Education and Crank Claims: Special Relativity

Done

The general expression, which you have been given before, has \(\gamma[tex] being a function of [tex]|| \mathbf{v}||\), which treats the three directions in the same manner, precisely as I said.

You were provided this before but obviously you didn't understand then.

No one ever said it wasn't. You were given the link to the general case multiple times, so trying to pretend you're saying something new is just a denial of reality.

SR doesn't assume what you said it does. SR derives the fact that the direction which is contracted is parallel to the direction of the boost.

In Rpenner's example that direction is \((v_{x},0,0)\). In your example that direction is \((v_{x},v_{y},v_{z})\). In Rpenner's example there are two directions which will not experience contraction, they are orthogonal to x and therefore can be chosen to be y and z. In your example you will find the same. Pick two vectors which are orthogonal to \((v_{x},v_{y},v_{z})\) and are not parallel to one another and you'll find neither of them get altered by the Lorentz transform you've just given. So all you've done is just rotate the vector, only 1 direction is transformed still.

Since you no doubt lack the understanding to work out two vectors orthogonal to \(\mathbf{v} = (v_{x},v_{y},v_{z})\) I'll do it for you. An easy one is \(\mathbf{u} = (v_{y},-v_{x},0)\). Clearly these are orthogonal. Once you have 2 non-parallel vectors in \(\mathbb{R}^{3}\) it's easy to find a third mutually orthogonal vector, use the cross product, ie \(\mathbf{w} = \mathbf{u} \times \mathbf{v} = (-v_{x}v_{z} , -v_{y}v_{z} , v_{y}^{2} + v_{x}^{2} ) \). Now we have a triplet of orthogonal vectors which form a basis for \(\mathbb{R}^{3}\), ie \(\mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{w} = \mathbf{w} \cdot \mathbf{u} = 0\). Check that for yourself.

I claim that the Lorentz transform you have provided doesn't alter vectors in the directions \(\mathbf{u}\) or \(\mathbf{w}\), you will just do a shift and contraction in the direction of \(\mathbf{v}\). More specifically, the Lorentz image of \(A\mathbf{u} + B\mathbf{w}\) at time T will be \(A\mathbf{u} + B\mathbf{w} - \gamma T \mathbf{v}\) at times \(\gamma T\).

Let's see if this is true. Here is the Lorentz transform in question, call it \(\Lambda\). The 3-vectors define 4-vectors by \((T,\mathbf{u})\) and \((T,\mathbf{w})\). Let's apply that \(\Lambda\) to \(\mathbf{u}\). We'll set c=1 so \(\beta_{x} \to v_{x}\) etc.

We end up with the 4-vector \((\gamma T , -v_{x}\gamma - v_{y} , -v_{y}\gamma + v_{x} , -v_{z}\gamma ) = (\gamma T , \mathbf{u}-\gamma \mathbf{v})\).

Do the same with \(\mathbf{w}\) and the get the same result. Thus the Lorentz transform you have provided (when you get all the components right) still leaves 2 directions unchanged, shifting and contracting only the direction parallel to the velocity vector \(\mathbf{v}\) and dilating the time.

So the expressions you've given aren't warping 3 dimensions, they are warping 1 in an unpleasant choice of basis. The expression is easily obtained by doing a rotation on the version Rpenner gave. It's really very simple stuff. Tell me, have you ever actually studied relativity? You seem extremely poor at basic mathematical methods, unfamiliar with well known results and you're always damn sure of yourself only to be knocked down by the next post by someone who knows some relativity. You obviously thought I'd make some excuse, as if your posts are unanswerable. I haven't had the slightest trouble knocking this post up, I sat down and typed it out.

You cranks always think you're stretching the understanding of people like myself or Rpenner or anyone who actually has a physics qualification. You aren't. You think you are because you use yourself as a measuring stick, that if you're struggling then we must be. It just shows how naive you are. You really have no idea just how massively unqualified you are for that Caltech job. It's like someone thinking they can do brain surgery because they own a butter knife.

 

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Based on the "Standard Configuration" of Special Relativity (SR) it assumes that the axis of motion is elastic/deformable while the other two axes remain fixed/rigid.

Standard Configuration -- Distance

\( \Delta{x'} \,\,\, = \,\,\, \frac{{\Delta{x} - v_x \, \Delta{t}}}{sqrt{1 \,\, - \,\, \frac{v^2_x}{c^2_{Light}}} \)

\( \Delta{y'} \,\,\, = \,\,\, \Delta{y} \)

\( \Delta{z'} \,\,\, = \,\,\, \Delta{z} \)

The below "Non-Standard Configuration" maintains that the axis of motion in this case the x-axis, y-axis, and z-axis are all elastic/deformable.

Please stop denying this, it makes you look contrary and not truly seeking the right answers.

Non-Standard Configuration -- Distance

\( \Delta{x'} \,\,\, = \,\,\, \frac{{\Delta{x} - v_x \, \Delta{t}}}{sqrt{1 \,\, - \,\, \frac{v^2_x}{c^2_{Light}_x}} \)

\( \Delta{y'} \,\,\, = \,\,\, \frac{{\Delta{y} - v_y \, \Delta{t}}}{sqrt{1 \,\, - \,\, \frac{v^2_y}{c^2_{Light}_y}} \)

\( \Delta{z'} \,\,\, = \,\,\, \frac{{\Delta{z} - v_z \, \Delta{t}}}{sqrt{1 \,\, - \,\, \frac{v^2_z}{c^2_{Light}_z}} \)

I don't disagree with this statement, in fact I agree with this statement.

However, based on "Physics" of the natural world, most objects that are in uniform motion relative to an inertial frame of reference, have motion in all three directions (x, y, & z); not just one direction. And which means that motion is occurring along parallel axes to all three directions (x, y, & z) simultaneously. Which would yield a Boost" in all three directions.

