Anamitra Palit 9 Posted January 16 Share Posted January 16 (edited) Four Acceleration vector \begin{equation}\left(c\frac{d^2 t}{d\tau^2},\frac{d^2 x}{d\tau^2},\frac{d^2 y}{d\tau^2}, \frac{d^2 z}{d\tau^2}\right)\end{equation} (1) [[ct]=[x]=[y]=[z]] Let \begin{equation}c^2N=c^2\left(\frac{d^2 t}{d\tau^2}\right)^2-\left(\frac{d^2 x}{d\tau^2}\right)^2-\left(\frac{d^2 y}{d\tau^2}\right)^2-\left( \frac{d^2 z}{d\tau^2}\right)^2\end{equation} (2) We consider the metric \begin{equation}c^2d\tau^2=c^2dt^2-dx2-dy^2-dz^2 \end{equation} (4) \begin{equation}\Rightarrow c^2=c^2\left(\frac{dt}{d\tau}\right)^2-\left(\frac{dx}{d\tau}\right)^2-\left(\frac{dy}{d\tau}\right)^2-\left(\frac{dz}{d\tau}\right)^2\end{equation} (5) Differentiating (5) with respect to proper time we have, \begin{equation} c^2\frac{dt}{d\tau}\frac{d^2 t}{d \tau^2}-\ frac{dx}{d\tau}\frac{d^2 x}{d \tau^2}-\ frac{dy}{d\tau}\frac{d^2 y}{d \tau^2}-\ frac{dz}{d\tau}\frac{d^2 z}{d \tau^2}=0\end{equation} (6) By applying the Cauchy Schwarz inequality we have, \begin{array}{l}\left(\ frac{dx}{d\tau}\ frac{d^2 x}{d \tau^2}+\ frac{dy}{d\tau}\frac{d^2 y}{d \tau^2}+\ frac{dz}{d\tau}\frac{d^2 z}{d \tau^2}\right)^2 \\ \le \left(\left(\ frac{dx}{d\tau}\right)^2+\left(\ frac{dy}{d\tau}\right)^2+\left(\ frac{dz}{d\tau}\right)^2\right)\\ \times \left(\left(\ frac{d^2 x}{d\tau^2}\right)^2+\left(\ frac{d^2 y}{d\tau^2}\right)^2+\left(\ frac{d^2 z}{d\tau^2}\right)^2\right)\end{array} (7) \begin{equation}\left(c^2\left(\ frac{dt}{d \ tau}\right)^2-c^2\right)\left(c^2\left( \ frac {d^2 t}{d \tau^2}\right)^2-c^2N\right) \ge\left( c^4\ \frac {d^2 t}{d\tau^2}\right)^2\left(\frac{dt}{d\tau}\right)^2\end{equation} \begin{equation}\left(\left(\frac{dt}{d \tau}\right)^2-1\right)\left(\left( \frac {d^2 t}{d \tau^2}\right)^2-N\right) \ge \left( \frac {d^2 t}{d\tau^2}\right)^2\left(\frac{dt}{d\tau}\right)^2\end{equation} (8) \begin{equation} \left( \frac {d^2 t}{d\tau^2}\right)^2 \left(\frac{dt}{d\tau}\right)^2-N\left(\frac{dt}{d \tau}\right)^2-\left( \frac {d^2 t}{d \tau^2}\right)^2+N\ge \left(\frac{ \frac {d^2 t}{d\tau^2}\right)^2\left(\frac{dt}{d\tau}\right)^2\end{equation} \begin{equation} \Rightarrow -{N}\left(\frac{dt}{d \tau}\right)^2-\left( \frac {d^2 t}{d \tau^2}\right)^2+{N}\ge \0\end{equation} (9) \begin{array}{N}\left({{1}-\left\frac{dt}{d\tau}\right)^2}\right)\ge \\ \left( \frac {d^2 t}{d \tau^2}\right)^2\end{array} \begin{equation}N\left(1-\gamma^2\right)\ge \left( \frac {d^2 t}{d \tau^2}\right)^2\end{equation}(10) The right side of (10) is always positive or zero. Therefore N<=0 ,gamma[Lorentz factor] being always positive[> unity]. N cannot be positive unless the particle is moving uniformly. If N is negative then from (1) we have \begin{equation}c^2\left(\frac{d^2 t}{d\tau^2}\right)^2\le \left(\frac{d^2 x}{d\tau^2}\right)^2-\left(\frac{d^2 y}{d\tau^2}\right)^2-\left( \frac{d^2 z}{d\tau^2}\right)^2\end{equation} For a particle at rest (spatially) and N<0, \begin{equation}\left(\frac{d^2 t}{d\tau^2}\right)^2\le 0\end{equation}(11) Equation (11) will not hold, the left side being a [perfect square and hence positive or zero]unless \begin{equation}\frac{d^2 t}{d\tau^2}=0\end{equation} that is unless \begin{equation}\frac{dt}{d\tau}=constant \Rightarrow \gamma=constant\end{equation} that is unless the particle is moving with a constant velocity. An accelerating particle will not cater toy N<0. For N=0 we have from (10) \begin{equation}\left(\frac{d^2x}{d\tau^2}\right)^2\le 0 \end{equation}(12) Equation (12) is not a valid on unless the particle moves with a constant velocity. We conclude that the norm square of the acceleration vector c^2N cannot be positive, negative or zero unless the particle is moving uniformly that is unless it moves with a constant speed. Edited January 16 by Anamitra Palit For Correcting Latex codes Link to post Share on other sites

