User talk:GHT153

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Talk:Lorentz transformation‎
Replied, happy editing. M&and;Ŝc2ħεИτlk 23:45, 23 May 2013 (UTC)

Article talk pages are for discussing the article, not the subject
Welcome to Wikipedia and thank you for your contributions. I am glad to see that you are discussing a topic. However, as a general rule, talk pages such as Talk:Lorentz transformation are for discussion related to improving the article, not general discussion about the topic. If you have specific questions about certain topics, consider visiting our reference desk and asking them there instead of on article talk pages. Thank you. - DVdm (talk) 08:29, 28 May 2013 (UTC)

Please stop using talk pages such as Talk:Lorentz transformation for general discussion of the topic. They are for discussion related to improving the article; not for use as a forum or chat room. If you have specific questions about certain topics, consider visiting our reference desk and asking them there instead of on article talk pages. See here for more information. Thank you. - DVdm (talk) 07:53, 27 January 2014 (UTC)

Please stop your disruptive editing. If you continue to use talk pages for inappropriate discussion, as you did at Talk:Lorentz transformation, you may be blocked from editing. - DVdm (talk) 07:41, 22 May 2014 (UTC)

Note about your user page
Hi Gerd, I was just reading your user page User:GHT153 and I notice a little mistake. Per our wp:talk page guidelines we can of course not discuss this on article talk pages, but we can do it safely here, if you like. Perhaps the following helps.

Instead of 
 * Substituting $$x=x_{+}=ct$$ for the first light front we get
 * $$x_{+}'=\gamma\left(x_{+}-vt\right)=\gamma\left(ct-vt\right)=\gamma\left(1-\frac{v}{c}\right)ct=\gamma\left(1-\frac{v}{c}\right)x_{+}$$
 * $$t'=\gamma\left(t-\frac{vx_{+}}{c^2}\right)=\gamma\left(t-\frac{vct}{c^2}\right)=\gamma\left(t-\frac{vt}{c}\right)=\gamma\left(1-\frac{v}{c}\right)t$$
 * with the common factor
 * $$\gamma\left(1-\frac{v}{c}\right)=\frac{1-\frac{v}{c}}{\sqrt{1 - \frac{v^2}{c^2}}}=\frac{1-\frac{v}{c}}{\sqrt{(1 + \frac{v}{c})(1 - \frac{v}{c})} }=\frac{\sqrt{1-\frac{v}{c}}}{\sqrt{1 + \frac{v}{c}}}<1$$
 * From these equations special relativity tells us that, for a given time $$t$$, the distance the light front travels is shorter than in the system $$S(x,t)$$, and time runs slower by the same factor, resulting in a constant speed of light $$\Delta x/\Delta t=\Delta x'/\Delta t'=c$$ in both systems.

you should write: 
 * From these equations special relativity tells us that, for a given time $$t$$, as the origin of the system $$S'(x',t')$$ is chasing the front of the ("positive") light signal, the distance the light front travels is shorter than in the system $$S(x,t)$$, and it takes less time by the same factor for the signal front to cover that shorter distance, resulting in a constant speed of light $$\Delta x/\Delta t=\Delta x'/\Delta t'=c$$ in both systems.
 * From these equations special relativity tells us that, for a given time $$t$$, as the origin of the system $$S'(x',t')$$ is chasing the front of the ("positive") light signal, the distance the light front travels is shorter than in the system $$S(x,t)$$, and it takes less time by the same factor for the signal front to cover that shorter distance, resulting in a constant speed of light $$\Delta x/\Delta t=\Delta x'/\Delta t'=c$$ in both systems.

And of course, likewise, instead of 
 * Now substituting $$x=x_{-}=-ct$$ for the second light front we get
 * $$x_{-}'=\gamma\left(x_{-}-vt\right)=\gamma\left(-ct-vt\right)=-\gamma\left(1+\frac{v}{c}\right)ct=\gamma\left(1+\frac{v}{c}\right)x_{-}$$
 * $$t'=\gamma\left(t - \frac{vx_{-}}{c^2}\right)=\gamma\left(t+\frac{vct}{c^2}\right)=\gamma\left(t+\frac{vt}{c}\right)=\gamma\left(1+\frac{v}{c}\right)t$$
 * with the common factor
 * $$\gamma\left(1+\frac{v}{c}\right)=\frac{1+\frac{v}{c}}{\sqrt{1 - \frac{v^2}{c^2}}}=\frac{1+\frac{v}{c}}{\sqrt{(1 + \frac{v}{c})(1 - \frac{v}{c})} }=\frac{\sqrt{1+\frac{v}{c}}}{\sqrt{1-\frac{v}{c}}}>1$$
 * From these equations special relativity tells us now that, for a given time $$t$$, the distance the light front travels is longer than in the system $$S(x,t)$$, and time runs faster by the same factor, resulting in a constant speed of light $$\Delta x/\Delta t=\Delta x'/\Delta t'=-c$$ in both systems.

you should write: 
 * From these equations special relativity tells us that, for a given time $$t$$, as the origin of the system $$S'(x',t')$$ is receding from the front of the ("negative") light signal, the distance the light front travels is longer than in the system $$S(x,t)$$, and it takes more time by the same factor for the signal front to cover that longer distance, resulting in a constant speed of light $$\Delta x/\Delta t=\Delta x'/\Delta t'=-c$$ in both systems.
 * From these equations special relativity tells us that, for a given time $$t$$, as the origin of the system $$S'(x',t')$$ is receding from the front of the ("negative") light signal, the distance the light front travels is longer than in the system $$S(x,t)$$, and it takes more time by the same factor for the signal front to cover that longer distance, resulting in a constant speed of light $$\Delta x/\Delta t=\Delta x'/\Delta t'=-c$$ in both systems.

Imagine that the light signals each trigger an explosion, the positive signal at event $$(x_{+},t)$$ and the negative signal at event $$(x_{-},t)$$, both as measured in the system $$S(x,t)$$. As these explosion events happen at different locations and are simultaneous in system S, they are of course not simultaneous in system S'. That does not mean that "time runs at different rates at the same time". It just means that the explosion events are not simultaneous in system S'. See Special relativity. Hope this helps. - DVdm (talk) 14:35, 5 May 2015 (UTC)


 * Hi Dirk, the equations are very simple. No need to play around with light sensitive explosives.


 * Following special relativity a light front, described by $$x=x_{+}=ct$$ in the system S, travels in the same time interval a shorter distance, i.e. $$x_{+}'=\gamma\left(1-v/c\right)ct$$, and time runs slower by the same factor, i.e. $$t'=\gamma(1-v/c)t$$. ...


 * That's what the equations say and nothing else. The "little mistake" is at your side. GHT153 (talk) 22:22, 20 May 2015 (UTC)


 * Well, I did my very best to carefully explain the meanings of the variables in the equations and the physics behind them. Apparently that didn't help much, so never mind. Sorry for perhaps having wasted your time. - DVdm (talk) 03:56, 21 May 2015 (UTC)

Final warning
You may be blocked from editing without further warning the next time you use talk pages for inappropriate discussions, as you did at Talk:Relativistic Doppler effect. Referring to this comment. - DVdm (talk) 06:59, 30 June 2016 (UTC)

Renewal of final warning
Please see. Also note that you are not allowed to make edits in logged-out mode as you did as with this edit, and as IP  with this edit. See wp:sock puppetry. - DVdm (talk) 16:48, 26 July 2017 (UTC)