The Lorentz Boost is a trick for the mathematicians. But the physicists of the world like to see this derived using geometry; the way Einstein did it. Plus a geometric derivation typically has more credence for the physicist.

Are you trolling again?

First, I have not rotated any "vector" or "frame of reference".

Second, it is well known that according to the Laws/Rules of the "Dot Product" that different vectors that are perpendicular (orthogonal) the dot product of the two vectors is the zero (0). And these rules are much more powerful than Lorentz.

However, I have been describing velocity vectors that are parallel to each other in every direction dimension.

You are using math descriptions here, that I don't particularly agree with! You are discussing three motion vectors including a (w) vector when there should only be two vectors (u & v) not counting the speed of light vector (c).

Now, this statment above I agree with. And this vector (\(\mathbf{v}\)) above is contracted where there is motion in the directions (vx, vy, vz).

It is obvious that you can do "Abstract Math" and likewise very clear that you don't know a thing about "Physics." I think that you might want to brush up by taking a “First Years" physics class again.

I would like to see you "derive" \( \Delta{y'} \,\,\, = \,\,\, \Delta{y} \,\,\, , \,\,\, \Delta{z'} \,\,\, = \,\,\, \Delta{z}\) of the "Standard Configuration" of Special Relativity (SR) using geometry or any other method other than the "Lorentz Boost" method; but I know you can't.

 

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I'm explaining to your the specifics, you just don't wan to listen.

The vector \((v_{x},v_{y},v_{z})\) is a single direction, it just has components in all 3 bases. There are directions perpendicular to that vector and I've given two such vectors to give an orthogonal basis for \(\mathbb{R}^{3}\), ie

\(\mathbb{R}^{3} = \textrm{span}\Big\{ \mathbf{v} \,,\, \mathbf{u} \,,\, \mathbf{w} \Big\} = \textrm{span} \Big\{ (v_{x},v_{y},v_{z}) \,,\, (-v_{y},v_{x}, 0 ) \,,\, (-v_{x}v_{z} \,,\, -v_{y}v_{z} , v_{x}^{2}+v_{y}^{2} ) \Big\}\)

Your formulation allows you to put in any values you want, while Rpenner's example he has already set \(v_{x}=1\), \(v_{y}=0\), \(v_{z}=0\).

Let's true yours with some actual numbers. Try \(v_{x}=v{y}=v_{z}=1\). Then \(\mathb{v} = (1,1,1)\). I claim there are 2 directions orthogonal to this and by putting in the values in the formulas I gave we get \(\mathbf{u} = (-1,1,0)\) and \(\mathbf{w} = (-1,-1,2)\). They are both orthogonal to (1,1,1).

All you've done is pick one general vector. The fact it has components in all directions doesn't mean 3 dimensions are altered, it means one general direction is altered and the two orthogonal to it are left unchanged, precisely as in Rpenner's example.

Ah, you're back to the "That isn't physics, it's maths!". You really do need to open a textbook on this stuff. I did a theoretical physics PhD and I was in a better position at the start because I'd done a maths degree than those people in my year who'd done a physics degree. Mathematics is the language of theoretical physics. Your attempt to complain a proof of something isn't valid because it's algebraic rather than geometrical is laughable. They both lead to the same thing, just doing the algebra is easier to work with.

This is stuff anyone whose done research in GR should know. Penrose? Mathematics degree. Hawking? Mathematics degree. Green? Mathematics degree. Perry? Mathematics degree. Dirac? Mathematics degree. The theoretical physics and general relativity research groups at Cambridge are in the mathematics department. They form a significant chunk of DAMTP. That's what the T and P stand for, theoretical physics.

You're really showing your lack of familiarity with the material by saying such things.

Yes, you have. If you work out a rotation Lorentz transform which rotates \((v,0,0)\) to \((v_{x},v_{y},v_{z})\) where \(v^{2} = v_{x}^{2}+v_{y}^{2}+v_{z}^{2}\) you'll be able to use it to turn Rpenner's version into your version.

You can see hints of this from the fact \(\gamma\) depends on \(v^{2} = v_{x}^{2}+v_{y}^{2}+v_{z}^{2}\), which is invariant under rotations and the same expression appears in the time component of each transform.

It would seem to me you're just copying equations Rpenner or Wiki have provided and not really understanding them. We've seen you do this in the past with things like arc lengths, the SC metric, tensors components, electromagnetic tensors. This is why I didn't give you the benefit of the doubt, you don't deserve it.

Actually the manner of combining elements in a vector space depends on a metric, which is an additional piece of information. That's why Riemannian manifolds are denoted as (M,g), not just M, because they can have multiple metrics.

The metric which defines the dot product is the Euclidean metric and it has symmetry group SO(N). In this case SO(3). SO(3) is a subgroup of SO(1,3) and the Euclidean 3d metric is an induced metric on spatial hyper-surfaces of Minkowski space. So the Lorentzian formalism on space-time is actually more general than the Euclidean formalism on space.

But then I suppose you don't know about induced metrics. Would you like me to provide some lecture notes from my 4th year about such things? It's a course on black holes.

Except that I've shown you can construct a triple of mutually orthogonal vectors, including your vector, which under Lorentz transforms transform in the same way as Rpenner's.

Let's look at that again. We had \(\mathbb{R}^{3} = \textrm{span}\Big\{ \mathbf{v} \,,\, \mathbf{u} \,,\, \mathbf{w} \Big\} = \textrm{span} \Big\{ (v_{x},v_{y},v_{z}) \,,\, (-v_{y},v_{x}, 0 ) \,,\, (-v_{x}v_{z} \,,\, -v_{y}v_{z} , v_{x}^{2}+v_{y}^{2} ) \Big\}\).

In Rpenner's case he picks \(v_{x} = v\) and the rest 0. Then you end up with a triplet \(\textrm{span} \Big\{ (v,0,0) \,,\, (0,v, 0 ) \,,\, (0 \,,\, 0 , v^{2} ) \Big\}\), basically vectors individually in the x, y and z directions.

