Einstein's Relativity

Discussion in 'Physics & Math' started by RJBeery, Mar 31, 2010.

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  1. RJBeery Natural Philosopher Valued Senior Member

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    Technically speaking, according to his description, both observers would concur on the simultaneity of the lightning strikes since they are taking the measurement when observer X sees them (and, by his own description, observer Y is co-located with X at that time). I am not questioning SR, I am complaining about his explanation of it, which others have pointed out may be lacking due to the fact that he was the first to describe it.
     
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  3. AlphaNumeric Fully ionized Registered Senior Member

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    Excellent response. You demand I give 'math proofs' and when I do so you don't retort any of them.

    Explain to me what precisely you didn't get about my previous posts comments about space-time intervals. I asked you a direct question and you ignored it.

    I just provided it. Obviously you need to have your hand held through this one.

    Frame 1 has coordinates \((t,\underline{x}})\). Light is associated with null 4-vectors \(p^{\mu}\) such that \(p^{\mu}p_{\mu} = 0\). For the space-time interval you get \(ds^{2} = dx^{\mu}dx_{\mu} = dx^{\mu}dx^{\nu}\eta_{\mu\nu} = -dt^{2} + d\underline{x}\cdot d\underline{x}\).

    Frame 2 has coordinates \((t',\underline{x}'})\). Light is associated with null 4-vectors \(q^{\mu}\) such that \(q^{\mu}q_{\mu} = 0\). For the space-time interval you get \((ds')^{2} = (dx')^{\mu}(dx')_{\mu} = (dx')^{\mu}(dx')^{\nu}\eta_{\mu\nu} = -(dt')^{2} + d\underline{x}'\cdot d\underline{x}'\).

    Let Frame 2 be moving with velocity \(\underline{v}\) with respect to Frame 1. Thus to go from Frame 1 coordinates to Frame 2 coordinates we apply a Lorentz boost in the direction \(\frac{\hat{v}}{|v|}\). We denote this by the matrix \(\Lambda^{\mu}_{\nu}\) and sync the origins of the two frames such that \((t,x) = (0,0) = (t',x')\) coincide initially. We will make use of the definition of \(\Lambda\), that \(\Lambda^{\top} \cdot \eta \cdot \Lambda = \eta\) or in a given set of coordinates \(\Lambda^{\rho}_{\mu} \eta_{\rho\sigma} \Lambda^{\sigma}_{\nu} = \eta_{\mu\nu}\).

    A null 4-vector in Frame 1, \(p^{\mu}\), transforms to \(P^{\mu} \equiv \Lambda^{\mu}_{\nu}p^{\nu}\). We compute its norm, \(P^{\mu}P_{\mu}\). Inserting its definition we have \(P^{\mu}P_{\mu} = P^{\mu} \eta_{\mu\nu} P^{\nu} = \Lambda^{\mu}_{\tau}p^{\tau} \eta_{\mu\nu} \Lambda^{\nu}_{\sigma}p^{\sigma} = p^{\tau} \Big( \Lambda^{\mu}_{\tau}\eta_{\mu\nu} \Lambda^{\nu}_{\sigma} \Big) p^{\sigma} \). By definition of the Lorentz transformations the terms in the brackets simplify : \( \Lambda^{\mu}_{\tau}\eta_{\mu\nu} \Lambda^{\nu}_{\sigma} = \eta_{\tau\sigma}\). Therefore \(P^{\mu}P_{\mu} = p^{\tau}\eta_{\tau\sigma}p^{\sigma} = p^{\mu}p_{\nu}\).

    Therefore \(p^{\mu}\) and \(P^{\mu}\) have the same norm. Therefore if \(p^{\mu}p_{\mu} = 0\) then \(P^{\mu}P_{\mu} = 0\). If \(p^{\mu}\) is null then its Lorentz transformation image, \(P^{\mu}\) is also null, aka on the light cone. I haven't had to restrict what particular Lorentz transformation I'm considering since they all obey the same defining property, they leave the metric unchanged. Thus I've proven that regardless of what Lorentz transformation you apply a null vector maps to a null vector aka a point on the light cone goes to a point on the light cone.

    Now no doubt you're struggling to follow this algebra so I'll do a particular case. We'll consider a 2d setup ie Frame 1 has coordinates (t,x) and Frame 2 has coordinates (t',x') and they are related by a Lorentz boost of speed v along the x axis. Therefore we have the standard formula for the coordinates : \(x' = \gamma(x-vt)\) and \(t' = \gamma(t - vx)\) (having set c=1 for less cluttered algebra.

