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

[Dale]

OK, belliott. I am willing to engage in a discussion about GR and the equivalence principle, but first I want you to address directly the following:

In my example I purposfully set up a scenario without gravitation so that there would be no confusion about if a given force was real or inertial. So, in the context of my example (i.e. flat spacetime, no gravity) I have performed an analysis demonstrating a clear violation of Newton's 3rd law in an accelerated reference frame. You are claiming that there is no such violation, so where did I make my error? Can you demonstrate that Newton's 3rd law is, in fact, satisfied in my scenario?

 

[belliott4488]

OK, belliott. I am willing to engage in a discussion about GR and the equivalence principle,

Nah, I don't think we need to go into GR - I haven't gotten the impression we have any disagreement there, unless you think it's all a crock, but you don't impress me as a crackpot , so I think we're fine just staying on-topic.

... but first I want you to address directly the following:

In my example I purposfully set up a scenario without gravitation so that there would be no confusion about if a given force was real or inertial. So, in the context of my example (i.e. flat spacetime, no gravity) I have performed an analysis demonstrating a clear violation of Newton's 3rd law in an accelerated reference frame. You are claiming that there is no such violation, so where did I make my error? Can you demonstrate that Newton's 3rd law is, in fact, satisfied in my scenario?

I agree this is the crux of our disagreement, but I don't see where you have demonstrated a "clear violation of Newton's 3rd law." You did show how the laws didn't work if the spaceman attempted to do Physics as if he were in an inertial frame, but then you showed how the 1st and 2nd laws did work if he invoked the fictitious inertial force. About the 3rd law, you simply said, "... all of the inertial forces act in the same direction, so they do not form 3rd law pairs. The inertial forces violate the 3rd law because there are no equal and opposite inertial forces."

This is what I've already responded to - I don't understand why you say this. First, the inertial forces are in the direction opposite to the acceleration, right? so if there were no other forces at work any objects under the influence of those forces would all be moving in that direction, which they aren't - at least, not until the spaceman drops the rock.

Let me go through it one more time, and tell me what it is that you object to (I've already been through this twice, but maybe you haven't been reading all the posts). If the spaceman stands in his rocket ship holding a rock, he feels that the rock exerts a "downward" force of ma on his hand. As long as he holds the rock motionless in his frame, he sees that he's exerting an equal and opposite force upward, ie. ma in the "upward" direction. We would describe it differently from outside, in our inertial coordinates, but you know how to do that and it doesn't matter for this discussion.

Now if he let's go of the rock and sees it fall to the floor, he will, of course, see it accelerate at a in the "downward" direction - again, we outside observers would say that it was moving inertially, but in his coordinates he would say it accelerates. If you now ask, "where's the force equal and opposite to the force he sees that causes this perceived acceleration?" the answer is essentially the same as it is when a student asks about the force equal and opposite to the gravitational force on a falling apple. In that case, the Earth is accelerated upward by a minuscule amount; in this case the rocket ship is accelerated "upward" when the spaceman releases the rock. In his frame, this is simply due to the equal and opposite force pulling "up" on the floor (which he can detect). In our frame, we see the additional acceleration on the ship because the rocket motor is now producing a greater acceleration of a' = F/M instead of the original acceleration a = F/(M+m), where F is the thrust of the rocket motor, m is the mass of the rocket, and M is the total remaining mass being accelerated by the motor.

Once the rock lands on the floor, the motor must now accelerate it as well (according to our inertial frame), so the acceleration we see returns to its original value, and the spaceman no longer feels the additional acceleration. (As an interesting side-note, his description would most likely be that he initially feels a downward force on his body, which he might call his "weight", due to the fictitious force, but when he drops the rock and feels the additional force, he would call that an inertial reaction force since he sees the floor (and himself) accelerating :"upward" in response to the equal and opposite force from the rock. In our inertial frame, we'd say it's all inertial reaction forces, just two different values for the two different acceleration magnitudes.)

That's not super rigorous, but do you really want me to go through the algebra? The reason I keep referring to the Equivalence Principle is just that I'm assuming you've seen this before and you know that the calculations in the accelerated frame are identical to the calculations in a constant gravitational field, so I'm not bothering to copy them here. I will, if you insist - or if you think there really is a subtlety there that I'm missing but will show up in the calculations.