Anamitra Palit 9 Posted January 16 Author Share Posted January 16 (edited) The initial posting may be understood with difficulty. I would repost the Latex form at my earliest convenience. In the mean time the viewer may consider the uploaded file. Norm 2.pdf Edited January 16 by Anamitra Palit Link to post Share on other sites

Markus Hanke 487 Posted January 16 Share Posted January 16 4 hours ago, Anamitra Palit said: We conclude that the norm square of the acceleration vector c^2N cannot be positive, negative or zero unless the particle is moving uniformly that is unless it moves with a constant speed. The combination of coordinate chart and metric you are using as an ansatz along with your constituent relation \[\gamma=\frac{dt}{d\tau}\] already implies inertial motion from the beginning, so the end result is hardly surprising. If you want to allow for the possibility of uniformly accelerated motion, you need to use a Rindler chart; and if you want to consider arbitrarily accelerated frames, you need to use a Fermi chart to cover your spacetime, along with the appropriate constituent relationships between the components of the velocity and acceleration vectors in each chart. Link to post Share on other sites

Anamitra Palit 9 Posted January 16 Author Share Posted January 16 (edited) With Improved Latex as mentioned in the last posting[One may require to refresh the page for proper viewing] Four Acceleration \begin{equation}\left(c\frac{d^2 t}{d\tau^2},\frac{d^2 x}{d\tau^2},\frac{d^2 y}{d\tau^2}, \frac{d^2 z}{d\tau^2}\right)\end{equation} (1) Let \begin{equation}c^2N=c^2\left(\frac{d^2 t}{d\tau^2}\right)^2-\left(\frac{d^2 x}{d\tau^2}\right)^2-\left(\frac{d^2 y}{d\tau^2}\right)^2-\left( \frac{d^2 z}{d\tau^2}\right)^2\end{equation} (2) We consider the metric \begin{equation}c^2d\tau^2=c^2dt^2-dx^2-dy^2-dz^2 \end{equation} (4) \begin{equation}\Rightarrow c^2=c^2\left(\frac{dt}{d\tau}\right)^2-\left(\frac{dx}{d\tau}\right)^2- \left(\frac{dy}{d\tau}\right)^2-\left(\frac{dz}{d\tau}\right)^2 \end{equation} (5) Differentiating (5) with respect to proper time we have, \[ c^2\frac{dt}{d\tau}\frac{d^2 t}{d \tau^2}- \frac{dx}{d\tau}\frac{d^2 x}{d \tau^2}-\frac{dy}{d\tau}\frac{d^2 y}{d \tau^2}-\frac{dz}{d\tau}\frac{d^2 z}{d \tau^2}=0\] (6) By applying the Cauchy Schwarz inequality we have, \[\left(\frac{dx}{d\tau}\frac{d^2 x}{d \tau^2}+\frac{d y}{d\tau}\frac{d^2y}{d \tau^2}+\frac{dz}{d\tau}\frac{d^2 z}{d \tau^2}\right)^2 \\ \ge \left(\left(\frac{d x}{d\tau}\right)^2+\left(\frac{dy}{d\tau}\right)^2+\left(\frac{dz}{d\tau} \right)^2\right)\left(\left(\frac{d^2 x}{d \tau^2}\right)^2+\left(\frac{d^2 y}{d \tau^2}\right)^2+\left(\frac{d^2 z}{d \tau^2}\right)^2\right)\](7) or, \[\left(c^2\left(\frac{d t}{d \tau}\right)^2-c^2\right)\left(c^2\left( \frac {d^2 t}{d \tau^2}\right)^2-c^2N\right) \ge\left( c^2 \frac {d^2 t}{d\tau^2}\right)^2\left(c^2\frac{d t}{d\tau}\right)^2\] \[\left(\left(\frac{dt}{d \tau}\right)^2-1\right)\left(\left( \frac {d^2 t}{d \tau^2}\right)^2-N\right) \ge \left( \frac {d^2 t}{d\tau^2}\right)^2\left(\frac{dt}{d\tau}\right)^2\] (8) \[ \left( \frac {d^2 t}{d\tau^2}\right)^2\left(\frac{dt}{d\tau}\right)^2-N\left(\frac{dt}{d \tau}\right)^2-\left( \frac {d^2 t}{d \tau^2}\right)^2+N\ge \left( \frac {d^2 t}{d\tau^2}\right)^2\left(\frac{dt}{d\tau}\right)^2\] \[ \Rightarrow-N\left(\frac{dt}{d \tau}\right)^2-\left( \frac {d^2 t}{d \tau^2}\right)^2+N\ge 0\] (9) \[N\left(1-\left(\frac{dt}{d\tau}\right)^2\right)\ge \left( \frac {d^2 t}{d \tau^2}\right)^2\] \[N\left(1-\gamma^2\right)\ge \left( \frac {d^2 t}{d \tau^2}\right)^2\](10) The right side