Principle of relativity in the proof of invariance of interval

[DrGreg]

First frame is stationary frame. Second frame is moving frame.

Only from one point of view.

The first postulate implies that:

"Second frame is stationary frame. First frame is moving frame."​

is also an equally valid point of view.

 

[Mike_bb]

Only from one point of view.

The first postulate implies that:

"Second frame is stationary frame. First frame is moving frame."​

is also an equally valid point of view.

Let's consider two cases for these statements:
1. First frame is stationary frame and second frame is moving frame.
Let's agree that we will see second frame from first frame. We obtain ${ds_2}^2=k(V){ds_1}^2$
2. First frame is moving frame and second frame is stationary frame.
As we agreed above we will see second frame from first frame but we obtain the same result: ${ds_2}^2=k(V){ds_1}^2$

Where did I go wrong?

 

[PeroK]

Let's consider two cases for these statements:
1. First frame is stationary frame and second frame is moving frame.
Let's agree that we will see second frame from first frame. We obtain ${ds_2}^2=k(V){ds_1}^2$
2. First frame is moving frame and second frame is stationary frame.
As we agreed above we will see second frame from first frame but we obtain the same result: ${ds_2}^2=k(V){ds_1}^2$

Where did I go wrong?

You forgot to swap the frames in the second equation.

 

[Sagittarius A-Star]

Ok. "The second system looks from the first is indistinguishable from the first system from the second"
I wrote about it when I provided drawings for this statement. It doesn't work as it's written. Where did I go wrong in posts #15 and #19?

Each picture in #15 and #19 contains a clock and a coordinate system, that move relative to each other.

But for a visualization of the reciprocity of the two frames a picture is helpful, that contains two coordinate systems that move relative to each other.

​

If the directions of the x- and z- axes in both frames $S$ and $S'$ are reversed, this does not effect the equation
$c^2dt'^2-dx'^2-dy'^2-dz'^2 = k(V) (c^2dt^2-dx^2-dy^2-dz^2)$
but it interchanges the roles of $S$ and $S'$. Then the following picture holds with primed and unprimed symbols interchanged:

Source:​

http://www.scholarpedia.org/article/Special_relativity:_kinematics​

Then with the principle of relativity follows also:
$c^2dt^2-dx^2-dy^2-dz^2 = k(V) (c^2dt'^2-dx'^2-dy'^2-dz'^2)$.

 

[Mike_bb]

Each picture in #15 and #19 contains a clock and a coordinate system, that move relative to each other.

But for a visualization of the reciprocity of the two frames a picture is helpful, that contains two coordinate systems that move relative to each other.

​

If the directions of the x- and z- axes in both frames $S$ and $S'$ are reversed, this does not effect the equation
$c^2dt'^2-dx'^2-dy'^2-dz'^2 = k(V) (c^2dt^2-dx^2-dy^2-dz^2)$
but it interchanges the roles of $S$ and $S'$. Then the following picture holds with primed and unprimed symbols interchanged:

View attachment 356489​

Source:​

http://www.scholarpedia.org/article/Special_relativity:_kinematics​

Then with the principle of relativity follows also:
$c^2dt^2-dx^2-dy^2-dz^2 = k(V) (c^2dt'^2-dx'^2-dy'^2-dz'^2)$.

Am I right that after the changes it will look like this?

 

[Sagittarius A-Star]

Am I right that after the changes it will look like this?
View attachment 356492

The changes would also include to reverse directions of the x- and z- axes including the related arrows.

 

[Mike_bb]

The changes would also include to reverse directions of the x- and z- axes.

I changed x- and -z- axes on the picture. Is it true?

 

[Mike_bb]

Sagittarius A-Star, big thanks!! You helped me a lot!

 

[Mike_bb]

Hello!

