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

[Eidos]

Incidentally, that is how you can get a shadow to move faster than the speed of light. If I take an object and put a light source behind it and rotate the light source about the object, the shadow that the object makes very far away can move faster than the speed of light.

$$\omega=\mathbf{r}\times\mathbf{v}=|\mathbf{r}||\mathbf{v}|\sin{\theta}$$
then
$$\exists\, r: |\mathbf{r}||\mathbf{v}|\sin{\theta} > c$$

 

[newTonn]

That has no physical meaning. In order for something to move it must have a velocity. Please answer the question: How do you propose to measure the velocity of the vacuum?

If there is no way to physically measure its velocity then there is no physical meaning of assigning it a velocity. The vacuuum does not in any physical sense move, it is not an entity to which velocity may be assigned. All that happens is that the particles of the clock move as do any particles in the environment. Where there are no particles there is empty space or vacuum. The vacuum itself is not a thing to which a velocity may be assigned, it is simply the absence of things.

As i clearly mentioned,The vacuum inside a vacuum chamber is moving from one place to another.Physically you can measure the initial position and final position of vacuum with a measuring tape(if the chamber is shifted only for a few meters).

 

[Dale]

It is OK, newTonn, I thought about this off and on for about 5 years before I figured it out, so don't feel bad that you don't get it right away.

What you have described is measuring the position of the container, not the velocity of the vacuum. To simplify things, imagine that you are adrift in intergalactic space. How do you measure the velocity of the vacuum that surrounds you?

 

[Eidos]

Incidentally, that is how you can get a shadow to move faster than the speed of light. If I take an object and put a light source behind it and rotate the light source about the object, the shadow that the object makes very far away can move faster than the speed of light.

$$\omega=\mathbf{r}\times\mathbf{v}=|\mathbf{r}||\mathbf{v}|\sin{\theta}$$
then
$$\exists\, r: |\mathbf{r}||\mathbf{v}|\sin{\theta} > c$$

To be clear, a shadow is not a 'thing', it is the absence of things.

 

[newTonn]

It is OK, newTonn, I thought about this off and on for about 5 years before I figured it out, so don't feel bad that you don't get it right away.

What you have described is measuring the position of the container, not the velocity of the vacuum. To simplify things, imagine that you are adrift in intergalactic space. How do you measure the velocity of the vacuum that surrounds you?

I am talking about the vacuum inside the container(clock) not the surrounding vaccum.I never mentioned that universe is enclosed.

 

[Dale]

I am talking about the vacuum inside the container(clock) not the surrounding vaccum.I never mentioned that universe is enclosed.

That is an interesting response. I can think of a few different ways to measure the velocity of air. All of them work just as well if the air is enclosed or not. Why would it be more difficult to measure the velocity of interstellar vacuum than the velocity of the vacuum in a bottle? Are you sure your proposed measurement technique is measuring the velocity of the vacuum?

 

[newTonn]

That is an interesting response. I can think of a few different ways to measure the velocity of air. All of them work just as well if the air is enclosed or not. Why would it be more difficult to measure the velocity of interstellar vacuum than the velocity of the vacuum in a bottle? Are you sure your proposed measurement technique is measuring the velocity of the vacuum?

Dalespam as far as analogy of sound clock to light clockis considered,we must not be bothered about the velocity of surrounding.Anyhow l am attaching some of my doubts and reasoning about the moving clock.
Can you please clear the doubts?

 

[Dale]

Dalespam as far as analogy of sound clock to light clockis considered,we must not be bothered about the velocity of surrounding.

I agree completely, but don't you see that is exactly what you are doing. You are focusing on the clock that surrounds the vacuum and atributing the surrounding clock's velocity to the vacuum. You are not considering the velocity of the vacuum itself. The vacuum has no velocity regardless of its surroundings.

I was not trying to move to interstellar vacuum in order to consider surroundings, but in order to get you to consider the vacuum itself rather than its surroundings.

Anyhow l am attaching some of my doubts and reasoning about the moving clock.
Can you please clear the doubts?

Sure, you have a right triangle with two legs of length:
L
v t
and a hypotenuse of length
c t

by Pythagorean theorem

L^2 + (vt)^2 = (ct)^2

Which is one equation in one unknown. Solve for t to get the time that it takes light to make the trip one way.

 

[newTonn]

I agree completely, but don't you see that is exactly what you are doing. You are focusing on the clock that surrounds the vacuum and atributing the surrounding clock's velocity to the vacuum. You are not considering the velocity of the vacuum itself. The vacuum has no velocity regardless of its surroundings.

I was not trying to move to interstellar vacuum in order to consider surroundings, but in order to get you to consider the vacuum itself rather than its surroundings.