Since \(\{ \mathbf{v},\mathbf{u},\mathbf{w}\}\) is a basis for \(\mathbb{R}^{3}\) an vector can be written as a unique linear combination of them, \(\mathb{V} = v_{x}\mathbf{e}_{x} + v_{y}\mathbf{e}_{y} + v_{z}\mathbf{e}_{z} = A\mathbf{v} + B\mathbf{u} + C\mathbf{w}\). As I demonstrated in my last post the effect of a Lorentz boost in the general \(\mathbf{v}\) direction ends up sending this to \(\tilde{A}\mathbf{v} + B\mathbf{u} + C\mathbf{w}\) for some \(\tilde{D}\) dependent on time and \(||\mathbf{v}||\). The two directions orthogonal to \(\mathbf{v}\) are left unchanged.

You are mistaking the fact \(\mathbf{v}\) has components in all 3 entries of its vector for the idea all possible directions are changed. The vector (1,2,3) has non-zero entries in all components but its still one direction, there are still 2 directions orthogonal to it.

I know you don't agree with them, you don't understand them. As for the number of vectors, firstly the speed of light isn't a vector, it's a scalar. There you go again with your inability to understand the difference between scalars and vectors. Secondly \(\mathbb{R}^{3}\) is three dimensional. To span it you need 3 vectors who are not pair-wise parallel (which is equivalent to having non-zero triple vector product). You give \(\mathbf{v}\) and I demonstrate there's a \(\mathbf{u}\) and \(\mathbf{w}\) which form a orthogonal basis for \(\mathbf{R}^{3}\).

This stuff isn't even relativity, it's basic linear algebra. If there's N directions in a space you need N different vectors at least to span it. \(\mathbb{R}^{3}\) is 3 dimensional, understand?

I suppose you did struggle with this before, you thought \(\frac{1}{\mathbb{R}^{N}}\) meant something, so perhaps you still haven't learnt what the notation means.

Typical crank excuse. You haven't understood what I've said so you cannot retort it. Since I've used mathematics you claim I'm just doing maths and not physics. You should take a look in the physics section of www.arxiv.org, you'll see what I'm doing here is tame compared to that. As for 1st year courses, a 1st year relativity course in a physics degree is something I've taught.

Besides, you claim you're competent at general relativity. This sort of maths should be second nature to you. You should have seen this stuff all the time, you should know it's the bread and butter of the relativity community. To try to claim it isn't physics is laugable. This is fundamental linear algebra. You can't do electromagnetism, gravity, Lagrangian mechanics, fluid mechanics, anything like that without an understanding of this. Again, I can provide you with books or lecture notes which will illustrate this material is central to those areas of physics. The fact you don't realise that exposes how little you know.

So you've been provided with one method and you don't understand it so you demand another one.

Please explain, demonstrating explicitly any claims, where the problem with using a Lorentz transform is. Lorentz transforms encode the geometry of Minkowski space-time into them. You ask about geometry, geometry is encoded in the metric. The Lorentz transforms are defined by the metric, so using Lorentz transforms is a way of describing the metric.

The fact you ask these things, while claiming to be up to speed on general relativity really shows the level of contradiction in your claims. You should have plenty of experience using symmetries to model geometry, it's how relativity works.

I've explicitly demonstrated that the effect of any Lorentz transform is to warp one direction in space, while leaving the orthogonal directions unchanged. I've explicitly demonstrated your claim that Lorentz transform warps all directions to be false. If you wish to contest that then show explicitly where I've made a mistake. Where is the algebra wrong? Asking for another derivation is pointless if you haven't refuted the first.

 

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How do we detect the slowdown of clocks?

I posted this question:

I received this:

And a much longer one from OnlyMe which I won't quote, but hopefully am able to link to here: [post]2833263[/post]

The two responses seemed to be somewhat in disagreement, although perhaps not totally unreconcilable. Anyway, OnlyMe also posted this: [post]2833461[/post], which I really didn't entirely understand, but it seems to me that he generally agrees with funkstar.

I am of course happy to accept funkstar's answer, as he confirmed what I suspected. That what I described is essentially a Doppler effect, makes sense to me.

Both of the responses leads me then to this question:

Is the effect of relativity (the time dilation?) on the clock rate then something I should expect to come "in addition" to this one? In other words: should we - at least in principle - expect to be able to to detect *by viewing it* a greater slowdown of a distant clock than what can be deduced from its velocity (away from us) according to my description above?

(I know, that was a mess of a sentence. Hope you can understand it. I'll try to clarify if required.)

Thanks.

 

The problem with you "Mathematicians" is that you assume too much!!:bugeye:

Inertial Frame Velocity Vector

\( \vec{u} \,\,\, = \,\,\, {(u_x, u_y, u_z)} \) \( \,\,\,\, ---> \,\,\,\, \frac{m}{s} \)

Mass Object Velocity Vector

\( \vec{v} \,\,\, = \,\,\, {(v_x, v_y, v_z)} \)\( \,\,\,\, ---> \,\,\,\, \frac{m}{s} \)

Induced Velocity Vector

\( \vec{w} \,\,\, = \,\,\, {(w_x, w_y, w_z)} \)\( \,\,\,\, ---> \,\,\,\, \frac{m}{s} \)

Induction Process below

\( \vec{w} \,\,\, = \,\,\, \vec{v} \,\,\, x \,\,\, \vec{u} \)\( \,\,\,\, ---> \,\,\,\, \frac{m}{s} \)

Induced Velocity Components

\( \vec{w^2_x} \,\,\, = \,\,\, [{v_z} \,\, {u_y} \,\,\, - \,\,\, {v_y} \,\, {u_z}] \,\,\, = \,\,\, - {v_x} \,\, {u_z} \)\( \,\,\,\, ---> \,\,\,\, \frac{m^2}{s^2} \)