I have a question.
As was mentioned above, according to principle of relativity we have:

${ds_2}^2=K(V){ds_1}^2$ $(1^*)$
${ds_1}^2=K(V){ds_2}^2$ $(2^*)$

Let's consider these equations:

In equation $(1^*)$: ${ds_2}^2=K(V){ds_1}^2$ coefficient $K(V)$ appropriate to interval ${ds_1}^2$.
But in equation $(2^*)$: ${ds_1}^2=K(V){ds_2}^2$ coefficient $K(V)$ must appropriate to interval ${ds_2}^2$.
If we obtain this contradiction then there must be two different coefficients for $(1^*)$ and $(2^*)$.

I have a doubt about this reasoning. How to solve this question? Where did I go wrong?
Thanks!

 

[Sagittarius A-Star]

Hello!

I have a question.
As was mentioned above, according to principle of relativity we have:

${ds_2}^2=K(V){ds_1}^2$ $(1^*)$
${ds_1}^2=K(V){ds_2}^2$ $(2^*)$

Let's consider these equations:

In equation $(1^*)$: ${ds_2}^2=K(V){ds_1}^2$ coefficient $K(V)$ appropriate to interval ${ds_1}^2$.
But in equation $(2^*)$: ${ds_1}^2=K(V){ds_2}^2$ coefficient $K(V)$ must appropriate to interval ${ds_2}^2$.
If we obtain this contradiction then there must be two different coefficients for $(1^*)$ and $(2^*)$.

I have a doubt about this reasoning. How to solve this question? Where did I go wrong?
Thanks!

There is no contradiction. If you substitute $(1^*)$ into $(2^*)$, then you get
${ds_1}^2=K(V)K(V){ds_1}^2$

 

[Mike_bb]

There is no contradiction. If you substitute $(1^*)$ into $(2^*)$, then you get
${ds_1}^2=K(V)K(V){ds_1}^2$

I thought about it. Do you mean that $K(V)$ is suitable for $K(V){ds_1}^2$ as coefficient for interval ${ds_1}^2$?

 

[Sagittarius A-Star]

I thought about it. Do you mean that $K(V)$ is suitable for $K(V){ds_1}^2$ as coefficient for interval ${ds_1}^2$?

I mean that $(K(V))^2=1$.

 

[Mike_bb]

I mean that $(K(V))^2=1$.

I mean another thing. We know that coeffiecient doesn't depend on coordinate and time of intervals. But we don't know what coefficient is appropriate for different intervals and are there different coefficients or not.

 

[Sagittarius A-Star]

I mean another thing. We know that coeffiecient doesn't depend on coordinate and time of intervals. But we don't know what coefficient is appropriate for different intervals and are there different coefficients or not.

In posting #34 is described, why the same coefficient can be used.

 

[Mike_bb]

In posting #34 is described, why the same coefficient can be used.

Is there another way how to explain it? Wiki says nothing about way that mentioned in posting#34.

 

[Sagittarius A-Star]

Is there another way how to explain it? Wiki says nothing about way that mentioned in posting#34.

Maybe the following summary is better to understand:

Source:
Book "Unusually Special Relativity" from Andrzej Dragon.

https://www.amazon.com/Unusually-Special-Relativity-Andrzej-Dragan/dp/1800610882?tag=pfamazon01-20

 

[Mike_bb]

Maybe the following summary is better to understand:

Source:
Book "Unusually Special Relativity" from Andrzej Dragon.

Hello!
I returned to beginning because I have a doubt. Am I right that for different intervals we have different coefficient K1(x,y,z,t); K2(x,y,z,t)... or we have one coefficient for different intervals K(x,y,z,t), i.e. ${ds_2}^2=K1(x,y,z,t){ds_1}^2$ and so on? How does it work in this context?

P.S. My attempt to obtain ${ds_2}^2=K{ds_1}^2$ and ${ds_1}^2=K{ds_2}^2$ is:

${ds_2}^2=K1(x,y,z,t){ds_1}^2$
${ds_1}^2=K2(x,y,z,t){ds_2}^2$

K1(x,y,z,t) doesn't depend on spacetime coordinate then we obtain constant K1;
K2(x,y,z,t) doesn't depend on spacetime coordinate then we obtain constant K2;

${ds_2}^2=K1{ds_1}^2$
${ds_1}^2=K2{ds_2}^2$

But it differs from ${ds_2}^2=K{ds_1}^2$ and ${ds_1}^2=K{ds_2}^2$

 

[Sagittarius A-Star]

${ds_2}^2=K1{ds_1}^2$
${ds_1}^2=K2{ds_2}^2$

But it differs from ${ds_2}^2=K{ds_1}^2$ and ${ds_1}^2=K{ds_2}^2$

$K_1=K_2$ because you can interchange the roles of the primed and unprimed frame, as described in posting #34.