of (10) is always positive or zero. Therefore for N<=0 has to hold for the left side ,gamma[Lorentz factor] being always positive[> =unity]. N cannot be positive unless the particle is moving uniformly. If N is negative then from (1) we have \[c^2\left(\frac{d^2 t}{d\tau^2}\right)^2\le \left(\frac{d^2 x}{d\tau^2}\right)^2+\left(\frac{d^2 y}{d\tau^2}\right)^2+\left( \frac{d^2 z}{d\tau^2}\right)^2\] For a particle at rest (spatially) and N<0, \[\left(\frac{d^2 t}{d\tau^2}\right)^2\le 0\](11) Equation (11) will not hold, the left side being a [perfect square and hence positive or zero]unless \begin{equation}\frac{d^2 t}{d\tau^2}=0\end{equation} that is unless \begin{equation}\frac{dt}{d\tau}=constant \Rightarrow \gamma=constant\end{equation} that is unless the particle is moving with a constant velocity. An accelerating particle will not cater to N<0. For N=0 we have from (10) \[\left(\frac{d^2x}{d\tau^2}\right)^2\le 0 \](12) Equation (12) is not a valid on unless the particle moves with a constant velocity. We conclude that the norm square of the acceleration vector c^2N cannot be positive, negative or zero unless the particle is moving uniformly that is unless it moves with a constant velocity @ Markus Hanke: In special Relativity the frames of reference are inertial and hence they are in uniform relative motion. But the particles in a given inertial frame of reference[and hence in other frames of reference,inertial ]can be in a state of acceleration. The momentum of a particle is given by \[ p_i=m\frac{dx^i}{dt}=m_0\gamma\frac{dx^i}{dt}=m_0 \frac{dx}{d\tau}\] since \[\gamma=\frac {dt}{d \tau}\] The \[\gamma\] in the last line is the particle's gamma and it is expected to be a variable if the particle is in a state of acceleration. But in this article we see that the particle cannot accelerate. Edited January 16 by Anamitra Palit Link to post Share on other sites

Markus Hanke 487 Posted January 16 Share Posted January 16 (edited) 50 minutes ago, Anamitra Palit said: In special Relativity the frames of reference are inertial and hence they are in uniform relative motion. Not necessarily true. A region of spacetime being special relativistic means only that the Riemann tensor vanishes everywhere within that region, so that spacetime is flat. It does not mean that all frames necessarily need to be inertial, or that there is only uniform relative motion. You can have non-inertial frames in flat spacetime, so SR is perfectly capable of handling acceleration. 50 minutes ago, Anamitra Palit said: But the particles in a given inertial frame of reference[and hence in other frames of reference,inertial ]can be in a state of acceleration. Not true. If a frame is inertial, then by definition there is no proper acceleration. There can, however, be coordinate acceleration, but that is an artefact of how we choose to label events, and not something that a physical accelerometer would measure. 50 minutes ago, Anamitra Palit said: in the last line is the particle's gamma and it is expected to be a variable if the particle is in a state of acceleration. This is incompatible with (1) and (2) in your last post. An accelerated particle undergoes hyperbolic motion in Minkowski spacetime, so you can’t just make the gamma factor variable, but retain a Minkowski coordinate chart (with proper time) and its corresponding metric. I suggest what you should do is start with the correct metric ansatz - i.e. a Rindler or Fermi metric -, and work through it again, and see what result you then get. Edited January 16 by Markus Hanke Link to post Share on other sites