\( \vec{w^2_y} \,\,\, = \,\,\, [{v_x} \,\, {u_z} \,\,\, - \,\,\, {v_z} \,\, {u_x}] \,\,\, = \,\,\, - {v_y} \,\, {u_z} \)\( \,\,\,\, ---> \,\,\,\, \frac{m^2}{s^2} \)

\( \vec{w^2_z} \,\,\, = \,\,\, [{v_y} \,\, {u_x} \,\,\, - \,\,\, {v_x} \,\, {u_y}] \,\,\, = \,\,\, {v^2_y} \,\,\, + \,\,\, {u^2_x} \)\( \,\,\,\, ---> \,\,\,\, \frac{m^2}{s^2} \)

Where the Induced Velocity Vector now becomes

\( \vec{w} \,\,\, = \,\,\, {(||\mathf{w_x}|| , ||\mathf{w_y}||, ||\mathf{w_z}||)} \,\,\, = \,\,\, {(\sqrt{{v_x} \,\,\, {u_z}}, \,\,\, \sqrt{{v_y} \,\, {u_z}}, \,\,\, \sqrt{{v^2_y} \,\,\, + \,\,\, {u^2_x}})} \)\( \,\,\,\, ---> \,\,\,\, \frac{m}{s} \)


So this is what you get. You are once again mixing units and quantities!

\(\mathbb{R}^{3} = \textrm{span}\Big\{ \mathbf{v} \,,\, \mathbf{u} \,,\, \mathbf{w} \Big\} = \textrm{span} \Big\{ (v_{x},v_{y},v_{z}) \,,\, (-v_{y},v_{x}, 0 ) \,,\, (-v_{x}v_{z} \,,\, -v_{y}v_{z} , v_{x}^{2}+v_{y}^{2} ) \Big\}\)

This is what I get:

\(\mathbb{R}^{3} = \textrm{span}\Big\{ \mathbf{v} \,,\, \mathbf{u} \,,\, \mathbf{w} \Big\} = \textrm{span} \Big\{ (v_{x},v_{y},v_{z}) \,,\, (u_{x},u_{y}, u_{z} ) \,,\, \,\,\, {(\sqrt{{v_x} \,\,\, {u_z}}, \,\,\, \sqrt{{v_y} \,\, {u_z}}, \,\,\, \sqrt{{v^2_y} \,\,\, + \,\,\, {u^2_x}})} \Big\}\)

Also
Induced Square Velocity Metric

\( \vec{w^2} \,\,\, = \,\,\, {(\vec{u} \,\,\, - \,\,\, \vec{v})^2} \,\,\, = \,\,\, \vec{u^2} \,\,\, + \,\,\, \vec{v^2} \,\,\, - \,\,\, 2||\mathf{u}||||\mathf{v}||cos{\theta} \)\( \,\,\,\, ---> \,\,\,\, \frac{m^2}{s^2} \)

Inertial Frame Squared Velocity Metric

\( \vec{u^2} \,\,\, = \,\,\, {u^2_x} \,\,\, + \,\,\, {u^2_y} \,\,\, + \,\,\, {u^2_z} \) \( \,\,\,\, ---> \,\,\,\, \frac{m^2}{s^2} \)

Mass Object Velocity Vector Squared Velocity Metric

\( \vec{v^2} \,\,\, = \,\,\, {v^2_x} \,\,\, + \,\,\, {v^2_y} \,\,\, + \,\,\, {v^2_z} \) \( \,\,\,\, ---> \,\,\,\, \frac{m^2}{s^2} \)

where,

\( cos{\theta} \,\,\, = \,\,\, \frac{(\vec{u} \vec{v})}{||\mathf{u}||||\mathf{v}||} \)\( \,\,\,\, ---> \,\,\,\, unitless \)

\( sin{\theta} \,\,\, = \,\,\, \frac{(\vec{u} \,\,\, x \,\,\, \vec{v})}{||\mathf{u}||||\mathf{v}||} \)\( \,\,\,\, ---> \,\,\,\, unitless \)

Now above, I have introduced physics and geometry into the solution.

What you have presented is wrong; your quantities and units don't match below.

\(\mathbb{R}^{3} = \textrm{span}\Big\{ \mathbf{v} \,,\, \mathbf{u} \,,\, \mathbf{w} \Big\} = \textrm{span} \Big\{ (v_{x},v_{y},v_{z}) \,,\, (-v_{y},v_{x}, 0 ) \,,\, (-v_{x}v_{z} \,,\, -v_{y}v_{z} , v_{x}^{2}+v_{y}^{2} ) \Big\}\)


Now, if you want to be a real physicist; "Learn Physics"!!

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If you want to be a mathematician then you are doing fine. But if you want to learn physics, I suggest you read the book:

Super Principia Mathematica - The Rage to Master Conceptual & Mathematical Physics - By Robert Louis Kemp

Now that you have introduced a new induced velocity vector (w), the "physics"question now becomes what does the induced velocity (w) mean in Special Relativity? In introducing this new vector you have to explain what is going on along that new vector that you introduced into the problem?


Edit:

Initially posted:
Induced Square Velocity Metric

\( \vec{w^2_z} \,\,\, = \,\,\, {(\vec{u} \,\,\, - \,\,\, \vec{v})^2} \,\,\, = \,\,\, \vec{u^2} \,\,\, + \,\,\, \vec{v^2} \,\,\, - \,\,\, 2||\mathf{u}||||\mathf{v}||cos{\theta} \)\( \,\,\,\, ---> \,\,\,\, \frac{m^2}{s^2} \)

This should be:

Induced Square Velocity Metric

\( \vec{w^2} \,\,\, = \,\,\, {(\vec{u} \,\,\, - \,\,\, \vec{v})^2} \,\,\, = \,\,\, \vec{u^2} \,\,\, + \,\,\, \vec{v^2} \,\,\, - \,\,\, 2||\mathf{u}||||\mathf{v}||cos{\theta} \)\( \,\,\,\, ---> \,\,\,\, \frac{m^2}{s^2} \)

 

DonQuixote,

1) Can you safely just assume that "apparent" phenomenon has nothing to do with relativity?

