

AlphaNumeric · Apr 4, 2010

Jack_ said: ↑

The objective is to map the stationary light sphere such that at any one time in the stationary frame, the light sphere will be spherical in the moving frame.

Do not forget, if SR is true, it must be spherical in the moving frame when mapped.

As you mentioned and I agree, it is distorted by time after mapping. It must be spherical for the theory to be consistent without distortion.

That requires one time in the stationary frame for the mapping and also requires the mapping of the origin of the light sphere which is not the origin in the stationary sphere.

Good luck.
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I demonstrated the Lorentz image of the light sphere is a light sphere. Its in my post. Didn't you understand it? If you're quoting the post and then claiming I didn't do something I clearly did and which I said as much in words, then either you're not reading what I post or you don't understand what I post. Either way you can't claim I didn't provide it, its right there.

AlphaNumeric · Apr 4, 2010

Jack_ said: ↑

AN, are you goiing to at least attack my first post?
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We've already discussed specific physical setups with you. The problem is you fail to understand even basic simple cases which you need to understand to discuss things like you're first post.

I've posted the mathematical demonstration that light cones go to light cones. If you are unable to understand or accept this then discussing your post post is pointless since you lack the understanding to have an informed discussion on it.

Jack_ said: ↑

You all will attack me personally but will not respond to several proofs I have given.
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What the fuck do you think my last post was? I addressed several things you'd claimed like the light cone not mapping to the light cone. You're unwilling or unable to talk to me about it. I address your claims, you ignore any and all retorts. You're incapable of having an honest discussion.

Jack_ said: ↑

That post is sufficient to take out SR.

Personally I refuse to believe a failed theory.
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I'm still willing to put $1000 dollars on thye table. If you're so sure you're right then why aren't you taking what is basically free money from me? I address claims you make, you never, never, enter into a discussion about them. I've demonstrated light cones map to light cones. Can't you reply? Can't you tell me where my maths is wrong? I provide more and more mathematics, which is what you demand, and then you refuse to talk about it.

If you can't answer direct simple questions then you're just a hypocrite. Claiming "No one has retorted my first post" is just an ignorant delusional lie. You pretend you're taking some kind of higher ground but you're not. You just resort to lies and ignorance. If you're so sure you're right why are you arguing with 'primitive' posts like myself? Why aren't you talking to a journal?

Because you know you're talking bullshit.

AlphaNumeric · Apr 4, 2010

I thought I'd comment further on the nature of Lorentz transformations while I remember what I've just remembered (if you see what I mean). Part of the issue Jack has is that he thinks there's some kind of ambiguity in how the define the light cone if two frames will disagree on the location of the centre of the expanding sphere of light. This can be examined using a bit of vector analysis.

We consider the light cone of some point p. If vectors v and u, components $v = v^{\mu}\partial_{\mu}$ and u = u^{\mu}\partial_{\mu}[/tex], define the velocities of the two rigid spheres then they transform under a Lorentz transformation. There is a set of coordinates such that $v = \partial_{t}$, ie $v^{\mu} = (1,0,0,0)$ but if the spheres are in relative motion in these coordinates $u^{\mu} \neq (1,0,0,0)$. If you solve the geodesic equations for v you find that at time t the centre of the rigid sphere is at $(t,0,0,0)$. This coincides with the centre of the photon sphere. Conversely there are coordinates such that $u^{\mu} = (1,0,0,0)$ and it defines a path of the form $(t',0,0,0)$. This cooincides with the photon sphere's centre in the new coordinates.

Jack's issue is that since the path associated to v is different from that associated to u it seems like the photon sphere are no well defined centre and you can't use Lorentz transformations consistently as its ambiguous what the photon sphere is if you can't define its centre. What he fails to realise is you define photon spheres/cones by their apex, which is a point in space-time, not by their centre in space.

