The Problem of Time leads to a Problem of Energy for the Universe

Discussion in 'Pseudoscience Archive' started by Reiku, Jan 10, 2012.

  1. Reiku Banned Banned

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    Interestingly there is what I call a ''planck limit'' (has that terminilogy been used before, if so I apologize) for this equation. If the distance is a solution in a ''Planck Parameter'', then there is really a limit on the equation for the distance reducing to the planck length.

    \(\dot{\chi} = i(\frac{\partial \mathcal{L}}{\partial \dot{q_i}} \cdot d_) \lim_{d \rightarrow P_l} \nabla^2\)

    This slices our time and thus defines a distinct action in a certain slice of our worldline.
     
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  3. Reiku Banned Banned

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    If so the equation would reduce to

    \(\dot{M} = \frac{\hbar}{i} \nabla^2\)

    Which would be a quantized representation, when the limit

    \(\lim_{d \rightarrow Pl}\)

    is satisfied., where \(P_l\) is the Planck Length.
     
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  5. James R Just this guy, you know? Staff Member

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    Reiku:

    Continuing with the non-maths part of your post on (I think) page 1:

    No, I don't know about timelessness. Time is in Einstein's equations as a coordinate.

    It is one thing to claim that the flow of time is an illusion; it's another to claim that time isn't important in Einstein's equations or that time doesn't really exist.

    If the state of a system is X at time t and Y at time t + dt then I'd say the system has evolved.

    What symmetry?

    What is diffeomorphism invariance?

    I haven't ever heard of him in the context of revolutionising spin.

    Can you please explain his revolutionary ideas and how they have affected subsequent mainstream physics?

    Nobody has measured how long my big toe nail is either, but that doesn't mean that I can't have one.

    How? What of causation?

    Einstein's relativity preserves causation.

    Explain.

    If the past is not real, then why do memories change over time?

    What problems does relativity pose for causation?

    What makes you sure?
     
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  7. James R Just this guy, you know? Staff Member

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    In your equation here, you seem to have a variable on the left-hand side and an operator on the right-hand side.

    What does your equation mean, and how is it relevant to the thread topic?
     
  8. AlphaNumeric Fully ionized Registered Senior Member

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    You have done it again. You focus in on one of a lengthy list of highlighted errors and make some excuse about that and then conclude all of what I've said can therefore be ignored.

    I stand by everything I said, including the \(\dot{x}\) vs v thing. The fact you transcribed an error on Susskind's part shows you don't understand what he's writing. I looked at it and saw the error, you didn't.

    You ignored how the fact this is your only exposure to Lagrangians and Noether's theorem completely undermines your claims to understand the Dirac equation. You couldn't answer my questions. You couldn't answer anything if you couldn't pick out something from the YouTube video to parrot.

    This isn't a 'pick and mix' of accusations, it's a lengthy list which has considerable overlap with previous threads where you've spouted mathematics. You attempt to threaten me with "I won't let you forget this", as if I'm the one whose made a mistake. You're welcome to bring this thread up in the future, it is evidence you've been lying for a long time about various things.

    I know people who did physics or maths degrees or PhDs in several Scottish universities, their quality is pretty good.

    This is nonsense. If you picked a physics degree in say Edinburgh university you'd need maths and physics highers (or A Levels) to get in but then you'd not need to study biology or chemistry, just mathematics and physics. As you're from Scotland you'd not have any student fees either.

    You couldn't manage to do highers in biology, chemistry and physics simultaneously? Perhaps science isn't for you if you struggle with something so 'basic' (in the grand scheme of things).

    And its nice of you to admit you did only a 6 month preliminary maths course. You wouldn't answer my question about whether or not you admitted you lied about doing Riemannian curvature in college but you'll admit to someone else the level of maths you've done is 'preliminary'. I'll take that as an admission of dishonesty.

    Why don't you define each term for us, state your initial assumptions and work through to your conclusion. Prove that's more than a mish-mash of LaTeX symbols you've seen Susskind write down.

    Not in the slightest. I'll give you a chance, why is it easy to prove what you just said to be false.

