Analogies for relativistic physics

[studiot]

2 hours ago, Killtech said:

but size does not equal size.

Love it to bits.

 

 

[KJW]

26 minutes ago, Killtech said:

Of course. I haven't made any physicals assumption apart from the well established acoustic wave equation in the case of a stationary medium. Everything else in that post is merely a consequence of that. Know what? i repost it...

There was no need to repost the earlier post.

It seems to me that you are not understanding the point I'm making. If the acoustic wave equation is only valid for a stationary medium, then that means a different equation applies to a moving medium. And a different equation for a moving medium means that the acoustic wave equation is not invariant to "A'rentz" transformations. You did make the mathematical error of failing to recognise where the analogy between light and sound fails.

 

13 minutes ago, Killtech said:

Normally there isn't but if you read the paper, you will find that SR in that scenario has a preferred frame. there are enough reviewed papers on this, if you don't believe me.

Yes, i meant there is one frame where one ages more rapid, more rapid then in any other inertial frame, i.e. that frame becomes uniquely special by that property and is hence considered the preferred frame. 

It's not a preferred frame of reference. It's two non-equivalent frames of reference. And they are non-equivalent because only the travelling twin accelerates.

 

Edited November 20, 2023 by KJW

[Killtech]

19 minutes ago, KJW said:

There was no need to repost the earlier post.

It seems to me that you are not understanding the point I'm making. If the acoustic wave equation is only valid for a stationary medium, then that means a different equation applies to a moving medium. And a different equation for a moving medium means that the acoustic wave equation is not invariant to "A'rentz" transformations. You did make the mathematical error of failing to recognise where the analogy between light and sound fails.

sorry, i have to correct you there. a change of coordinates does not in any way change the original equation, but merely chooses a different representation of it. there is a clearly defined method how a change of variable must be performed exactly, so it has no effect on the predictions being made.

The form of the equation after a coordinate transformation is derived from know facts, not assumed. 

I have calculated a prediction using two ways, one using the straight forward calculation using most basic knowledge and only the other makes use of the A'rentz invariance. you then have to explain to me how it is possible that the result is the same, regardless of the method - with or without the use of invariance.

Edited November 20, 2023 by Killtech

[swansont]

1 hour ago, Killtech said:

if the math agrees with the experiment, what more do you want?

https://www.scienceforums.net/topic/132777-analogies-for-relativistic-physics/#comment-1253247

What experiment does this math agree with?

[KJW]

43 minutes ago, Killtech said:

I have calculated a prediction using two ways, one using the straight forward calculation using most basic knowledge and only the other makes use of the A'rentz invariance. you then have to explain to me how it is possible that the result is the same, regardless of the method - with or without the use of invariance.

You made a mistake with your arithmetic. You said γs=5/3 whereas it's γs=5/4

 

Edited November 20, 2023 by KJW

[Bufofrog]

52 minutes ago, Killtech said:

Yes, i meant there is one frame where one ages more rapid, more rapid then in any other inertial frame, i.e. that frame becomes uniquely special by that property and is hence considered the preferred frame.

Ah, so every single person is in their own preferred frame.  Another way to say the same thing is moving clocks tic slower and moving rulers are shorter.

[Killtech]

8 minutes ago, KJW said:

You made a mistake with your arithmetic. You said γs=5/3 whereas it's γs=5/4

I made that typo in my original post, and i corrected it in the repost. the value i used for \(\gamma_s^2\) is correct though and the calculation that follows is too. thanks for finally reading it.

11 minutes ago, Bufofrog said:

Ah, so every single person is in their own preferred frame.  Another way to say the same thing is moving clocks tic slower and moving rulers are shorter.

No, this is different. Each person has the means to determine if it is in the preferred frame or not. Locally the preferred frame changes nothing, so you cannot detect it by local means if your frame is in fact the preferred one or not. Only globally it makes a difference. it is different then in regular case of SR where the global detection isn't possible.

[swansont]

1 hour ago, Killtech said:

Normally there isn't but if you read the paper, you will find that SR in that scenario has a preferred frame. there are enough reviewed papers on this, if you don't believe me.

That’s not what is meant by a preferred frame.

[Killtech]

12 minutes ago, swansont said:

That’s not what is meant by a preferred frame.

https://en.wikipedia.org/wiki/Preferred_frame

that's the meaning and that is the case in that special topology.

[Bufofrog]

46 minutes ago, Killtech said:

No, this is different. Each person has the means to determine if it is in the preferred frame or not.

Well that's surprising.  What are the 'means' by which we can determine the preferred frame?

[studiot]

On 11/4/2023 at 7:54 PM, Killtech said:

Let's remember the linear acoustic wave equations is △p−c−2s∂p∂t2=0 in the rest frame of the medium.

