Abouzar Bahari Posted February 18 Share Posted February 18 Since the introduction of Special theory of Relativity (SR), Lorentz invariance has been a fundamental part of our description of nature. Over the past decades, this (Lorentz invariance) was tested and verified by many research groups and almost no Lorentz violation has been found. However, when we look at the Lorentz transformations, we can mathematically find that they are not symmetrical relations. In fact, the inertial frames are not symmetrical in relation to each other and as a result, there must be a reference frame. If the inertial frames are symmetrical in relation to each other, as the SR argues, some paradoxes and contradictions are happened. Hence, we need to find the misconceptions in this theory. Many researches up to now, focused on Zero-Point Field (ZPF) as a cause for many physical phenomena like mass, inertia, and gravity. One can argue that the Lorentz transformations can be applied only for the inertial frames, moving inside the ZPF and the inertial frames, being at rest in relation to the ZPF, must not bear the Lorentz transformations. In addition, we have found that the SR and General Relativity (GR) are not two separated concepts. They are one thing and the space-time of them are the same. In both cases, it is the ZPF, which curves around the accelerated body or a gravitational mass, causing changes in the body’s length, time, and mass. The full-length published paper can be available via the following link: Bahari, Abouzar. " Zero-Point Field is the Cause for the Lorentz Transformations and Leads to Find Misconceptions about the Special Relativity" BULETIN FISIKA [Online], Volume 25 Number 1 (5 September 2022), ISSN 2580-9733, https://ojs.unud.ac.id/index.php/buletinfisika/article/view/82042 -2 Link to comment Share on other sites More sharing options...

Markus Hanke Posted February 18 Share Posted February 18 (edited) 44 minutes ago, Abouzar Bahari said: However, when we look at the Lorentz transformations, we can mathematically find that they are not symmetrical relations. Oh yes, they are. The Lorentz transformation leaves the metric invariant: \[g_{\mu \nu } =\Lambda _{\sigma }^{\mu } \Lambda _{\rho }^{\nu } =g_{\sigma \rho }\] Rewrite this in matrix notation: \[\Lambda ^{T} g\Lambda =g\] Take determinant on both sides: \[det\left( \Lambda ^{T}\right) det( g) det( \Lambda ) =det( g)\] Since none of these determinants is ever zero, and since the determinant of the transpose equals the determinant of the original matrix, you get: \[det( \Lambda )^{2} =1 \] which implies that \(\Lambda\) is always invertible. Thus, inertial frames related via Lorentz transformations are always symmetric. 44 minutes ago, Abouzar Bahari said: The full-length published paper can be available via the following link ! Moderator Note It is against the rules of this forum to post personal theories into the main physics section, let alone onto an existing thread. If you wish to discuss this, you must open your own thread under “Speculations” and explain your thoughts there (don’t just give links). Edited February 18 by Markus Hanke 1 Link to comment Share on other sites More sharing options...

joigus Posted February 18 Share Posted February 18 @Markus Hanke is absolutely right. When one talks about something being symmetric or not, one must specify what is symmetric --the object-- with respect to what --change of POV, transformation, etc. What Markus has shown to you is that, assuming two observers assign respectively coordinates \( \left(t,x,y,z\right) \) and \( \left(t',x',y',z'\right) \), the metric --given by \( t^{2}-x^{2}-y^{2}-z^{2} \) doesn't change --it's the same in the primed coordinates and the unprimed ones. It might be that what you mean is that the law that user with primed coordinates uses to correlate his observations with those of user with unprimed coordinates is not the same with \( \boldsymbol{v} \) replaced by \( -\boldsymbol{v} \). But it is. Both relative velocities are obviously collinear, so, \[ x'=\frac{x-vt}{\sqrt{1-v^{2}/c^{2}}} \] \[ ct'=\frac{ct-vx/c}{\sqrt{1-v^{2}/c^{2}}} \] \[ y'=y \] \[ z'=z \] (simple Lorentz transformations in one direction, AKA 'boosts') Introducing the definitions, \[ \gamma=\frac{1}{\sqrt{1-\beta^{2}}} \] \[ \beta=v \] The reciprocal ones obviously are, \[ \gamma'=\gamma \] \[ \beta'=-\beta \] and you get, \[ \left(\begin{array}{cccc} \gamma & -\beta\gamma & 0 & 0\\ -\beta\gamma & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{array}\right)\left(\begin{array}{cccc} \gamma & \beta\gamma & 0 & 0\\ \beta\gamma & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{array}\right)=\left(\begin{array}{cccc} \gamma^{2}\left(1-\beta^{2}\right) & \beta\gamma-\beta\gamma & 0 & 0\\ \beta\gamma-\beta\gamma & \gamma^{2}\left(1-\beta^{2}\right) & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{array}\right)= \] \[ =\left(\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{array}\right) \] which more compactly reads, \[ \Lambda\left(-\boldsymbol{v}\right)=\Lambda^{-1}\left(\boldsymbol{v}\right) \] In what other sense you might want it to be more symmetrical, I don't know. 1 Link to comment Share on other sites More sharing options...