Anamitra Palit 9 Posted January 17 Author Share Posted January 17 (edited) In relation to the last comment made by Markus Hanke [One may have to refresh the page for viewing the formulas] We may have accelerating frames in Flat Space time.That is right.But in my discussion I used the relation(metric) \[c^2d\tau^2=c^2dt^2-dx^2-dy^2-dz^2\] (1) The metric represented by (1) follows from the Lorentz Transformations. We can also derive the Lorentz transformations from the metric that is from (1)[Steven Weinberg, Gravitation and Cosmology, Chapter 2: Special Relativity]. Equation (1) characterizes Special Relativity When I talked of Special Relativity , I meant the effect of the metric represented by(1). The Lorentz transformations consider frames in relative uniform motion. The metric represented by (1):has it got anything to do with frames in relative acceleration which are possible in flat space time?What are the transformation laws there? Special Relativity and flat space time are not identical objects. Flat space time may have reference frames with relative acceleration between hem. This is not possible with Special Relativity because Lorentz Transformations consider only such frames of reference as have uniform relative motion between them[only inertial frames are considered in Special Relativity] \[\frac{dt}{d\tau}\] considered in the article is not the gamma between the reference frames. It is the gamma for the particle. \[\gamma=\frac{1}{\sqrt{1-\frac{v_p^2}{c^2}}}\] (2)where v_p is the particle velocity and not the relative velocity between reference frames. A particles motion in a specified reference frame may be considered independent of other reference frames and the relative velocity between them. Recalling (1) and dividing both sides by dtau^2 we obtain \[c^2=c^2v_t^2-v_x^2-v_y^2-v_z^2\] (3) The velocities are variable in respect of time.The term v_t relates to energy and energy of an accelerating particle is not constant. \[\left(v_t,v_x,v_y ,v_z\right)\] are variable quantities and not constants \[E=mc^2=m_0 \gamma c^2\](3) gamma in (3) is a variable gamma corresponding to the motion of the particle in a given frame of reference. The Lorentz transformations involve the relative velocity of separation between the two frames of reference and a different gamma[different from the particle gamma]. Both disappear when we formulate (1) from them: \[c^2dt^2-dx^2-dy^2-dz^2= c^2dt’^2-dx’^2-dy’^2-dz’^2\] (4) Each side of (2) is denoted by \[c^2d\tau^2\] Again from (2),by considering the invariance of proper time we may derive the Lorentz transformations uniquely : and gamma[between the two reference frames as distinct from the particle gamma] pops up . On Particle gamma: From (1) \[c^2 \left(\frac {dt}{d\tau}\right)^2=c^2-\left(\frac{dx}{d\tau} \right)^2-\left(\frac{dy}{d\tau} \right)^2-\left(\frac{dz}{d\tau} \right)^2\] \[c^2 \left(\frac {dt}{d\tau}\right)^2=c^2-v_x^2-v_y^2-v_z^2\] \[ \left(\frac {dt}{d\tau}\right)^2=c^2-v^2 \] \[ \frac {dt}{d\tau}=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\](5) \[v^2=v_x^2+v_y^2+v_z^2\] is the particle velocity. The particle could at rest at the origin of one of the reference frames when is motion with respect to other reference frames in terms of dt,dx ,dy and dz when referred to (5) represents the gamma between the reference frames. Else it is the particle gamma.The invariance expressed by (4) does not sand in the way of the velocity components being variable.For example \[ dx’=\gamma\left(dx-vdt\right)\] \[ \Rightarrow \frac{dx’}{d\tau}=\gamma\left(\frac{dx}{d\tau}-v\frac{dt}{d\tau}\right)\] The constancy of v does not stand in the way of dx/dtau ,dt/dtau and dx’/dtau being variables. Edited January 17 by Anamitra Palit Link to post Share on other sites

Anamitra Palit 9 Posted January 17 Author Share Posted January 17 In relation to the last posting one should take note of the following facts: We may consider two close events (t,x,y,z) and (t+dt,x+dx,y+dy,z+dz) on the particle's world line[space time path] and apply the Lorentz transformation on the infinitesimals dt, dx, dy and dz [asides t,x,y and z]. The reference frames involved have to be inertial obviously with uniform relative motion between themselves for the application of the Lorentz transformation. For that it is not necessary that the quantities dt/dtau,dx/dtau,dy/dtau and dz/tau have to be constants.In general they are variables. Flat space time, of course ,is supportive to relative acceleration between reference frames.If Special Relativity is identified with the Lorentz transformations it does not support acceleration between reference frames [for the transformations] Link to post Share on other sites

md65536 383 Posted January 17 Share Posted January 17 10 hours ago, Markus Hanke said: Not true. If a frame is inertial, then by definition there is no proper acceleration. There can, however, be coordinate acceleration, but that is an artefact of how we choose to label events, and not something that a physical accelerometer would measure. I'm not following this. An accelerometer that is measuring proper acceleration can be described in the coordinates of an inertial observer. My reading of what you replied to, is "An inertial observer remains inertial as measured in any other inertial reference frame, but a particle can (properly) accelerate as measured in an inertial reference frame." It only mentions particles being able to accelerate. What are you referring to when you say "there is no proper acceleration"? Link to post Share on other sites