I was going to examine your 'mathematics'.

I can't make head not tail of this equation  unless you have nabla the wrong way up ?

If you have written Δp then the equation is nonsense.

On 11/13/2023 at 1:57 PM, Killtech said:

that is if we have a sound equation like ∂2xp−c(x)−2∂2tp=0

You seem to have corrected this few posts further on but a recognisable wave equation is stated and is indeed the equation you seem to have copied from Wikipedia.

 

Unfortunately you seem not to have read the text where Wikipedia clearly states that this is the equation of a standing wave.

My example involves a travelling wave and Wiki refers you to a simpler first order differential equation, which it call a one way wave equation in its own style.

 

Quote

n physics, the acoustic wave equation is a second-order partial differential equation that governs the propagation of acoustic waves through a material medium resp. a standing wavefield. The equation describes the evolution of acoustic pressure p or particle velocity u as a function of position x and time t. A simplified (scalar) form of the equation describes acoustic waves in only one spatial dimension, while a more general form describes waves in three dimensions. Propagating waves in a pre-defined direction can also be calculated using first order one-way wave equation.

 

https://en.wikipedia.org/wiki/One-way_wave_equation

 

20 minutes ago, Bufofrog said:

Well that's surprising.  What are the 'means' by which we can determine the preferred frame?

Note Wiki clearly states these 'preferred frames' to be hypothetical for the purposes of exploring what if there were such a frame.

It makes no guarantees that there is one and indeed states there is not such a frame in an inertial set of frames (as we all know).

Edited November 20, 2023 by studiot

[swansont]

1 hour ago, Killtech said:

https://en.wikipedia.org/wiki/Preferred_frame

that's the meaning and that is the case in that special topology.

“In theoretical physics, a preferred frame or privileged frame is usually a special hypothetical frame of reference in which the laws of physics might appear to be identifiably different (simpler) from those in other frames.”

It works for any inertial frame, though, so that one frame is not a preferred frame. The scenario has an accelerated frame, and accelerations are not relative. That’s the crux of the matter.

[Killtech]

15 minutes ago, studiot said:

I was going to examine your 'mathematics'.

I can't make head not tail of this equation  unless you have nabla the wrong way up ?

If you have written Δp then the equation is nonsense.

it's this one here: https://en.wikipedia.org/wiki/Acoustic_wave_equation#/media/File:Derivation_of_acoustic_wave_equation.png

the time derivative is also 2nd order but one character is missing. without a preview function i had some trouble with the latex.

15 minutes ago, studiot said:

Unfortunately you seem not to have read the text where Wikipedia clearly states that this is the equation of a standing wave.

My example involves a travelling wave and Wiki refers you to a simpler first order differential equation, which it call a one way wave equation in its own style.

As one can easily see the solutions to that equation are normal travelling waves. remember you posted a more general book on the topic and you can find the equation in your link as well (https://www.scienceforums.net/topic/132856-wave-equation-in-a-medium-with-smooth-nx-refractive-index/#comment-1253798) where travelling waves are studied. i am not sure i understand that remark on wiki mentioning the standing wavefield.

Here is also the slightly more general case used for traveling waves in a medium with refractive index of which this isa special case: https://wiki.seg.org/wiki/The_eikonal_equation

15 minutes ago, studiot said:

Note Wiki clearly states these 'preferred frames' to be hypothetical for the purposes of exploring what if there were such a frame.

It makes no guarantees that there is one and indeed states there is not such a frame in an inertial set of frames (as we all know).

Indeed the topology discussed in that paper is hypothetical since we never observed the universe to have a limited expanse in any direction so far. However, it that were the case, the existence of preferred frame is a consequence. The issue is that the finite expanse does not allow for the relativity principle to hold globally. 

9 minutes ago, swansont said:

It works for any inertial frame, though, so that one frame is not a preferred frame. The scenario has an accelerated frame, and accelerations are not relative. That’s the crux of the matter.

the interesting thing about that scenario is that the inertial frames are no longer perfectly equivalent and can be distinguished. in other words the relativity principle cannot not hold globally any longer and instead it defines an unique preferred inertial frame. That is the crux of the matter and it is discussed there e.g. here:

https://arxiv.org/abs/gr-qc/0101014
you can find other papers on the topic since it is a known case. 

the wiki article has a section about the special meaning of the preferred frame in aether theories and that is the meaning here as well.

[studiot]

38 minutes ago, Killtech said:

Indeed the topology discussed in that paper is hypothetical since we never observed the universe to have a limited expanse in any direction so far

How's that ?

 

 

42 minutes ago, Killtech said:

i had some trouble with the latex.