    You wouldn't get into a decent physics degree without a decent grade in mathematics. If you can't get whatever the highers version of A's in maths and physics are then a career in physics isn't really for you.

    It's considerably cheaper for a Scot to go to university in Scotland than any one else at any other university and student loans are the cheapest type of loan you'll ever have access to in your life.

    As for long distance, I've known students who used to do 60 miles each way commute each day. Someone I work with did a 2 hour commute each way every day for 2 years until he moved closer to the office. If people want something enough and have the drive then they'll find a way.

    In all honesty though if you're going to struggle to get the money together to go to university then in your case I'd advise to think long and hard about it. Your attitude to learning is the antithesis of what I'd expect from a good driven student. Last thing you want to do is 'rage quit' when you find all your YouTube parroting doesn't get you a passing grade and then you have nothing to show for your wasted time but a couple of grand debt. Assuming you could even get onto a physics degree course at a decent university, which I would be very surprised if you did.
     
  9. Reiku Banned Banned

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    Sorry, it had to do with a paper I read on treating the wheeler de witt equation in a new light which included matter fields \(\chi\) if this type.

    To be honest, it was just me thinking aloud. Just ignore it. Did you get back to the questions you were gonna ask, I haven't checked yet?
     
  10. Reiku Banned Banned

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    Well, actually some institutions here will not fund you alphanumeric if you are a part time student. Bursaries are not allowed for a full time student. This is why I made the mention about the three different courses.

    I am still trying to figure out what to do, yet I appreciate believe it or not, your input, even the negative light.
     
  11. Reiku Banned Banned

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    AN

    ''conclude all of what I've said can therefore be ignored.''

    It's not that I can't answer you, I have told you once before in this thread and more times before this than I even wish to remember, it is your attitude which stops me from answering you.
     
  12. James R Just this guy, you know? Staff Member

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    Reiku:

    See post #123
     
  13. Reiku Banned Banned

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    ''No, I don't know about timelessness. Time is in Einstein's equations as a coordinate.

    It is one thing to claim that the flow of time is an illusion; it's another to claim that time isn't important in Einstein's equations or that time doesn't really exist.''


    One solution which will help explain timelessness for you is the Wheeler deWitt equattion. It is a solution of relativity which the time derivative (it is a global time) vanishes. Wiki even has a bit on timelessness now. It didn't used to.

    ''If the state of a system is X at time t and Y at time t + dt then I'd say the system has evolved''

    Sure, but it's not always as simple as that.

    ''What is diffeomorphism invariance?''

    The Einstein equations are invariant under diffeomorphism's of the spacetime manifold - it is basically a set of operations which map spacetime points to other spacetime points

    Here is a great paper explaining timelessness and diffeomorphism invariance in the theory

    http://fqxi.org/data/essay-contest-files/Markopoulou_SpaceDNE.pdf

    ''I haven't ever heard of him in the context of revolutionising spin.

    Can you please explain his revolutionary ideas and how they have affected subsequent mainstream physics?''


    Him and his ''fysiks group'' actually saved physics according to David Kaiser which you can read about here

    http://en.wikipedia.org/wiki/Fundamental_Fysiks_Group

    ''Nobody has measured how long my big toe nail is either, but that doesn't mean that I can't have one''

    Your toe nail is different. It is a system which has already collapsed due to tightly bound particles which are more or less quantum entangled with themselves, a process called decoherence. Take a box with a few particles able to move about in. They will act as waves for a long time (the language which is the same as saying they are in a smear of probable states but no one state really exists, that is until the particles slowely decohered - the particles slowely observing each other until they settle into one final state. Now you toe nail is in one final state isn't it (for now atleast before you cut it

    Please Register or Log in to view the hidden image!

    )?

    ''Einstein's relativity preserves causation.''

    This is in reference to all your causality questions...

    Just the same problem as I said before. If timelessness is not preserved in Einstein's theory then surely causation is not fully understood either since that is a time-dependant phenomenon.

    ''Explain.''

    Well, the universe can be a bit like the Wheelers delayed choice experiment. It is not until someone make's the decision to observe a particle in the present state do they ''create'' some defined path for that particle, instead of being a smeared probability over all the path's it made in it's history. To have many histories is the same as the sum over histories.

