Eugene Shubert
Valued Senior Member
I prefer to phrase that differently. Physicists don't know how to distinguish between inertial motion and accelerated motion.
On the Definition of an Inertial Frame of Reference
- Thread starter Eugene Shubert
- Start date
I prefer to phrase that differently. Physicists don't know how to distinguish between inertial motion and accelerated motion.
I just told you: no. To my knowledge, all physical laws are generally covariant.
$$g_{\mu\nu} = \eta_{\mu\nu}$$ defines a particular class of coordinate systems that we like to call "inertial". It's not a physical law.
Eugene Shubert
Valued Senior Member
Guest,
Don't you realize that you've been fighting The Black Knight? Your joust with Sir Knight has been beautifully illustrated by Monty Python. Have you ever seen the skit? http://www.youtube.com/watch?v=zKhEw7nD9C4
Eugene Shubert
Valued Senior Member
Anything you say worthy adversary. You still don't know how to distinguish between inertial motion and accelerated motion.
Eugene Shubert
Valued Senior Member
Where did Guest say that should be covariant? You've given an expression specifically in a particular set of coordinates, it is not a tensor expression. It follows from Maxwell's equations, which can be written covariantly, when you pick particular coordinates but there's no reason to think the form would remain the same if you do some general valid transformation.
Tach, I've been slightly less blunt than I might have otherwise been since you said you've got more than 20 published papers and I was giving you the benefit of the doubt but it seems you're either lying or whatever you're published in isn't anything to do with the topic at hand. All you're doing is throwing ad homs and avoiding entering any kind of detailed discussion. If you bother to check other threads you'll see Guest is more than capable of doing high level mathematics. Its up to him whether he wants to reveal much information about himself but through various PMs and emails I can state with complete certainty he knows his stuff, being a PhD in a particular area of mathematics.
If you're incapable of having a discussion then no matter how many papers you have and how many citations its pointless you being here. Presently I see no reason to think you have the capabilities to have published 20+ papers in reputable journals, at least in nothing related to this topic.
I have given you the benefit of the doubt as well on this one since I know that you are a good guy otherwise but I can see that you two go to the same university and that you are both making asses of yourselves the same exact way on THIS issue. The fact that you are both good guys doesn't change the fact that you are both stuck on the tensor formalism which in the case of Shubert theory CANNOT be applied. Let me prove that to you.
The point is (and always was) that the Shubert crackpot transforms are NOT a set of "valid transformations". Rather than repeating mechanically the same idiocy over and over, think about it:
1. Shubert's crackpot theory has no metric.
2. Because there is no metric , there are no $$g_{ij}$$
3. Because there are no $$g_{ij}$$, there are no Christoffel symbols, $$\Gamma_{i,jk}$$
4. Because there are no Christoffel symbols, $$\Gamma_{i,jk}$$, covariant derivatives cannot be constructed in Shubert's crackpot theory. You need a valid set of transformations in order to do that. You need to go back to the basics and learn.
The only way of verifying whether Shubert crackpot transforms make Maxwell's equation covariant is the way I have been asking Guest to work it, by calculating the partial derivatives. So, I am going to ask you to do the same thing. Put up or shut up. You will need to work the problem the hard way, by calculating the second order partial derivatives, no hiding behind cutting and pasting the same tensorial expression that does not even apply to this problem will not cut it. Stop being lazy and start working the exercise.
Consider the possibility that you have just been schooled on the subject.
Why are you so defensive of him? He turned up to be a bigger fraud than even Shubert. Because you two are going to the same school? On this subject, the school's education was wasted on you both. You two kept repeating the same mantra without even thinking about what is the problem in front of you. So, is
$$\Bigg(\nabla^2 - { \mu\epsilon } {\partial^2 \over \partial t^2} \bigg) \mathbf{E}\ \ = \ \ 0$$
covariant wrt the Shubert transforms or not? Roll up your sleeves and start calculating. Are the Shubert transforms a "valid set of transforms" or not? Let's see you and/or Guest prove that you can calculate.
