I Principle of relativity in the proof of invariance of interval

[Mike_bb]

Hello!

I saw such interpretation of principle of relativity when I read proof of invariance of infinitesimally small interval:

"The second inertial frame of reference looks from the first in no way different from how the first inertial frame of reference looks from the second."

Proof of invariance of infinitesimally small interval:

The second inertial frame of reference looks from the first in no way different from how the first inertial frame of reference looks from the second.
If we go to the second inertial frame of reference from the first inertial frame of reference and then back we will have:

Doing such way we obtain K(V)=1 and we can prove invariance of interval.

But I can't understand what does this interpretation mean? How does it work in proof? Could anyone give example for this case?

Thanks!

 

[Sagittarius A-Star]

View attachment 356300

Doing such way we obtain K(V)=1 and we can prove invariance of interval.

But I can't understand what does this interpretation mean? How does it work in proof? Could anyone give example for this case?

Thanks!

The formulas you showed are reciprocal between both frames because inertial frames are homogeneous in spacetime and isotropic in space. The orientation of the rectangular axes in both frames can be arbitrary.

From for formulas you showed follows $K(V) = \pm 1$. The negative sign can be dismissed for $v \rightarrow 0$ and then argue with continuity.

when I read proof of invariance of infinitesimally small interval:

Where did you read this?

 

[Mike_bb]

The formulas you showed are reciprocal between both frames because inertial frames are homogeneous in spacetime and isotropic in space. The orientation of the rectangular axes in both frames can be arbitrary.

From for formulas you showed follows $K(V) = \pm 1$. The negative sign can be dismissed for $v \rightarrow 0$ and then argue with continuity.


Where did you read this?

I read this proof in Russian Wiki

 

[Mike_bb]

Does this interpretation work in experiment of time dilation using photon clock?

"The second inertial frame of reference looks from the first in no way different from how the first inertial frame of reference looks from the second."

If it works then could anyone explain how? Thanks.

 

[Sagittarius A-Star]

I read this proof in Russian Wiki

The validity of one equation you cited in the OP can be derived from the 2nd postulate of SR (invariance of the speed of light in vacuum), and from there for spacetime intervals unequal to zero:

Source (1960 book "Special Relativity" of Wolfgang Rindler, in §8 on page 16:):
https://www.amazon.com/-/de/dp/101342879X/?tag=pfamazon01-20

 

[Sagittarius A-Star]

Does this interpretation work in experiment of time dilation using photon clock?

"The second inertial frame of reference looks from the first in no way different from how the first inertial frame of reference looks from the second."


View attachment 356304


If it works then could anyone explain how? Thanks.

It works with the relativity of simultaneity. In the Lorentz transformation, $t'$ is a function not only of $t$, but also of $x$.

 

[Mike_bb]

The validity of one equation you cited in the OP can be derived from the 2nd postulate of SR (invariance of the speed of light in vacuum), and from there for spacetime intervals unequal to zero

My thread is about interpretation of principle of relativity rather than proof.
I try to understand how "The second inertial frame of reference looks from the first in no way different from how the first inertial frame of reference looks from the second"?

 

[Mike_bb]

How could this interpretation be apply to proof?

 

[Sagittarius A-Star]

My thread is about interpretation of principle of relativity rather than proof.
I try to understand how "The second inertial frame of reference looks from the first in no way different from how the first inertial frame of reference looks from the second"?

You can compare the elapsed proper times of the two light clocks in posting #4 only directly when they meet a second time (twin "paradox"). If the "moving" light clocks turns around to meet the "stationary" light clock a second time, then the scenario is asymmetrical. For calculating the time dilation in case of no turnaround, you need to involve the coordinate-time of the reference frame.

 

[Mike_bb]

You can compare the elapsed proper times of the two light clocks in posting #4 only directly when they meet a second time (twin "paradox"). If the "moving" light clocks turns around to meet the "stationary" light clock a second time, then the scenario is asymmetrical. For calculating the time dilation in case of no turnaround, you need to involve the coordinate-time of the reference frame.

We have two reference frames: first stationary reference frame and second reference frame moving with velocity V.
Does this interpretation hold for this example?

 

[Sagittarius A-Star]

We have two reference frames: first stationary reference frame and second reference frame moving with velocity V.
Does this interpretation hold in this example?

Yes. According to the principle of relativity, you are also allowed to regard the first frame as moving and the second as stationary.

 

[Mike_bb]

Yes. According to the principle of relativity, you are also allowed to regard the first frame as moving and the second as stationary.

Can this lead to following formulas?

 

[Sagittarius A-Star]

How could this interpretation be apply to proof?

