Time in Special Relativity

Jack, the lorentz transform isn't that hard a concept. Why are you being so dense?
If I'm doing it wrong, then why did you accept the table I drew up earlier in the thread?
Are you admitting you are incapable of transforming an event from one reference frame to another?

Here is an event in the frame of O':
(x',t')=(-r,0)

Transform it to the frame of O. easy.

 

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I do not know what to tell you.

You cannot have an event from 0 to -r with no time. Are you claiming a different start time or something?

Why did you get off task of the LT contradiction?

 

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:bugeye:
Jack, you're making no sense at all.

At time t'=0 in the frame of O', there is a clock at x'=-r.

Right?

That's an event.
t'=0
x'=-r


What's so hard to understand?

 

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Apparently, everything.

Jack_, do as he says: Transform the event at (x',t') = (-r,0) in coordinates of frame O' to the coordinates of frame O.

It's ridiculously easy, and doesn't even require any understanding of relativity. If you cannot, it's yet more evidence that you're trolling this subforum.

 

If you want to call that an event, be my guest. You cannot apply LT unless time has elapsed. I am not seeing an elapsed time.

 

Perhaps you can help me out with this difficult problem. It appears you understand something.

What do you think the correct answer is under LT. Remember, some motion must occur. LT is concerned with mapping vectors to vectors. These vectors require a change in position and change in time.

It is not a point mapper.

 

You have the strangest ideas. Where do you get them?

Jack, we're doing simple coordinate transformations. This is really simple algebra. You just substitute the values for x' and t' into the equations. What's stopping you?

Lorentz transformation:

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x'=-r
t'=0
t=?

I know you can do the algebra, so what's the problem?

Do you agree that plugging x'=-r and t'=0 into the above lorentz transform equation for t gives t = -γrv/c² ?

 

I have told you time and time again, LT is not for t'=0. We need some elapsed time to apply LT. t'>0 and then we can apply LT.

Are you messing or is it the case you really have no understanding of LT?

 

You can tell me all you like, Jack, but that doesn't make you right.

You had no problem earlier in the thread when we transformed event with x=0 and x'=0. So what's your problem now?

What's stopping you from plugging in x'=-r and t'=0 into the equations?

If you do plug in those values, do you agree that the result for t is t -γrv/c² ?

 

I never plugged in x'=x=0 unless t'=t=0.

You need an elapsed time to apply LT.

Your results are therefore, false.

Can you prove you can use LT for static points? I am betting you need vectors.

 

That's not what I meant.
I'm asking you to plug in x'=-r amd t'=0. x and t are unknowns.
Earlier in the thread, we transformed events with x=0, t<>0, and events with x'=0, t'<>0.

You agreed (post 23) that transforming (x',t')=(0,r/c) to (x,t)=(γvr/c, γr/c) was a correct application of the lorentz transform. Note that x'=0, t'<>0, x<>0, t<>0.

You agreed (post 40) that the transforms in the table were correct including six for x'=0, t'<>0, and six for x=0, t<>0.

You seem happy with plugging in x'=0, so why are you unable to plug in t'=0? Really, Jack, what is so hard about plugging values in to an equation?

No, you don't. Clearly, you just need the appropriate values to plug in to the equations.

Jack, look at the equations:

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We have x', we have t'. You can consider these to be spacetime coordinates, or you can consider them to be the components of a spacetime displacement vector. It works either way. Plug in the values, get out the answer.

Why are you unable to do this?

Look, it's really this simple:
x'=-r
t'=0

x = γ(x' + vt)
t = γ(t' + vx/c²)

Plug in x' and t', get out x and t. What's your problem?

 

What? No, that's not correct.

What? No, that's not correct.

I think AN's technical discussion has broken you.

It really is as simple as Pete's showing you. Plug-and-play. Shake-and-bake.

