Understanding Relativity: A Blind Man's Perspective on Time and Physics

[Dale]

Hai Dalespam,Lorentz tranform is not such a complicated thing.Only Debate is wheather the results (length contraction and time dilation) have a real physical meaning or it is only the perception of the observer.

That has been well http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html" .

Please don't come to a conclusion before reading carefully,what i am saying.In my drawing and my explanation clocks are moving perpendicular to their length 'L'.
when the moving observer turn back and see the other clock,he will see the time in that clock is moving fast.I think you will agree this.
For any observer velocity of light is 'c' a constant.
for him if L' = 1m =ct'
or t' = 1/c
if he see back and find L = 1m
then ct = 1m
so t = 1/c
Then he will not observe any time dilation.

No, I do not agree. That violates the first postulate. Since the laws of physics are the same in all inertial reference frames then if a moving clock is slow in one frame a moving clock must be slow in all frames. The symmetry is required by the first postulate.

From the Lorentz transform in units where c=1:
\left(<br /> \begin{array}{l}<br /> t&#039; \\<br /> x&#039; \\<br /> y&#039; \\<br /> z&#039;<br /> \end{array}<br /> \right)=\left(<br /> \begin{array}{llll}<br /> \gamma &amp; -v \gamma &amp; 0 &amp; 0 \\<br /> -v \gamma &amp; \gamma &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 1 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 1<br /> \end{array}<br /> \right).\left(<br /> \begin{array}{l}<br /> t \\<br /> x \\<br /> y \\<br /> z<br /> \end{array}<br /> \right)
In the first column and first row you see that the primed clock ticks slower by the factor γ in the unprimed frame.

Solving for the unprimed frame we obtain:
\left(<br /> \begin{array}{l}<br /> t \\<br /> x \\<br /> y \\<br /> z<br /> \end{array}<br /> \right)=\left(<br /> \begin{array}{llll}<br /> \gamma &amp; v \gamma &amp; 0 &amp; 0 \\<br /> v \gamma &amp; \gamma &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 1 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 1<br /> \end{array}<br /> \right).\left(<br /> \begin{array}{l}<br /> t&#039; \\<br /> x&#039; \\<br /> y&#039; \\<br /> z&#039;<br /> \end{array}<br /> \right)
In the first column and first row you see that the unprimed clock ticks slower by the factor γ in the primed frame.

 

[newTonn]

That has been well http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html" .
No, I do not agree. That violates the first postulate. Since the laws of physics are the same in all inertial reference frames then if a moving clock is slow in one frame a moving clock must be slow in all frames. The symmetry is required by the first postulate.

From the Lorentz transform in units where c=1:
\left(<br /> \begin{array}{l}<br /> t&#039; \\<br /> x&#039; \\<br /> y&#039; \\<br /> z&#039;<br /> \end{array}<br /> \right)=\left(<br /> \begin{array}{llll}<br /> \gamma &amp; -v \gamma &amp; 0 &amp; 0 \\<br /> -v \gamma &amp; \gamma &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 1 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 1<br /> \end{array}<br /> \right).\left(<br /> \begin{array}{l}<br /> t \\<br /> x \\<br /> y \\<br /> z<br /> \end{array}<br /> \right)
In the first column and first row you see that the primed clock ticks slower by the factor γ in the unprimed frame.

Solving for the unprimed frame we obtain:
\left(<br /> \begin{array}{l}<br /> t \\<br /> x \\<br /> y \\<br /> z<br /> \end{array}<br /> \right)=\left(<br /> \begin{array}{llll}<br /> \gamma &amp; v \gamma &amp; 0 &amp; 0 \\<br /> v \gamma &amp; \gamma &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 1 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 1<br /> \end{array}<br /> \right).\left(<br /> \begin{array}{l}<br /> t&#039; \\<br /> x&#039; \\<br /> y&#039; \\<br /> z&#039;<br /> \end{array}<br /> \right)
In the first column and first row you see that the unprimed clock ticks slower by the factor γ in the primed frame.

May be i couldn't convey properly what i want to say.I will try to explain it further.
The speed of light is 'c' constant for all observers.
1st case
If Observer A(regardless of which frame),found a clock ticking faster in other frame,that means the light in that clock has to travel a lesser distance than that of the length to be traveled in his clock to reach the opposite side.
2nd case
Similarly if he found a clock ticking slower in other frame,it means the light in the other clock has to travel more distance than that of his clock.
Otherwise we can say,in the first case,for observer A,light appears to be traveling with more than the speed 'c' and in the second case light appears to be traveling with less than the speed 'c',i think which is against the special relativity.

 

[Dale]

May be i couldn't convey properly what i want to say.I will try to explain it further.
The speed of light is 'c' constant for all observers.
1st case
If Observer A(regardless of which frame),found a clock ticking faster in other frame,that means the light in that clock has to travel a lesser distance than that of the length to be traveled in his clock to reach the opposite side.
2nd case
Similarly if he found a clock ticking slower in other frame,it means the light in the other clock has to travel more distance than that of his clock.
Otherwise we can say,in the first case,for observer A,light appears to be traveling with more than the speed 'c' and in the second case light appears to be traveling with less than the speed 'c',i think which is against the special relativity.