Markus Hanke 487 Posted January 17 Share Posted January 17 4 hours ago, md65536 said: I'm not following this. An accelerometer that is measuring proper acceleration can be described in the coordinates of an inertial observer. Yes, proper acceleration (unlike coordinate acceleration) is a quantity that everyone agrees on, so the coordinate basis does not matter. 4 hours ago, md65536 said: What are you referring to when you say "there is no proper acceleration"? I was referring to the fact that, if an accelerometer co-moving with an observer measures something other than zero, then that observer cannot be inertial by definition. I understood the OPs comment to say that a particle can be in a state of proper acceleration, yet still be inertial - hence my comment. 5 hours ago, Anamitra Palit said: has it got anything to do with frames in relative acceleration which are possible in flat space time?What are the transformation laws there? If you want to use transformation laws, then those can be found on this page (scroll down to the relations shown in red and blue). As mentioned before, accelerated particles will undergo hyperbolic motion in Minkowski spacetime, so the coordinate transformations use hyperbolic functions and their inverses. Alternatively, and that’s what I have been suggesting above, you can just adopt an appropriate coordinate chart and metric from the beginning. The Rindler metric is the most commonly used, but it is valid only for constant accelerations; for arbitrary accelerations you can use Fermi coordinates (e.g.). Link to post Share on other sites

md65536 383 Posted January 17 Share Posted January 17 14 hours ago, Markus Hanke said: I was referring to the fact that, if an accelerometer co-moving with an observer measures something other than zero, then that observer cannot be inertial by definition. I understood the OPs comment to say that a particle can be in a state of proper acceleration, yet still be inertial - hence my comment. That's not what OP literally said, and you've interpreted what OP wrote as nonsense, but still I don't know what OP really meant because it could be interpreted different ways. I don't think you two are talking about the same thing, at least in some cases like this, and I don't see how the problems can possibly be resolved if you're not even talking about the same things. I think OP needs to clarify first. Eg. in this case, Anamitra were you talking about general particles, as they are measured in various inertial frames of reference? Or particles in their own rest frames, being inertial in some inertial frames of reference and accelerating in other inertial frames of reference? Or something else? I think some consider a particle being "in" a given frame to mean that the frame is its rest frame, but I take it to mean the particle "as measured in" a given frame. Eg. a train can have a positive speed in the bank's frame. The train is in all the different frames, not just its rest frame. Is that at odds with your interpretation of OP's statement? Link to post Share on other sites

Markus Hanke 487 Posted January 18 Share Posted January 18 11 hours ago, md65536 said: That's not what OP literally said It’s how I understood it, based on the fact that the metric given is of a form that would generally be used by an inertial observer, so it is natural to assume that these are Minkowski coordinates, and not hyperbolic ones. The geodesics calculated from his metric ansatz are straight lines, not hyperbolas - unless the coordinate basis is not Cartesian, but the OP never indicated that. This is also consistent with the OP’s conclusion: “[...] we see that the particle cannot accelerate”, which is of course trivially true, based on that metric. Link to post Share on other sites

md65536 383 Posted January 18 Share Posted January 18 8 hours ago, Markus Hanke said: It’s how I understood it, based on the fact that the metric given is of a form that would generally be used by an inertial observer, so it is natural to assume that these are Minkowski coordinates, and not hyperbolic ones. The geodesics calculated from his metric ansatz are straight lines, not hyperbolas - unless the coordinate basis is not Cartesian, but the OP never indicated that. This is also consistent with the OP’s conclusion: “[...] we see that the particle cannot accelerate”, which is of course trivially true, based on that metric. Then I still don't understand. Particles *can* accelerate, and their motion can be described in the coordinates of an inertial observer. Why do you need an accelerating observer to describe accelerating particles? It can be described in Minkowski coordinates, why suggest Rindler? Not all particles are observers; to me it looks like where OP is talking about particles, you're describing observers. "in this article we see that the particle cannot accelerate." (emphasis mine) I understand to mean that we're talking about the case of constant gamma, ie. we're limiting ourselves to particles that don't accelerate. If you two are really on the same page and understanding each other, then I apologize. (Besides all that, a particle can accelerate and maintain constant speed relative to an inertial observer, thus constant gamma in that observer's inertial frame but only changing direction, while the particle undergoes proper acceleration, so I think I disagree with both of you however your statements are understood!) Link to post Share on other sites