Genady Posted February 18 Share Posted February 18 7 hours ago, Abouzar Bahari said: almost no Lorentz violation has been found almost? 7 hours ago, Abouzar Bahari said: when we look at the Lorentz transformations, we can mathematically find that they are not symmetrical relations When going from frame A to frame B, we use in the Lorentz transformation the speed of B relative to A. When going from frame B to frame A, we use in the Lorentz transformation the speed of A relative to B. These two statements are symmetrical. In fact, I've copied and pasted the first to make the second, and then have simply replaced A by B and B by A. How more symmetrical it can be? Link to comment Share on other sites More sharing options...

Bufofrog Posted February 18 Share Posted February 18 8 hours ago, Abouzar Bahari said: If the inertial frames are symmetrical in relation to each other, as the SR argues, some paradoxes and contradictions are happened. Just to add to what was already said, there are no actual paradoxes or contradictions. Link to comment Share on other sites More sharing options...

Abouzar Bahari Posted March 1 Author Share Posted March 1 Yes, there are many contradictions. Please read the paper first. Thanks On 2/18/2023 at 8:58 PM, Bufofrog said: Just to add to what was already said, there are no actual paradoxes or contradictions. This is not as easy as you said. Have a look at these transformations first. They are not symmetrical and I have demonstrated it by mathematics. Read the paper first. Thanks. On 2/18/2023 at 7:55 PM, Genady said: almost? When going from frame A to frame B, we use in the Lorentz transformation the speed of B relative to A. When going from frame B to frame A, we use in the Lorentz transformation the speed of A relative to B. These two statements are symmetrical. In fact, I've copied and pasted the first to make the second, and then have simply replaced A by B and B by A. How more symmetrical it can be? Your mathematics is not true. First, you mention that β=v, which is not right. then, you mention γ′=γ, which is meaningless. What is meaning of γ′ or β′? did Einstein himself applied such kinds of meaningless terms himself? On 2/18/2023 at 8:58 PM, Bufofrog said: Just to add to what was already said, there are no actual paradoxes or contradictions. On 2/18/2023 at 12:46 PM, Markus Hanke said: Oh yes, they are. The Lorentz transformation leaves the metric invariant: gμν=ΛμσΛνρ=gσρ Rewrite this in matrix notation: ΛTgΛ=g Take determinant on both sides: det(ΛT)det(g)det(Λ)=det(g) Since none of these determinants is ever zero, and since the determinant of the transpose equals the determinant of the original matrix, you get: det(Λ)2=1 which implies that Λ is always invertible. Thus, inertial frames related via Lorentz transformations are always symmetric. ! Moderator Note It is against the rules of this forum to post personal theories into the main physics section, let alone onto an existing thread. If you wish to discuss this, you must open your own thread under “Speculations” and explain your thoughts there (don’t just give links). On 2/18/2023 at 12:46 PM, Markus Hanke said: Oh yes, they are. The Lorentz transformation leaves the metric invariant: gμν=ΛμσΛνρ=gσρ Rewrite this in matrix notation: ΛTgΛ=g Take determinant on both sides: det(ΛT)det(g)det(Λ)=det(g) Since none of these determinants is ever zero, and since the determinant of the transpose equals the determinant of the original matrix, you get: det(Λ)2=1 which implies that Λ is always invertible. Thus, inertial frames related via Lorentz transformations are always symmetric. ! Moderator Note It is against the rules of this forum to post personal theories into the main physics section, let alone onto an existing thread. If you wish to discuss this, you must open your own thread under “Speculations” and explain your thoughts there (don’t just give links). Your formula only imply that Lorentz transformation leaves the metric invariant. Yes, this is true. However, it does not mean that the Lorentz transformations are symmetrical. You have to refer to the main formula not to the variant, covariant or determinant of these formula to find how they are not symmetrical. Read my paper first. Thanks. -2 Link to comment Share on other sites More sharing options...

Genady Posted March 1 Share Posted March 1 13 minutes ago, Abouzar Bahari said: Your mathematics is not true. First, you mention that β=v, which is not right. then, you mention γ′=γ, which is meaningless. What is meaning of γ′ or β′? did Einstein himself applied such kinds of meaningless terms himself? Your reply is not true. I didn't mention any of that. This is a meaningless interpretation of what I said, namely: On 2/18/2023 at 12:25 PM, Genady said: When going from frame A to frame B, we use in the Lorentz transformation the speed of B relative to A. When going from frame B to frame A, we use in the Lorentz transformation the speed of A relative to B. These two statements are symmetrical. In fact, I've copied and pasted the first to make the second, and then have simply replaced A by B and B by A. How more symmetrical it can be? Unless you were replying to somebody else, you're trolling. Link to comment Share on other sites More sharing options...

Abouzar Bahari Posted March 1 Author Share Posted March 1 21 minutes ago, Genady said: Your reply is not true. I didn't mention any of that. This is a meaningless interpretation of what I said, namely: Unless you were replying to somebody else, you're trolling. Sorry, I have replied to somebody else. But, the answer to you is that, This is not as easy as you said. Have a look at these transformations first. In the Lorentz equations, the direction of motion is not important. It means, whether the S’ is moving to the right or to the left, the S observer applies the Lorentz relations without changing the sign of 𝑣. Therefore, we could argue that 𝑣 is a scalar not a vector quantity in the Lorentz relations. Hence, when these relations are inverse, only the quantity of 𝑣 is important not its direction and we must not change the sign of 𝑣. Thence, because the inverse equations are different with the main Lorentz equations, we must not consider them as symmetrical with each other. 23 minutes ago, Genady said: Your reply is not true. I didn't mention any of that. This is a meaningless interpretation of what I said, namely: Unless you were replying to somebody else, you're trolling. Please look at the images I have attached to this post from the main paper. Thanks. -2 Link to comment Share on other sites More sharing options...

