# Interpretations of QM

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Well trust me  I can relate when I first started studying gauge groups I didn't realize the above and kept getting lost and mislead. Once I realized the reasoning of gauge group axioms life got a whole lot simpler in understanding numerous relations described by the group. In particular the (U(1), SU(2) and SU(3) groups including the corresponding SO(N) groups. As SO(N) groups are a double cover of the SU(N) groups.

Edited by Mordred

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16 hours ago, Lorentz Jr said:

But it's often presented as a "why" explanation.

Yes, unfortunately you are right. I see a lot of problems with the way physics is presented in various media. Sadly this appears to be true across the board, including the other sciences too.

16 hours ago, Lorentz Jr said:

Sean Carroll and Lee Smolin have both said relativity can be "explained" by a "change of intuition"

I think they meant explaining the model itself, rather than any underlying ontology.

16 hours ago, Lorentz Jr said:

The only way you can "explain" anything in relativity is by changing reference frames.

Well, relativistic effects are always relationships between reference frames, so this is not really a surprise. But the true power and beauty lies in the exact opposite - that relativity allows us to write the laws of physics such that they do not in any way depend on which reference frame is chosen. It’s about the fact that nature appears to be generally covariant (within the classical domain) and thus does not care about observers at all. That’s a powerful symmetry.

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@Joigus something to keep in mind in topological spaces you will have connected and disconnected sets. For any topological space dealing with the Feymann path integrals. You will be dealing with connected graphs (sets) where momentum conservation is applied and the graph is oriented. Just an FYI to help better understand path integrals.

A way to help is think of any point on a graph or if you prefer manifold/geometry etc. In Feymann path integrals that point is a vertex. In GR its a reference frame, In Feymann Integrals that vertex can be connected to other vertexes or disconnected. If its disconnected its assigned a valency of zero. If its connected to another vertex it will have a valency of 1 or greater depending on the number of connections.

so your path integral has a valency of 1 it has 2 vertexes joined by an edge.(the two vertexes are the initial state/event and the final state/event. The edge being the path defined by the Euler Langrangian for the extremum of the action integral which will be the geodesic. Which will also apply the affine connections through the Christoffels with the use of parallel transport. This then gets into the metric connection which that parallel transport of two vectors must stay parallel along a curve in metric space. (covariant derivative of the above gauge groups).

Hopefully the correlation for 4d spacetime will help with regards to the various topological space connections. You can readily see how they may apply in degrees of freedom

Edited by Mordred
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3 hours ago, Markus Hanke said:

the true power and beauty lies in the exact opposite - that relativity allows us to write the laws of physics such that they do not in any way depend on which reference frame is chosen. It’s about the fact that nature appears to be generally covariant (within the classical domain) and thus does not care about observers at all. That’s a powerful symmetry.

Yes, it's powerful and beautiful and symmetric, but it doesn't make sense. There's no reason the apparent laws of physics shouldn't be affected by the observer's state of motion at all. It only seems reasonable based on our primitive intuition of space as a substanceless geometric void, because how can nothingness have a reference frame?

On the other hand, how can a substanceless void support fields? And if the vacuum has some kind of substantive existence (which GR says it does), why wouldn't it have its own reference frame?

If "explaining" SR is simply a matter of describing the math behind it, then SR is inconsistent with any deterministic mechanism whatsoever  to explain why time dilation occurs in the twins paradox. According to SR, nothing happens in the spaceship any differently from the way things happen on Earth, and yet somehow less time passes on the spaceship. That's fundamentally unscientific. As I was saying earlier, the relativistic "explanation" relies on the same kind of frame-changing errors that perpetual-motion machines are based on (and that Einstein, as a competent patent clerk, must have been intimately familiar with).

It's a kindergarten-level problem in control theory: There has to be some mechanism that causes the difference in elapsed time, and there has to be some difference between Earth and the spaceship that the mechanism can detect so it acts on them differently.

To make a simple analogy, you can't drive your car around town unless it has an engine in it, and you can't start the engine unless you do something (like turning the key in the ignition) that it can detect. And you can't make the car go faster or slower unless you interact with the engine though the gas pedal. There has to be a mechanism, and there has to be something that controls it.

Edited by Lorentz Jr
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5 minutes ago, Lorentz Jr said:

difference between Earth and the spaceship

The difference is that the spaceship accelerates while Earth does not.

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The difference is that the spaceship accelerates while Earth does not.

The acceleration breaks the mathematical symmetry of the paradox, but it doesn't provide a physical mechanism that can cause the aging difference. This can be disproven with two counterexamples:

If one performs the experiment around the Sun instead of, say, Alpha Centauri, with exactly the same accelerations (which would have to be very high in this case, given the relatively short distance to the sun), the difference in elapsed times is a matter of minutes rather than years.

There's a three-observer variant of the experiment (which I like to call the "triplets paradox" ) that involves no acceleration at all but has the same result. One spaceship flies by Earth and travels to Alpha Centauri, and the other one gets up to speed near AC, heads toward Earth, crosses paths with the first one, and flies by Earth.