If we omit the $y$ and $z$ coordinates, then the invariance of the spacetime interval is
$$c^2\Delta t^2-\Delta x^2 = c^2\Delta t'^2-\Delta x'^2$$If a clock rests in the primed frame, then between two tick-events $\Delta x' =0$ and the clock's proper time is $\Delta \tau' = \Delta t'$.
It follows:
$c^2\Delta t^2 (1-{\Delta x^2 \over \Delta t^2})= c^2\Delta \tau'^2-0$

$\Rightarrow \Delta \tau'^2 = \Delta t^2 (1-v^2/c^2)$

You could repeat the calculation for the reciprocal case $\Delta x=0$ and $\Delta \tau = \Delta t$ for a clock that rests in the unprimed frame.

 

[Sagittarius A-Star]

Can this lead to following formulas?

View attachment 356307

Yes, but only in combination with i.e. what I wrote in posting #5.

 

[Mike_bb]

Hello, Sagittarius A-Star!

If we omit the $y$ and $z$ coordinates, then the invariance of the spacetime interval is
$$c^2\Delta t^2-\Delta x^2 = c^2\Delta t'^2-\Delta x'^2$$If a clock rests in the primed frame, then between two tick-events $\Delta x' =0$ and the clock's proper time is $\Delta \tau' = \Delta t'$.
It follows:
$c^2\Delta t^2 (1-{\Delta x^2 \over \Delta t^2})= c^2\Delta \tau'^2-0$

$\Rightarrow \Delta \tau'^2 = \Delta t^2 (1-v^2/c^2)$

You could repeat the calculation for the reciprocal case $\Delta x=0$ and $\Delta \tau = \Delta t$ for a clock that rests in the unprimed frame.

Thanks. I obtained the same results.
But I have a doubt about representation on picture #2 where inertial frame with photon clock is stationary and another inertial frame is moving with velocity V to the left side.

1.

2.

No matter how hard I tried, I couldn't obtain

How to obtain this equation?
Thanks!

 

[PeroK]

My thread is about interpretation of principle of relativity rather than proof.
I try to understand how "The second inertial frame of reference looks from the first in no way different from how the first inertial frame of reference looks from the second"?

If space and time are isotropic and homogeneous, then we can argue as follows:

Suppose the origin of one inertial reference frame (frame B) is moving relative to another (frame A). Because of our assumption, we can choose coordinates where the origin of the one frame $O_B$ is moving along the x-axis of frame A. We can formalise this and say that the position of $O_B$ is $vt_A$, where $v$ is some constant (representing the velocity of the origin of frame B, as measured in frame A).

Note that this is something we do naturally when we tackle any kinematics problem. We choose a suitable time and space origin and choose our coordinate axes in the most suitable way. If time were not homogeneous, we would not be able to do this - there would be an absolute $t=0$ and space and/or time would be different in some way at other times. Likewise, the isotropy and homogeneity of space allows us to choose our spatial origin and x-axis as the direction of motion etc.

We can also take $v$ to be positive and, using our normal convention, if we draw a diagram of this scenario, then $O_B$ is moving along the x-axis away to the right from $O_A$, with the two origins coinciding at time $t_A = 0$ and $t_B = 0$.

We can see that the motion of $O_A$ as measured in frame B must be $-v't_B$, where $v'$ is some positive constant that represents the speed of $O_A$, as measured in frame B.

Moreover, if $v' \ne v$, then we must have either $v' < v$ or $v' > v$. But, the only variable we have is that $O_B$ is moving to the right. If we now change the direction of our x-axis, we find that $O_A$ is moving to the right, as measured in frame B. So, we must have $v < v' < v$ or $v > v' > v$, corresponding to the two cases above. This is a contradiction, so we must have $v' = v$.

And now we see that there is complete symmetry. The motion of $O_B$ as measured in frame A is $vt_A$. And, the motion of $O_A$ as measured in frame B is $-vt_B$. Finally, by changing the orientation of our x-axis in the second case, we can equally describe the motion as $vt_B$. And that is exactly the same description of frame A as frame A's description of frame B.

That gives us the required conclusion that whatever A measures to be the case in frame B, frame B (by an appropriate choice of origin and orientation of the axes) can measure to be the case in frame A.

This is the fundamental symmetry argument that leads us to the conclusion that, for inertial reference frames in isotropic and homogeneous space and time, there is a single parameter $v$ that defines the relative motion between them; and that the motion of each is physically identical as measured by the other.

 

[martinbn]

My thread is about interpretation of principle of relativity rather than proof.
I try to understand how "The second inertial frame of reference looks from the first in no way different from how the first inertial frame of reference looks from the second"?