Lorentz Jr Posted March 1 Share Posted March 1 (edited) v is defined as the speed of S' relative to S, given the convention that positive values are to the right (i.e. toward [math]+\infty[/math]) and negative values are to the left (toward [math]-\infty[/math]). There's nothing in the proof that requires v to be positive. 59 minutes ago, Abouzar Bahari said: we could argue that 𝑣 is a scalar not a vector quantity in the Lorentz relations. Hence, when these relations are inverse, only the quantity of 𝑣 is important not its direction and we must not change the sign of 𝑣. Nonsense. There's no law against scalars being negative. EDIT: More accurately, v is a component of a vector, and vector components are allowed to be negative. Edited March 1 by Lorentz Jr Link to comment Share on other sites More sharing options...

Abouzar Bahari Posted March 1 Author Share Posted March 1 On 2/18/2023 at 12:46 PM, Markus Hanke said: Oh yes, they are. The Lorentz transformation leaves the metric invariant: gμν=ΛμσΛνρ=gσρ Rewrite this in matrix notation: ΛTgΛ=g Take determinant on both sides: det(ΛT)det(g)det(Λ)=det(g) Since none of these determinants is ever zero, and since the determinant of the transpose equals the determinant of the original matrix, you get: det(Λ)2=1 which implies that Λ is always invertible. Thus, inertial frames related via Lorentz transformations are always symmetric. ! Moderator Note It is against the rules of this forum to post personal theories into the main physics section, let alone onto an existing thread. If you wish to discuss this, you must open your own thread under “Speculations” and explain your thoughts there (don’t just give links). Please see the images I have attached from the main paper below to this topic. Thanks. 7 minutes ago, Lorentz Jr said: v is defined as the speed of S' relative to S, given the convention that positive values are to the right (i.e. toward +∞ ) and negative values are to the left (i.e. toward +∞ ). There's nothing in the proof that requires v to be positive. No, dear. you just use the main Lorentz equations, regardless the direction of S' motion. It means, if the S' moves to the right, you use 𝑥 = 𝛾(𝑥′ - 𝑣𝑡′), and if the S' moves to the left, you use 𝑥 = 𝛾(𝑥′ - 𝑣𝑡′), again. the amount of V is important, not its direction. That is the key point. -2 Link to comment Share on other sites More sharing options...

Markus Hanke Posted March 1 Share Posted March 1 1 hour ago, Abouzar Bahari said: However, it does not mean that the Lorentz transformations are symmetrical. You have to refer to the main formula not to the variant, covariant or determinant of these formula to find how they are not symmetrical. Lorentz transformation matrices are always invertible: \[\Lambda(v) \Lambda(-v)=\Lambda \Lambda^{-1}=I\] What I have shown you in my post is one of the standard methods to formally proof this; there are many other ways to provide the same proof. Therefore, all Lorentz transformations are necessarily symmetrical. 1 hour ago, Abouzar Bahari said: They are not symmetrical and I have demonstrated it by mathematics. No you have not. Link to comment Share on other sites More sharing options...

Lorentz Jr Posted March 1 Share Posted March 1 12 minutes ago, Abouzar Bahari said: No, dear. you just use the main Lorentz equations, regardless the direction of S' motion. It means, if the S' moves to the right, you use 𝑥 = 𝛾(𝑥′ - 𝑣𝑡′), and if the S' moves to the left, you use 𝑥 = 𝛾(𝑥′ - 𝑣𝑡′), again. the amount of V is important, not its direction. That is the key point. Your key point was that the transformations are asymmetric. That point is false. Link to comment Share on other sites More sharing options...

Markus Hanke Posted March 1 Share Posted March 1 46 minutes ago, Abouzar Bahari said: Therefore, we could argue that 𝑣 is a scalar not a vector quantity in the Lorentz relations. It’s neither a scalar not is it a vector - it’s a parameter of the transformation matrix, and as such it can be positive or negative. However, this is totally irrelevant, since you need only show that the matrix itself is invertible, which is what I have done already. Link to comment Share on other sites More sharing options...

Genady Posted March 1 Share Posted March 1 (edited) 53 minutes ago, Abouzar Bahari said: Therefore, we could argue that 𝑣 is a scalar not a vector quantity in the Lorentz relations. Hence, when these relations are inverse, only the quantity of 𝑣 is important not its direction and we must not change the sign of 𝑣. Yes, v is not a vector, but it has a sign. The two observers have the same x axis. So, for one observer, the x coordinate has increased, e.g., in 1 sec from 5 to 7. The speed, v = 2. For another observer, the x coordinate has decreased in 1 sec from 4 to 2. The speed v = -2. Edited March 1 by Genady Link to comment Share on other sites More sharing options...

