Is there anything faster than light?

I mean, look at this:

http://apod.nasa.gov/apod/ap140615.html

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Have you ever seen anything more glorious?

The YECs no doubt hate this image because:

1) it does not contain a CENTRAL LOCUS of any kind where G-d might sit on his almighty cosmic throne like he/she/it evidently did at the precision nanosecond of the Big Bang or beginning of Inflation which they have co-opted for themselves to identify the moment of creation, and
2) G-d him/her/itself has not photobombed this image.

G-d and science damn, rather than bless the YEC creationist cosmologists. They don't like anything resembling relativity either, because G-d is relative only to us.

 

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Tell us again why you can't bring yourself to just write God or even god, Dan. Are you being selectively 'anti-fundamentalist' here?

 

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It's a tradition of Judaism that no one actually knows the name of G-d. This prohibits our tradition from anything like using the name of G-d in a blasphemous manner.

 

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OK - I believe in respecting folks beliefs. But given such reverential care, I take it you believe that a deity is in fact 'out there somewhere' right? No I don't want a sidetracking into religious debate, just pointing out that it might pay to lay low on making fun of other's religious ideas in an ostensibly physics related sub-forum.

 

I'm not making fun of a religious idea, Q-reeus. I'm making fun (actually, hard criticism, not fun) of YEC's ideas about making science into a religion. It isn't fun. It doesn't serve science, and it doesn't serve religion either.

If you think your holy scripture (whichever tradition) is the unerring word of G-d, and you worship it as if it were an idol, then it is an idol. In Judaism, the Torah is revered as though it were a person, but NOT as though it were G-d. To assure that translation bias does not enter into the tradition, religious services and text are never translated from the original Hebrew or Aramaic. Even with these measures, some Helenistic polytheism has insinuated itself into the Torah from the days of Alexander the Great, but for the most part, the corruption was successfully contained.

And for thousands of years before anyone self-identified as a follower of Christ or of Islam, the rabbinical debate commemorated in the Talmud argued over the account of Genesis and did not call any of it blasphemy. Nor does modern Judaic tradition have the slightest idea that G-d, or religious leaders should prohibit or interfere the advancement of science not associated in any manner with religious tradition or faith.

There is a reason there is a first commandment, and it means exactly what it says. Don't make idols of artifacts, religious leaders, or religious texts. Anyone who thinks that pursuing science for its own sake is idolatry would either be an idiot, or else someone more interested in benefiting from the exploitation of religious devotion for personal gain.

 

Have a look at this article by Professor Douglas Scott. Note my bolding:

"How come we can tell what motion we have with respect to the CMB? Doesn't this mean there's an absolute frame of reference? The theory of special relativity is based on the principle that there are no preferred reference frames. In other words, the whole of Einstein's theory rests on the assumption that physics works the same irrespective of what speed and direction you have. So the fact that there is a frame of reference in which there is no motion through the CMB would appear to violate special relativity! However, the crucial assumption of Einstein's theory is not that there are no special frames, but that there are no special frames where the laws of physics are different. There clearly is a frame where the CMB is at rest, and so this is, in some sense, the rest frame of the Universe. But for doing any physics experiment, any other frame is as good as this one. So the only difference is that in the CMB rest frame you measure no velocity with respect to the CMB photons, but that does not imply any fundamental difference in the laws of physics."

 

Let me ask you: what do you mean exactly by "scalar curvature"? You seem to be referring to the Ricci scalar - the trace of the Ricci curvature tensor, upon which it clearly depends totally.

But I cannot see that this justifies the use of the term "scalar curvature". Please explain

PS Sorry to be so late on the case

PPS We need to know that the trace of a rank 2 tensor is precisely the inner (or scalar) product of those vectors that enter into the tensor product that defines it. Once we know this, we can see that taking the trace, finding the inner product and "contracting tensors" are equivalent operations. Maybe a new thread may be in order??

 

I used terminology in the Wikipedia article on the EFE, which calls it a scalar curvature. The first use of the term "scalar curvature" in the EFE article points to an article titled, "Scalar curvature." That second article confirms your claim it is the trace of the Ricci curvature tensor. I have been told by people who demonstrated knowledge of both differential geometry and relativity that this usage is one of the peculiarities of relativity terminology, and that the more common term in mathematics (as opposed to relativity) is Ricci scalar, so I guess that might be where the confusion lies.