    ''If the past is not real, then why do memories change over time?''

    Memories don't change over the past however. Memories are experienced always in the present time. We do no such thing as jumping into a past state upon remembering an experience. Our experiences are bound, just as much as we remember them in the present state. The thing we call the past is just a bunch of memories which neither they or the person experiencing them have ever really been in.

    ''What makes you sure? ''

    This is what physics and relativity says. As Einstein ellucidated, and as Dinosaur previously said, anyone who need a definition of a ''before'' and ''after'' cannot understand time.
     
    Last edited: Jan 16, 2012
  14. Reiku Banned Banned

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    Here's a bunch of links to some reading material.

    http://arxiv.org/PS_cache/gr-qc/pdf/9812/9812027v1.pdf
    http://www.ipod.org.uk/reality/reality_wheeler_dewitt.asp
    Since the Wheeler de Witt wave function is a global case, there is a suggestion that an incomplete wave function might still describe the universe
    http://books.google.co.uk/books?id=...esnum=8&ved=0CEoQ6AEwBzgK#v=onepage&q&f=false
    more on the wave function
    http://arxiv.org/PS_cache/gr-qc/pdf/0308/0308029v1.pdf
    Before there was some question of an equation I wrote. The reasons for it came from this paper:
    http://arxiv.org/PS_cache/hep-th/pdf/9503/9503073v2.pdf
    Which allows one to treat a matter field \(x\) which is given as x, but I call \(\chi\) as being used to measure time \(\tau\). It takes the matter field into a minisuperspace model. The equation of interest is
    \((-\frac{1}{4} \frac{\partial^2}{\partial \alpha^2} + \alpha^2 - g^2\alpha^4 + \frac{1}{4} \frac{\partial^2}{\partial x^2}) \Psi(\alpha, x)\)
    Which can be subjected to a seperation of variables leading to two equations, one describing the scale factor \(\alpha\) and the matter field \(x\).
     
  15. Reiku Banned Banned

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    \(\dot{\chi} = i(\frac{\partial \mathcal{L}}{\partial \dot{q}} \cdot d_) \lim_{d \rightarrow P_l} \nabla^2\)

    So how did I get

    \(\frac{\partial \mathcal{L}}{\partial \dot{q}}\) [1]

    part? First we need to assume worldlines, our job is compute the action. First

    \(d \tau^2 = dt^2 - d\vec{x}^2\) [2]

    There are some \(c^2\) parts in there which we will omit for now. First [2] concentrating on the right hand side, can be redressed as

    \(-M \int \sqrt{\frac{dt^2}{dt^2} - \frac{dx^2}{dt^2}} dt\) [3]

    Where the integral ''cuts up'' our worldline into small fragments. This is just the same as

    \(-M \int \sqrt{1 - \dot{x}} dt\) [4]

    The canonical momentum takes the appearance of \(\frac{\partial \mathcal{L}}{\partial \dot{x}}\) and thus the momentum can be given as \(\frac{\partial \mathcal{L}}{\partial\dot{q}}\) just as a change of notation. Differentiating [4] in our new notation leads to

    \(\frac{ \partial \mathcal{L}}{\partial \dot{q}} = \frac{+M2\dot{q}}{2\sqrt{1 - \frac{v^2}{c^2}}\) [5]

    Just in case no one knows what is being done here, differentiating the square root puts it in the denominator with an extra factor of 2, then we differentiate the arguement of the square root in reference to \(\dot{q}\) which gives another minus sign changing the \(-M\) into a positive mass, and thus simplifying [5] leads to the relativistic

    \(\frac{\partial \mathcal{L}}{\partial \dot{q}} = \frac{M V_x}{\sqrt{1 - \frac{v^2}{c^2}}\) [6]

    Which is how [1] is obtained. This is multiplied by the distance to obtain the physical action on our worldline

    \(\frac{\partial \mathcal{L}}{\partial \dot{q}}\cdot d\) [7]

    The operator \(\nabla^2\) turns out to make the dimensions of the equation mass over time, making the equation describe mass flow for some matter field \(\chi\). That makes our final equation then

    \(\dot{\chi} = i(\frac{\partial \mathcal{L}}{\partial \dot{q}} \cdot d_) \lim_{d \rightarrow P_l} \nabla^2\)

    Assuming I have done everything right. The limit comes from slicing your systems world line to the infinitessimal Planck Limits, where the equation reduces to \(\frac{\hbar}{i}\nabla^2\) which will describe the mass flow.
     