Sure, you have a right triangle with two legs of length:
L
v t
and a hypotenuse of length
c t

by Pythagorean theorem

L^2 + (vt)^2 = (ct)^2

Which is one equation in one unknown. Solve for t to get the time that it takes light to make the trip one way.

you didnt get my point .solve t by the equation and put v*t for the distance traveled by clock.You will see the clock is still ahead of light.

 

[Dale]

you didnt get my point .solve t by the equation and put v*t for the distance traveled by clock.You will see the clock is still ahead of light.

That is incorrect:
ct > vt
for all t>0 and c>v

So since ct>vt in what sense does the clock get ahead of the light?

You might want to show the derivation of your figures. I don't really understand what they mean. It seems like you are trying to set up an infinite sum and then claim that light never reaches the mirror. I hope you are aware that many infinite sums do converge.

 

[newTonn]

That is incorrect:
ct > vt
for all t>0 and c>v

So since ct>vt in what sense does the clock get ahead of the light?

You might want to show the derivation of your figures. I don't really understand what they mean. It seems like you are trying to set up an infinite sum and then claim that light never reaches the mirror. I hope you are aware that many infinite sums do converge.

But L=ct ; so the triangle will remains open always.

 

[Dale]

But L=ct ; so the triangle will remains open always.

No.

L^2 + (vt)^2 = (ct)^2

So L=ct iff v=0.

 

[newTonn]

No.

L^2 + (vt)^2 = (ct)^2

So L=ct iff v=0.

Does it means there is a length expansion in the direction perpendicular to the motion when v>0.?

 

[newTonn]

i would like to add that the L is shortened when v>0 and still it will give you an open triangle.

 

[Dale]

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

Since c>v the light will eventually catch up with the detector. Is that clear to you? If so, then we can define t to be the time between when the light was released from the source to when it was received by the detector. During time t the detector traveled a distance vt and the light traveled a distance ct. So L, vt, and ct form a right triangle with ct being the hypotenuse. So

L^2 + (vt)^2 = (ct)^2

is just the Pythagorean theorem.

 

[Antenna Guy]

...During time t the detector traveled a distance vt and the light traveled a distance ct. So L, vt, and ct form a right triangle with ct being the hypotenuse. So

L^2 + (vt)^2 = (ct)^2

is just the Pythagorean theorem.

Let ct=1, and your run-of-the-mill space-time diagram becomes a velocity diagram; wherein all velocities are normalized with respect to the velocity of light.

Regards,

Bill

 

[newTonn]

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

Since c>v the light will eventually catch up with the detector. Is that clear to you? If so, then we can define t to be the time between when the light was released from the source to when it was received by the detector. During time t the detector traveled a distance vt and the light traveled a distance ct. So L, vt, and ct form a right triangle with ct being the hypotenuse. So

L^2 + (vt)^2 = (ct)^2

is just the Pythagorean theorem.

let us consider t' as the time solved for the moving observer.For him L =ct'.When he looks back to the other clock,which is not moving, he will see the L of other clock is shortened and hence time in his clock moving fast .For example if v =0.9c,and L'=1m for the moving observer, t' = 7.65E-9 seconds.
When he looks back to other clock,he will see the light in that clock has traveled 2.29 times of L in the time interval t'.So for him L = 1/2.29 = 0.436 m.
Similarly,the observer who is not moving will see L = 1m and L'=2.29m.
Now stop the moving frame x'y'z' and call it x1y1z1(sudden stop can be allowed in a thought experiment)
.Then L= L1= 1 m as observed from both frame and As observed from xyz there will be a lag of time in x1y1z1.Now let the other clock move with 0.9c in the same direction.
(frame x''y''z'').He will see L'' = 1m and the L1 = 0.436m(the clock which stopped moving) and the clock at x1y1z1 ticks faster.Until finaly when they meet ,both of them will agree same time.
So physically nothing is changed,in due course.If they didnt observe each other in this journey,time.length and all the physical measurements are exactly same for both.

 

[neopolitan]

let us consider t' as the time solved for the moving observer.For him L =ct'.When he looks back to the other clock,which is not moving, he will see the L of other clock is shortened and hence time in his clock moving fast .For example if v =0.9c,and L'=1m for the moving observer, t' = 7.65E-9 seconds.
When he looks back to other clock,he will see the light in that clock has traveled 2.29 times of L in the time interval t'.So for him L = 1/2.29 = 0.436 m.
Similarly,the observer who is not moving will see L = 1m and L'=2.29m.
Now stop the moving frame x'y'z' and call it x1y1z1(sudden stop can be allowed in a thought experiment)
.Then L= L1= 1 m as observed from both frame and As observed from xyz there will be a lag of time in x1y1z1.Now let the other clock move with 0.9c in the same direction.
(frame x''y''z'').He will see L'' = 1m and the L1 = 0.436m(the clock which stopped moving) and the clock at x1y1z1 ticks faster.Until finaly when they meet ,both of them will agree same time.
So physically nothing is changed,in due course.If they didnt observe each other in this journey,time.length and all the physical measurements are exactly same for both.