This can be seen by considering a more formal construction. Given the space-time manifold M its tangent space at point p is $T_{p}M$ and the union over M of these is the tangent bundle, $\bigcup_{p}T_{p}M = TM$. The vectors v and u are defined in a specific tangent space, $T_{p}M$ while the velocities of the spheres for t,t'>0 are not in the same tangent space, they are elsewhere in the tangent bundle. Usually a vector is denoted in terms of components like $X^{\mu} = (a,b,c,d)$ etc. This is a point in the fibre of $TM$ at a given point. 4 dimensional Minkowski has dibres morphic to $\mathbb{R}^{4}$. As such a vector in TM is not written as just $\mathbb{R}^{4}$ elements but in the form $M \times \mathbb{R}^{4}$, for a local trivialisation of the bundle, ie $TM \to^{\pi} M$ such that $\pi^{-1}(p) \sim \mathbb{R}^{4}$. In such a trivialisation a point P in TM has coordinates $(p,X)$ where p defines the point in M where the vector X is defined and $\pi(P) = p$.

Lorentz transforms map vectors to vectors pointwise, ie generically $\Lambda : TM \to TM$ but for vectors defined at p we have $\Lambda_{p} : T_{p}M \to T_{p}M$. Hence in order to change frames you have to specify a point in M at which you're defining your Lorentz transformations. The very definitions of the vectors in TM associated to the velocities of the rigid spheres or vectors u and v carry with them the location p in M where they are defined, ie $V \in TM$ such that $\pi(V) = p$.

You cannot naively compare vectors are different points in M, if V,W in TM then V+W or V-W or any other binary operation of this form is only well defined if $\pi(V) = \pi(W) \in M$. You must use things like parallel transportation for such things.

As such by defining the vectors of Frame 2 in terms of vectors of Frame 1 both constructions have the location of the apex of the light cone. Since each point in M is associated to one and only one light cone there is no contradiction or ambiguity or issue with the light cone. All Frame 1 needs to tell Frame 2 is the point p (which is definable without reference to specific coordinates, see any decent book on manifolds) and Frame 2 then knows the light cone in question. Since the Lorentz images of the vector velocities of the rigid spheres carry with them this information via the projection $\pi$ then if Frame 2 knows what vectors to consider then it knows the light cone.

Jack doesn't like the fact that the geodesics defined from p and the vectors transform non-trivially but the light cone only depends on the choice of p, not on any vectors at p. This only happens for the light cone. To work out the motion of one of the rigid spheres in Frame 2 you need both the point p and the Lorentz image of the velocity in Frame 1, as the geodesics are not invariant. This is two bits of information. To work out the light cone you need only p.

In terms of physics this is equivalent to the fact light's motion doesn't depend on the motion of the emitter, an experimental fact. The rigid spheres being in relative motion is immaterial, when their centres coincide and they both make a flash the two resultant photon spheres will remain coincident with one another, ie overlap. Instantaneously at the moment of emission you can boost from the vector velocity of one sphere to the vector velocity of the other sphere but nothing will happen to the light sphere, it is 'blind' to these vectors as it must be if all frames are to see the speed of light being the same. Given a vector V you know the point its defined at, $\pi(V)$. The light cone only cares where the vector is defined, not what the vector's actual value is. Jack claims that since different vectors lead to the same light cone but different geodesics (aka paths swept out by centres of some rigid sphere with that initial velocity) then its a contradiction. No, its an inevitable result from light and only light being Lorentz invariant and is required in physics if light's motion is to be independent of its emitter's motion.

Simply because you have the freedom to boost to a frame where a particular geodesic is associated to the stationary vector $\partial_{t}$ doesn't mean its inconsistent. All frames agree on the causal structure and no geodesics become degenerate under a Lorentz transformation, since all elements of SO(1,3) are invertible due to unit determinant).