    Last edited: Jan 17, 2012
  16. James R Just this guy, you know? Staff Member

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    That's not the action.

    That's not even an equation, so it can't be a "re-dress" of [2].

    Why? How?

    And what is all this supposed to prove again?
     
  17. Reiku Banned Banned

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    I don't want you to focus on this too much. I wrote it very quickly yesterday to show alphanumeric that I wasn't simply rehashing the langrangian arguement from that lesson of susskind. I am sick of his hollow accusations.

    You're right though, that is an expression, and what I meant to say to begin with was that we would be given after the first few processes. I did not intend to mean it was the action per se. And you ask what it was to prove, well itt proves very little, only that I am trying to prove a point with AN. In short though, I'd like to think it describes a mass flow for some certain slice of a worldline. You will notice in the link, it tackles to deal with the time problem by introducing a matter field like mine as a reparametrization of the WDW equation.
     
  18. Reiku Banned Banned

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    I fixed those quibbles James.

    Anyway, are you finally settling in believing their is an existing time problem then?
     
  19. AlphaNumeric Fully ionized Registered Senior Member

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    No, it very much is. If you could answer my questions and demonstrate understanding I'd not be so blunt with you. People like Rpenner and Guest demonstrate such things and I get on fine with them. If instead all you display is dishonest then you're hardly going to earn my respect. If you want to be treated like the informed and honest person you believe yourself to be then you should act like it and demonstrate the knowledge you claim to have.

    It's a little more specific than that. Come on, you're posting plenty of LaTeX when you want to, what's stopping you giving the proper definition?

    That's a completely exaggerated thing to say. Firstly no one has ever 'saved physics'. Saved it from what? Why isn't his name well known? I've never heard his name anywhere other than your like of him.

    As for the mathematics you post in the post which followed that, it's clear you are once again copying from some source and mangling it. James has picked up on a number of simple mistakes you make, such as where you write down a space-time interval and call it an action. No, the space-time action is very closely related to the GR action but they aren't literally the same thing. You then skip a few lines in whatever you're copying from and state something you get once you convert the space-time interval into an action. After James pointed some of them out you've updated it a bit but still not entirely correct.

    Let's be more specific.....

    Obviously the right hand side of [2] cannot be written as [3] because you've magically pulled M from nowhere. You mentioned worldlines but you've skipped the bit from wherever you're getting this where it explained how to go from [2] to [3].

    The action of a particle moving through space-time is taken to be proportional to the length of it's worldline, with the proportionality constant involving the mass of the particle.

    The length of the worldline is obtained from the space-time interval, \(d \tau^2 = dt^2 - d\vec{x}^2\) via

    \(L(C) = \int_{C} \textrm{d}\tau = \int_{C} \sqrt{ \textrm{d}t^2 - \textrm{d}\vec{x}^2}\)

    This is 1 dimensional, since we're talking about world-lines and not the motion of something higher dimensional. Thus we can parameterise it by a single parameter. Due to the nice nature of most worldlines you can decide to use t by the chain rule of \(d \xi = \frac{\textrm{d}\xi}{\textrm{d}t}\textrm{d}t\) we have

    \(L(C) = \int_{C} \textrm{d}\tau = \int_{t_{0}}^{t_{1}} \sqrt{ \left(\frac{\textrm{d}t}{\textrm{d}t}\right)^2 - \frac{\textrm{d}\mathbf{x}}{\textrm{d}t}\cdot \frac{\textrm{d}\mathbf{x}}{\textrm{d}t}} \textrm{d}t = \int_{t_{0}}^{t_{1}} \sqrt{ 1 - \frac{\textrm{d}\mathbf{x}}{\textrm{d}t}\cdot \frac{\textrm{d}\mathbf{x}}{\textrm{d}t}} \textrm{d}t\)