We all have trouble with the TEX of one sort or another.

I have a ridiculously expensive commercial generator called Mathtype.

Alternatively you can use free sites Codecogs or Sciweavers to have a TEX editor and copy paste from it.

Another possibility (though limited by the inability to produce almost any sort of fraction) is to combine this SF super and subscript and use windows charmap to pick out special characters such as greek letters. You can actually achieve quite a lot this way.

[swansont]

1 hour ago, Killtech said:

the interesting thing about that scenario is that the inertial frames are no longer perfectly equivalent and can be distinguished. in other words the relativity principle cannot not hold globally any longer and instead it defines an unique preferred inertial frame. That is the crux of the matter and it is discussed there e.g. here:

https://arxiv.org/abs/gr-qc/0101014
you can find other papers on the topic since it is a known case. 

That’s valid for a compact space, a restriction not specified for the general case.

 

 

[joigus]

14 hours ago, studiot said:

I can't make head not tail of this equation  unless you have nabla the wrong way up ?

That surely is the Laplacian operator, which is nabla dot nabla.

"dot" meaning the 3D "dot".

4D nabla dot 4D nabla is called the d'Alembertian, and it's a square (at least in physics).

https://en.wikipedia.org/wiki/Laplace_operator

https://en.wikipedia.org/wiki/D'Alembert_operator

[studiot]

1 hour ago, joigus said:

That surely is the Laplacian operator, which is nabla dot nabla.

"dot" meaning the 3D "dot".

4D nabla dot 4D nabla is called the d'Alembertian, and it's a square (at least in physics).

https://en.wikipedia.org/wiki/Laplace_operator

https://en.wikipedia.org/wiki/D'Alembert_operator

Thanks, Joigus. +1

I think you may well be right, although I try to avoid these more exotic non mainstream symbols.

You have done better than the author who had the opportunity to this is what he meant, but chose not to for some reason.

 

If nabla squared is meant then in my opinion that is what should be said.

The square is often called the box operator but it suffers from the disadvantage that I have seen it confused (used for in lecturres) with the Hodge star operator and also is presented by some parsers as a general symbol when they come across something not in their vocablulary. Word is bad in this respect.

 

Also remaining is the mathematical demonstration that the second order differential equation obeys the Lorenz transformation, using the speed of sound as an invariant.
The simple fact is that to an inertial observer moving differently from the air, the speed of sound appears different.

[joigus]

20 minutes ago, studiot said:

Also remaining is the mathematical demonstration that the second order differential equation obeys the Lorenz transformation, using the speed of sound as an invariant.
The simple fact is that to an inertial observer moving differently from the air, the speed of sound appears different.

Absolutely. I reacted to your pointing this out from the very beginning.

On 11/4/2023 at 12:31 PM, studiot said:

Sound waves do not obey Lorenz transformations.

This is the key result implied but not explicitly stated (you have to think about it to find it)  in the first postulate in Einstein's theory of relativity.

I elaborated (or tried to) a little more on that. But to no avail.

The funny thing is, when I first saw this thread, I interpreted the title literally, "analogies for relativistic physics". I said to myself. "oh, that's interesting". Nothing further from the truth. Not interesting, not enlightening, not even funny at all or in any sense.

--funny as in,

"That's funny, this member doesn't seem to understand the principle of special relativity: It has two parts; the second one destroys his reasoning".

[Killtech]

On 11/21/2023 at 12:26 AM, swansont said:

That’s valid for a compact space, a restriction not specified for the general case.

If you cared to read the conversation you were responding to, you would know you don't add anything already said.

But if you put it that way, the "general" case does not apply so generally, if there are counterexamples against it. note that the compact space case is just the easiest example to illustrate the issue, so do not make the fallacy to assume its a necessary a condition.

On 11/20/2023 at 10:47 PM, studiot said:

I can't make head not tail of this equation  unless you have nabla the wrong way up ?

On 11/21/2023 at 1:09 PM, joigus said:

 

4D nabla dot 4D nabla is called the d'Alembertian, and it's a square (at least in physics).
https://en.wikipedia.org/wiki/Laplace_operator

https://en.wikipedia.org/wiki/D'Alembert_operator

oh, sorry it didn't cross my mind that anyone on these forums wouldn't know these, so i didn't understand your question.

On 11/21/2023 at 3:41 PM, joigus said:

I elaborated (or tried to) a little more on that. But to no avail.

The funny thing is, when I first saw this thread, I interpreted the title literally, "analogies for relativistic physics". I said to myself. "oh, that's interesting". Nothing further from the truth. Not interesting, not enlightening, not even funny at all or in any sense.