Abouzar Bahari Posted March 2 Author Share Posted March 2 16 hours ago, Genady said: Yes, v is not a vector, but it has a sign. The two observers have the same x axis. So, for one observer, the x coordinate has increased, e.g., in 1 sec from 5 to 7. The speed, v = 2. For another observer, the x coordinate has decreased in 1 sec from 4 to 2. The speed v = -2. If a parameter is scalar like mass, density, ... it is characterized by magnitude and have no corresponding direction. They are physical quantities that are represented solely by magnitude as well as size. So, it has not a sign (+ or -). if a parameter is a vector, like velocity, acceleration, ... it has a sign. You mentioned the true statement. Here, in the Lorentz formulas, the velocity is not a vector. So, the inverse equations for S' observer is not the same as the main Lorentz equations. in inverse equations you will have +v instead of -v. That is why they are not symmetrical equations. 16 hours ago, Lorentz Jr said: Your key point was that the transformations are asymmetric. That point is false. No, the key point is that the velocity is a scalar parameter, not a vector here. -2 Link to comment Share on other sites More sharing options...

Genady Posted March 2 Share Posted March 2 22 minutes ago, Abouzar Bahari said: Lorentz equations. in inverse equations you will have +v instead of -v. This math is not a matter of Lorentz equation or inverse equation, just geometry. The same in Galilean transformation, x' = x - vt If S' moves to the right relative to S, v is positive. If S' moves to the left, v is negative. 1 Link to comment Share on other sites More sharing options...

Markus Hanke Posted March 2 Share Posted March 2 (edited) 3 hours ago, Abouzar Bahari said: No, the key point is that the velocity is a scalar parameter, not a vector here. The parameter v is not a scalar, nor is it a vector, since it does not transform like either of those kinds of objects. It’s simply a real-valued parameter of the transformation matrix, which can take either positive or negative values. To see why, you need only consider the geometrical meaning of the general Lorentz transformation - it’s simply a combination of a boost and a hyperbolic rotation. As such, the transformation parameter can also be expressed as a hyperbolic angle (called rapidity) - and since a rotation about a point of origin can always be either clockwise or counterclockwise, the rotation angle can and does carry a sign. So it’s really simple - you start at a point A, and hyperbolically rotate your coordinate system by some angle ϕ to arrive at a new point B; you then perform the same rotation in the opposite direction, ie by the angle −ϕ , and arrive back at A. That’s just what it means for a linear transformation to be invertible (=symmetric), and that’s exactly what the Lorentz transformation does in spacetime. This is all just elementary linear algebra. 3 hours ago, Abouzar Bahari said: That is why they are not symmetrical equations. Several people here have already shown you that they are symmetrical - including a formal mathematical proof. If you choose not to believe us here, you can find different proofs of their invertibility in pretty much any decent textbook on Special Relativity; here is another online one. And here you will find a long list of experimental results that show that Lorentz invariance does indeed hold in the real world. So where do we stand with this thread? We have explained to you why the transformations are symmetrical; we have shown you formal proofs that they are symmetrical; and we have provided experimental evidence that the whole theory matches up with real-world experimental data. I think we’re done here. Edited March 2 by Markus Hanke 1 Link to comment Share on other sites More sharing options...

joigus Posted March 2 Share Posted March 2 β=v in any system of units such as light-years per year, light-seconds per second, etc. That is, any system of units in which c=1 . I thought you understood that, @Abouzar Bahari. Link to comment Share on other sites More sharing options...