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Shubert,
You have a couple of simple exercises to do, you sttill haven't managed the very simple one on speed composition. Instead of trolling Youtube, you need to start answering the questions. Or are you starting to realize that your theory is a worthless piece of crap?
what Shubert has been showing is that if $$g(\mu_{ij}) = \frac{\sqrt{1+k\mu_{ij}^{2}}}{\mu_{ij}}$$ is a valid solution to his equations then $$g(\mu_{ij}) = \epsilon \pm \frac{\sqrt{1+k\mu_{ij}^{2}}}{\mu_{ij}}$$ is also a valid solution since it differs from the first one by a constant that gets canceled in the subtraction. Sad to see that you have been schooled by Shubert, the pathetic crackpot. Pay better attention to his paper next time and you'll avoid making an ass of yourself.
Eugene Shubert
Valued Senior Member
Sadly,
But don't get your hopes too high, your theory is still worthless crap. The fact that AlphaNumeric wasted so much time in pursuing an imagined error doesn't mean that your theory is not full of holes.
No, I'm not even in university any more.
I don't think either of us claimed Eugene's work is valid or could have consistent formulations using tensors. I have been making (and I think Guest has too) broader points.
Which, as I've said many times, is beside the point.
You seem to have completely missed the point Guest and I are making. How many times did I say that Eugene's work is not what we're talking about?
Maxwell's equations, when written in the form $$\nabla^{a}F_{ab} = 0$$ etc are covariant, full stop. By definition of covariant you are only considering valid coordinate transformations. Whether Eugene's are valid is beside the point of whether Maxwell's equations are expressibly covariantly.
Again, I've said this repeatedly.
Did you not see him use covariant derivatives, which make Maxwell's equations covariant, regardless of which valid coordinate setup you wish to use. IF Eugene's transformations have valid Jacobians then when you apply them to Maxwell's equations written in the covariant tensor formalism then you'll get the same form of the tensor equations. If Eugene's transformations are not valid coordinate transformations then all bets are off, which is something which none of us have denied.
How many times are you going to need to have this explained? If the Jacobian of the transformations is non-singular then the transformation will leave the tensor form of Maxwell's equations unchanged, no explicit calculation beyond the Jacobian needs to be done. You have given a specific term in the tensor expression and said "Do the transformation on this!!", which is not individually going to be covariant. That's why things are written in the tensor form and not individually component by component. Gauss and Stokes and Maxwell used to do everything component by component but in the centuries since we've gotten more powerful mathematics to use. Like the differential forms which Guest mentioned.
You have failed to get anything we've said it seems.
No, though I happen to know which university he got his education from, including a PhD, and which university currently employs him and he knows likewise about me (though I'm no longer at university, I'm employed by a private research company).
I 'defend' him because you are just throwing out bullshit and I know for a fact your claims about him are false.
Our doctorates and our pay cheques from people who employ us because of our research histories would say otherwise.
Well done, you just demonstrated you completely failed to understand my previous response on this matter. You've given a single equation, in a particular coordinate system. The form of Maxwell's equations which Guest and I have been referring to are tensors, a family of equations which relate to one another and which, under coordinate transformations, mix together in a specific way. A single component out of a tensor expression will transform in a very nasty way in most cases. Its only when you consider how the other terms in a tensor contribute do you get nicer things.
Have you never seen how covariant and contravariant indices transform? They transform in opposite ways and a single component becomes a linear combination of components via $$v_{a} \to \frac{\partial \tilde{x}^{b}}{\partial x^{a}}v_{b}$$. Notice how the set of partial derivatives form a matrix which transforms the components in a linear combination? That means that a general valid coordinate transformation will need you to consider the components as a set, not just single out a specific component to consider, as you've now asked several times and which I've already explained your mistake on.
How many times have I explained that's beside the point we're making? Is English not your first language?
I commented on how the method he derived one solution failed to provide him with the solution he ultimately used. He assumes things like an odd function then gets a solution which is not odd.
And do I have any problem admitting I didn't spend much time reading his paper? No, because its obviously nonsense. You try to insult me for supposedly supporting him and now you're complaining I haven't given him enough attention?