My understanding is that both the Ricci curvature tensor (first term of the EFE) and the scalar curvature (part of the second term) are inherent curvature parameters of the Riemann manifold upon which GRT is defined, and that the metric tensor is the non-inherent curvature (part of both the second and third terms)(perhaps due to mass-energy? I hesitate because it is not clear to me that the Ricci curvature tensor and scalar curvature are not affected by the mass-energy contents of the manifold). This understanding is probably vastly oversimplified.

Just so we're clear about what I mean when I talk about a "term,"

\(R_{\mu \nu}\) First term, Ricci curvature tensor term
\({1 \over 2}g_{\mu \nu}\,R\) Second term, scalar curvature term
\(g_{\mu \nu} \Lambda\) Third term, cosmological term
\({8 \pi G \over c^4} T_{\mu \nu}\) Resulting stress-energy tensor term, multiplied by the constants, G and c

The whole EFE:
\(R_{\mu \nu} - {1 \over 2}g_{\mu \nu}\,R + g_{\mu \nu} \Lambda = {8 \pi G \over c^4} T_{\mu \nu}\)

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Perhaps. I have had parts of matrices and tensors explained to me before but it never quite "took;" I can't relate it to other things I know, so it's just a bunch of odd rules that I can't memorize, and can't derive from other things I know. Maybe you could connect those dots for me in a way that I will understand better.

 

Odd rules like that are technically called "conventions" just like EM's right hand rule.

There is a mathematical convention in Lorentz transforms of length contraction and Minkowski rotations that both occur from the geometric centers of contracting lengths. The trouble with those two mathematical ones I just mentioned as opposed to the EM conventions of the right hand rule and the agreed upon direction of electric current is that they are completely, totally arbitrary with no real basis in physical reality. Either end of a contracting object could have been chosen, but the calculated physical result would not really have been equivalent, useful, or correct. The choice of origin of a coordinate system may be arbitrary, but if it is going to be useful for anything physical, the units are not. Stretching of space is not something to be determined by means of mathematical convention either.

 

That's not a convention, it's a change in terminology; whereas a convention defines a particular basis under which all calculations will be performed, like conventional current vs. electron flow, or the sign convention -+++ in GRT, that defines time as negative and the three space dimensions as positive; it is possible to choose another basis, viz., electron flow or +---.

It is possible to use hyperbolic trig to define the Lorentz transform rather than algebra as it is usually presented, and get all the same results, and those results have been confirmed by experiment, viz. the muon lifetime increasing in the frame of the observer as the muon speed increases in that same frame. When using this form of the Lorentz transform the direct physical meaning of the transform is explicitly treated as a rotation in which part of the space coordinate(s) along the vector of motion (in the frame of the observer) is transformed into extra duration in the time coordinate. This is the direct unambiguous measurement of the rotation, just as when an object is rotated by some angle in space alone, what was X (for example) is turned into some combination of less X, and more Y and/or Z. All that's been done is to add a fourth coordinate, time, and use the same law that causes a stick that was normal to your line of sight appear shorter when it is rotated to some other angle.

Try the Physics FAQ page on adding velocities in SRT here, and look at the section titled, "How can that be right?" You will see the derivation of hyperbolic trig from the more common algebraic form of the Lorentz transform.

 

Surely you not being so rude as suggest that my posts are only "correct" if they are confirmed by Wikipedia?

Then I am sorry. I simply do not understand how anyone can claim knowledge of some of the more recent advances in physics without understanding the mathematics that were used to develop them.

Still less do I understand how those who seek to "improve" or "better explain" current physics reject the use of the relevant mathematics.

I guess I may be in the wrong place.....

 

Never. I merely cite it in support of your point and as an aid to understanding both between us and for any spectators.

By listening to/reading the explanations of those who have demonstrated knowledge, and repeating them back in different terms to see if they agree not only with my explanation of what they said, but of what others said. Which is what I'm trying to do with you.