Abouzar Bahari Posted March 2 Author Share Posted March 2 17 hours ago, Markus Hanke said: Lorentz transformation matrices are always invertible: Λ(v)Λ(−v)=ΛΛ−1=I What I have shown you in my post is one of the standard methods to formally proof this; there are many other ways to provide the same proof. Therefore, all Lorentz transformations are necessarily symmetrical. No you have not. 20 hours ago, Markus Hanke said: Lorentz transformation matrices are always invertible: Λ(v)Λ(−v)=ΛΛ−1=I What I have shown you in my post is one of the standard methods to formally proof this; there are many other ways to provide the same proof. Therefore, all Lorentz transformations are necessarily symmetrical. No you have not. In mathematics, this term you mentioned : Λ(v)Λ(−v)=ΛΛ^{−}^{1}=I does not mean that Λ is symmetrical. For instance, [1/8] is the reverse of [8] in 1D matrix and [1/8].[8]= I. But 1/8 is not symmetrical with 8. Λ is symmetrical when Λ(v)= Λ(−v), which is not in Lorentz equations. See the term "Parity" in physics. In 1 dimension: 𝑥′ = 𝛾(𝑥 - 𝑣𝑡) but the reverse equation is 𝑥 = 𝛾(𝑥' +𝑣𝑡′). They are not equal to each other. Therefore, they are not symmetrical. we can not change the sign of v, when we reverse the equation. Why? Because it is not a vector. It is a scalar parameter. Even if you consider it as a vector, its sign in both of above-mentioned equation remains +v because we defined it as the velocity of S’ direction to the S direction which is to the right (+). it must be mention that we are talking about physical parameters like velocity and it must be a scalar or vector. they are not just mathematics. 50 minutes ago, Markus Hanke said: The parameter v is not a scalar, nor is it a vector, since it does not transform like either of those kinds of objects. It’s simply a real-valued parameter of the transformation matrix, which can take either positive or negative values. To see why, you need only consider the geometrical meaning of the general Lorentz transformation - it’s simply a combination of a boost and a hyperbolic rotation. As such, the transformation parameter can also be expressed as a hyperbolic angle (called rapidity) - and since a rotation about a point of origin can always be either clockwise or counterclockwise, the rotation angle can and does carry a sign. So it’s really simple - you start at a point A, and hyperbolically rotate your coordinate system by some angle ϕ to arrive at a new point B; you then perform the same rotation in the opposite direction, ie by the angle −ϕ , and arrive back at A. That’s just what it means for a linear transformation to be invertible (=symmetric), and that’s exactly what the Lorentz transformation does in spacetime. This is all just elementary linear algebra. Several people here have already shown you that they are symmetrical - including a formal mathematical proof. If you choose not to believe us here, you can find different proofs of their invertibility in pretty much any decent textbook on Special Relativity; here is another online one. And here you will find a long list of experimental results that show that Lorentz invariance does indeed hold in the real world. So where do we stand with this thread? We have explained to you why the transformations are symmetrical; we have shown you formal proofs that they are symmetrical; and we have provided experimental evidence that the whole theory matches up with real-world experimental data. I think we’re done here. Your proof is completely wrong, as the others made this mistake. I have provided an easy and understandable of asymmetry of these equations. Please read my paper completely and learn about many contradictions in today's Special Relativity interpretation. I have worked on this subject more than 10 years and know all matters you have mentioned before. Please read my paper. Kind regards, and thanks for your discussion. 3 hours ago, Genady said: This math is not a matter of Lorentz equation or inverse equation, just geometry. The same in Galilean transformation, x' = x - vt If S' moves to the right relative to S, v is positive. If S' moves to the left, v is negative. No, in both of those cases you mentioned, you will use 𝑥′ = 𝛾(𝑥 - 𝑣𝑡), when you want to use Lorentz boost, without applying + or - for v sign, or else, instead of length contraction and time dilation, you will achieve length elongation and faster timing. -4 Link to comment Share on other sites More sharing options...

Genady Posted March 2 Share Posted March 2 27 minutes ago, Abouzar Bahari said: No, in both of those cases you mentioned, you will use 𝑥′ = 𝛾(𝑥 - 𝑣𝑡), when you want to use Lorentz boost, without applying + or - for v sign, or else, instead of length contraction and time dilation, you will achieve length elongation and faster timing. No, I will achieve the correct result in both cases. You are mistaken. I will not try to prove a simple arithmetic to you. Goodbye. Link to comment Share on other sites More sharing options...

Abouzar Bahari Posted March 2 Author Share Posted March 2 3 minutes ago, Genady said: No, I will achieve the correct result in both cases. You are mistaken. I will not try to prove a simple arithmetic to you. Goodbye. Do not claim. Prove it, if you can Mr. mathematician. 1 hour ago, joigus said: β=v in any system of units such as light-years per year, light-seconds per second, etc. That is, any system of units in which c=1 . I thought you understood that, @Abouzar Bahari. I knew this, but please note that c is a physical parameter, and is not a mathematical one. So, it can be lessen to 1 in mathematical formulas or Minkowski space-time, metrics, etc, but it should not be ignored or lessen to 1, when we discuss about physics or physical parameters. -3 Link to comment Share on other sites More sharing options...

Lorentz Jr Posted March 2 Share Posted March 2 (edited) 3 hours ago, Abouzar Bahari said: I have provided an easy and understandable of asymmetry of these equations. You have provided incoherent nonsense. 3 hours ago, Abouzar Bahari said: instead of length contraction and time dilation, you will achieve length elongation and faster timing. You're boring us, Mr. Bahari. Please post something funnier. 😄 Length contraction: [math]L' = {x'}_2 − {x'}_1 = \gamma(x_2 − vt)−\gamma(x_1 − vt)=\gamma(x_2 − x_1)=\gamma L[/math] Time dilation: [math]\displaystyle{\Delta t' = {t'}_2 − {t'}_1 = \gamma\left({t_2} −\frac{vx}{c^2}\right)−\gamma\left({t_1}−\frac{vx}{c^2}\right)=\gamma\left(x_2 − x_1\right)=\gamma\Delta t}[/math] No dependence on the sign of v. 😉 Edited March 2 by Lorentz Jr Link to comment Share on other sites More sharing options...