    Writing time derivatives as dots gives

    \(L(C) = \int_{t_{0}}^{t_{1}}\sqrt{ 1 - ||\dot{\mathbf{x}}||^{2}} \textrm{d}t\)

    The Lagrangian density is then taken to be the integrand multiplied by \(\pm M\) (depending on preferences), so using your choice we get \(\mathcal{L} = -M \sqrt{ 1 - ||\dot{\mathbf{x}}||^{2}} \). To convert this into the usual notation for the coordinates of a Lagrangian we use \(q_{i} = x_{i}\) and so \(\dot{q}_{i} = \dot{x}_{i}\), so \(\mathcal{L}(q_{i},\dot{q}_{j}) = -M \sqrt{ 1 - (\dot{q}_{1}^{2} +\dot{q}_{2}^{2}+\dot{q}_{3}^{2}) } = -M \sqrt{ 1 - \dot{\mathbf{q}}\cdot \dot{\mathbf{q}} } \)

    That's the explanation. If you're going to attempt to explain something to someone who doesn't know how to do it cutting corners doesn't help them. It also helps you avoid making silly mistakes like dropping the square inside the square root in your final expression.

    Just like any other kind of integration.


    You mix notation, a habit Susskind has of doing and which I've commented on before, by reverting to writing the denominator in terms of v, not \(\dot{\mathbf{q}}\). It's sloppy.

    And you do the same again, mixing notation. In fact you actually get the answer wrong because of this muddling of notation.

    You actually have 3 standard coordinates, \(q_{1},q_{2},q_{3}\), because the particle is in 3 dimensional space, a fact I highlighted further up. As such you need to be careful when differentiating because you should differentiate with respect to one of these or their time derivatives. It's trivial to do this general if you're competent at index notation but if not it can be confusing.

    Clearly where you are getting this from (probably the Susskind YouTube video, I haven't watched all of it) decided to differentiate with respect to \(q_{1}\), or in a different notation, \(q_{x}\) because the velocity term you've got has a little x on. Writing it properly and not with bad notation it should be, if we set c=1, \(M \frac{q_{x}}{\sqrt{1- \dot{\mathbf{q}}\cdot \dot{\mathbf{q}}}}\). This is what we get if we computed \(\frac{\partial \mathcal{L}}{\partial \dot{q}_{x}}\). Notice the little index on the denominator's \(\dot{q}\)? It's because I'm specifying which coordinate to differentiate with respect to, since it's a multicoordinate system.

    Just to do it explicitly in general index notation, \(\frac{\partial \mathcal{L}}{\partial \dot{q}_{i}} = -M\frac{\partial}{\partial \dot{q}_{i}}\sqrt{1- \dot{\mathbf{q}}\cdot \dot{\mathbf{q}}} = -M \frac{1}{2} \frac{1}{\sqrt{1- \dot{\mathbf{q}}\cdot \dot{\mathbf{q}}} } \frac{\partial}{\partial q_{i}}(-\dot{\mathbf{q}}\cdot \dot{\mathbf{q}})\)

    That's using the chain rule. Now to compute the right hand term,

    \(\frac{\partial}{\partial q_{i}}(-\dot{\mathbf{q}}\cdot \dot{\mathbf{q}}) = -\frac{\partial}{\partial q_{i}} \left( \dot{q}_{j}\dot{q}_{j} \right) = -\delta_{ij}\dot{q}_{j} - \dot{q}_{j}\delta_{ij} = -2\delta_{ij}\dot{q}_{j} = -2\dot{q}_{j}\)

    and so we have \(\frac{\partial \mathcal{L}}{\partial \dot{q}_{i}} = -M \frac{1}{2} \frac{1}{\sqrt{1- \dot{\mathbf{q}}\cdot \dot{\mathbf{q}}} } -2\dot{q}_{i} = \frac{M\dot{q}_{i}}{\sqrt{1- \dot{\mathbf{q}}\cdot \dot{\mathbf{q}}} }\)

    Done properly and written correctly.