I reject nothing; I merely invited you to explain it another way. A good place to start would be the derivation of matrices, from the POV of someone comfortable with algebra, Euclidean geometry, trigonometry, and some calculus (as much of it as I remember which is mostly differential calculus, derivatives, limits, and the fact that integration is the opposite of differentiation- I just don't use integration enough to be facile in it). Then you can show how matrices relate to tensors. If you can do it.

You'll have to be the judge of that.

 

Some kind of new discovery http://forum.nasaspaceflight.com/index.php?PHPSESSID=vp35t4k2p4lupqf0md86f2tih5&topic=36313.0

 

Well of course I can do it. It's just a matter of notation......

Suppose a 3-vector space \(V\) and further suppose \(v,\,\,w \in V\). I write \(v= (a^1+a^2+a^3)e_j\) and similarly \(w=(b^1+b^2+b^3)e_k\) where the \(a^i,\,\,b^i \) are just numbers - scalars - and the \(e_i\) are basis vectors i.e a subspace of \(V\).

Define the tensor product \(v\otimes w\) as \((a^1b^1+a^1b^2+a^1b^3+a^2b^1+a^2b^2+........+a^3b^3)e_j\otimes e_k\)

Then rearrange the scalar components as \(\begin{pmatrix}a^1b^1 &a^1b^2& a^1b^3\\a^2b^1&a^2b^2&a^2b^3\\a^3b^1&a^3b^2&a^3b^3 \end{pmatrix}e_j \otimes e_k\) and call this "a tensor"

One easily sees that the trace of this matrix is \(a^1b^1+a^2b^2+a^3b^3\) - just a Real number - and this completely coincides with the definition of the inner product of the 2 vectors we started with.

So there - easy

 

I'll look that over, I'll have questions.

I should note that I know what a vector is and see why the interval from \(e_j\) to \(e_k\) is a vector in \(V\).

My first observation is that this is a 3-vector space and GRT uses a 4-vector space, and that I think the signs in \(V\) are are +++ whereas the fourth vector in relativity has a negative sign, so -+++. I think that's the sign convention for \(R_{\mu \nu}\) in the EFE, too, and also the other tensors (I think they all have to have the same sign convention for the tensor sums and products to come out right).

So my first question has two parts, first, is this a correct understanding, and second, how does this affect a 4d tensor and make it different from the way your 3d tensor works?

My second question is, what is a basis?

 

Actually, no, I don't think this is quite correct. First, the concept of an "interval" requires a metric, which is not yet defined. Second, even it it had been, the only vector that describes the interval between 2 or more basis vectors is the zero vector. This is because (more as a matter of convenience rather than strict mathematical necessity) the basis vectors are taken to be linearly independent

Yes, this is one common convention. It does have some interesting consequences, though

It does, but pretty much only in the inner product as I defined it. In my last, we may assume that the inner product (the trace of the matrix I gave) is either definitely positive or definitely zero. One says thereby that this inner product is "positive-definite".

In the Minkowski case you mention, the inner product is not definitely anything, so it is called an "indefinite metric". (Hmm. We need to work a little on this - give me a while to cook up a post on the metric tensor

Given a vector space \(V\) a basis \(\{e_i\}\) is a subspace (of vectors, obviously) such that any other \(v \in V\) can uniquely be written as a linear combination of the form \(\sum\nolimits_j \alpha^j e_j\)

 

OK, that's going to take some digesting. It appears I have stumbled over a piece of terminology, "interval." Also I need to understand which is the vector and what the other elements are. So I'll think about that a while.

OK, I'll keep an eye out for that.

Given a vector space \(V\) a basis \(\{e_i\}\) is a subspace (of vectors, obviously) such that any other \(v \in V\) can uniquely be written as a linear combination of the form \(\sum\nolimits_j \alpha^j e_j\)[/QUOTE]What's \(\alpha^j\)?

 

[QUOTE="Schneibster, post: 3375297, member: 284370"[/QUOTE]

They are scalars - numbers. In the present simple case they are Real numbers.

In other words, vectors have the property they can be arithmetically added to give a new vector, and scalar multiplied also to give a new vector. Hence \(v = \alpha^1e_1+\alpha^2e_2+.....+\alpha^ne_n\) since the \(e_i\) are also vectors.