joigus Posted March 2 Share Posted March 2 1 hour ago, Abouzar Bahari said: I knew this, but please note that c is a physical parameter, and is not a mathematical one. I don't know what you mean by "it's a physical parameter, and not a mathematical one." A physical parameter, in the usual sense of the term, most definitely it is not. A physical parameter is any quantity that we can vary either freely, or subject to some specified conditions. Eg, the magnetisation of a medium of given magnetic susceptibility, etc. In the context of relativistic physics, c is a universal constant, not a parameter. Theoretically, it is derived from principles of electromagnetismf. Experimentally, it is measured. If you mean otherwise, you should say so. Because I've been studying these things in excruciating detail for many years, I can tell you you're using the poor-man's version of boosts. The grown-up version of it is, \[ \boldsymbol{x}_{\Vert}'=\frac{\boldsymbol{x}_{\Vert}-\boldsymbol{v}t}{\sqrt{1-v^{2}/c^{2}}} \] \[ ct'=\frac{ct-\boldsymbol{v}\cdot\boldsymbol{x}/c}{\sqrt{1-\left\Vert \boldsymbol{v}\right\Vert ^{2}/c^{2}}} \] \[ \boldsymbol{x}_{\bot}'=\boldsymbol{x}_{\bot} \] Where you have to decompose position 3-vector \( \boldsymbol{x} \) as, \[ \boldsymbol{x}=\boldsymbol{x}_{\Vert}+\boldsymbol{x}_{\bot} \] \[ \boldsymbol{x}_{\Vert}=\frac{\boldsymbol{x}\cdot\boldsymbol{v}}{\boldsymbol{v}\cdot\boldsymbol{v}}\boldsymbol{v} \] \[ \boldsymbol{x}_{\bot}=\boldsymbol{x}-\boldsymbol{x}_{\Vert} \] So the expression in the numerator is actually not a positive 3-scalar, but a 3-vector projection in some inertial frame. You don't understand anything, and what's worse, you don't ask. So Markus's noble attempt to help you, my attempt to close down possible loopholes, and other members' attempts to walk you through the logic of Lorentz transformations, is --most unfortunately-- to no avail. Pitty. Good day. Link to comment Share on other sites More sharing options...

Markus Hanke Posted March 3 Share Posted March 3 (edited) 19 hours ago, Abouzar Bahari said: In mathematics, this term you mentioned : Λ(v)Λ(−v)=ΛΛ^{−}^{1}=I does not mean that Λ is symmetrical. It means that Λ is invertible, which implies that the aforementioned frames A and B are symmetric, see above and below. 19 hours ago, Abouzar Bahari said: For instance, [1/8] is the reverse of [8] in 1D matrix and [1/8].[8]= I. But 1/8 is not symmetrical with 8. “Symmetry” means that you apply a transformation to an object in order to obtain a new object; and then apply the inverse transformation to the new object; you end up again with the original object. That’s exactly what you have demonstrated here - ⅛ x 8 = identity. Thank you for confirming this for us (once again). Likewise in physics - you Lorentz-transform a frame A into a frame B; and then you reverse-transform B back into A using the inverse of the original transformation matrix. That’s how symmetry is defined. Physically, it means that all inertial frames experience the same laws of physics, irrespective of their state of relative motion. 19 hours ago, Abouzar Bahari said: Λ is symmetrical when Λ(v)= Λ(−v) No, your just repeating this nonsense does not make it any less wrong. Two frames A and B are symmetric iff B=Λ(v)A A=Λ(−v)B=Λ(−v)Λ(v)A=IA Physically speaking, this means simply that all inertial frames experience the same laws of physics, irrespective of relative motion. Once again, here is the experimental evidence for that, which is clear and unambiguous. You are perfectly entitled to your own misconceptions, but not your own physical facts. 19 hours ago, Abouzar Bahari said: Your proof is completely wrong What I have shown you is elementary linear algebra, and it is not in contention by anyone except people who have agendas that are incompatible with actual science. 19 hours ago, Abouzar Bahari said: I have worked on this subject more than 10 years Then you have wasted your time, because evidently you don’t even understand simple linear algebra - or, more likely, you don’t want to understand it. That being the case, you are really not in any position to argue about SR. 18 hours ago, Abouzar Bahari said: Do not claim. Prove it, if you can Mr. mathematician. The proof has already been provided. Several times, in fact. So far as anti-relativity sentiments are concerned, this was a very underwhelming and childish attempt, I have to say. With this kind of approach you will never be taken seriously by anyone who has even just cursory knowledge of the subject matter. Needless to say you have utterly failed to convince anyone here on this forum. And honestly, given the overwhelming amount of experimental and observational evidence for SR (a small selection of which I have linked above), I will never understand why people like yourself are even wasting your time with this. You might as well argue that a round shape isn’t in fact the best shape for the wheels on a car - this debate has been settled long ago. You aren’t making any kind of valuable contribution to science, you know. Had you used those 10 years you mentioned to actually learn real physics and maths, you might have been able to contribute something of value. It’s a missed opportunity. We really are done here now. Good luck to you. Edited March 3 by Markus Hanke 3 Link to comment Share on other sites More sharing options...