He hasn't 'schooled me', just like you haven't. If his work is not wrong in the way I commented, so be it. How does that mean he's 'schooled me'? Does his work somehow become more valid? No. Does any of it negate my comments to you? No. It seems to me you're just trying to find anything you can to throw at people, regardless of whether or not you actually retort what they said. You failed to respond to me comments about your use of specific coordinates and instead simply repeated the same mistake! Maybe you should pay a little more attention too? At least in my case I failed to pay enough attention to a crackpot. Oh no, I'm so ashamed.
To think this week I spent less time reading Eugene's work and more time reading Landau and Lifshitz's 'Fluid Mechanics'. Yes, I obviously squandered my time, to think I've been learning about turbulent jet formation when I should have been reading Eugene's work. Wow, I'm so embarrassed to have learnt some fluid mechanics.
Eugene Shubert
Valued Senior Member
The reality is that Sir Knight put his sword through your head and you have been completely and thoroughly discredited: http://www.youtube.com/watch?v=dhRUe-gz690
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Wrong, Maxwell's equations are covariant INDEPENDENT of notation. Heck, all the macroscopic laws of physics are covariant.
The point that you and Guest have missed (since you both appear not to understand the notion of covariance correctly) is that the fact that since the macroscopic laws of physics are covariant they can be used in verifying if a set of transforms or a metric are valid. For example, the Lorentz transforms preserve covariance , the Galilei, Tangherlini, Shubert crackpot transforms do not. The precise equation I gave you guys to work on has been used by Lorentz to prove that Galilei transforms do not preserve covariance while developing the transforms named after him (that do preserve covariance). You learned this early on in your SR class but you have obviously forgotten.
You two guys are so brainwashed by the tensor notation that you missed the basics.
Good, you are starting to remember the basics.
No, this is the whole point of the thread, to prove that the Shubert crackpot transforms ARE NOT VALID since they DO NOT preserve covariance.
Are you daft? I have just proven to you that you cannot construct covariant derivatives for the Shubert theory. The notion of covariant derivative does NOT exist in the framework of Shubert's theory.
Good, so the UK government should ask for the money back it wasted on you two. You need to relearn the basics.
You should do the honorable thing and pay the money back.
This is a form just as valid as any. This is the form used to prove Lorentz covariance (and lack of Galilei covariance, remember?)
Sure, PROVIDED that the theory that you are examining has a valid metric allowing you to calculate the Christoffel symbols. Shubert's doesn't.
But you CANNOT construct the tensor formulation in the Shubert formalism because it has no valid metric. You need to go back to your books and re-read the chapter on general covariance.
And he points out to you that he simply added a constant to the solution in order to get a more general form to his solution. You need to learn how to read.
No, what I am criticizing you is for being shallow. You look at things for a few minutes, you cu and paste something that you remember from your school days and, when challenged to get deeper, you cut and paste some more of the same tripe.
He's exposed your shallowness.
True, Shubert's work is the same pile of crap, we agree on that. But when you and Guest254 show up with red herring claims of his theory being general covariant, it doesn't help the thread.
You can't use generalized coordinates in the Shubert framework, you dork. His theory doesn't have a metric.
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You mean that you can't.
By contrast, physicists can and do. It is really trivial, do you know how? (hint: you should have been taught this, if not in high school then during your year at UCSD).
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Eugene Shubert
Valued Senior Member
No Sir Knight. It is not trivial for physicists. It must be done within a mathematical model. Physicists are confused by this challenge.
So, you have nu clue. Figures.
You're becoming like Terry, linking to YouTube videos rather than making a relevant point.
Yes, Nature is independent of how we formalise our description, just like Nature doesn't care whether you and I communicate in English or French or Sindar.
But that also means that if you have an expression which is not invariant under coordinate transformations then its not formalised in the proper way or its incomplete. Picking out a single component of Maxwell's equations or from Maxwell's tensor will be such an example. Under Lorentz transforms you can convert electric fields into magnetic and vice versa, which means that if you didn't know about magnetic fields and you only considered electric you'd have an incomplete view of the physics you're attempting to describe, ie the choice of coordinate ends up defining a particular form of the electric field component of the underlying things you're examining.
