It's not that I can't answer you
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.
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
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?
Him and his ''fysiks group'' actually saved physics according to David Kaiser which you can read about here
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.....
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]
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.
Where the integral ''cuts up'' our worldline into small fragments.
Just like any other kind of integration.
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]
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 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]
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.
Which is how [1] is obtained.
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.
This is multiplied by the distance to obtain the physical action on our worldline
$$\frac{\partial \mathcal{L}}{\partial \dot{q}}\cdot d$$ [7][/tex]
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.
The operator $$\nabla^2$$ turns out to make the dimensions of the equation mass over time
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?
making the equation describe mass flow for some matter field $$\chi$$. That makes our final equation then
$$\dot{\chi} = i(\frac{\partial \mathcal{L}}{\partial \dot{q}} \cdot d_) \lim_{d \rightarrow P_l} \nabla^2$$
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.
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.
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.