Define a vector that transfers the reference systems to the future

[Renato Iraldi]

TL;DR Summary

We define a vector that transfers the reference systems to the future.

Suppose a stationary reference frame, any other reference frame will have a clock that advances at t'= t/gand is moving away at velocity v. g is the relativistic factor.

we can write:

C d t' = Cdt/g and

d x = v dt

this can be written

(Cdt',Cdx)=(C/g,C v) dt ;

since 1/g2+ v2 = 1-v2+ v2 = 1

We can write:

(Cdt',Cdx)= ( C cos a , C sin a) dt ;

v= sina ; and 1/g= cosa .

We see that all system go to the future at speed C. C is a vector xof magnitude the velocity oh ligth.
The translation speed is a component of the speed towards the future, the speed C is the only speed, for this reason it cannot be exceeded.
In a stationary system all the transfer toward the future is in time, for a photon all the transer toward his future is traslational.

 

[Ibix]

You seem to be groping towards the concept of four vectors, which are a good way of understanding relativity. You need hyperbolic functions rather than trigonometric for them - your maths is too messy to work out how you've ended up with trigonometric ones, although I suspect you're just doing things kind of backwards and have ended up with frame dependant conclusions instead of the more useful invariant ones.

The translation speed is a component of the speed towards the future, the speed C is the only speed, for this reason it cannot be exceeded.
In a stationary system all the transfer toward the future is in time, for a photon all the transer toward his future is traslational.

This bit is not at all a helpful avenue to go down. In particular, attempting to use forms of the Lorentz transforms that only apply between rest frames is always going to lead to confusion and self contradiction if you try to deduce anything about things travelling at the speed of light, since they do not have rest frames.

 

[PeterDonis]

We see that all system go to the future at speed C.

No, this is not correct. Light does not. See below.

C is a vector xof magnitude the velocity oh ligth.

Yes, and a timelike vector can be normalized to this magnitude. But a null vector (describing how light moves) cannot; a null vector has magnitude zero.

The translation speed is a component of the speed towards the future, the speed C is the only speed, for this reason it cannot be exceeded.

This is not correct.

In a stationary system all the transfer toward the future is in time, for a photon all the transer toward his future is traslational.

This is not correct either, although you will find something like it in some pop science books, even ones by physicists who should know better (such as Brian Greene).

 

[Renato Iraldi]

You seem to be groping towards the concept of four vectors, which are a good way of understanding relativity. You need hyperbolic functions rather than trigonometric for them - your maths is too messy to work out how you've ended up with trigonometric ones, although I suspect you're just doing things kind of backwards and have ended up with frame dependant conclusions instead of the more useful invariant ones.

This bit is not at all a helpful avenue to go down. In particular, attempting to use forms of the Lorentz transforms that only apply between rest frames is always going to lead to confusion and self contradiction if you try to deduce anything about things travelling at the speed of light, since they do not have rest frames.

[Mentor note: the text below has been edited to use Latex. Renato Iraldi, your posts will be much easier to read and will attract more helpful answers if you format the equations properly.]

I have shown that from the point of view of a stationary system all clocks are described by $x=vt$, and $t' =t/\gamma$. This is correct.
that $v^2+(1/\gamma)^2 =1$ therefore it can be written:
$v= \sin a$ and $1/\gamma = \cos a$.
We use C=1.
for all watches
$(dx,dt')= (\sin a, cos a) dt.$ Vectorial equation.
the coordinates of all the clocks advance with the same type of equation according to a vector of magnitude 1 equal to say C.
It's math.

 

[Ibix]

(dx,dt')= (sin a, cos a) dt.

Oh, I see. What you're doing is mixing quantities from different frames. That's not a particularly helpful thing to do, although it's not necessarily wrong and seems to have been done correctly here. It's more helpful to try to express things in terms of invariants, or at least quantities relevant to one frame only, since that eventually lets you see what things you are writing are always true and which ones depend on your frame choice.

A more generally useful approach is to observe that the four velocity $(\cosh\psi,\sinh\psi)$, where $\mathrm{tanh}\ \psi =v/c$ is the rapidity, is a vector. That immediately lets you add them and scale them, giving you access to easier and more general coordinate transforms and vectors such as the four momentum and then tensors.

It also allows you to generalise to null vectors properly. Your approach is not valid, since it depends on the quantity $t'$ being defined for objects moving at the speed of light, which is not correct.

 

[Renato Iraldi]

Oh, I see. What you're doing is mixing quantities from different frames. That's not a particularly helpful thing to do, although it's not necessarily wrong and seems to have been done correctly here. It's more helpful to try to express things in terms of invariants, or at least quantities relevant to one frame only, since that eventually lets you see what things you are writing are always true and which ones depend on your frame choice.

A more generally useful approach is to observe that the four velocity $(\cosh\psi,\sinh\psi)$, where $\mathrm{tanh}\ \psi =v/c$ is the rapidity, is a vector. That immediately lets you add them and scale them, giving you access to easier and more general coordinate transforms and vectors such as the four momentum and then tensors.

It also allows you to generalise to null vectors properly. Your approach is not valid, since it depends on the quantity $t'$ being defined for objects moving at the speed of light, which is not correct.

I thank you for the interest shown-
I know the treatment in hyperbolic geometry, however I believe that it can be done in geometry of circles, I have not understood your last sentence. t' is the time marked by a moving clock, the dilated time.
I have made a model of a block of events universe with time as a sweep at speed C, which with elementary geometry can be deduced the Lorentz transformations. I can send if e mail.
I thank you again.

 

[Ibix]

I have not understood your last sentence. t' is the time marked by a moving clock, the dilated time.

It is not possible to define this in the case $v=c$, so your approach is not valid in that case.

I'm also not at all sure that your approach is actually vector maths, as you claim. Your use of $(dx,dt')$ might be better written as $(dx,d\tau)$ where $\tau$ is the proper time of the moving object. In that case it's clearer that you seem to be thinking in terms of space-proper time diagrams, also called Epstein diagrams. They aren't terribly popular, although people do use them, largely because they don't have the nice geometrical properties of Minkowski diagrams - for example, worldlines crossing on such diagrams does not mean that they actually meet.

 

[Renato Iraldi]

It is not possible to define this in the case $v=c$, so your approach is not valid in that case.

I'm also not at all sure that your approach is actually vector maths, as you claim. Your use of $(dx,dt')$ might be better written as $(dx,d\tau)$ where $\tau$ is the proper time of the moving object. In that case it's clearer that you seem to be thinking in terms of space-proper time diagrams, also called Epstein diagrams. They aren't terribly popular, although people do use them, largely because they don't have the nice geometrical properties of Minkowski diagrams - for example, worldlines crossing on such diagrams does not mean that they actually meet.

I agree with you that it is not possible to deduce what the photon sees, (as some popularizers do) making a limit with v = C because it is not possible to speak of a photon reference system.
because for every system the physical laws are the same but not for the photon system.
What can be said is that the passage of time has a magnitude C, one component is velocity, and the other is the passage of time, as higher speeds are seen, the component orthogonal to velocity decreases.
It can also be said in the twin paradox that one of them has to change the angle of the direction of the vector that pushes it into the future, the longest time is when the angle is zero. the systems that change the angle end up advancing less toward the future.

 

[PeterDonis]

I know the treatment in hyperbolic geometry, however I believe that it can be done in geometry of circles

I'm not sure what you mean by "geometry of circles", but I suspect you mean something like ordinary Euclidean geometry. That won't work because there are no timelike or null vectors.