    In [1] you don't give the above expression anyway, you just give it as \(\frac{\partial \mathcal{L}}{\partial \dot{q}_{i}}\). Why are you now bothering to explicitly state something you never stated originally? That entire derivation of a form for \(\frac{\partial \mathcal{L}}{\partial \dot{q}_{i}}\) for a particle is entirely irrelevant to motivating [1].

    This is what happens when you don't know what you're copying and you attempt to explain something you don't know how to explain.

    Why? Why should you multply in distance? Firstly you give a dot product in your definition but distance is a scalar. Looks like you've made that mistake again.

    Of course I am asking questions I already know the answer to and I'll explain what it was you were supposed to say. What you've computed is actually a set of expressions, because you can differentiate the Lagrangian density with respect to any of the 3 \(\dot{q}_{i}\) it depends on.

    In much the same way \(\nabla \phi\) for scalar \(\phi\) is a vector and you can dot product it with things you can do the same with this set of expressions by treating \(\frac{\partial}{\partial \dot{\mathbf{q}}}\) in the same manner, ie \(\frac{\partial \mathcal{L}}{\partial \dot{\mathbf{q}}} \equiv \left( \frac{\partial \mathcal{L}}{\partial \dot{q}_{1}} , \frac{\partial \mathcal{L}}{\partial \dot{q}_{2}} , \frac{\partial \mathcal{L}}{\partial \dot{q}_{3}} \right)\). Then dotting this with a vector, say \(\mathbf{X}\) is going to give a scalar quantity \(\mathbf{X} \cdot \frac{\partial \mathcal{L}}{\partial \dot{\mathbf{q}}} = \sum_{i} X_{i} \frac{\partial \mathcal{L}}{\partial \dot{q}_{i}}\).

    The question is what do you dot it with to make a scalar. It depends entirely upon the context and what you're trying to do. Usually it's something to do with \(\mathbf{q}\) or \(\dot{\mathbf{q}}\) but not always. It certainly isn't 'distance' because that's a scalar.

    Firstly \(\nabla^{2}\) hasn't come into anything yet so you have to motivate why you're going to consider it. While it's true that if you dot \(\frac{\partial \mathcal{L}}{\partial \dot{\mathbf{q}}}\) with something with units of length and then multiply by \(\nabla^{2}\) you get something with units of mass per unit time there's infinitely many different operators or quantities with the same units (inverse length squared) as \(\nabla^{2}\). You haven't motivate your choice at all.

    This is yet another piece of evidence that you're copying all of this from here and there and you're trying to cut corners so it's not blindingly obvious (yet it still is) you're lifting it from somewhere. In doing so what you post here doesn't define terms and picks results from nowhere or skips several important steps or refers to things incorrectly.

    Is this going to be like earlier where if I check the Susskind YouTube video it turns out all these expressions were there, in the sequence you've given them, with the same little notational quirks, just like you had with the potential stuff earlier in the thread?

    Now you've magically pulled this matter field from somewhere and proclaimed it to have that equation of motion. It cannot, for multiple reasons. Firstly the fact you dot d with something implies it's a vector, yet the limit in the equation implies its a scalar. Contradiction. Secondly the quantity you're limiting isn't inside your limit, the limit does nothing! Thirdly you've picked up a factor of i from nowhere which calls into question what exactly the matter field is. Fourthly it's not an actual equation of motion, since the \(\nabla^{2}\) isn't acting on anything. It's like saying \(\frac{dr}{dt} = \frac{d}{dx}\). It's meaningless. Now it's possible that the action of the thing on the left on some object is the same as the action of the thing on the right on the same object but that doesn't mean they are equal in the sense you've written down, in the same way a matrix A can satisfy \(A \cdot \mathbf{X} = \lambda \mathbf{V}\) for some vector \(\mathbf{V}\) and scalar \(\lambda\) but it would be incorrect to say \(A = \lambda\).

    As it stands the expression is mathematically meaningless in several ways and physically without context or justification.