I'll add one more thing, which is just to reinforce that I agree completely that the fictitious forces in non-inertial frames screw up the calculations. This is particularly evident in rotating frames, where we can't even attribute the centrifugal or Coriolis force to an apparent gravitational field. Nonetheless, we can write down expressions for what these forces look like in the coordinates of the rotating frame - in all their ugliness - and then do Physics using Newton's laws. This is a pain, but as far as I know, it works - do you disagree?

Again, my whole point boils down to this: whatever the origin of the forces, if you can write down expressions for them, then you can use Newton's laws to describe the resulting motion in the coordinates of the accelerated frame. I don't think you've actually denied this, except to assert that there aren't any equal and opposite forces. Can you show where this is true and you would be unable to describe the motion in the accelerated frame?

If you still think so, then how about we try the simple case of the spaceman tossing the rock in the air and recording its trajectory (just one dimension is fine, of course)? I predict that his application of Newton's laws will correctly describe the motion he observes. I'll gladly go through it in detail if you like, but I'd ask that you do the same so that we can see where our calculations diverge.

 

[Dale]

I agree this is the crux of our disagreement, but I don't see where you have demonstrated a "clear violation of Newton's 3rd law." You did show how the laws didn't work if the spaceman attempted to do Physics as if he were in an inertial frame, but then you showed how the 1st and 2nd laws did work if he invoked the fictitious inertial force. About the 3rd law, you simply said, "... all of the inertial forces act in the same direction, so they do not form 3rd law pairs. The inertial forces violate the 3rd law because there are no equal and opposite inertial forces."

This is what I've already responded to - I don't understand why you say this.

In SI units the inertial force on the rocket is (-2000,0,0) and the inertial force on the ball is (-0.2,0,0). I didn't explicitly mention the inertial force on the exhaust, but if the exhaust masses 1 kg then the inertial force on the exhaust is (-2,0,0). These are all of the inertial forces that exist in the example, they are all of different magnitudes and in the same direction so none of them are equal nor opposite to each other, so they cannot form 3rd law pairs. There are no missing forces and no missing bodies so the 3rd law is violated in the accelerated frame.

First, the inertial forces are in the direction opposite to the acceleration, right?

Yes.

so if there were no other forces at work any objects under the influence of those forces would all be moving in that direction, which they aren't - at least, not until the spaceman drops the rock.

I don't understand your comment here.

Let me go through it one more time, and tell me what it is that you object to (I've already been through this twice, but maybe you haven't been reading all the posts). If the spaceman stands in his rocket ship holding a rock, he feels that the rock exerts a "downward" force of ma on his hand. As long as he holds the rock motionless in his frame, he sees that he's exerting an equal and opposite force upward, ie. ma in the "upward" direction. We would describe it differently from outside, in our inertial coordinates, but you know how to do that and it doesn't matter for this discussion.

Now if he let's go of the rock and sees it fall to the floor, he will, of course, see it accelerate at a in the "downward" direction - again, we outside observers would say that it was moving inertially, but in his coordinates he would say it accelerates.

Yes.

If you now ask, "where's the force equal and opposite to the force he sees that causes this perceived acceleration?" ... in this case the rocket ship is accelerated "upward" when the spaceman releases the rock. In his frame, this is simply due to the equal and opposite force pulling "up" on the floor (which he can detect). In our frame, we see the additional acceleration on the ship because the rocket motor is now producing a greater acceleration of a' = F/M instead of the original acceleration a = F/(M+m), where F is the thrust of the rocket motor, m is the mass of the rocket, and M is the total remaining mass being accelerated by the motor.

This is where you need to run the numbers. I understand what you are saying here, but since you have not run the numbers you don't realize that you are wrong in your conclusions: the resulting forces in this slightly more detailed scenario will still not satisfy Newton's 3rd law. Do you want to do the math or do you want me to do it?

That's not super rigorous, but do you really want me to go through the algebra?

Yes. I honestly believe that is the best way to understand non-inertial frames and inertial forces.

The reason I keep referring to the Equivalence Principle is just that I'm assuming you've seen this before and you know that the calculations in the accelerated frame are identical to the calculations in a constant gravitational field, so I'm not bothering to copy them here. I will, if you insist - or if you think there really is a subtlety there that I'm missing but will show up in the calculations.

I will be glad to discuss GR later, but before we can have a productive conversation about the equivalence principle and curved spacetime we need to come to a mutual understanding about non-inertial reference frames and ficticious forces in flat space.