 

This is wrong. I have explained to you several times that LT transform vectors, not points and that the reason they are expressed in terms of coordinate in common discussion is that you can easy obtain the associated integral curve through space-time as the space-time is flat. General relativity does not assume flat space-time and thus when you apply a Lorentz transformation you do not express it in terms of coordinates. In fact the method of demonstrating Lorentz invariance in GR explicitly requires you to consider sets of vectors at a point, not sets of coordinates. If the space-time is curved then you cannot replace the vector components with space-time coordinates. If you actually understand the posts I make then you should easily know how to prove this to yourself, you should be aware of what integral curves are and their relationship to geodesics.

When you transform from one frame to another you do it by picking a point in the space-time, p, and then applying a Lorentz transformation \(\Lambda : T_{p}M \to T_{p}M\). The frames, by definition, agree on the point p. If you let time run for any non-zero amount of time, in any frame for an object whose velocity is any non-zero 4-vector, then you are no longer considering the same space-time point, call it q. The Lorentz transforms \(T_{p}M \to T_{p}M\) are not those of \(T_{q}M \to T_{q}M\). You might be fooled into thinking they are because the flat space-time of SR means that they have the same matrix representations, the same general components etc but formally they are entirely seperate spaces. This is the essence of the 'fibre' in 'fibre bundle', you have a set (a 'bundle') of spaces, one space (ie space of tangent vectors) for each point in your set (ie your space-time). If you were doing your analysis in full generality then you'd have to consider the impact on a particular \(\Lambda(p) : T_{p}M \to T_{p}M\) when you parallelly transport the base space point p to q. This is measured in terms of curvature (I can run through an explicit example in regards to vectors if needs be) and for SR that's zero so the assumption that you can work in terms of coordinates is justified. This is not the case generally.

And because I'm showing off and in order to head you off before you make some comment about me not understanding this kind of thing, the notion of transporting vectors and indeed any tensor or spinor through space-time forms the basis of my PhD and unlike you I've demonstrated my knowledge and understanding to professors. And to top it off the professor who did my PhD viva emailed yesterday to say he's accepted my amended thesis and thus I'm now a Dr. Its amazing what a little bit of effort and intellectual honesty will get you, you should try it some time Jack.

No, this is something I've explained to you before. You no longer have an excuse of ignorance, since you have been told it several times, you're now simply being dishonest (well, more so than before). Lorentz transformations map vectors defined at some point p to vectors also defined at p. I explained this more formally when I went through things like \(\pi(T_{p}M) = p\) with you. Would you like me to provide a link again to that post? Or perhaps you'd like another book recommendation? Or a link to some lecture notes on the matter?

You have fallen into the trap of not understanding the difference between a manifold M and its tangent space fibre \(T_{p}M \sim V\). The local bundle trivialisation of the patch \(\mathcal{U} \in M\) is \(T\mathcal{U} \equiv \mathcal{U} \times V \sim \mathbb{R}^{n} \times \mathbb{R}^{n}\), where n is the dimensional of a basis of V. Lorentz transformations transform the second \(\mathbb{R}^{n}\), not the first. The local trivialisation provides \(u = (p,v)\) for \(v \in V\) and a general point \(u \in T\mathcal{U}\) contains sufficient information to solve the second order geodesic equation to obtain an integral curve through \(T\mathcal{U}\), \(\gamma(s)\) such that \(\gamma(0) = u\). By varying s you move through the curve to other points in \(T\mathcal{U}\) and due to the fact \(\pi(\gamma(s))\) is injective (ie the integral curve defines a section of some \(T\mathcal{U}'\)) you obtain a new set of space-time coordinates as you move along the curve, ie \(\pi(s_{1}) = \pi(s_{2}) \Rightarrow s_{1}=s_{2}\) (as you should be aware of, injectivity is defined in this manner). The Lorentz transform didn't change p, it changed the part of u which is associated to V and thus provides a new set of integral curves.

Once again you have tried to give the impression you're well read when it comes to vectors and points etc but all you've done is demonstrate how incredibly naive you are about both the level of work and amount of material done in regards to these things by the scientific community and also how massively you over estimate yourself while under estimating others. Don't you get tired of posting things you don't you don't have a clue about? Don't you get tired of lying and having those lies exposed? Do you have some kind of compulsion to lie again and again?