No. And Yes.

2) and that the slowing down of clocks in SR is something else entirely?

Yes. And No.

I see your confusion. When discussing the subject "relativity" and different frames of reference (inertial or non-inertial) and someone uses the statements "as viewed from" or "as seen from" or "as judged from" you want to begin understanding by removing the concept of "seeing", like to look out into the horizon type of "seeing".

When the terms "as viewed from" or "as seen from" or "as judged from" are being used, all of these terms can be substituted with the term to "make measurements from".

So what you are doing is "making measurements" from or in different frame of reference.

Now the question becomes, why is "making measurements from/in" a better term? Because making measurements encompases measuring:
1) Sonic Doppler Effects (hearing)
2) Optical Doppler Effects (seeing)
3) Time Measurements (Clocking)
4) Distance Measurements (Apparent Geometry)

Imagine that you are standing on the podium at the "train station" making any one or more of the various measurements above; that is considered an inertial frame.

Now, imagine that someone else "AlphaGeneric" is riding in one of the box cars of the "moving train", making any one or more of the various measurements above; that is also considered an inertial frame.

Now in your inertial frame on the "station podium" you get your own measurements. And in "AlphaGeneric's" "moving train" inertial frame of reference he is also getting his own measurements.

Now the "paradox" comes in when you compare the measurements in your "station podium" inertial frame of reference, with AlphaGeneric's measurements in the "moving train" inertial frame of reference. You naturally assume that you and AlphaGeneric should get the same values for your frame measurements, but you don't.

It turns out that adding motion to the experiments, and in this case the motion of the moving train it will cause different observers, to measure different values, in their different frames, for the various experiments described above.

Best

 

I haven't assumed anything. I've explicitly walked you through your mistakes and misconceptions.

Wow, just wow.

You've just shown you don't know how to do the cross product. Despite me giving the calculations explicitly in my previous post.

If \(X = (x_{1},x_{2},x_{3})\) and \(Y = (y_{1},y_{2},y_{3})\) are given and \(Z = (z_{1},z_{2},z_{3})\) is defined by \(Z = X \times Y\) then we get

\(z_{1} = x_{2}y_{3}-x_{3}y_{2}\)
\(z_{2} = x_{3}y_{1}-x_{1}y_{3}\)
\(z_{3} = x_{1}y_{2}-x_{2}y_{1}\)

Note the lack of a squared on the \(z_{i}\) terms. You also didn't even use the right symbol for the cross product, using \(x\) instead of \(\times\). Haven't you ever done the cross product before? It comes up all over the place in electromagnetism, which you claim to know about.

Little things like that reveal how little experience you have with this stuff, even if you hadn't just shown you don't know the cross product's actual formula. Hell, you could have just copied the results from my post or looked it up on Wikipedia (it's not like you haven't copied from Wikipedia before!).

Here's a tip. If you're going to try to correct someone who knows what he's talking about on a matter you don't remember or never knew in the first place it's a good idea to check your formulae are right.

Firstly that isn't an induced metric. An induced metric would be the metric on a subset of the space formed by the ambient metric restricted to the space. I've even gone over the induced metric with you before, with the equation of a sphere and the spherical metric. You clearly didn't understand it before, afterwards or now.

Secondly you got your results wrong. If \(\mathbf{u} = (-v_{y},v_{x},0)\) and \(\mathbf{v} = (v_{x},v_{y},v_{z})\) then \(||\mathbf{u}-\mathbf{v} ||^{2} = ||(-v_{y}-v_{x},v_{x}-v_{y},-v_{z}) ||^{2} = (-v_{y}-v_{x})^{2} + (v_{x}-v_{y})^{2} + (-v_{z})^{2} = v_{y}^{2} + 2v_{x}v_{y}+v_{x}^{2} + v_{x}^{2} - 2v_{x}v_{y} + v_{y}^{2} + v_{z}^{2} = 2v_{x}^{2} + 2v_{y}^{2} + v_{z}^{2}\), which is not equal to \(w_{z}^{2} = (v_{x}^{2}+v_{y}^{2})^{2}\) (or even your incorrect expression for it).

Are you struggling to expand out brackets now?

Well done, you've managed to quote a few dot product identities which are known to 17 year olds. Such a shame you couldn't actually use them correctly.

You've shown you can quote formula but you can't use them. Seriously this stuff is high school level. And you thought you were in the running for a CalTech position?!

The vector cross product is an identity relating components in a vector space, there isn't an attempt to match units. Besides, the components can be viewed as describing the multiples of some unitful quantity, so \(v_{x}\) etc can have no units. Alternatively just work with unit vectors of dimensionless quantities.

Nice true though. Such a shame the relevant issue is the direction. I've shown there are 2 directions which are orthogonal to each other and \(\mathbf{v}\) which are left unaltered by your Lorentz boost. Although your vector has all non-zero components that doesn't mean all 3 dimensions are warped, only that the 1 direction which is warped isn't parallel to any of the three bases. If you work in a coordinate frame which has \(\hat{\mathbf{v}},\hat{\mathbf{u}},\hat{\mathbf{w}}\) as it's basis then you'll end up with exactly the same Lorentz boost as Rpenner did. All Rpenner has to do to convert his version into yours is apply a rotation which takes \((v,0,0)\) to \((v_{x},v_{y},v_{z})\). That's all there is too it.

I've demonstrated a better conceptual understanding of the scenario than you. I've demonstrated a more indepth understanding of the mathematics than you. I've demonstrated a more accurate understanding of the mathematics than you. Every single step of the way I've out done you.