I'll add one more thing, which is just to reinforce that I agree completely that the fictitious forces in non-inertial frames screw up the calculations. This is particularly evident in rotating frames, where we can't even attribute the centrifugal or Coriolis force to an apparent gravitational field. Nonetheless, we can write down expressions for what these forces look like in the coordinates of the rotating frame - in all their ugliness - and then do Physics using Newton's laws. This is a pain, but as far as I know, it works - do you disagree?

Newton's 1st and 2nd laws I agree. Newton's 3rd law I disagree. In fact, you can make an arbitrarily accelerating reference frame with arbitrarily ugly inertial forces and you can use those inertial forces to correctly predict the outcome of traditional (non-relativistic) physics experiments in the non-inertial frame. But those inertial forces will always violate the 3rd law.

Again, my whole point boils down to this: whatever the origin of the forces, if you can write down expressions for them, then you can use Newton's laws to describe the resulting motion in the coordinates of the accelerated frame. I don't think you've actually denied this, except to assert that there aren't any equal and opposite forces. Can you show where this is true and you would be unable to describe the motion in the accelerated frame?

I never made such a claim. I have been very consistent with my statements.

In a non-inertial reference frame objects experiencing no real forces accelerate. This is a violation of Newton's 1st and 2nd law, however it is a "repairable" violation in the sense that we can introduce inertial forces. These inertial forces modify the normal laws but can be used to correctly predict the motion of objects in the non-inertial frame, but they do not satisfy Newton's 3rd law. This last is a "non-repairable" violation.

If you still think so, then how about we try the simple case of the spaceman tossing the rock in the air and recording its trajectory (just one dimension is fine, of course)? I predict that his application of Newton's laws will correctly describe the motion he observes. I'll gladly go through it in detail if you like, but I'd ask that you do the same so that we can see where our calculations diverge.

Sounds good. I will do that, and I predict that his application of Newton's first two laws, including the inertial forces, will correctly describe the motion he observes and that the inertial forces will not satisfy the 3rd law.

So that we can be consistent let's use the following:
mass of the ball = .1 kg
mass of the rocket (including ball) = 1000 kg
mass of the exhaust at t=1 s (if needed) = 1 kg
thrust of the engine = 2000 N
transformation equation between inertial and non-inertial frame: x' = x-t²
ball released gently at t=t'=1s

If I missed something just specify it explicitly.

 

[belliott4488]

In SI units the inertial force on the rocket is (-2000,0,0) and the inertial force on the ball is (-0.2,0,0). I didn't explicitly mention the inertial force on the exhaust, but if the exhaust masses 1 kg then the inertial force on the exhaust is (-2,0,0). These are all of the inertial forces that exist in the example, they are all of different magnitudes and in the same direction so none of them are equal nor opposite to each other, so they cannot form 3rd law pairs. There are no missing forces and no missing bodies so the 3rd law is violated in the accelerated frame.

Sorry, I do think you're missing some forces here, specifically the "real" forces that cause the objects to accelerate and hence to give rise to the inertial reaction forces in the first place. In fact, I believe inertial reaction forces are often defined as the equal and opposite forces applied by massive objects back on whatever is applying the forces on them. Here are the specifics for your case:

1. The ball: In the accelerating frame the ball is subject to an inertial force of -2 N, as you've said. It has this because it is being accelerated along with the rocket (relative to the inertial frame), which means it is motionless in the accelerated frame. So, if it is subect to this force in this frame, why isn't it moving? Simply because the spaceman is holding it in his hand, exerting an equal and opposite force of +2 N. Better yet, he can place it on a spring scale, which will show the "weight" of the ball as 2 N, which of course means that the spring is exerting a +2 N force "upward" on the ball.

2. The rocket: In the accelerating frame the spaceman notices that the ball has a "weight" (2 N), he has a weight (his mass X the acceleration), and he quickly deduces that the rocket itself must have weight, so something must be holding it up, since it is motionless in his frame. Looking out his port hole, he sees the rocket plume and realizes that the rocket motor is supplying thrust in the "upward" direction, which must be exactly equal and opposite to the "weight" of the rocket+spaceman+ball. In other words, the rocket is "hovering" in his frame due to the thrust force, which is exactly equal and opposite to the inertial force on the rocket. We know that this must be, because the thrust of the motor is exactly what is needed to accelerate this same mass in the inertial system to the acceleration that produces the "weight" in the accelerated system.

