Transformation equations presented in a different way

[bernhard.rothenstein]

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 energy 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

 

[Meir Achuz]

Never use "it is obvious" in pedagogy. That is like the old mathematics professor joke.

 

[nakurusil]

E=g(u')g(V)E(0)(1+Vu'/cc)

Don't you think that $$E=\gamma(V)*E(0)$$ is much cleaner?

 

[Hootenanny]

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

 

[bernhard.rothenstein]

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?

 

[bernhard.rothenstein]

transformation equations

Don't you think that $$E=\gamma(V)*E(0)$$ is much cleaner?

Thanks. Yes it is but I think that the Forum could offer a simple equation editor.

 

[cristo]

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 $$[ /tex] tags (without the space) around the equations.$$

 

[bernhard.rothenstein]

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

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.

 

[nakurusil]

m=g(u')g(V)E(0)(1+Vu'/cc)=g(u')g(V)m(0)(1+Vu'/cc).

$$\gamma(u')\gamma(V)(1+Vu'/c^2)=\gamma(u)$$ , so the above reduces the the much cleaner, well known :

$$m(u)=\gamma(u)*m(0)$$

 

[robphy]

The identity (rearranged in a minor way)
$$\gamma(u')\gamma(V)(1+u'V/cc)=\gamma(u)$$
is more recognizable in terms of rapidities:
$$\begin{align*}<br /> \gamma(u')\gamma(V)(1+u'V/cc)<br /> &=<br /> \cosh{\theta'}\cosh{\phi}(1+c\tanh{\theta'}\ c\tanh{\phi}/cc)\\<br /> &=<br /> \cosh{\theta'}\cosh{\phi}+\sinh{\theta'}\sinh{\phi}\\<br /> &=<br /> \cosh{(\theta'+\phi)}\\<br /> &=<br /> \cosh{\theta}\\<br /> &=<br /> \gamma(u)<br /> \end{align*}$$
where u$=c\tanh{\theta}$ 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.

 

[bernhard.rothenstein]

transformations

The identity (rearranged in a minor way)
$$\gamma(u')\gamma(V)(1+u'V/cc)=\gamma(u)$$
is more recognizable in terms of rapidities:
$$\begin{align*}<br /> \gamma(u')\gamma(V)(1+u'V/cc)<br /> &=<br /> \cosh{\theta'}\cosh{\phi}(1+c\tanh{\theta'}\ c\tanh{\phi}/cc)\\<br /> &=<br /> \cosh{\theta'}\cosh{\phi}+\sinh{\theta'}\sinh{\phi}\\<br /> &=<br /> \cosh{(\theta'+\phi)}\\<br /> &=<br /> \cosh{\theta}\\<br /> &=<br /> \gamma(u)<br /> \end{align*}$$
where u$=c\tanh{\theta}$ 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.

 

[bernhard.rothenstein]

The identity (rearranged in a minor way)
$$\gamma(u')\gamma(V)(1+u'V/cc)=\gamma(u)$$
is more recognizable in terms of rapidities:
$$\begin{align*}<br /> \gamma(u')\gamma(V)(1+u'V/cc)<br /> &=<br /> \cosh{\theta'}\cosh{\phi}(1+c\tanh{\theta'}\ c\tanh{\phi}/cc)\\<br /> &=<br /> \cosh{\theta'}\cosh{\phi}+\sinh{\theta'}\sinh{\phi}\\<br /> &=<br /> \cosh{(\theta'+\phi)}\\<br /> &=<br /> \cosh{\theta}\\<br /> &=<br /> \gamma(u)<br /> \end{align*}$$
where u$=c\tanh{\theta}$ 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$$m(0)gamma(u0=m(0)gamma(u')gamma(V)(1+u'V/cc$$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.