If you can't do the cross product, understand the difference between vector and scalar, don't grasp rotations, have no understanding of Lie groups and Lie algebras, don't grasp orthonormal bases, don't know how to do matrix exponentiation, misunderstand what an induced metric is, think rotations are more general than Lorentz transforms and can't expand out \((\mathbf{u}-\mathbf{v})^{2}\) properly (and all of that is just this thread!!) why should anyone think reading 3 volumes of your thoughts will teach them anything other than the depths of self delusion some people will dive to.

Oh and I'll tell you now, if you mention your book at all in this forum again for any reason than someone else bringing it up you'll get a warning. You've done enough self advertising and you only got away with it to the extend you have because the forum's moderator hasn't been around much. I am now a moderator in this forum so consider this your final warning.

The \(\mathbf{w}\) I've given is a direction which is analogous to \(\mathbf{e}_{z}\) in Rpenner's case, just as \(\mathbf{u}\) is analogous to \(\mathbf{e}_{y}\). \(\hat{\mathbf{v}},\hat{\mathbf{u}},\hat{\mathbf{w}}\) for an orthonormal basis, as do the x,y,z unit vectors. I've demonstrated you are nothing but a rotation away from Rpenner's set up. It's a shame you don't grasp that but none the less that is the case. If you care to address my mathematics, rather than quoting high school level expressions you got from Wikipedia at me then perhaps we could discuss it and you could learn somethimg.

 

Yes.

Doppler compression and relativistic doppler effect are separate effects, and thus the actual light you see coming from rapidly moving objects is the result of both effects. Experiments have demonstrated this; objects moving quickly are more red-shifted than classical doppler theory alone would account for.

 

There I would be educating PhDs, just like I am educating you here on this forum!

Don't you threaten me again you little "pip-squeak"!!:bugeye:

First, I was giving a reference to a published work. I am "legally" able to reference any published work that has an ISBN and is sold on Amazon.com; whether that be my personal work, or your grandmother's published work. That is not advertising. If you don't know the difference:

a) You should not be a "moderator"
b) You need "moderator" training.

Being a moderator does not give you "God Powers"!!!

Do your homework; referencing any published work that has an ISBN is legal; and is not advertising. Learn the difference between "advertising" and making a "reference"; no matter how jealous you are!

Do not let your personal jealousy get in the way of your new job!!:bugeye:

Second, this is pure entertainment for me. I do not have to contribute here. I am "not" tied to this place for any reason. I am currently looking for a new and fun place to discuss physics with some cool people, anyway. I will now, speed up my search, since you are "now" in charge!!

 

The correct expression of it is! Get it right!

\(w_{z}^{2} = (v_{x}^{2}+v_{y}^{2})\)

No. What I am struggling to understand, like an instructor that ask his student; why did you write that?

Why did you write that?

\( u_{x} = -v_{y} \)
\( u_{y} = v_{x} \)
\( u_{z} = 0 \)

What I don't comphrend, is why do you start with these initial "assuming" conditions for the velocity of the inertial frame of reference?

This is starting from assumptions. I know that mathematicians do this a lot!

Oh, you give complements so so generously.

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I agree with this totally!

I agree with this totally!

 

Congratulations!

Great, do you need any help?

 

So, Origin what of your beginnings?

 

DonQuixote,

First as funkstar's post quoted below (bold emphasis mine) says, the Doppler effect has nothing to do with SR, I might add, at least not as far as clock rates are involved. The Doppler effect involves a change in the wavelength of light, observed either as a redshift or blue shift. I did not understand that to be what you were thinking or describing and it would not result in a distant clock appearing to run fast or slow, just redder or bluer than it would at rest.

There are actually three aspects of SR where two clocks in inertial frames of reference (A and B) stationary or moving uniformly relative to one another, can appear to or actually run slow, from the frame of reference of the other.

The first is simple time of light delay. If A and B are one light minute apart an observer in either location will see the clock at the the other location one minute slow compared to their own. It takes light that long to move between them. This time of light delay is involved whether A and B are at rest or moving relative to one another. If they are moving apart the time of light delay will be increasing and both clocks when viewed from the opposite frame of reference will appear not only slow but slowing as the time of light distance increases. If they move toward one another they will both appear to the other as slow but the rate will be decreasing as they move closer the time of light delay will also be decreasing. This is not time dilation but it is still part of the special theory of relativity.

The second two involve Time Dilation.

First, a clock that is moving will run slow when compared to a clock at rest. This is only apparent with very accurate clocks and/or high velocities. This is a real time dilation and has been experimentally demonstrated.

SR states that the laws of physics are the same for all inertial frames of reference and that all inertial frames of reference are equally valid. This and the time dilation mentioned above lead to the twin paradox(es).

Assuming that A and B are moving relative to one another, A can assume it is at rest and B is moving and B can assume that it is at rest and A is moving. Both are equally valid under SR. Both would see the other as experiencing length contraction and time dilation. In the twin paradox(es) it is only when the two twins or in our case A and B return to the same location that they can compare their clocks and determine who's clock was "really" running slow, compared to the other. It only when brought back to gather that which was actually moving relative to the other can be determined.

I am not sure this addressed your question(s). The inclusion of the Doppler effect suggests perhaps not. So once more, the Doppler effect can occur both as a function of both SR and GR, but it has nothing time dilation, it involves changes in the wavelengths of light.

 

And once again, Robert Louis Kemp demonstrates that he is a complete fraud who doesn't understand even the most elementary concepts in physics.

 

Except that you don't have a PhD Mr Kemp. You lack the qualifications for the positions you have applied for. You haven't done any published research ever.

As for educating me, funny how I've proven you wrong and you haven't refuted me.

No, I'm enforcing the rules of this forum. Self advertising for books which are self published and cost money is not allowed.