Markus Hanke 487 Posted January 19 Share Posted January 19 14 hours ago, md65536 said: If you two are really on the same page and understanding each other, then I apologize. I am uncertain what it is really is that the OP is trying to show, so I can’t guarantee this. What I presented is my own understanding of what he has posted. 14 hours ago, md65536 said: Particles *can* accelerate, and their motion can be described in the coordinates of an inertial observer. Yes of course, but that isn’t what the OP has been doing. That is kind of my point. He’s essentially using an inertial coordinate system to show that there is no acceleration - which is trivially true. 14 hours ago, md65536 said: "in this article we see that the particle cannot accelerate." (emphasis mine) I understand to mean that we're talking about the case of constant gamma, ie. we're limiting ourselves to particles that don't accelerate. Ok, but then, what is his point? He starts with a metric in Cartesian coordinates, then manipulates it using relations that imply an inertial observer, and ends up with the conclusion that there is no acceleration...? 14 hours ago, md65536 said: It can be described in Minkowski coordinates, why suggest Rindler? I’m not sure I understand your question. Both charts cover the same spacetime, so the difference is merely one of coordinate basis. Rindler is what you naturally get when you apply the proper transformations that arise in a frame with non-zero constant acceleration (since such frames are not related by Lorentz transformations): \[ct\rightarrow \ \left(\frac{c^{2}}{a} +x’\right) sinh\left(\frac{at’}{c}\right) \] \[x\rightarrow \ \left(\frac{c^{2}}{a} +x’\right) cosh\left(\frac{at’}{c}\right) -\frac{c^{2}}{a} \] The advantage here is that the world line of such a uniformly accelerated particle is one of constant x (in this coordinate basis!) as t “ticks along”, so in some sense a particle in this coordinate frame is “at rest”, even though it is uniformly accelerating. It’s the most natural choice for uniformly accelerated motion. Very simply put, the line element \(ds^2=...\) does exactly what it says on the tin - it is an infinitesimal section (element) of a curve in spacetime. To obtain the entire geometric length of a curve C, you integrate: \[L=\int _{C} ds\] To describe a uniformly accelerated particle, you can do one of two things: 1. Keep the line element in Minkowski coordinates, but make C a hyperbola to describe the motion 2. Use the line element in Rindler coordinates, which makes C a “straight line” in these coordinates Of course, the result will be the same (since this integral is an invariant), so both are valid, but the computational effort differs - it turns out that option (2) is very much easier to do. It’s simply the more natural choice - in much the same way as (e.g.) spherical coordinates are the natural choice to describe the surface of a sphere, rather than Cartesian coordinates. 1 Link to post Share on other sites

md65536 383 Posted January 19 Share Posted January 19 9 hours ago, Markus Hanke said: He’s essentially using an inertial coordinate system to show that there is no acceleration - which is trivially true. No acceleration of what? It originally mentions and later clarifies that the acceleration of "particles" is being discussed, and that's not trivially true. If you're assuming it's reference frames being accelerated, where is that stated? This topic is confusing from the start, with seemingly some explanation or details missing?? It's difficult for me to follow if I first have to guess at the same assumptions being made, even if they're reasonable. 9 hours ago, Markus Hanke said: Ok, but then, what is his point? He starts with a metric in Cartesian coordinates, then manipulates it using relations that imply an inertial observer, and ends up with the conclusion that there is no acceleration...? Yes, that seems to explain what is going on, thanks. I did not get that from the discussion so far. Basically is it: specify a particle that has no acceleration, conclude that it cannot be changing speed? (I'm guessing then the main flaw in reasoning is that OP starts with equations of inertial motion, but is treating them as "the equations of general motion in SR"?) Link to post Share on other sites

Markus Hanke 487 Posted January 20 Share Posted January 20 13 hours ago, md65536 said: No acceleration of what? It originally mentions and later clarifies that the acceleration of "particles" is being discussed, and that's not trivially true. If you're assuming it's reference frames being accelerated, where is that stated? As I said previously, my response was based upon my own understanding of the OP. If it completely missed the point, then it is the OP’s job to clarify things. Unfortunately this poster seems to be in the habit of opening threads, and then abandoning them after a couple of comments are made. 13 hours ago, md65536 said: I'm guessing then the main flaw in reasoning is that OP starts with equations of inertial motion, but is treating them as "the equations of general motion in SR"? Yes that’s pretty much it. Accelerated motion involves a little more than simply allowing gamma to vary. Link to post Share on other sites

