I Derivation of SR's time-dilatation in 1d?

[Sagittarius A-Star]

Without the second postulate I don't think I can show that $X^2-T^2 = X'^2-T'^2$ on H. Wouldn't that be too strong?

I didn't write "without the second postulate". MacDonald wrote in his first line "Assume: (A) The speed of light is the same in all inertial frames. (Take c = 1.)". The following equation are already based on postulate 2.

Thus on H $$ X + T = e_1 (X'+T') $$ $$ X - T = e_2 (X'-T') $$ for some constants $e_1,e_2.$ If I could show that $e_1 e_2 = 1$ then that would be enough, sadly I can't do that at this point.

Now assume that c, the speed of light is 1 in both frames.

I should better re-write them without $c=1$:
$x + ct = e_1 (x'+ct')$
$x - ct = e_2 (x'-ct')$
That the 2nd postulate was invoked can be seen from the fact, that they contain $ct'$ instead of $c't'$.

You could now take the first equation, invoke the 1st postulate to switch primed and unprimed coordinates, if you also describe for symmetry reasons a light beam, that moves in the opposite direction:
$x'-ct' = e_1 (x - ct)$
By putting the unprimed quantities on the left side, you get your new second equation in:
$$ x + ct = e_1 (x'+ct') $$$$ x - ct = \frac{1}{e_1} (x'-ct') $$

 

[Pony]

I didn't write "without the second postulate". MacDonald wrote in his first line "Assume: (A) The speed of light is the same in all inertial frames. (Take c = 1.)". The following equation are already based on postulate 2.

Not how I wrote it, I think I never used postulate 2 before I said it. I only used postulate 1/Galilean relativity. So does the Macdonald paper BTW, it uses postulate 2 first when it refers it. (In a relativist mindset it is hard to imagine that time and space could exist without something relating them. But then Newtonian kinematics, where time is measured with wristwatches and distance is measured with measuring tapes, and just assuming that light follows newtonian mechanics too, is consistent.)

That the 2nd postulate was invoked can be seen from the fact, that they contain $ct'$ instead of $c't'$.

You can't change what I write and point out that you use postulate 2 in that.

It doesn't really matter tho.

 

[Sagittarius A-Star]

Not how I wrote it, I think I never used postulate 2 before I said it.

How are the units in your $X+T$ in posting #46 consistent without $c=1$?
And after you say that you define $c=1$ you show the same equations.

 

[Pony]

Both are real numbers.

And after you say that you define you show the same equations.

This may seem confusing, but I believe it works the way I wrote it, that is, we can not use an equation, and derive some results in a more general way, then restrict ourselves later. "Now we assume c=1, which means some additional restrictions for our T and X functions, but the previous facts still remain true, and we even can keep using the same symbols."

So those equations are true in the Newtonian setting as well (I think at least), and I hoped for a nice looking spacetime interval formula that is agnostic whether we assume postulate 2 or don't.

 

[Sagittarius A-Star]

This may seem confusing

Yes. I think it would be less confusing to use from begin to end the same unit system. But yes, I also think it works the way you wrote it.

 

[Pony]

Well, okay. I am sorry about that.

I am not interested talking about aesthetics anymore.

Yes. You didn't write to which value you defined $c$ before you defined $c=1$.
You are adding a distance to a time interval, which makes no sense with an undefined $c$.

The way I wrote does not need it, and is a bit more general. But you can assume that Galileo worked with seconds and light-seconds, and so do I until step 1. Then Einstein came and noticed that light moves with velocity = 1 lightsec/sec in every frame, and so I postulated this as well before step 2. (But not sooner.)

But it really is super agnostic to measurement units, and later only assumes that c=1.

 

[DAH]

The way I see it in SR $c=1$ but it still has units $\text{m/s}$ so that $ct$ has units of meters, the same as the spatial coordinates. The relative velocities in the transformations will always be less than 1 or a fraction.

For time dilation in one dimension, if two ticks of a watch are separated by $(\delta t, 0)$ in the frame of the watch, where $\delta t=1$ second. Then in a frame that is moving at $v=\text{d}x^{\prime}/\text{d}t^{\prime}$ relative to the watch, the same ticks are separated by $(\delta t^{\prime}, \delta x^{\prime})$.