    A null space-time interval is \(ds^{2} = -dt^{2} + dx^{2}\) in Frame 1 and in Frame 2 \((ds')^{2} = -(dt')^{2} + (dx')^{2}\). We have (t',x') in terms of (t,x) so we'll consider how the second interval changes. To do that we need to work out the dx' and dt' using the Lorentz transformations.

    \(dx' = \frac{\partial x'}{\partial x}dx + \frac{\partial x'}{\partial t}dt} = \gamma dx - \gamma v dt\)
    \(dt' = \frac{\partial t'}{\partial x}dx + \frac{\partial t'}{\partial t}dt} = -\gamma v dx + \gamma dt\)

    Putting them into \((ds')^{2} = -(dt')^{2} + (dx')^{2}\) we have :

    \((ds')^{2} = -( -\gamma v dx + \gamma dt)^{2} + ( \gamma dx - v\gamma dt)^{2} = -\Big( \gamma^{2}v^{2} dx^{2} + \gamma^{2}dt^{2} - 2v\gamma^{2} dx dt \Big) + \Big( \gamma^{2} dx^{2} + v^{2}\gamma^{2}dt^{2} - 2\gamma^{2} dx dt \Big) = -\gamma^{2}\Big( 1 - v^{2} \Big) dt^{2} + \gamma^{2} \Big( 1 - v^{2} \Big) dx^{2} = -dt^{2}+dx^{2}\)

    The final equality makes use of \(\gamma^{-2} = 1-v^{2}\). Thus the specific Lorentz transforms used, ie a boost along the x axis by speed v, leave the space-time interval invariant, irrespective of what v is. Thus a null vector maps to a null vector and a point on the light cone maps to a point on the light cone.

    QED.

    By the way, this exercise is exactly one which 1st year physics students in my university covered in their first term. It's little more than just algebraic substitutions. It's that simple. And yet you haven't managed to grasp it.

    Again, who precisely are you trying to convince. This stuff is so basic and so simple to work through that anyone doing physics covers it and even people who don't do physics can work through the algebra (provided you give them the dx' and dt' expressions as they require the person to know differentiation otherwise).

    You claim there's a contradiction between although the light cone of Frame 1 is mapped to a light cone in Frame 2 their centres do not map to one another. That is not a contradiction. In fact its a requirement else all velocities would be frame independent, which really would be a contradiction.

    I've 'put up' before. I've provided plenty of explanations, diagrams and algebra. Now I've done it again. You really seem to be struggling to grasp that this stuff is easy. Its second nature to anyone whose done a physics degree. You make the mistake of assuming that because many of these results are new or confusing to you then everyone else is in the same boat. You haven't done some amazing new result which no one has ever thought of, you've just done a bit of algebra everyone whose done physics has done and you've come up against a result which confuses you. Happens to everyone. I have had it happen plenty of times to me. But unlike you I went away and read and read and then asked people for explanations which I actually listened to and thought about. Each and every time I've realised what my mistake was. Its a shame you can't just swallow your pride and accept you don't understand something. It'd have been a lot easier to do before you started declaring to all and sundry you've toppled relativity. It now seems you've got too much pride invested in your easily refutable claims to be willing to man up and say "I'm wrong".
     
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  5. Jack_ Banned Banned

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    This is all nice.

    This is about the train experiment.

    Let's see that proof of yours from the stationary train frame.

    Now, let me help you. The light emission points in the train frame will not be A and B.

    They will be A' and B' stationary to M' or the SR clock sync method fails.

    Once you correctly determine the correct space time interval, you will find M' must be struck by the light simultaneously.

    Here is the problem, when you take a frame stationary, here is what you must do.

    Let us take a system of co-ordinates in which the equations of Newtonian mechanics hold good.2 In order to render our presentation more precise and to distinguish this system of co-ordinates verbally from others which will be introduced hereafter, we call it the ``stationary system.''
    http://www.fourmilab.ch/etexts/einstein/specrel/www/

    Do you understand this?
     
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  7. AlphaNumeric Fully ionized Registered Senior Member

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    Well done on not responding to anything I said. What precisely do you not understand about my demonstration that null vectors map to null vectors under a LT? It demonstrates that the light cone goes to the light cone. This is counter to a claim you have made. I'd like you to respond to it. All you're doing is trying to change the subject. I've quoted bits of your replies and I've responded to them. Can't you do the same? Is all you're capable of just avoiding anything anyone says?