    Completely without justification, not to mention Planck lengths are not infinitesimal, that's what makes them so important! Infinitesimals is what can be used to construct the mathematical objects used in calculus, ie derivatives and integrals. If you make your space-time discrete then you have to be very careful about how you use differential operators. If you can make your divisions of a region arbitrarily small then you can construct differentials. If you're stuck at a non-zero length, like the Planck length, then you can't use the limiting process. You've actually worked backwards, you've assumed all your calculus is fine, which means you can divide your worldline as much as you like and take meaningful differential limits, and then declared you're somehow working with larger lengths. Even the definition of the worldline and its action is altered if you discretise space-time so the very construction of \(\frac{\partial \mathcal{L}}{\partial \dot{\mathbf{q}}}\) is brought into question. Hell, your very space-time interval could be called into question! If the worldline is to be cut up into an explicitly finite number of pieces things should be done with sums, not integrals.

    Either you've pulled your final conclusion from your backside or the place you're lifting this from contains an awful lot more explanation and elaboration on where all of these terms originate and the motivation behind them. You said in a previous post "I started to conjecture such a field and came up with" and then stated this equation. If you really have put this together yourself and not just lifted (and mangled) something from somewhere else then you're just showing how little you understand even basic equations in calculus or even how to take limits. You obviously want to be seen to be 'playing physicist' but you're just making nonsense up. And not just "Opps, I forgot a factor of 2" but "This expression is 5 kinds of bull****". If you can't afford to go play physicist at university (assuming you'd even get in) you should be spending your time a little better than wasting it deluding yourself.
     
    Last edited: Jan 19, 2012
  20. Reiku Banned Banned

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    I will get back to this later. The dot ''d'' is simply a multiplication, by the way. Nothing fancy like the dot product.
     
  21. Reiku Banned Banned

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    And it is not that I have pulled a matter field from somewhere. You are actually forgetting the context I am using it is. The upper dot makes the equation mass over time, or a mass flux rate. It just so happens that the physical action multiplied by our operator satisfies the same dimensions. But as I said, I will get back to it later.
     
  22. AlphaNumeric Fully ionized Registered Senior Member

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    Your equation is still gibberish for multiple reasons.

    So? I can write down as many nonsense equations with correct units as I like. For example \(E = 4mc^{2}\). Units work. In fact, since E, \(mc^{2}\) and pc all have the same units the ratios \(\frac{E}{mc^{2}}\) and \(\frac{E}{pc}\) are dimensionless. Therefore given ANY function \(f(\frac{E}{mc^{2}},\frac{E}{pc}) = 0\) whose terms are dimensionless will have valid units. For example, f(a,b) = a^{2} \sin \frac{a}{b} - \exp(a) - \ln \ln b[/tex] will give an equation relating E, m and p with correct units but it'll be physically nonsense.

    Matching units is a necessary but not sufficient condition for something being physically valid.

    The equation is meaningless and you don't define the physical system you're talking about. You construct \(\mathcal{L}\) from a particle's worldline but the equation of the particle is something completely different, it has nothing to do with \(\chi\). You haven't given any justification for your equation, even ignoring the fact it's gibberish.

    Is it really so hard for you to say "Looks like what I wrote was nonsense. I made a number of mistakes."? I've outlined in detail the problems. You can't reinterpret your way out of it, the equation you give as your conclusion is nonsense.
     
  23. Reiku Banned Banned

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    Hold right there. I spoke out load one evening after returning home from a drunken night out. Saw some stuff I had been scribbling earlier and decided to post it. Ok... then, the next day I think you accused me of just using the Susskind arguements for the langrangian. To show you it was not just based on the video's I have watched by him I provided you a link on the matter fields (which if you haven't read then maybe you have a bit of a hypocritical nature about you again concerning the fact you slagged me off for not reading your paper earlier). I had seen the dimensions working in my head before I even wrote the full equation out. The distance, in my mind was important because a mass flow over some area would almost certainly invoke the idea of distances. The operator, I simply used that because I wanted to settle the dimensions on both sides. If it is gibberish, it's gibberish, I really don't care.

    I admitted a long while back, I didn't want anyone to focus on the equation. I was thinking aloud when I wrote it here I said to James, but in typical alphanumeric behaviour, you seized the chance to debate it as soon as you could.

    Lastly, susskind is not my only source of information. I have loads. It just so happens I am pro-susskind, I prefer his teaching methods over any one.
     

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