Edited by Lorentz Jr
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29 minutes ago, Lorentz Jr said:

The acceleration breaks the mathematical symmetry of the paradox

I think it is more than this, it determines who ages more. Let's say A is on Earth and B is in the spaceship. When B reaches the turning point, it doesn't turn, but instead A accelerates to a faster relative speed and catches up with B somewhere behind that point. Then it will be B who has aged more, right?

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it determines who ages more.

That depends on what you mean by "determines". "Breaking the symmetry" means making it possible to mathematically determine who ages more, but my counterexamples show that the acceleration itself can't directly cause a slowdown of aging.

When B reaches the turning point, it doesn't turn, but instead A accelerates to a faster relative speed and catches up with B somewhere behind that point. Then it will be B who has aged more, right?

Yes, that's right. Because the Lorentz transformations are nonlinear.

Edited by Lorentz Jr
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9 minutes ago, Lorentz Jr said:

acceleration itself can't directly cause a slowdown of aging

Right. The slowdown of aging is not physical as long as they are in transit. Only when they meet again, their ages can be physically compared. The result is the consequence of the entire history between their separation and their meeting. There are two different worldlines with two different lengths. The acceleration is just an indication of the difference. If the histories are different, why to expect the outcome to be the same?

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40 minutes ago, Lorentz Jr said:

That depends on what you mean by "determines". "Breaking the symmetry" means making it possible to mathematically determine who ages more, but my counterexamples show that the acceleration itself can't directly cause a slowdown of aging.

Yes, that's right. Because the Lorentz transformations are nonlinear.

They are linear.

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If the histories are different, why to expect the outcome to be the same?

Because the principles that govern their behavior are independent of their histories (according to SR).

10 minutes ago, joigus said:

They are linear.

Don't be silly. $\gamma$ is nonlinear in the velocity.

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16 minutes ago, joigus said:

They are linear.

The transformations are linear.

The formulae involved are not.

58 minutes ago, Lorentz Jr said:

Yes, that's right. Because the Lorentz transformations are nonlinear.

7 minutes ago, Lorentz Jr said:

Don't be silly. γ is nonlinear in the velocity.

Edit since composing this I see a non productive exchange of red points.

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2 minutes ago, studiot said:

The transformations are linear.

The formulae involved are not.

The formulae in question are the Lorentz transformations.

The Lorentz transformations are nonlinear in the relative velocity.

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9 minutes ago, Lorentz Jr said:

Because the principles that govern their behavior are independent of their histories (according to SR).

Don't be silly. γ is nonlinear in the velocity.

Why do you act so childishly? Saying they are non-linear is not the standard way of referring to them, and certainly not the appropriate way. It's at best confusing. If you wish to say they are non-linear in velocities, you should specify that. That's not the way we refer to them, among other things because once you are in the domain of SR, 3-velocity is not a vector quantity on which to define linear / non-linear transformations. It's just a set of 3 parameters.

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16 minutes ago, Lorentz Jr said:

the principles that govern their behavior are independent of their histories (according to SR)

This is not SR. This is a wordplay.

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1 minute ago, joigus said:

Why do you act so childishly?

Why do you play "Gotcha" with people, making critical comments without providing any more information to explain your criticism? The context was Genady's comment "A accelerates to a faster relative speed and catches up with B".

20 minutes ago, studiot said:

Edit since composing this I see a non productive exchange of red points.

I didn't start it.

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18 minutes ago, Lorentz Jr said:

The formulae in question are the Lorentz transformations.

The Lorentz transformations are nonlinear in the relative velocity.

I was rather hoping you might read my post and say to yourself:

"That's interesting I wonder why he said that and made the distinction between the formula and the transformation.
Perhaps there is something I can learn here."

And then ask what a professional mathematician meant when he said

21 minutes ago, studiot said:

The transformations are linear.

The formulae involved are not.

For instance the sine function is nonlinear, but the fourier series and transform is a linear application of it.

This is a linear polynomial

$a{x^5} + b{x^4} + c{x^3} + d{x^2} + ex$

although only the last term is specified by a  linear formula.

Edited by studiot
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6 minutes ago, Lorentz Jr said:

Why do you play "Gotcha" with people, making critical comments without providing any more information to explain your criticism? The context was Genady's comment "A accelerates to a faster relative speed and catches up with B".

@Lorentz Jr, you'd be well advised to refrain from your usual hissy fits, stop using neg-reps as some kind of covering fire for your arguments, and stick to the topic.

If I didn't understand your point, say it nicely and we can all get back to business. I can assure you I, and many others, think you make a valuable contribution to the forums.

Lorentz transformations don't act on the v's, they're parametrised by them. That's whay I got confused.

I'm sorry for having ruffled your feathers. Jeez!

I've just reverted my red point.

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35 minutes ago, Lorentz Jr said:

The formulae in question are the Lorentz transformations.