md65536 383 Posted January 21 Share Posted January 21 On 1/20/2021 at 1:23 AM, Markus Hanke said: As I said previously, my response was based upon my own understanding of the OP. If it completely missed the point, then it is the OP’s job to clarify things. Okay, but I'm trying to understand a statement that you made, because it makes no sense to me and I'm trying to figure out where the misunderstanding is. You wrote, "He’s essentially using an inertial coordinate system to show that there is no acceleration - which is trivially true." That makes NO sense to me, because you can describe a particle that's accelerating, using an inertial coordinate system. The coordinate system has nothing to do with whether the particle is properly accelerating or not. Eg. Consider an inertial observer on a train bank. A train accelerates from rest at a rate of 1 m/s^2, relative to the observer. The observer's coordinate system doesn't trivially show that there is no acceleration. Am I using the term "coordinate system" incorrectly? What I don't understand is if one can refer to a particle, and its own coordinate system, as the same thing, or if you and Anamitra are talking about different things. If I have an inertial observer and a train accelerates, I don't think I can sensibly describe it as "an accelerating coordinate system". I'm trying to find out what you're saying is trivially true. Link to post Share on other sites

Markus Hanke 487 Posted January 22 Share Posted January 22 20 hours ago, md65536 said: That makes NO sense to me, because you can describe a particle that's accelerating, using an inertial coordinate system. Yes, you can indeed - the result is the hyperbolic transformation I gave earlier. This just doesn’t seem to be what the OP was doing or getting at - and if it is, then the conclusion he reaches is of course meaningless (“in this article we see that the particle cannot accelerate”). It is, however, correct and fully consistent (albeit trivial) with the absence of proper acceleration in an inertial frame. Hence my reading. Link to post Share on other sites

Kino 8 Posted January 22 Share Posted January 22 (edited) Hasn't the OP just proved (in a rather long-winded way) that the inner product of four acceleration and four velocity is zero and, hence, that if the four velocity is \( V^t=c \), \( V^x=V^y=V^z=0 \) then the four acceleration has \( A^t=0 \) (and that the four acceleration is spacelike if it isn't zero)? They just appear not to have noticed that in general \( \frac{dA^t}{d\tau} \neq 0 \), so the conclusion that \( \gamma=\mathrm{const} \) does not follow. The fact that a function is instantaneously zero does not mean that either its integral or its derivative need be zero. To me, the argument looks similar to saying that in circular motion in the x,y plane there comes a point where \( (v_x,v_y)=(v,0) \), and at that point \( a_x=0 \), and hence \( v_x \) can never change and circular motion is impossible. Obviously that's nonsense, and realising that \( \frac{da_x}{dt}\neq 0 \) is a part of understanding why it's nonsense. Edited January 22 by Kino LaTeXnichal errors and other minor corrections 3 Link to post Share on other sites

joigus 370 Posted January 23 Share Posted January 23 (edited) 1 hour ago, Kino said: Hasn't the OP just proved (in a rather long-winded way) that the inner product of four acceleration and four velocity is zero and, hence, that if the four velocity is Vt=c , Vx=Vy=Vz=0 then the four acceleration has At=0 (and that the four acceleration is spacelike if it isn't zero)? They just appear not to have noticed that in general dAtdτ≠0 , so the conclusion that γ=const does not follow. The fact that a function is instantaneously zero does not mean that either its integral or its derivative need be zero. To me, the argument looks similar to saying that in circular motion in the x,y plane there comes a point where (vx,vy)=(v,0) , and at that point ax=0 , and hence vx can never change and circular motion is impossible. Obviously that's nonsense, and realising that daxdt≠0 is a part of understanding why it's nonsense. You may be right. I'm not following this thread very closely, and I'm not sure if what you say is what the OP is trying to prove. But here's a more standard proof: This is obvious, but let's do a check. And gammas, of course, are in general time-dependent. SR can deal with accelerations, as Markus said. The 4-vectors are, and their 4-product is, It's necessary to keep in mind that, The derivative of the gamma is, So the 4-product is indeed identically zero: As Markus also said, the concept that in SR supersedes constant acceleration is that of hyperbolic motion. He also has been very careful to distinguish flat space-time from prescription to adopt inertial frames. Indeed, the Minkowski spacetime can be studied in terms of Rindler charts --hyperbolically accelerated frames--. It is not difficult to show that when the motion is completely collinear (spacial 3-vectors of velocity, acceleration, and force). It's in wikipedia, although the proof is not complete, and I can provide a completion, if anyone's interested. As I said, I'm not completely sure that what I'm saying is relevant to the discussion. It is the standard, reliable, mainstream formalism that we know and love. Edited January 23 by joigus minor correction 1 Link to post Share on other sites