--funny as in,

"That's funny, this member doesn't seem to understand the principle of special relativity: It has two parts; the second one destroys his reasoning".

Yeah, you misread the original post so much, i didn't even bother to try to correct you. But fair enough, especially for physicists, i suppose it is very easy to forget that there is not just one Lorentz group since only the one specific to the speed of light is used in physics. You forgot that if you take the Lorentz group and replace c by any other constant, you get another group with identical structure - and it is also still commonly referred to Loretz group, because in math c play no special role. However, the group acts quite differently on the space, and in particular combining elements of different Lorentz groups end up being elements of neither.

So in this case this creates a little ambiguity and i guess i cannot entirely blame someone for stopping reading at "sound wave equation is Lorentz-invariant" and therefore missing out that a different Lorentz group was meant, which should have become clearer when reading further. 

Since you mentioned the d'Alembert operator, lets go back to the linear acoustic wave equation and notice how we are talking about the same structure - where only the constant is a different one. One could just write \(\Box p=0\) instead but that would omit the material specific propagation constant within, which just fuels the confusion you fell for.

But since i apparently do not understand the concepts of relativity, would you please be so kind to enlighten me and the other people in the conversation on the invariances of the d'Alembert operator (or any other operator with the identical structure) and name the 3 subgroups of transformations of the Poincaré group it is invariant under?

[joigus]

8 minutes ago, Killtech said:

Yeah, you misread the original post so much, i didn't even bother to try to correct you. But fair enough, especially for physicists, i suppose it is very easy to forget that there is not just one Lorentz group since only the one specific to the speed of light is used in physics. You forgot that if you take the Lorentz group and replace c by any other constant, you get another group with identical structure - and it is also still commonly referred to Loretz group, because in math c play no special role. However, the group acts quite differently on the space, and in particular combining elements of different Lorentz groups end up being elements of neither.

I told you this since the very beginning there are as many "Lorentz" groups as you like just by re-defining the parameter. Studiot implied it too. Thanks for expressing what I told you and you had missed as if it were my fault. Your comments are easily the silliest I've found in these forums so far.

The good thing is --apparently-- you finally understood. I can only hopel

9 minutes ago, Killtech said:

oh, sorry it didn't cross my mind that anyone on these forums wouldn't know these, so i didn't understand your question.

It was studiot who asked for clarification, not me. Every time you introduce an abbreviation you must clarify what you mean. That's what professionals do. It's obvious you're not one.

You seem to have problems following up on the comments. Take more time to answer instead of embarrassing yourself so much.

Bye.

PS: Re-read the criticism. All the answers you need are there.

[Killtech]

2 minutes ago, joigus said:

I told you this since the very beginning there are as many "Lorentz" groups as you like just by re-defining the parameter. Studiot implied it too. Thanks for expressing what I told you and you had missed as if it were my fault. Your comments are easily the silliest I've found in these forums so far.

On 11/14/2023 at 4:08 PM, joigus said:

So is it clear now that the sound equation is not Lorentz invariant, as v_s is not a universal constant, nor is it a Lorentz scalar, or are we still discussing that?

English isn't my native language, so please help me understand how that only quote of yours in this thread implies there is more then one Lorentz group? I have to admit i still cannot find an interpretation of it that would conform with what you just said. Or was one of your responses maybe deleted?

Also where did you get the impression that i mixed up the Lorentz groups? i have tried to point out quite a lot that this is an analogy, i.e. not the same thing, and i tried to stress that out a lot by prefixing the concepts with the word "analog" to make it apparent it is not the same as in relativity, but rather a mathematical analog with the constant replaced by another.

[swansont]

43 minutes ago, Killtech said:

But if you put it that way, the "general" case does not apply so generally, if there are counterexamples against it. note that the compact space case is just the easiest example to illustrate the issue, so do not make the fallacy to assume its a necessary a condition.

Do we exist in a compact space?

[Killtech]

21 minutes ago, swansont said:

Do we exist in a compact space?

Is it a neccessary condition? Actually, aren't some big bang models of the universe compact (honestly i don't know)?

what you need for comparisons are geodesics that repeatedly meet.

Edited November 25, 2023 by Killtech

[swansont]

7 minutes ago, Killtech said:

Is it a neccessary condition? 

It's a condition of the paper you cited, so if you're going to rely on that for an example, yes, it's absolutely required. Otherwise the conclusions don't apply to the real world, and it's not valid to use it.

[Killtech]

2 minutes ago, swansont said:

It's a condition of the paper you cited, so if you're going to rely on that for an example, yes, it's absolutely required. Otherwise the conclusions don't apply to the real world, and it's not valid to use it.

The question was, what would happen if we could meaningfully compare the length of a meter and that of a second in between inertial frames.