We know the spacetime separations are $\delta s=\delta s^{\prime}$, so then $(c\delta t)^2=(c\delta t^{\prime})^2-(\delta x^{\prime})^2$.
If we divide by $(c\delta t^{\prime})^2$ we get:
\begin{align*}
\frac{c^2(\delta t)^2}{c^2(\delta t^{\prime})^2}&=1-\frac{(\delta x^{\prime})^2}{c^2(\delta t^{\prime})^2}\\\\
\left(\frac{\delta t}{\delta t^{\prime}}\right)^2&=1-\left(\frac{\text{d}x^{\prime}}{c\text{d}t^{\prime}}\right)^2=1-\frac{v^2}{c^2}\\\\
\frac{\delta t}{\delta t^{\prime}}&=\sqrt{1-\frac{v^2}{c^2}}
\end{align*}
Which can be rearranged to give $$\delta t^{\prime}=\gamma \delta t$$
This is basically the same as @Dale 's post #2

 

[Pony]

I don't think that the symmetry approach in #38 is working anymore. It is too vague to point out a mistake, but I don't think that the assumptions imply the conclusion.

A better approach can be orienting the x axes against each other from the start, then we get $$ T+X = e_1 (T'-X')$$ $$ T-X = e_2 (T'+X')$$ equations which are true for every event ( a light ray has +1 and -1 velocities in the frames when we extend these equations from the trajectories to every event ) and in both of the two frames, with the same $e_1,e_2$ constants (because the frames now have a truly symmetric role). So $$ T'+X' = e_1 (T-X)$$ $$ T'-X' = e_2 (T+X)$$ also hold, which means
$$ T + X = e_1(T'-X') = e_1e_2 (T+X) $$ which means $e_1 e_2 = 1.$ Thus $T^2 - X^2 = T'^2 -X'^2$ and SR follows easily.

 

[Sagittarius A-Star]

A better approach can be orienting the x axes against each other from the start
$$T+X = e_1 (T'-X')$$

Then you can define time as that, what a standardized clock shows and directly conclude with the two postulates and symmetry from ... $$T+X = e_1 (T'-X')$$$\Rightarrow$ (swap frames and satisfy descriptions in both frames of a light beam in opposite direction)$$T'+X' = e_1 (T-X)$$$\Rightarrow$$$\begin{cases}T+X = e_1 (T'-X')\\
T- X = \frac{1}{e_1} (T'+X')\end{cases}$$Then multiply those.

These are essentially calculations in light-cone coordinates. If you rotate the axes of a normal spacetime coordinate system $(X, T)$ counter-clockwise by $45°$, than you get the axes of a light-cone coordinate system $({L_+}, {L_-})$.$$\begin{cases}{L_+} = X \cos(45°) + T \sin(45°) = (T+X) \frac{1}{\sqrt{2}}\\
{L_-} = -X \sin(45°) + T \cos(45°) = (T- X) \frac{1}{\sqrt{2}}\end{cases}$$Lorentz transformation (with x axes in same direction):$$\begin{cases}{L_+}=k {L_+}' \\{L_-}=\frac{1}{k} {L_-}'\end{cases}$$

Calculating $k$ on the worldline of an object at $X' = 0 \ \ \ \Rightarrow \ \ \ {L_+}'={L_-}'$:​

$$k^2 = \frac{{L_+}}{{L_-}} = \frac{1+v}{1-v} = \gamma^2 (1+v)^2 = \frac{1}{\gamma^2 (1-v)^2}$$$$\begin{cases}T+X = \gamma (1+v) (T'+X') \ \ \ \ \ (1)\\
T- X = \gamma (1-v) (T'-X') \ \ \ \ \ (2)\end{cases}$$

Add and subtract Eqs. (1) and (2):​

$$\begin{cases}T= \gamma (T'+v X') \ \ \ \ \ (3)\\X = \gamma (v T'+X') \ \ \ \ \ (4)\end{cases}$$​

Invariance of the spacetime interval:
$${L_+}{L_-} = {L_+}'{L_-}'$$

 

[Pony]

Who's clock shows, where, what do you mean 'shows'? What does the first equation mean, and how do I conclude that?

 

[Sagittarius A-Star]

Who's clock shows, where, what do you mean 'shows'?

See posting #40. That definition of time separates for example SR from Lorentz ether theory.

What does the first equation mean, and how do I conclude that?

I copied the first equation from your posting #58.