You provided the wave equation applied to an electric field. Under a Lorentz transformation that electric field can be made into a linear combination of electric and magnetic fields, so clearly something is amiss and that something is what I previously said, by picking out a single component in a particular coordinate system you've extracted part of a system and your choice of coordinates has defined that 'part'. Unless you include the other parts you can't expect to have a covariant formalism because your choice of coordinates determines the part you're considering so you must have all parts.
Hence why tensor expressions for Maxwell's work (and the non-abelian extensions into Yang Mills theory) involve all components. $$nabla^{1}F_{12}$$ is not covariant, but $$\nabla^{a}F_{ab}$$ as a whole is.
Your comment about Nature being covariant doesn't contradict anything I've said, it actually gels perfectly with my point that your choice of coordinates when you give that expression for the electric field wave equation is violating the "Coordinate choices don't matter" principle.
I can't speak for Guest but it happens that general relativity was my best course in my 4th year and my PhD was in extra dimensional formalisms in string theory. Coordinates and the structures they have was my area of research. Clearly people whose job it is to know whether I know my stuff disagree with you.
Galilean transformations include as a subgroup rotations, which do preserve the Lorentz structure.
There's a difference between those transformations which leave the metric invariant and those which change its form. This is a point you previously demonstrated you didn't get when you said there's no metric involved. If Eugene is talking about Euclidean space then the metric is Euclidean. If he's talking about Minkowski space in the SR formalism then its the Minkowski metric. Either way if his space has a notion of distance then unless its pathological there'll be a metric.
A rotation in Euclidean space or a Lorentz transform in Minowski space-time leave the metric unchanged, ie $$g \to M \cdot g \cdot M^{\top} = g$$. If g is Euclidean metric then $$M \in SO(N)$$ and if $$g = \eta$$ then $$M \in SO(N-1,1)$$. Notice how both of these are linear operations represented by a matrix? But what about something else like going from Cartesians to polars? That isn't a linear operation, you can't write it as a matrix and the metric changes. In the 2d Cartesian to polars you have $$ds^{2} = dx^{2}+dy^{2} = dr^{2} + r^{2}d\theta^{2}$$. This is a transformation on the coordinates which doesn't alter anything, it is just a different way of describing things. From this point of view the change to polars is valid, despite it not being a transformation which leaves the metric's form invariant. Instead all covariant and contravariant indices transform with powers of the Jacobian.
Changing from Cartesians to polars in Maxwell's equations in this context is entirely valid, despite the metric also changing. From this point of view, which is what Guest and I have been talking about, Maxwell's equations are covariant. From this point of view you can change coordinates with a Galilean transform and Maxwell's equations keep their form, because the Jacobian is valid. That is what covariance means. It is distinct from the space of transformations which are expressible as linear operators and which leave some particular choice of inner product, typically the metric form $$g(v_{1},v_{2})$$, invariant.
For instance if you do a Lorentz transform from $$(t,x) \to (t',x')$$ then it preserves the space-time interval, $$ds^{2} = -dt^{2} + dx^{2} = -(dt')^{2} + d(x')^{2}$$, you just as ' to all your labels. This is a very special kind of coordinate transformations, precisely for that reason. do a change of coordinates of the Euclidean 2d space interval to polars, $$(x,y) \to (r,\theta)$$ and you don't get an invariant expression, $$ds^{2} = dx^{2}+dy^{2} = dr^{2} + r^{2}d\theta^{2}$$. Still utterly valid in terms of covariance. You can write Maxwell's equations in polar coords and you'll find you can still express them as $$\nabla^{a}F_{ab} = 0$$ for a,b being $$r,\theta$$ etc, rather than $$x,y$$ etc. That is what covariance means, its a much much larger set of transformations than those which leave the inner product associated to the vector space your space-time's tangent spaces define.