Yes, these are both correct. However, only the 2nd case ever happens. The distance that the light travels is always greater in the frame where the clock moves than in the frame where the clock is stationary.

This is a direct consequence of the Pythagorean theorem where the hypotenuse is always longer than either of the two legs (one leg being the length of the clock and the other being the distance the clock travels during one tick).

 

[newTonn]

Yes, these are both correct. However, only the 2nd case ever happens. The distance that the light travels is always greater in the frame where the clock moves than in the frame where the clock is stationary.

This is a direct consequence of the Pythagorean theorem where the hypotenuse is always longer than either of the two legs (one leg being the length of the clock and the other being the distance the clock travels during one tick).

So in fact you agree that the length is contracted in the perpendicular direction of motion also.(1st case)
What does this means?.The moving observer will see the other objects in a smaller scale,propotional to his relative velocity.
Do this observation prove that the object is physicaly contracted or is it only the perspective of the observer?

 

[Dale]

So in fact you agree that the length is contracted in the perpendicular direction of motion also.(1st case)

No I do not agree. I don't know how you could come to that conclusion when I have clearly and consistently stated that there is no length contraction in the perpendicular direction:

only the 2nd case ever happens.

There is no length contraction in the direction perpendicular to the relative motion.

From your drawing I though that you were considering clocks that were moving perpendicular to their length. If so then there is no length contraction involved.

What I agreed with is that the distance, d, traveled by light in a certain amount of time, t, in any inertial frame is always d=ct. This does not imply length contraction for a clock oriented in the perpendicular direction, only time dilation.

I have been very clear about my position and for you to make this assertion seems a deliberate attempt to pretend misunderstanding. When I have not understood something you posted I have specifically identified it, please have the courtesy to do the same.

 

[newTonn]

No I do not agree. I don't know how you could come to that conclusion when I have clearly and consistently stated that there is no length contraction in the perpendicular direction:

What I agreed with is that the distance, d, traveled by light in a certain amount of time, t, in any inertial frame is always d=ct. This does not imply length contraction for a clock oriented in the perpendicular direction, only time dilation.

I have been very clear about my position and for you to make this assertion seems a deliberate attempt to pretend misunderstanding. When I have not understood something you posted I have specifically identified it, please have the courtesy to do the same.

I am really sorry if you feel that as a deliberate attempt.
Now coming back to the point
Let us say 'd' is the distance between the mirrors(or reflectors or sensors) and 't' is the time taken by the light beam to reach from one mirror to other for both the clocks, when there is no relative velocity.
d= ct for both observers.
Now one of the clock moved perpendicular to the clock with a relative velocity near to that of light.
The observer ,A in the moving frame will not see any difference in his clock
So still,for his clock, d=ct-----(1)
when he turns back to other clock there is a time difference in the other clock.let us say time 't1'.
if he still see the distance between the mirrors of other clock as d
then equation for the other clock from his frame will be
d = ct1----(2)
since left hand side of the equations are same,equating RHS we will get

ct1=ct
Since c is constant,t=t1 (no time dilation?)
if he observes a time dilation then d in the other clock should be either > or < d in his clock and never will be equal to d.That means a length contraction.

 

[Dale]

if he still see the distance between the mirrors of other clock as d
then equation for the other clock from his frame will be
d = ct1----(2)
since left hand side of the equations are same,equating RHS we will get

Here is the key mistake. d is the distance between the mirrors of the other clock, but d is not the distance that the light travels. The light travels a distance equal to the hypotenuse of a right triangle with one leg being d and the other leg being (v t1).

d^2 + (v t1)^2 = (c t1)^2

 

[newTonn]

Here is the key mistake. d is the distance between the mirrors of the other clock, but d is not the distance that the light travels. The light travels a distance equal to the hypotenuse of a right triangle with one leg being d and the other leg being (v t1).

d^2 + (v t1)^2 = (c t1)^2

But the observer will always see the light traveling in a straight line perpendicular to the mirrors,because there is no relative motion of the mirrors(in x,y and z direction) at any point of time.
So his conclusion will be either the light is traveling at a lesser or more speed than c or the length is shortened or elongated.(triangle is a third party observation-or else result of assigning an absolute space )

 

[JesseM]

But the observer will always see the light traveling in a straight line perpendicular to the mirrors,because there is no relative motion of the mirrors(in x,y and z direction) at any point of time.
So his conclusion will be either the light is traveling at a lesser or more speed than c or the length is shortened or elongated.(triangle is a third party observation-or else result of assigning an absolute space )

What observer are you talking about here? An observer at rest relative to the mirrors, or one who sees the mirrors moving in the x direction? If the latter, naturally he must see the light moving diagonally, not perpendicular to the axis of the mirrors, assuming the light is aimed straight up in the rest frame of the mirrors.

 

[newTonn]

What observer are you talking about here? An observer at rest relative to the mirrors, or one who sees the mirrors moving in the x direction? If the latter, naturally he must see the light moving diagonally, not perpendicular to the axis of the mirrors, assuming the light is aimed straight up in the rest frame of the mirrors.