 

My hair is ruffled... did anyone else hear a "whoosh"?

 

What hair?

I hope he won't mind mind my saying so, but I think Alpha is being just a little obtuse, possibly for his own entertainment in baiting Jack, as much as anything. All he said was correct, as far as I am qualified to judge, but a little hard to follow for those who know little or nothing about manifolds, tangent spaces and bundles.

I feel sure that if you, or anyone else for that matter, wants a breakdown of his excellent post(s) there are some here who would be willing to oblige; I count myself as one, provided it is done in good faith.

Mind you, it would be an extremely long session.....

 

Yep, I'm being deliberately over the top with my explanations.

This is partly because a full explanation of my points in simpler terms would have been an even longer post and possibly require a number of diagrams*; But mostly because Jack claims to have a good understanding of vectors, space-time, coordinates etc, believes he's doing insightful new advanced physics (which he's gracious enough to teach us...) and has previously implied he understands posts of mine where I start talking about bundles (I can provide a link if needed). He complains how 'primitive' I am and constantly claims to have a wish to discuss his 'work' and so one might think that he'd welcome my attempts at raising the level of discussion to something a little more advanced, particularly given they strip away the cumbersome and tedious nature of doing explicit Lorentz transformations.

Of course he and I (and anyone else who reads his posts) knows that he doesn't have a clue what I'm talking about. I'm doing a slightly modified crank tactic; cranks often dial up the level of their buzzword usage to frighten off any pseudo-laymen who might be willing to stand up to them (hence Jack's love of 'decidable'). This allows them to control the discussion more and basically bluff their way through, rather than provided justification for their claims. I'm simply using the same tactic, to a sufficiently large extent that even with a bit of furious Wikipedia searching its extremely hard to even come up with some vaguely coherent and relevant reply, never mind an attempt at a retort. The difference in my case is that I actually know the terminology, rather than simply being someone who picks out buzzwords from Wikipedia.

Yes, I'm more than happy to dial it down for anyone else. Its important to go in at the deep end, reducing as much as possible a crank piecing some vague semblence of an understanding by Wiki'ing bits of a simpler explanation and then claiming they understand the physics being discussed. I fully expect Jack to either not reply at all or to dismiss my comments as long winded or simply to complain I haven't retorted his claims, all the while professing knowledge of bundles. No doubt as soon as I (or someone else) provides a simpler explanation he'll get his foot in the door and become a little more vocal, backed up with newly gathered 'Google Intelligence'. GI is like AI, it simply matches words, rather than provide understanding and reasoning.

* I asterix'd something above and what I wanted to say is also a response to this comment. Often, particularly when having to talk a crank through some bit of mainstream material I often think "This could be done in 10 minutes if we were face to face and next to a black board!", particularly when diagrams are needed. I lurch from thinking "I like typing it out" through to "Oh god its going to take bloody ages to explain this, I need a lot of diagrams and tons of equations".

 

LOL, I'll but all you do is surf.

 

The events require action in the universe. In LT construction, Einstein used the clock synch method in the moving frame and then took the partial derivative of Tau with respect to t since obviously the position is also changing.

Thus, events are mapped not simply by mapping but because light is "action" and it transfers in a finite fashion.

As such Minkowsky space-time is generated by the action of light.

Now, to prove your case, you will need to prove LT construction was not derived by taking the partial derivative of Tau with respect to t against the clock synch method in the moving frame.

Let me know.

See section 3
http://www.fourmilab.ch/etexts/einstein/specrel/www/

 

The reason I will not accept any t or t'=0 is because when frames are in realtive motion, to determine a position, one needs time.

LT is a dynamic application and you, not knowing what you are doing, are attempting to caste it a static space to space mapping. This indicates you do not understand SR and since James T agrees with you, he does not either.

I note both all of you have conceeded that my time dilation and light sphere argument has refuted LT.

Thanks.