Markus Hanke 487 Posted January 23 Share Posted January 23 (edited) 7 hours ago, Kino said: Hasn't the OP just proved (in a rather long-winded way) that the inner product of four acceleration and four velocity is zero and, hence, that if the four velocity is Vt=c , Vx=Vy=Vz=0 then the four acceleration has At=0 (and that the four acceleration is spacelike if it isn't zero)? Quite possibly so! To be honest, I am still struggling to understand what it is the OP is actually trying to do. It also seems he has abandoned this thread, just like all the other ones he opened before this. 7 hours ago, Kino said: They just appear not to have noticed that in general dAtdτ≠0 , so the conclusion that γ=const does not follow. The fact that a function is instantaneously zero does not mean that either its integral or its derivative need be zero. To me, the argument looks similar to saying that in circular motion in the x,y plane there comes a point where (vx,vy)=(v,0) , and at that point ax=0 , and hence vx can never change and circular motion is impossible. Obviously that's nonsense, and realising that daxdt≠0 is a part of understanding why it's nonsense. Indeed - very well summarised! +1 Edited January 23 by Markus Hanke 1 Link to post Share on other sites

joigus 370 Posted January 23 Share Posted January 23 (edited) Yeah, nice insight. \( \gamma^{-1} \frac{d\gamma}{d\tau} \) doesn't have to be zero, even though \( \gamma^{-1} \frac{d\gamma}{d\tau} - \gamma^{-1} \frac{d\gamma}{d\tau} \) is identically zero. Welcome to the forums, @Kino. Edited January 23 by joigus 1 Link to post Share on other sites

Kino 8 Posted January 23 Share Posted January 23 (edited) 3 hours ago, joigus said: Welcome to the forums, @Kino. Thanks! I think the easiest way to make the point is actually to extend your one liner. \( c^2=\eta_{\mu\nu}U^\mu U^\nu \) by definition, and hence differentiating with respect to \( \tau \) we get \( 0=\eta_{\mu\nu}A^\mu U^\nu+\eta_{\mu\nu}U^\mu A^\nu=2\eta_{\mu\nu}U^\mu A^\nu \). This equation is obviously satisfied if \( A \) is a zero vector. If it's not a zero vector, we simply note that for any \( U \) there exists a frame in which it represents a state of rest, and hence in that frame the only non-zero component is \( U^t=c \). In that frame \( 0=\eta_{\mu\nu}U^\mu A^\nu \) simplifies to \( A^t=0 \), but the spacelike components of \( A \) are not restricted. Thus in this frame \( \eta_{\mu\nu}A^\mu A^\nu=-((A^x)^2+(A^y)^2+(A^z)^2) \) is obviously negative, or zero iff the vector is a zero vector and, since the left hand side is manifestly covariant, it's negative in any coordinate system. I think this is everything useful that @Anamitra Palit expressed in equations 1-11 and the un-numbered equation immediately afterwards, just derived with much less mess. But you can differentiate \( 0=\eta_{\mu\nu}U^\mu A^\nu \) with respect to \( \tau \) again, giving \( 0=\eta_{\mu\nu}(A^\mu A^\nu+U^\mu J^\nu) \), where \( J^\nu=dA^\nu/d\tau=d^3t/d\tau^3 \) is the "four jerk". In the frame where \( U \) represents "at rest", this simplifies to \( cJ^t=\eta_{\mu\nu}A^\mu A^\nu \), which is manifestly non-zero if the four-acceleration is non-zero. So the OP is simply wrong when they state in the un-numbered equation before (12) than \( \gamma = \mathrm{const} \). That would only follow from \( d^2t/d\tau^2=0 \) if that were true for all \( \tau \) (which it isn't in general) or if \( d^nt/d\tau^n=0 \) for all \( n\geq 2\) (which isn't true in general). 8 hours ago, Markus Hanke said: It also seems he has abandoned this thread, just like all the other ones he opened before this. I see what you mean. The ones where I can see the maths all seem to have an error, too. By the way, is there any way to make LaTeX render in the preview? I still see the source, although it seems to render when posted. Edited January 23 by Kino Link to post Share on other sites

joigus 370 Posted January 23 Share Posted January 23 55 minutes ago, Kino said: By the way, is there any way to make LaTeX render in the preview? I still see the source, although it seems to render when posted. Not that I know of. You do have a section to test your LaTeX beforehand, as you already know: https://www.scienceforums.net/forum/99-the-sandbox/ 1 Link to post Share on other sites

Kino 8 Posted January 23 Share Posted January 23 (edited) 5 minutes ago, joigus said: Not that I know of. You do have a section to test your LaTeX beforehand, as you already know: https://www.scienceforums.net/forum/99-the-sandbox/ Thanks. It looks like you get an hour to edit, so it's not a big problem. Edited January 23 by Kino Link to post Share on other sites

Markus Hanke 487 Posted January 24 Share Posted January 24 14 hours ago, Kino said: By the way, is there any way to make LaTeX render in the preview? Unfortunately not, at least not that I know of. It’s a little bug bear of mine too, but not a massive issue. Link to post Share on other sites

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