 

[Mike_bb]

$K_1=K_2$ because you can interchange the roles of the primed and unprimed frame, as described in posting #34.

Thanks. It's one of possible ways.
But you wrote at the beginning of this topic: "The formulas you showed are reciprocal between both frames because inertial frames are homogeneous in spacetime and isotropic in space."

Coefficient doesn't depend on coordinate and time because specetime is homogenoeous. Could you explain in more details which of coordinate and time are involved ? Thanks!

 

[Mike_bb]

$K_1=K_2$ because you can interchange the roles of the primed and unprimed frame, as described in posting #34.

I found answer here:

https://en.wikipedia.org/wiki/Deriv..._and_Proof_of_Proportionality_of_ds2_and_ds′2

On what may

depend? It may not depend on the positions of the two events in spacetime, because that would violate the postulated homogeneity of spacetime.

But I can't understand how "violate the postulated homogeneity of spacetime"?

 

[Sagittarius A-Star]

I found answer here:

https://en.wikipedia.org/wiki/Derivations_of_the_Lorentz_transformations#Rigorous_Statement_and_Proof_of_Proportionality_of_ds2_and_ds′2


But I can't understand how "violate the postulated homogeneity of spacetime"?

From homogeneity of spacetime and isotropy of space follows that all coefficients in a transformation between primed and unprimed coordinates must be constants. Therefore, if you transform the coordinates of the spacetime interval formula i.e. from unprimed to primed coordinates, then you must get a polynomial of second order. A theorem of algebra states, that two irreducible polynomials of the same degree (i.e. quadratic), which share all their zeros (invariance of the speed of light!), can differ by at most a constant factor $K$.

 

[Mike_bb]

Hello, Sagittarius A-Star!

1.
View attachment 356460
2.
View attachment 356458

See post

I decided to add more details to equations. Now equation contain coefficients a1 and a2 that depend on coordinates and time:

$$c^2\Delta t^2-\Delta x^2 = a1(v\Delta t , L, \Delta t)c^2\Delta t'^2-\Delta x'^2$$
$$c^2\Delta t'^2-\Delta x'^2 = a2(v\Delta t , L, \Delta t)c^2\Delta t^2-\Delta x^2$$

If $$a1(v\Delta t , L, \Delta t) = a2(v\Delta t , L, \Delta t)$$ then $$a1(v\Delta t , L, \Delta t) = a2(v\Delta t , L, \Delta t) = K$$

So:

$$c^2\Delta t^2-\Delta x^2 = K(c^2\Delta t'^2-\Delta x'^2)$$
$$c^2\Delta t'^2-\Delta x'^2 = K(c^2\Delta t^2-\Delta x^2)$$

Is it right?

Thanks!

 

[Mike_bb]

Sagittarius A-Star

In previous post I meant that coordinates and time are the same for two intervals in different coordinate systems (by Principle of Relativity). Am I right?

 

[bquuune]

Greetings Mike_bb,

I just saw the reply you made at https://math.stackexchange.com/ques...-infinitesimal-be-non-integer/5069096#5069096

I don't understand the proof you wrote there.
If you can prove
(1) $c^2\Delta t^2-\Delta x^2 = K(c^2\Delta t'^2-\Delta x'^2)$ and
(2) $c^2\Delta t'^2-\Delta x'^2 = K(c^2\Delta t^2-\Delta x^2)$

Then you can plug the second equation into the first and get
$c^2\Delta t^2-\Delta x^2 = K K(c^2\Delta t^2-\Delta x^2)$
So $K=1$ or $-1$.