'Legally'? This forum is not legally required to let you do anything. You, me and everyone else here posts at the discretion of the owners. If the owner came along tomorrow and said "Stuff it, you're all banned" we'd have to accept it. If the moderators and I decided you deserved to be banned then we can ban you. I cannot ban you on a whim, there are rules which you have to have broken for me to ban you. One of those rules is no self advertising.

This is the maths and physics forum, for reputable science. Pseudo science and unpublished alternative theories have their own subforums. Your work is not published in any reputable journal or nor is it based on work which is published in a reputable journal. As such your work does not qualify as 'published' in that sense. Simply paying someone to print your work on paper and bound it into a book doesn't count as 'published work' in the academic sense. The fact you try to confuse the two illustrates you either don't understand the important difference or you're dishonest.

For example, the Bible is published. As is The Lord of The Rings. Should we count those as valid references in a science debate? No. Clearly they have not met scientific standards and one is an admitted work of fiction (and the other an unadmitted work of fiction). Your work has not meet those standards. Furthermore it is not freely available. My work, which has been published in reputable journals, is freely available. As such I am a published researcher. If it is relevant I am allowed, within the rules of this forum, to reference my work as it holds scientific weight. You are not. You charge people money for work which has not been assessed for its scientific merit. Worse, what has been looked at has been found to be false.

As such citing it in a scientific discussion is not a valid way of justifying a claim about science. Citing Hawking in a discussion about general relativity is valid, his work has been reviewed and examined and passed said examination. Your books have not, so they cannot be trusted and thus are not relevant to discussion.

'Legal' has nothing to do with it. You do not have freedom of speech here, the forum is not required to let you post anything and everything, just as a newspaper is not required to print every single letter which is sent to it. If you start your own newspaper then you can pretty pretty much anything, just as starting your own forum means you can post pretty much anything. You've been to Farsight's forum, that's what he did. He was laughed off every science forum he could find so he started his own.

Do your homework.

What do I have to be jealous of you for? I have more peer reviewed published work, I have more qualifications and a better research job.

Then leave.

 

We have 2 vectors \(\mathbf{v} = (v_{1},v_{2},v_{3})\) and \(\mathbf{u} = (-v_{2},v_{1},0)\). The formula for the cross product gives that the components of \(\mathbf{w} = \mathbf{v} \times \mathbf{u}\) are \((-v_{2}v_{3},-v_{1}v_{3},v_{1}^{2}+v_{2}^{2})\).

The cross product has the property that the end result is orthogonal to both \(\mathbf{v}\) and \(\mathbf{u}\), ie \(\mathbf{v}\cdot(\mathbf{v}\times \mathbf{u}) = 0 = \mathbf{u} \cdot (\mathbf{v}\times \mathbf{u})\).

But what do we get if \(\mathbf{w} = (\sqrt{v_{2}v_{3}},\sqrt{v_{1}v_{3}},\sqrt{v_{1}^{2}+v_{2}^{2}})\). Clearly \(\mathbf{v}\cdot(\mathbf{v}\times \mathbf{u}) = 0 = \mathbf{u} \cdot (\mathbf{v}\times \mathbf{u})\) fails.

Thus your expression cannot be right. You wanted a geometric analysis. The geometric meaning of the cross product is that it gives a vector at right angles to the 2d plane defined by \(\mathbf{v}\) and \(\mathbf{u}\). The reason it gives quadratic expressions is that its length is proportional to the length of each of the input vectors. By the sin rule \(|| \mathbf{v} \times \mathbf{u} || = || \mathbf{v}|||| \mathbf{u} || \sin \alpha\). Since they are, by construction, at right angles to one another \(\alpha = \frac{\pi}{2}\) and so \(|| \mathbf{v} \times \mathbf{u} || = || \mathbf{v}|||| \mathbf{u} || \). If you view the components as just dimensionless values, ie its the \(\mathbf{e}_{i}\) which carry the units then it isn't a problem.

For example, if \(\mathbf{v} = (1,1,1)\) then \(\mathbf{u} = (-1,1,0)\) and \(\mathbf{w} = (-1.-1,2)\). No problem, no units. Your obsession with units is blinding you to the real structure of the problem, because you don't actually know how to deal with units in any more than the most trivial way.

If you struggle to accept this (likely) then the simplest way around it is to make everything a unit vector, so you end up with not just an orthogonal basis but an orthonormal one.

\(\hat{\mathbf{v}} = \frac{1}{\sqrt{v_{1}^{2}+v_{2}^{2}+v_{3}^{2}}}(v_{1},v_{2},v_{3})\)
\(\hat{\mathbf{u}} = \frac{1}{\sqrt{v_{1}^{2}+v_{2}^{2}}}(-v_{2},v_{1},0)\)
Putting these into the cross product formula gives
\(\hat{\mathbf{w}} \equiv \hat{\mathbf{v}} \times \hat{\mathbf{u}} = \frac{1}{\sqrt{v_{1}^{2}+v_{2}^{2}+v_{3}^{2}}} \frac{1}{\sqrt{v_{1}^{2}+v_{2}^{2}}} (\mathbf{v} \times \mathbf{u}) = \frac{1}{\sqrt{v_{1}^{2}+v_{2}^{2}+v_{3}^{2}}} \frac{1}{\sqrt{v_{1}^{2}+v_{2}^{2}}} (-v_{1}v_{3},-v_{1}v_{2},v_{1}^{2},v_{2}^{2})\)

The third vector is a unit vector. This can be seen by direct computation or by noting \(|| \mathbf{X} \times \mathbf{Y}|| = || \mathbf{X} ||||\mathbf{Y}|| \) if \(\mathbf{X}\cdot \mathbf{Y}\) and if X,Y are unit vectors so is their cross product. This happens with the x,y,z axes, \(\mathbf{e}_{z} = \mathbf{e}_{x} \times \mathbf{e}_{y}\).