The Schrodinger equation is simple to understand in Cartesians but to do the Hydrogen atom its better to change into spherical polars. The result is a very different $$\Delta$$ form but its perfectly valid because the end results are the same.
You need to realise there's a difference between inner product invariance and covariance. Inner product invariance involves acting on elements of the (typically) Hilbert space with linear operators and then deducing which space of linear operators leaves the inner product invariant (this is how you get orthogonal, unitary and symplectic matrices). Covariance involves a non-singular Jacobian. For spaces with 'nice' inner products (like Euclidean or Lorentzian ones) you automatically get a valid Jacobian associated with the linear transform because you have $$|J| = det(M)$$ where J is the Jacobian and M the linear transform. I did this for the Lorentz group in a previous post but it obviously passed you by
.
And somehow it didn't come up when I submitted papers for publication in reputable journals or in my PhD viva. Weird that, its almost like people who know what they're talking about think I know what I'm talking about.
So you're saying Eugene isn't working in either Euclidean or Lorentzian space? A change of coordinates has nothing to do with a fundamental change in the metric. Defining coordinates doesn't have anything to do with defining a metric, only how you define the metric's components in those coordinates.
A good course in GR will cover things like how $$g : TM \times TM \to \mathbb{R}$$, ie $$g(X,Y) \in \mathbb{R}$$ for $$X,Y \in TM$$ is a coordinate free definition. Only when you pick coordinates do you get components $$g_{ab} = g(e_{a},e_{b})$$ for $$X = X^{a}e_{a}$$ and likewise for Y. Doing a change in coordinates $$e_{a} \to f_{a}$$ induces a change in the components but the overall expression $$g(X,Y)$$ is unchanged. Eugene is saying he wants to use different coordinates to the usual ones, which has no impact on whether or not the metric exists or whether $$g(X,Y)$$ exists.
Firstly a tensor can be defined without reference to a metric, which is why covariance is a larger notion than transformations which leave the metric unchanged. A tensor is a multilinear operator, it doesn't mention 'metric' in the definition. The whole covariant and contravariant concepts are definable without a metric being known or even existing. When you go eventually get to metric spaces and the like then you can define a smaller class of linear operators which satisfy $$g_{ab} = M^{c}_{a}g_{cd}M^{d}_{b}$$, which is much more restrictive than the general way a (2,0) tensor would transform (look up tensor densities).
You've gotten these two concepts jumbled together.
So I'm shallow for not giving a crank much attention, a crank you also agree is a crank? I guess the hypocrisy you're displaying is lost on you given you'd concluded within a dozen or two posts from joining this forum that I'm a hack and Guest is a crank?
You're welcome to search for threads I've started, you'll notice they tend to be a touch higher level than the average one and I display working understanding of the relevant concepts. I don't post my work specifically, I only ask questions related to it when I think someone here will have more first hand experience with things which I might find of interest but if you want a discussion on the specifics of say the last paper I got published in a reputable journal (of which I was the only author) we could start another thread.
Irony meter just went off the scale....
I've been saying that assuming his transformation has a non-singular Jacobian then things like Maxwell's equations in their well known covariant tensor formalism would be unchanged in their tensor formulation when you go from your favourite coordinate system to his. Whether or not they can be written as a linear operator and that linear operator leaves whatever metric might be relevant invariant is entirely another irrelevant matter.
Here's a question to think about which should help you realise your mistake : Where in the definition of the coordinate transformation in 2D of Cartesians to polars is a metric required or even mentioned?
Eugene Shubert
Valued Senior Member
What's wrong with me answering unjust criticisms and tailoring my response so it becomes understandable to this audience? The Quintessence of Axiomatized Special Relativity Theory is remarkably pertinent but you couldn't discern its validity. With one extraordinarily superficial look at my paper, you immediately jumped to the conclusion you wanted and then attacked me ferociously. To that ferocity you added outrageous misrepresentation and slander. Why aren't you hiding in shame? You are a disgrace to science. You are simply not credible in anything you say.
In the generally covariant formulation, $$\frac{\text{D}}{\text{d}\lambda} \, \dot{\gamma} \,=\, 0$$ ... ?
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