Abouzar Bahari Posted March 3 Author Share Posted March 3 22 hours ago, Lorentz Jr said: You have provided incoherent nonsense. You're boring us, Mr. Bahari. Please post something funnier. 😄 Length contraction: L′=x′2−x′1=γ(x2−vt)−γ(x1−vt)=γ(x2−x1)=γL Time dilation: Δt′=t′2−t′1=γ(t2−vxc2)−γ(t1−vxc2)=γ(x2−x1)=γΔt No dependence on the sign of v. 😉 span widget When the S is stationary and S’ is moving, what you are saying is that the term x'=(x-vt) in the Lorentz formula is in the Galilean-Newtonian mechanics and does not relate to SR. OK, this is true. In Newtonian mechanics, when we use the negative quantities for v, you will use x'=(x+vt), instead. Example: x= 10 m t= 2 s v (S’ velocity) = 3 m/s x'=(x-vt)= 10-(3)(2)=4 m For v= -3 m/s --> x'=(x+vt)=10+(3)(2)=16 m However, in Lorentz transformations, we are not going to use x'=γ (x+vt) and t'=γ(t+vx/c^{2}) in the case that S’ is moving with the speed v to the left (for instance v = -10^{8} m/s) and in any papers, texts, books, etc about the SR, nobody use such terms for Lorentz equations, in the case that S’ is moving to the left. Not Lorentz, nor Einstein, nor any other people. All the times, everybody use the main formulas, whether the S’ frame is moving to the right or to the left. Have you asked of yourself, why? That is because the Lorentz formulations have been invented to consider the light speed to be constant to c when it is propagated spherically for both S and S’ frame: metrics invariance for both S and S’. Before I make an example, I will derive the Lorentz equations when a light signal is propagated spherically : 𝑥 ^{2} + 𝑦 ^{2} + 𝑧 ^{2} − 𝑐 ^{2} 𝑡^{2} = 𝑥 ′^{2} + 𝑦 ′^{2} + 𝑧 ′^{2} − 𝑐 ^{2} 𝑡 ′^{2} = 0 --> x=ct , x'=ct' x'=γ (x-vt) --> ct'=γ (ct-vt) x=γ (x'+vt' ) --> ct=γ (ct'+vt') 𝑦 = 𝑦’ 𝑧 = 𝑧’ t=γt' (1+v/c) t'=γt (1-v/c) t'=γ^2 t' (1-v/c)(1+v/c) γ=1/√(1-v^{2}/c^{2} ) For instance, suppose the S frame is stationary and the S’ frame is moving to the left with a constant velocity v = 10^{8} m/s. When the center of coordinates (zero point) of these frames is coincided with together, a light signal is propagated from this point spherically (suppose in the figure 1, S' is moving to the left instead of moving to the right). At this moment, the clocks of both S and S’ observers which have already adjusted together, start to work. We want to calculate the light coordinate in both S and S’ frames after 2 ms. In the S frame: t=2 ms x=ct=3×10^{8}×2×10^{-3}=6×10^{5 }m =600 km In the S’ frame: with employing the Lorentz transformations, we obtain: x'=γ (x-vt)=1.06 (600-10^{8}×2×10^{-3})=424 km t'=γ(t-vx/c^{2} )=1.41 s Graphically, you can find that the quantities we have achieved with these formulas for the light signal coordinate for S’ is not true quantities for positive axes of x’. However, since the light is propagated spherically, for negative axes of x’, it becomes true. I mean when the x=ct=-600 km --> x'=-424 km Therefore, we can neglect the sign of x and v in Lorentz transformations and always use the main formulas, whether S’ is moving to the right or left. But if you persist to use x'=γ (x+vt), when the S’ is going to the left, it is OK. However, when you reverse the formula to x=γ (x'-vt'), you must not to apply v reversal, as Dr. Rindler says, to make them symmetrical. If so, you would find γ=1/(1+v/c) which is less than 1 and so, your length would be elongated and time becomes faster. Now, you can go and find something funnier. 23 hours ago, joigus said: I don't know what you mean by "it's a physical parameter, and not a mathematical one." A physical parameter, in the usual sense of the term, most definitely it is not. A physical parameter is any quantity that we can vary either freely, or subject to some specified conditions. Eg, the magnetisation of a medium of given magnetic susceptibility, etc. In the context of relativistic physics, c is a universal constant, not a parameter. Theoretically, it is derived from principles of electromagnetismf. Experimentally, it is measured. If you mean otherwise, you should say so. Because I've been studying these things in excruciating detail for many years, I can tell you you're using the poor-man's version of boosts. The grown-up version of it is, x′∥=x∥−vt1−v2/c2−−−−−−−−√ ct′=ct−v⋅x/c1−∥v∥2/c2−−−−−−−−−−√ x′⊥=x⊥ Where you have to decompose position 3-vector x as, x=x∥+x⊥ x∥=x⋅vv⋅vv x⊥=x−x∥ So the expression in the numerator is actually not a positive 3-scalar, but a 3-vector projection in some inertial frame. You don't understand anything, and what's worse, you don't ask. So Markus's noble attempt to help you, my attempt to close down possible loopholes, and other members' attempts to walk you through the logic of Lorentz transformations, is --most unfortunately-- to no avail. Pitty. Good day. First, the light speed is a universal constant c = 3*10^{8} m/s. It is not equals to 1. So, when you use the c in the equations, if you want to simplify your mathematical equations, you can use c=1 instead. However, when you want it as a physical constant paprameter, you are not allowed to use c=1. For instance, in the Lorentz equations β= v/c is a number between 0 to 1. But if you put c=1, β= v/c is a number equals to v which could be much larger than 1. Therefore, it is wrong to use c=1 in the Lorentz formulas. Second, as a physicist, you are supposed to simplify the equations not to make them more complicated by copy and paste other derivations of them for electrons inside the magnetic fields and we have from the Internet. Yes, absolutely there are other derivations of the Lorentz transformations in 4 dimensions. But, I am using the simple and completely applicable original Lorentz equations only for x direction, intentionally to explain why these equations are not symmetrical: You have the equation x'=γ (x-vt), when you reverse it, you will find x=γ (x'+vt) , these two equations are not the same, so they are not symmetrical. You have the equation t'=γ(t-vx/c^{2} ), when you reverse it you will find t=γ(t'+(vx')/c^{2}), these two equations are not the same, so they are not symmetrical. We are not allowed to change the v sign, when we reverse the equation. Because we are performing just an arithmetic job, not changing the observer. Third, after a PhD in nuclear physics and Ms. And Bs. In engineering and writing many outstanding papers that published in the ISI journals and teaching several years in the university and 10 years of hard studying about the SR (at least 200 papers and 20 books) and deep thinking and learning, yes I can not understand the bullshits you add to my comments like "Λ(−v)Λ(v)A=IA means the symmetry of two equations" or γ' or other wrong statements of yours. But, what I understand is that I am talking to some young boys who ever in BS or lesser period and try to discuss with others without even study and learn their papers. This is crazy. I taught I am discussing with some literate people not some dogmatic people who try to speak fast and loudly without even studying and hearing the words of the person in front of them. If I knew, I did not discuss with you never. 