Jack, I'll say it again; you haven't come up with some amazing new result or seen something no one else has. I've drawn a few worldline/light cones diagrams and the result is little more than the fact given a time-like vector X you can pick a frame such that $X^{\mu} = (1,0,0,0)$. Lorentz transformations are the residual coordinate freedoms left after causal structure is defined. Causal structures are defined by light cones and then you can pick freely what vector at the apex of the light cone you wish to be 'constant'.

If you simply repeat "OMG they disagree on the centres!" you're not demonstrating a contradiction. I've repeatedly explained how this isn't a contradiction as the centre of the light sphere has no physical meaning/properties. All objects in the space whose worldlines transform interact consistently.

If you haven't already done so please confirm that you understand a Lorentz transform leaves the space-time interval invariant, ie $p^{\mu}p_{\mu} \to (\Lambda\cdot p)^{\mu}(\Lambda \cdot p)_{\mu} = p^{\mu}p_{\mu}$ and thus null worldlines map to null worldlines, aka light cone to light cone. If you do not understand or do not accept this result please explain specifically where you disagree with my previous lengthy post where I demonstrate it.

Neddy Bate · Apr 4, 2010

Jack_ said: ↑

AN, are you goiing to at least attack my first post?

How about any of you?

You all will attack me personally but will not respond to several proofs I have given.

If you all are intelligent, I assume you would focus on my presentations.

If it is the case that the mods here felt this is just off science, it would seem you all could do something to prove something to stop for example the twins contradiction.

Here is the bottom line.

That post is sufficient to take out SR.

Personally I refuse to believe a failed theory.

Looks like I am different froms others.

So, who here can refute the first post in this thread?
Click to expand...

I think Pete did that in post #2, but I'll have a go at it.

You start with two twins together at the same place. You accelerate Twin#1 first, while Twin#2 stays inertial. This creates reciprocal time dilation where each twin thinks the other twin is younger. To resolve the reciprocality, the twins must be brought together to the same reference frame again. This is can be accomplished by accelerating either one of the twins to the same frame as the other. Whichever twin is accelerated second, will be younger, because the other one will age drastically during the acceleration.

Neddy Bate · Apr 4, 2010

Jack_ said: ↑

The objective is to map the stationary light sphere such that at any one time in the stationary frame, the light sphere will be spherical in the moving frame.
Click to expand...

Let there be two small light bulbs, one stationary in each frame. When the bulbs are very close to each other, they both flash, creating one light sphere. Each frame can think of their own light bulb as the center of their own light sphere. They don't have to care about the moving bulb.

Jack_ · Apr 4, 2010

Neddy Bate said: ↑

I think Pete did that in post #2, but I'll have a go at it.

You start with two twins together at the same place. You accelerate Twin#1 first, while Twin#2 stays inertial. This creates reciprocal time dilation where each twin thinks the other twin is younger. To resolve the reciprocality, the twins must be brought together to the same reference frame again. This is can be accomplished by accelerating either one of the twins to the same frame as the other. Whichever twin is accelerated second, will be younger, because the other one will age drastically during the acceleration.
Click to expand...

No, this is not correct. The acceleration phases are symmetric leaving on the relative motion phases.

I provide a detailed calculation of both acceleration phases here.

based on this paper which is the standard.

Jack_ · Apr 4, 2010

Neddy Bate said: ↑

Let there be two small light bulbs, one stationary in each frame. When the bulbs are very close to each other, they both flash, creating one light sphere. Each frame can think of their own light bulb as the center of their own light sphere. They don't have to care about the moving bulb.
Click to expand...

OK, in the stationary frame, after time t, what does the clock read at the moving light bulb?

If you say t/γ, then you are correct.

If the clock in the stationary frame reads rγ/c, then the clock at the bulb in the moving frame reads r/c.

If a clock reads r/c at the origin of light emission, then light is a distance r in all directions.

All of this is SR logic.

Now, run LT on t= rγ/c, and x = rγ. You will find x' ≠ r the moving frame which is a constradiction.

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