If U is a 4-vector, then the formula for a Lorentz transformation from the original frame to some primed frame is written as

$U’=\Lambda(v)U$

where $$\Lambda$$ is a square matrix, the Lorentz transformation matrix, and v is some parameter of the transformation (not necessarily speed!). I invite you to verify yourself that

$\Lambda(v+u)=\Lambda(v)+\Lambda(u)$

and (c=const.)

$\Lambda(cv)=c\Lambda(v)$

by whatever means you find most convenient. The above two relations define the property of linearity in the context of matrix transformations.

So yes, the Lorentz transformations are indeed very much linear - as of course they have to be, since they map lines into lines, ie inertial frames into inertial frames. This is pretty trivial tbh.

P.S. Cross posted with joigus, studiot and Grenady! Had my reply open on screen some time before hitting “Submit”.

Edited by Markus Hanke
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To put it mathematically

If LT represents a Lorentz transformation is represents a transformation of something.

That something is one of the four coordinates, x,y,z,t.

It is easy to show that aLT (x)  = LT (ax) ;  where a is some coefficient.

and the same for the other three coordinates.

As joigus noted, LT does not act directly on v, the relative velocity.

which is a condition that forms part of the definition of linear in mathematics.

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12 minutes ago, Markus Hanke said:

If U is a 4-vector, then the formula for a Lorentz transformation from the original frame to some primed frame is written as

U=Λ(v)U

where Λ is a square matrix, the Lorentz transformation matrix, and v is some parameter of the transformation (not necessarily speed!). I invite you to verify yourself that

Λ(v+u)=Λ(v)+Λ(u)

and (c=const.)

Λ(cv)=cΛ(v)

by whatever means you find most convenient. The above two relations define the property of linearity in the context of matrix transformations.

So yes, the Lorentz transformations are indeed very much linear - as of course they have to be, since they map lines into lines, ie inertial frames into inertial frames. This is pretty trivial tbh.

P.S. Cross posted with joigus, studiot and Grenady! Had my reply open on screen some time before hitting “Submit”.

Thank you, Markus. +1. As usual, you helped to take the discussion back on its track. And that's right. They are linear.

11 minutes ago, studiot said:

To put it mathematically

If LT represents a Lorentz transformation is represents a transformation of something.

That something is one of the four coordinates, x,y,z,t.

It is easy to show that aLT (x)  = LT (ax) ;  where a is some coefficient.

and the same for the other three coordinates.

As joigus noted, LT does not act directly on v, the relative velocity.

which is a condition that forms part of the definition of linear in mathematics.

I must confess I must go back to @Genady and @Lorentz Jr's parallel discussion and understand the point. The argument has spilt over into different aspects of physics, and it's hard to keep up.

Edited by joigus
minor correction
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40 minutes ago, Markus Hanke said:

If U is a 4-vector, then the formula for a Lorentz transformation from the original frame to some primed frame is written as

U=Λ(v)U

where Λ is a square matrix, the Lorentz transformation matrix, and v is some parameter of the transformation (not necessarily speed!). I invite you to verify yourself that

Λ(v+u)=Λ(v)+Λ(u)

and (c=const.)

Λ(cv)=cΛ(v)

I would recommend against using the letter  $v$ for clarity since not only does it come out different in the standard font used on this site but it also can represent velocity, as can  $u$, and so can easily be confused with 'parameters'.

Addition of velocities is decidedly non linear in SR.

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Yes, I don't know what all of this has to do with interpretations of QM, but composition of collinear boosts is not linear in velocities, although it is in rapidities[?*] (which are essentially hyperbolic-angle arguments.) That's probably what @Markus Hanke meant.

The part where I get lost, as I say, is what any of this has to do with QM and its interpretations.

I do not wish to keep talking about this, as it has no bearing on the particular aspects I'm more interested in, but suffice it to say that linear or non-linear depends on the arguments on which the object is considered to be acting. Eg, and totally analogous to collinear boosts, rotations about the same axis are linear in the angle, but not in the trig functions the rotation depends on.

* Is that the word?

Edited by joigus
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2 minutes ago, joigus said:

what any of this has to do with QM and its interpretations

Nothing. The discussion between @Lorentz Jr and @Markus Hanke has moved to GR and then to SR some time ago.

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16 hours ago, studiot said:

I would recommend against using the letter  v for clarity since not only does it come out different in the standard font used on this site but it also can represent velocity, as can  u , and so can easily be confused with 'parameters'.

Addition of velocities is decidedly non linear in SR.

Ok, point noted - though I did make it quite clear that this refers to general transformations between frames, and not just composition of velocities.

14 hours ago, joigus said:

That's probably what @Markus Hanke meant.

They were just meant to be general Lorentz transformations, and I left the parameter unspecified. I don’t think it matters what kind of parameter you choose - speed, angle, gamma factor etc -, since the transformation itself always remains linear; it acts on 4-vectors and general tensors, not the parameter itself.

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