What is the statement of the principle of relativity that you use? Because to me this is not an interpretation it is just a particular formulation of the principle.

 

[Sagittarius A-Star]

No matter how hard I tried, I couldn't obtain View attachment 356459
How to obtain this equation?
Thanks!

You can obtain your light clock drawings from the 2nd postulate (the speed of light is the same in both inertial frames, which are moving relative to each other). I propose that you use in the 2nd drawing primed coordinates because that is a different coordinate system.

Form the 2nd postulate you can also derive the formula you mentioned.

 

[Mike_bb]

I propose that you use in the 2nd drawing primed coordinates because that is a different coordinate system.

Could one drawing be enough?

From this drawing:

1. From 1st inertial frame we can see 2nd inertial frame (moving clock with velocity V) and this lead to ${ds_2}^2=k(V){ds_1}^2$
2. From 2nd inertial frame we can see 1st inertial frame but we can't conclude that ${ds_1}^2=k(V){ds_2}^2$ because initially we agree to consider 2nd inertial frame relatively to 1st inertial frame and not 1st relatively to 2nd.

 

[PeroK]

Could one drawing be enough?

View attachment 356465

From this drawing:

1. From 1st inertial frame we can see 2nd inertial frame (moving clock with velocity V) and this lead to $ds^2=k(V)ds^1$
2. From 2nd inertial frame we can see 1st inertial frame but we can't conclude that $ds^1=k(V)ds^2$ because initially we agree to consider 2nd inertial frame relatively to 1st inertial frame and not 1st relatively to 2nd.

The proposition that I justified in post #16 is not directly derived from this particular scenario. Nor, in fact, does it depend on the invariance of the speed of light: the argument I made in post #16 applies equally in Newtonian space and time. Note that, as an aside, Special Relativity and Newtonian space and time are essentially the only two possibilities for homogeneous and isotropic spacetime.

I suggest that the authors have invoked this principle of equivalence of all inertial reference frames in addition to the invariance of $c$. You may be able to justify it from the invariance of the speed of light alone - although, if you could, there would be no need for the first postulate of SR. I suspect, therefore, that you cannot justify the proposition from an analysis of a light clock alone.

In fact, my argument in post #16 is really based on the first postulate - as one of the direct implications of the first postulate is the isotropy and homegeneity of space and time.

 

[Mike_bb]

The proposition that I justified in post #16 is not directly derived from this particular scenario. Nor, in fact, does it depend on the invariance of the speed of light: the argument I made in post #16 applies equally in Newtonian space and time. Note that, as an aside, Special Relativity and Newtonian space and time are essentially the only two possibilities for homogeneous and isotropic spacetime.

I suggest that the authors have invoked this principle of equivalence of all inertial reference frames in addition to the invariance of $c$. You may be able to justify it from the invariance of the speed of light alone - although, if you could, there would be no need for the first postulate of SR. I suspect, therefore, that you cannot justify the proposition from an analysis of a light clock alone.

In fact, my argument in post #16 is really based on the first postulate - as one of the direct implications of the first postulate is the isotropy and homegeneity of space and time.

I need to lead to

without using only 2nd postulate.

 

[Sagittarius A-Star]

I read this proof in Russian Wiki

Can you please provide the link?

 

[Mike_bb]

Can you please provide the link?

Proof

 

[Sagittarius A-Star]

Proof

View attachment 356467

They argue in the Russian Wiki with the 1st postulate. English translation:

 

[Sagittarius A-Star]

Could one drawing be enough?

View attachment 356465

From this drawing:

1. From 1st inertial frame we can see 2nd inertial frame (moving clock with velocity V) and this lead to ${ds_2}^2=k(V){ds_1}^2$
2. From 2nd inertial frame we can see 1st inertial frame but we can't conclude that ${ds_1}^2=k(V){ds_2}^2$ because initially we agree to consider 2nd inertial frame relatively to 1st inertial frame and not 1st relatively to 2nd.

No. The mentioned formula cannot be derived from the light clock drawings, but from the two postulates as shown in the Russian Wiki.

 

[Mike_bb]

No. The mentioned formula cannot be derived from the light clock drawings, but from the two postulates as shown in the Russian Wiki.

What does this mean roughly speaking? :

 

[Sagittarius A-Star]

What does this mean roughly speaking? :
View attachment 356469

It means that you can interchange the roles of system 1 and system 2.

 

[Mike_bb]

It means that you can interchange the roles of system 1 and system 2.

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?

 

[PeroK]

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?

What special about the first frame? Or, it is the second frame that's special?

 

[Mike_bb]

What special about the first frame? Or, it is the second frame that's special?

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

 

[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"?