Transformation equations presented in a different way

  • #1
Consider a particle that moves with speed u relative to the inertial reference frame I and with speed u' relative to the inertial reference frame I'. Let g(u), g(u') and g(V) be the orresponding gamma factors (V the relative speed of I and I'). m(0) stands for its rest mass, E(0) for its rest energy, p and E for its momentum and enegy measured by observers from I. It is obvious that
p=g(u')g(V)m(0)(V+u')=g(u')g(V')E(0)(V+u')cc
E=g(u')g(V)E(0)(1+Vu'/cc)
I consider that such a presentation presents some (pedagogical) advantages showing clearly what observers from the I frame measure in the case when u'=0 and when u' and V are both equal to zero.
Even if I know that the concept of relativistic mass is persona non grata on the Forum I would also suggest for the relativistic mass
m=g(u')g(V)E(0)(1+Vu'/cc)=g(u')g(V)m(0)(1+Vu'/cc).
The oppinion of those who teach or learn special relativity theory is highly appreciated of course in the spirit of
sine ira et studio
 

Answers and Replies

  • #2
Meir Achuz
Science Advisor
Homework Helper
Gold Member
3,529
110
Never use "it is obvious" in pedagogy. That is like the old mathematics professor joke.
 
  • #3
329
0
Last edited:
  • #4
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,621
6
Even if I know that the concept of relativistic mass is persona non grata on the Forum [...]
As an aside, persona non grata means 'an unwelcome person' and is not usually used in reference to a concept. Absit invidia :wink:
 
  • #5
transformation equation

Never use "it is obvious" in pedagogy. That is like the old mathematics professor joke.
Thanks. I think that there is a big difference between posting on the Forum, where the participants can easily transform the usual transformations in those I propose and presenting them in all its steps.
I know very much joks with teachers of physics and mathematics. Which of them do you mean?
 
  • #6
transformation equations

Don't you think that [tex]E=\gamma(V)*E(0)[/tex] is much cleaner?
Thanks. Yes it is but I think that the Forum could offer a simple equation editor.
 
  • #7
cristo
Staff Emeritus
Science Advisor
8,107
73
Thanks. Yes it is but I think that the Forum could offer a simple equation editor.
The forum does! LaTex is very easy to learn, and it is easy to use on this forum too; simply put [tex] [ /tex] tags (without the space) around the equations.
 
  • #8
latina ginta est Regina

As an aside, persona non grata means 'an unwelcome person' and is not usually used in reference to a concept. Absit invidia :wink:
Thanks. My first language is close to Latin. I thought that physicists are able to extrapolate from persona non grata to relativistic mass which is there non grata. I end with
absit invidia which is shorter and more adequate then sine ira et studio I used so far.
 
  • #9
329
0
m=g(u')g(V)E(0)(1+Vu'/cc)=g(u')g(V)m(0)(1+Vu'/cc).
[tex]\gamma(u')\gamma(V)(1+Vu'/c^2)=\gamma(u)[/tex] , so the above reduces the the much cleaner, well known :

[tex]m(u)=\gamma(u)*m(0)[/tex]
 
Last edited:
  • #10
robphy
Science Advisor
Homework Helper
Insights Author
Gold Member
5,700
991
The identity (rearranged in a minor way)
[tex]\gamma(u')\gamma(V)(1+u'V/cc)=\gamma(u)[/tex]
is more recognizable in terms of rapidities:
[tex]
\begin{align*}
\gamma(u')\gamma(V)(1+u'V/cc)
&=
\cosh{\theta'}\cosh{\phi}(1+c\tanh{\theta'}\ c\tanh{\phi}/cc)\\
&=
\cosh{\theta'}\cosh{\phi}+\sinh{\theta'}\sinh{\phi}\\
&=
\cosh{(\theta'+\phi)}\\
&=
\cosh{\theta}\\
&=
\gamma(u)
\end{align*}
[/tex]
where u[itex]=c\tanh{\theta}[/itex] is the spatial-velocity obtained by "spatial-velocity-composition of u' and V".

In addition, in terms of rapidities, one can immediately transcribe the calculation into a spacetime diagram, which provides a hopefully more intuitive interpretation of what is happening physically [and mathematically].

So, it's not clear to me if anything is gained in the proposed formula, except maybe for a particular type of problem.
 
Last edited:
  • #11
transformations

The identity (rearranged in a minor way)
[tex]\gamma(u')\gamma(V)(1+u'V/cc)=\gamma(u)[/tex]
is more recognizable in terms of rapidities:
[tex]
\begin{align*}
\gamma(u')\gamma(V)(1+u'V/cc)
&=
\cosh{\theta'}\cosh{\phi}(1+c\tanh{\theta'}\ c\tanh{\phi}/cc)\\
&=
\cosh{\theta'}\cosh{\phi}+\sinh{\theta'}\sinh{\phi}\\
&=
\cosh{(\theta'+\phi)}\\
&=
\cosh{\theta}\\
&=
\gamma(u)
\end{align*}
[/tex]
where u[itex]=c\tanh{\theta}[/itex] is the spatial-velocity obtained by "spatial-velocity-composition of u' and V".

In addition, in terms of rapidities, one can immediately transcribe the calculation into a spacetime diagram, which provides a hopefully more intuitive interpretation of what is happening physically [and mathematically].

So, it's not clear to me if anything is gained in the proposed formula, except maybe for a particular type of problem.
Thanks. My intention is to present the transformation equations in such a way that theirs right sides contain only a proper physical quantity and velocities reducing the long discussions related to the concept of relativistic mass. That is the direction in which I hope our discussions will evolve.
Reading my lines please take into account that English is not my first language.
 
  • #12
The identity (rearranged in a minor way)
[tex]\gamma(u')\gamma(V)(1+u'V/cc)=\gamma(u)[/tex]
is more recognizable in terms of rapidities:
[tex]
\begin{align*}
\gamma(u')\gamma(V)(1+u'V/cc)
&=
\cosh{\theta'}\cosh{\phi}(1+c\tanh{\theta'}\ c\tanh{\phi}/cc)\\
&=
\cosh{\theta'}\cosh{\phi}+\sinh{\theta'}\sinh{\phi}\\
&=
\cosh{(\theta'+\phi)}\\
&=
\cosh{\theta}\\
&=
\gamma(u)
\end{align*}
[/tex]
where u[itex]=c\tanh{\theta}[/itex] is the spatial-velocity obtained by "spatial-velocity-composition of u' and V".

In addition, in terms of rapidities, one can immediately transcribe the calculation into a spacetime diagram, which provides a hopefully more intuitive interpretation of what is happening physically [and mathematically].

So, it's not clear to me if anything is gained in the proposed formula, except maybe for a particular type of problem.
Thank you for having brought the formula to a more transparent shape. Consider the concept of proper mass m(0) and multiply both sides of with it. It leads to[tex]m(0)gamma(u0=m(0)gamma(u')gamma(V)(1+u'V/cc[/tex]An exercised eye will recognise in the left side of the equation the expression of the relativistic mass in I in the left side its expresion as a function of phyhsical quantities measured in I. Do you consider that such a presentation is time saving, transparent and convincing for the fact that conservation laws are not compulsory in the derivation. I do not convinced that the equation will appear correctly in myh message.
 

Related Threads on Transformation equations presented in a different way

Replies
34
Views
2K
Replies
38
Views
1K
  • Last Post
2
Replies
48
Views
8K
  • Last Post
Replies
21
Views
858
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
4
Views
1K
Replies
35
Views
4K
  • Last Post
Replies
13
Views
2K
Top