No.He is observing light beam and mirror moving together in x direction.So the relative position of light beam and the mirrors will be such that it will be always on a line which is perpendicular with the mirror.He will always see the light beam moving perpendicular to the mirror.I will agree that It is traveling diognaly with respect to fixed space.
This is true with galelian relativity also.If ship is the only reference,the observer at land will see the ball falling verticaly from mast to the deck of ship.If there is a bird or anything in the background which have no relative motion with ship,he can see a parabolic path to the ball with respect to the bird.

 

[JesseM]

No.He is observing light beam and mirror moving together in x direction.So the relative position of light beam and the mirrors will be such that it will be always on a line which is perpendicular with the mirror.He will always see the light beam moving perpendicular to the mirror.I will agree that It is traveling diognaly with respect to fixed space.

Of course at any given instant the position of the photon is always along the line between the two mirrors, but because the mirrors are moving in his coordinate system, the direction of motion of the photon in his coordinate system is not perpendicular to the line between the two mirrors.

 

[newTonn]

Of course at any given instant the position of the photon is always along the line between the two mirrors, but because the mirrors are moving in his coordinate system, the direction of motion of the photon in his coordinate system is not perpendicular to the line between the two mirrors.

Then he can easily say that the other clock is wrong,instead of telling the time is dilated.
For example,You have a sand clock(the old one),which is working purely with gravity.You designed it to work at ground level.If you take it to a high altitude, there is no logic in expecting it to work as it was in the ground level .
Similarly a light clock designed for a rest frame will not work same in a moving frame,isn't it the whole thing? or there is anything more?

 

[Dale]

But the observer will always see the light traveling in a straight line perpendicular to the mirrors,because there is no relative motion of the mirrors(in x,y and z direction) at any point of time.

No, all observers in all inertial frames will always see the light traveling in a straight line between the event where it leaves one mirror and the event where it reaches the other mirror. That line will only be perpendicular to the mirrors in the clock's rest frame. If the light were to travel perpendicular to the mirrors in any other frame then it would miss the other mirror.

 

[newTonn]

No, all observers in all inertial frames will always see the light traveling in a straight line between the event where it leaves one mirror and the event where it reaches the other mirror. That line will only be perpendicular to the mirrors in the clock's rest frame. If the light were to travel perpendicular to the mirrors in any other frame then it would miss the other mirror.

The diognal movement of light will be identified by the observer only if there is a reference object in the background,which is still in the observers frame.
In the absense of this third reference,the observer will see a perpendicular motion of light beam with respect to mirrors.
if the observer,measure the distance from him to the beam at each minute fraction of a second and plot the position of the beam,he will get a diognal line.
But then he should be honest enough to tell that a light clock designed in a still frame will not work properly in a moving frame,because the principle- basis of measurement in this clock,depends on the distance traveled by the light.
So in a moving frame time will be misunderstood as dilated,instead of understanding correctly that the light has to travel more distance(and hence more time) to reach the other mirror.

 

[Dale]

The diognal movement of light will be identified by the observer only if there is a reference object in the background,which is still in the observers frame.
In the absense of this third reference,the observer will see a perpendicular motion of light beam with respect to mirrors.

You are really stretching here. No "reference object" is required. The light leaving one mirror is one event, the light reaching the other mirror is another event, the light travels a straight world line between the two events. No external reference object is required.

The distance between the two events is, in general, not d.

 

[newTonn]

You are really stretching here. No "reference object" is required. The light leaving one mirror is one event, the light reaching the other mirror is another event, the light travels a straight world line between the two events. No external reference object is required.

The distance between the two events is, in general, not d.

If i agree with your last statement,If the distance is not d,why should i say time is dilated,instead of saying the basic principle of measurement of time in the moving clock is wrong?

 

[Dale]

If i agree with your last statement,

Do you agree with the last statement? I don't want to start a new discussion when we still disagree about the geometry. I already feel like we left the vacuum discussion prematurely.

 

[JesseM]

The diognal movement of light will be identified by the observer only if there is a reference object in the background,which is still in the observers frame.
In the absense of this third reference,the observer will see a perpendicular motion of light beam with respect to mirrors.

Are you completely unfamiliar with the notion of inertial coordinate systems in relativity? Each observer is assumed to use a network of rulers and clocks at rest in their own frame to assign coordinates to events, and it is only relative to these rulers and clocks that statements like "light moves at c in every frame" are meaningful.

 

[newTonn]

Are you completely unfamiliar with the notion of inertial coordinate systems in relativity?

You have ignored part of my post where i stated"if the observer,measure the distance from him to the beam at each minute fraction of a second and plot the position of the beam,he will get a diognal line." to show that i am unfamiliar with the co-ordinate systems.

Each observer is assumed to use a network of rulers and clocks at rest in their own frame to assign coordinates to events, and it is only relative to these rulers and clocks that statements like "light moves at c in every frame" are meaningful.

Does this directly implies that there is no physical change,but only the difference in rulers and clocks?