7 hours ago, Markus Hanke said: It means that Λ is invertible, which implies that the aforementioned frames A and B are symmetric, see above and below. “Symmetry” means that you apply a transformation to an object in order to obtain a new object; and then apply the inverse transformation to the new object; you end up again with the original object. That’s exactly what you have demonstrated here - ⅛ x 8 = identity. Thank you for confirming this for us (once again). Likewise in physics - you Lorentz-transform a frame A into a frame B; and then you reverse-transform B back into A using the inverse of the original transformation matrix. That’s how symmetry is defined. Physically, it means that all inertial frames experience the same laws of physics, irrespective of their state of relative motion. No, your just repeating this nonsense does not make it any less wrong. Two frames A and B are symmetric iff B=Λ(v)A A=Λ(−v)B=Λ(−v)Λ(v)A=IA Physically speaking, this means simply that all inertial frames experience the same laws of physics, irrespective of relative motion. Once again, here is the experimental evidence for that, which is clear and unambiguous. You are perfectly entitled to your own misconceptions, but not your own physical facts. What I have shown you is elementary linear algebra, and it is not in contention by anyone except people who have agendas that are incompatible with actual science. Then you have wasted your time, because evidently you don’t even understand simple linear algebra - or, more likely, you don’t want to understand it. That being the case, you are really not in any position to argue about SR. The proof has already been provided. Several times, in fact. So far as anti-relativity sentiments are concerned, this was a very underwhelming and childish attempt, I have to say. With this kind of approach you will never be taken seriously by anyone who has even just cursory knowledge of the subject matter. Needless to say you have utterly failed to convince anyone here on this forum. And honestly, given the overwhelming amount of experimental and observational evidence for SR (a small selection of which I have linked above), I will never understand why people like yourself are even wasting your time with this. You might as well argue that a round shape isn’t in fact the best shape for the wheels on a car - this debate has been settled long ago. You aren’t making any kind of valuable contribution to science, you know. Had you used those 10 years you mentioned to actually learn real physics and maths, you might have been able to contribute something of value. It’s a missed opportunity. We really are done here now. Good luck to you. · Please search in the internet and find the symmetry means. If you can prove that Λ(−v)Λ(v)A=IA means the symmetry of two equations, I will accept your words, or else, please shut your mouth and first, study, then talk. · My attempt is not anti-relativity completely. I have accepted the invariant of the Lorentz equations, but, I do not agree with that they are symmetrical. · All the experimental tests agree with the above-mentioned claims. For instance, the atomic clock inside the moving high-speed airplane is dilated with the gamma factor, but the stationary atomic clock on the earth is not dilated. The researchers test the Lorentz formulas for the moving frames and get results for the Lorenz invariance of the inertial frames. However, they have ignored to test the Lorentz formulas from the view of the moving observer who thinks he is at rest according to what the SR says. Therefore, what they actually achieved is the Lorentz invariance not the symmetry of the inertial frames. · My paper after 5 month from its publication by a refereed journal was taken into account much more than I expected. In your forum, you mentors do not allow the others to speak. You just want to speak what you have learnt already in your books. But the real researchers are open- minded persons who do not think that all the matters in the books are true. They are finding the false theories. Just search how many famous scientists are completely or partially against relativity (you can find it in my paper). · About what you said I do not want to understand the negative value in the equation, read my answers to other. · But, whenever, some dogmatic people are dominant on the world of science, they could not be successful to talk and make their theories worldwide spreading. I do not want to say the journals should publish bullshits. But, I want to say the new theories if they are elaborated scientifically, must be published. Recently a published research showed that the velocity of the progress of science decreased rapidly in recent years. The reason is exactly those dogmatic people in the world of science today. Unfortunately, the people like you, when they have not enough logic in their hands (mathematical or physical logic), try to humiliate the person against them. That is the method of not literate and not civilized persons and I am so familiar with such people. I again recommend you to study more and then come and discuss with honorable people. Just search my name then open your mouth and say bullshits. -5 Link to comment Share on other sites More sharing options...

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