    You keep saying "Oh my points still stand, SR is dead" but if all you do is avoid replying to any retorts people make then you're in denial and its plain for anyone following your posts to see.

    You keep demanding people answer your questions and when I do you ignore my questions and ignore anything I've said. I respond directly to points you raise. Can't you do the same? If you're so confident shouldn't you have nothing to hide? I've offered to help you submit your work to a journal. I've offered to raise the level of discussion beyond Calculus 101. I've provided pictures, equations, explanations, you refuse to discuss any of them. Instead you want people to reply so you can monologue a bit more. You aren't interested in what people have to say about your claims, you just want to claim some more.

    If you can't reply to direct questions don't expect other people to reply to yours. You claim you've killed SR but if you're unwilling to reply to anything anyone says and you're always changing the subject whenever anyone treds too close to things you don't understand.

    AGAIN, if you think I'm 'too primitive' then why do you want my comments? Either you think I'm intelligent enough to be worth listening to or you're simply wanting more attention. Please answer this question. If you don't then I'll simply keep repeating it until you do answer it. As I said, don't expect people to answer your questions if you can't answer theirs. Rational discourse is the foundation of science but you wouldn't be aware of that....
     
  8. QuarkHead Remedial Math Student Valued Senior Member

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    Very nice explanation Alpha. However the words "pearls" and "swine" spring to mind
     
  9. CptBork Valued Senior Member

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    When a guy comes along and states that he knows Relativity is wrong a priori because "it's in his way", you know he has no interest in any sort of rational dialogue; it becomes more like preaching some sort of silly cult religion, I guess we can call it the Church of Jack.
     
  10. RJBeery Natural Philosopher Valued Senior Member

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    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?
     
  11. AlphaNumeric Fully ionized Registered Senior Member

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    Thanks. You'd think that Jack would respond to the maths he keeps demanding people provide. He claims to have taught vector calculus to undergraduates but I'm dubious about that so I thought I'd provide both the general tensor derivation and a specific case in terms of coordinates that the light cone is Lorentz invariant. No response, no comment, no retort.

    And then he has the laughable hypocrisy to demand people answer his questions, retort his comments. :shrug:

    He still hasn't answered my repeated question of why he wants to have me reply to his posts when he considers me 'too primitive' and obviously not a 4.0 student like him (so he says) and that I don't understand relativity (despite having more qualifications and more published work than him and wanting to raise the discussion to the level of fibre bundles). If I'm so ignorant and crap at physics why does he want me to comment on his 'work'? He's ignored it previously and yet he craves it. Sounds a bit of an attention whore.
     
  12. QuarkHead Remedial Math Student Valued Senior Member

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    You see what I fail to understand about this forum is this. Say I read a pop-sci book about, say, relativity or string theory, to take two recent examples. I believe I am sufficiently intelligent to know when things are being horribly over-simplified, even though I do not not know enough about either subject to detect what it is exactly.

    So I might (not likely, in my case) come here and ask for clarification. This does happen on this forum, I grant you, but not often enough to my taste.

    Rather, those with a poor understanding of what they have read come and read the riot act to the professionals here. This is OK, I believe, as far as it goes, provided ONLY they are willing to acknowledge there are those here who are professionals in the subject they are pontificating about and thus are willing to become "students" (as far as this makes sense on an internet chat room)

    BUT, what my gets my goat, and it seems that of some others, is when these people pretend to knowledge, skills and education that they manifestly don't posses. I cannot for the life of me see why anyone would want to do this. I can think of no member here "of good standing" (whatever that might mean) who has not had to ask a question and been grateful for the responses. I know that I have.

    Our Jack seems to be a case in point, but I can think of several others in the same category, of those who are quite clearly a little confused about the mathematical niceties of, say, coordinate transformations in flat (or otherwise) spacetime, and refuse to acknowledge that relativity has a very great deal more to it than trains and embankments.

    Alpha showed us this quite beautifully, and yet I doubt our Jack even read it, and if he did, has not the skills to critique it. And yet again he persists on what he terms "calling Alpha to put up".

    Likewise, I recall there was a thread entitled something like "I question string theory", but no real grounds were given, certainly no math was offered as I remember. So what's the beef? Just don't like it? Tough: I don't "like" QM, but there it is.

    /end rant

    Ben, you may delete or trash this as off-topic, at your discretion
     
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