I could have arrived at \(\hat{\mathbf{w}}\) directly from my original results by just normalising \(\mathbf{w}\), they all give the same results because \(\hat{\mathbf{w}}\) and \(\mathbf{w}\) point in the same direction.

Now you have three vectors which point in the same directions are the vectors I gave before except now they are all unit lengths and form an orthonormal basis. Does it make any difference to my analysis? No, which is why your whining about it is completely irrelevant but I wanted to show you how easily it is dismissed. The entire point I was making, which you failed to grasp, was that given a \(\mathbf{v}\) vector to do a Lorentz boost along in \(\mathbb{R}^{3}\) there is a 2 dimensional space spanned by the vectors orthogonal to \(\mathbf{v}\) which is left unaltered by the boost. In Rpenner's case this was the space spanned by \(\mathbf{e}_{y},\mathbf{e}_{z}\). In general it is the space spanned by \(\mathbf{u},\mathbf{w}\). The reason your whining was pointless is because

\(\textrm{span}\{ \mathbf{u},\mathbf{w} \} = \textrm{span}\{ \hat{\mathbf{u}},\hat{\mathbf{w}}\} \)

They are the same two dimensional subspace of \(\mathbb{R}^{3}\). This argument easily generalised to \(\mathbb{R}^{N}\), the boost will leave an N-1 dimensional subspace orthogonal to the boost vector invariant. This principle is well known to any mathematician or physicist, because every vector defines a 1 dimensional subspace and an N-1 dimensional subspace. The 1 dimensional subspace is the space spanned by the vector and the N-1 dimensional subspace is the one spanned by all vectors orthogonal to it. It's the classic dual structure in the set of hyperplanes in \(\mathbb{R}^{N}\).

You claim to know a lot about geometry and relativity but you don't understand this. As I've repeatedly said, I'll happily provide you with lecture notes which go over this stuff, to illustrate I'm not making this up or trying to present some obscure result. This is required learning to anyone doing general relativity at university. Well, perhaps I'm extrapolating too much, clearly it wasn't required learning at the University of Phoenix, just a research isn't a required component of a lecturing job there.

My aim was to construct a set of 3 vectors, including \(\mathbf{v}\) which are mutually orthogonal (at right angles to one another, since you might not know what that means). If \(\mathbf{v} = (v_{x},v_{y},v_{z})\) then clearly \((-v_{y},v_{x},0)\) is orthogonal to it. There isn't an assumption, it can be checked, \((-v_{y},v_{x},0) \cdot (v_{x},v_{y},v_{z}) = -v_{x}v_{y}+v_{x}v_{y} + 0 = 0\). You keep thinking I'm making assumptions, when in fact I'm explaining my methodology and then applying procedures I'd expect a 1st year to be familiar with. Why is it you never seem to have a grasp of any undergraduate physics or maths?f course it isn't the only vector orthogonal to \(\mathbf{v}\). However, in \(\mathbb{R}^{N}\) once you construct N mutually orthogonal vectors you have an orthogonal basis so you can uniquely write any vector in \(\mathbb{R}^{N}\) in terms of the set.

The fact you are unfamiliar with this process, which is very common, shows how little experience you have with linear algebra. As I said, a vector defines an N-1 dimensional orthogonal hyperplane and I wanted to construct a basis for that hyperplane, which I did so. A basis is not unique, any non-zero determinant linear transformation will convert a basis into another basis but the important thing is the N-1 dimensional basis is always orthogonal to the original vector.

You seem not to have even grasped what it is I'm doing. Are cross products and scalar products too much for you to follow?

Then you accept that your Lorentz boost does not warp all directions, there are N-1 orthogonal directions, which are orthogonal to your velocity vector, which are left unchanged by the Lorentz boost. All you have done is apply a rotation to Rpenner's case.

If you disagree with this you need to show where I have made a mistake in my algebra. And remember, check on Wikipedia whether the formula you're using is right, clearly you don't remember how to do the cross product.

 

Ah, that's beautiful. The delusion knows no bounds.

 

Just to let everyone know, I've been working out a derivation of the Lorentz transformations directly from the basic postulates, including a careful and thorough treatment of the axes orthogonal to the relative motion (the \(y,\ z,\ y'\) and \(z'\) axes). Without making any assumptions about the speed of light, I've completed the derivation of the most general possible coordinate transforms, which then become either the Galilean transforms with classical assumptions, or the Lorentz transformations when the speed of light postulate is added.

However, instead of adding in the lightspeed postulate and completing the derivation, I'm working on determining the transforms by demanding that Maxwell's equations in vacuum will hold in all inertial frames (determining how charges, currents and EM forces transform can come afterwards). Seems to be going well so far; I'm pretty sure I will have enough conditions to find all the undetermined constants and thus derive the Lorentz transformations in the process, in such a way that the transformation is entirely unique. If I'm successful, this should deal with any attempted counter-argument that two different observers are tracking two different EM pulses when they both measure their pulses moving at \(c\).

Then next on the list is to try and deal with Prometheus' questions about Wilsonian renormalization in a different thread, as I promised I would.

 

Go to Google.
Enter physics forums.
Find one where you have not been banned.
Register.
Stay there.

 

My rationale for thinking observed slowdown of clock rate is due to Doppler effect might be this:

Sound, light and clocks have somethig in common - frequency.

The frequency of sound, light or clocks will therefore show the Doppler effect.

We might, for the sake of simplicity, say that a clock that moves away from us at the speed of 100 000 km./sec. has a frequency of 1 tic per second. (That, I have learned today is one Hertz, one cycle per second)
Let's imagine it blinks one time per second. When it has moved for one second, it is 100 000 kilometers away, so when the local clock where we are and the distant clocks tic *simultaneously* it will still take 1/3 of a second for the light signal from the distant clock to arrive here. That makes it appear as if the frequency of the distant clock is 1/3 greater than the local clock. For every second of the local clock, the distant clock will lose 1/3 of a second.