I think I have addressed something like this in a different thread, where I pointed out that there is no real "twins paradox". The reason I am unsure is that it is difficult to work out precisely what you are saying.

There is a consistency in both frames, but I don't think that amounts to clocks in different frames being synchronised except in very special circumstances. In the scenario above, if I have it right you have one clock moving away from the other with a relative velocity. Then that clock stops suddenly and the other clock suddenly moves towards the other with the same relative velocity. If the clocks were initially synchronised, then they will be synchronised when they meet again. That is true enough.

But I don't think the clocks would be synchronised if the relative velocities of separation and approach were different. You can use three clocks and a long ruler to work that out.

You have a clock that just sits on the ruler (is at rest relative to the ruler). The ruler has a rest length of L. In your scenario, it measures the time for one clock to move from one end of the ruler to the other at a velocity v1. Let's make that velocity c/a where a>0. Then that same clock which remains at rest relative to the ruler), immediately measures the time it takes for the second clock to move from one end of the ruler to the other at a velocity v2. Let's make that velocity c/b where b>0.

The total time expired according to this clock at rest relative to the ruler is
t3 = L/v1 + L/v2 = aL/c + bL/c

If the first clock moves, then stops suddenly, the total time expired is
t1 = L'1/v1 + L/v2 = aL/c * sqrt (1 - (c/(a/c))^2) + bL/c = L/c * sqrt (a^2 - 1) + bL/c

If the second clock waits, then starts moving suddenly, the total time expired is
t2 = L/v1 + L'2/v2 = aL/c + bL/c * sqrt (1 - (c/(b/c))^2) = aL/c + L/c * sqrt (b^2 - 1)

The times t1 and t2 will be equal when

sqrt (a^2 - 1) + b = a + sqrt (b^2 - 1)

and as far as I can tell, the only solution to this is where a=b.

And ... the time discrepancy between the clock at rest relative to the ruler and the other two clocks is irreducible unless a=b=1, which means your clock must have some sort of photonic construction.

cheers,

neopolitan

 

[Dale]

When he looks back to the other clock,which is not moving, he will see the L of other clock is shortened and hence time in his clock moving fast .

You are really getting confusing here. 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.

When you are learning something you should always stick with the simplest thing that you don't understand. There is absolutely no point in trying to analyze a complicated situation when you already don't understand a far simpler one.

Analyze only a single light clock moving inertially in a direction perpendicular to the length. Once you understand that, then analyze a single light clock moving inertially in a direction parallel to the length. Use what you learned from those two analyses to understand the Lorentz transform. Only go to complicated scenarios with multiple clocks or non-inertial clocks after you fully understand the Lorentz transform.

 

[newTonn]

.

When you are learning something you should always stick with the simplest thing that you don't understand. There is absolutely no point in trying to analyze a complicated situation when you already don't understand a far simpler one.

Analyze only a single light clock moving inertially in a direction perpendicular to the length. Once you understand that, then analyze a single light clock moving inertially in a direction parallel to the length. Use what you learned from those two analyses to understand the Lorentz transform. Only go to complicated scenarios with multiple clocks or non-inertial clocks after you fully understand the Lorentz transform.

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.

.You are really getting confusing here. 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

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.

 

[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:
$$\left( \begin{array}{l} t' \\ x' \\ y' \\ z' \end{array} \right)=\left( \begin{array}{llll} \gamma & -v \gamma & 0 & 0 \\ -v \gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right).\left( \begin{array}{l} t \\ x \\ y \\ z \end{array} \right)$$
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:
$$\left( \begin{array}{l} t \\ x \\ y \\ z \end{array} \right)=\left( \begin{array}{llll} \gamma & v \gamma & 0 & 0 \\ v \gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right).\left( \begin{array}{l} t' \\ x' \\ y' \\ z' \end{array} \right)$$
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:
$$\left( \begin{array}{l} t' \\ x' \\ y' \\ z' \end{array} \right)=\left( \begin{array}{llll} \gamma & -v \gamma & 0 & 0 \\ -v \gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right).\left( \begin{array}{l} t \\ x \\ y \\ z \end{array} \right)$$
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:
$$\left( \begin{array}{l} t \\ x \\ y \\ z \end{array} \right)=\left( \begin{array}{llll} \gamma & v \gamma & 0 & 0 \\ v \gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right).\left( \begin{array}{l} t' \\ x' \\ y' \\ z' \end{array} \right)$$
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.