If you now ask, "where's the force equal and opposite to the force he sees that causes this perceived acceleration?" the answer is essentially the same as it is when a student asks about the force equal and opposite to the gravitational force on a falling apple. In that case, the Earth is accelerated upward by a minuscule amount; in this case the rocket ship is accelerated "upward" when the spaceman releases the rock. In his frame, this is simply due to the equal and opposite force pulling "up" on the floor (which he can detect). In our frame, we see the additional acceleration on the ship because the rocket motor is now producing a greater acceleration of a' = F/M instead of the original acceleration a = F/(M+m), where F is the thrust of the rocket motor, m is the mass of the rocket, and M is the total remaining mass being accelerated by the motor.

This is where you need to run the numbers. I understand what you are saying here, but since you have not run the numbers you don't realize that you are wrong in your conclusions: the resulting forces in this slightly more detailed scenario will still not satisfy Newton's 3rd law. Do you want to do the math or do you want me to do it?

Okay, let's do this more carefully, then. First, let's look at it in inertial coordinates, where we know we have no disagreement. Before the spaceman releases the ball, the rocket thrust is accelerating the entire system of rocket + ball, so the acceleration is a_1=F/(M+m), where, as before, F is the thrust of the motor, m is the mass of the ball (I think it was a rock last time ...), and M is the remaining accelerated mass. After he releases the ball, the thrust is applied only to the remaining mass, so the acceleration is now a_2 = F/M. After the ball hits the floor and comes to rest, the acceleration will return to its initial value.

What is the difference in the acceleration: it's simply
\delta a = a_2 - a_1 =\frac{F}{M} - \frac{F}{(M+m)} = \frac{F}{(M+m)} * \frac{m}{M}

But since the original acceleration is a_1=\frac{F}{(M+m)} we have,
\delta a = a_1 * \frac{m}{M}

Now, what does the spaceman see in the accelerated frame? He feels the ball with its "weight" of ma, and when he releases it, he sees it accelerate "downward" with an acceleration of a. He also feels the floor accelerate upward, however, due to the equal and opposite force exerted by the ball on the floor. What acceleration does he detect? Well, the force is given by ma, so that is the force "upward" on the floor. That force must accelerate the entire rocket (minus ball), so it produces an acceleration
a = \frac{F}{M} = a*\frac{m}{M}
which is the same acceleration we saw from the inertial point of view. Therefore by assuming a force equal and opposite to the inertial force he felt on the ball, he predicts exactly the same resulting acceleration as we did in the inertial frame.

I didn't explicitly plug any specific numbers in there, but since I have the same result for both frames, plugging in arbitrary numbers isn't going to prove anything.

For some reason the forum produces a database error if I post my whole response, so I'll leave this one here, and then post the rest in a subsequent response.

 

[belliott4488]

This is the remainder of my response, which the Forum wouldn't let me post in one piece:

Sounds good. I will do that, and I predict that his application of Newton's first two laws, including the inertial forces, will correctly describe the motion he observes and that the inertial forces will not satisfy the 3rd law.

So that we can be consistent let's use the following:
mass of the ball = .1 kg
mass of the rocket (including ball) = 1000 kg
mass of the exhaust at t=1 s (if needed) = 1 kg
thrust of the engine = 2000 N
transformation equation between inertial and non-inertial frame: x' = x-t²
ball released gently at t=t'=1s

If I missed something just specify it explicitly.

Okay, let's go through this. I'm not sure I really see the point, since we're just applying the 1st and 2nd laws, which you've already agreed work fine in the accelerated system, but since if you believe the error in my reasoning will be revealed, then on we go ...

Let's define some variables so that the equations aren't just a mash-up of numbers:

mass of ball: m_b = 0.1 \,\text{kg}
mass of rocket (including ball and spaceman): m_r = 1000.0 \,\text{kg}
thrust of rocket motor: T=2000.0 \,\text{N}

We're using primed coordinates, velocities, and accelerations for the accelerating system, unprimed for the inertial system.
I'm going to override your specification that the ball is released at t=1 s (gently - that's good! :-) we don't need any extraneous forces, here!), since there's nothing important happening 1 s before then to start our clocks. I'll just say that t=0 at the moment of release and avoid unnecessary extra terms that look like (t - 1s); if you've gone ahead with your definition, then there will be a lot of "t-1" factors that we can easily correct for.

I'll also say that the origins of both coordinate systems are aligned with the point of release of the ball at t=0, so that point is x = x' = 0 as well.

Then we have initial velocities, all at time t = 0:
rocket initial velocity in accel. stytem: v'_{r0} = 0\,\text{km/sec}
rocket initial velocity in inertial system: v_{r0} = 10.0 \,\text{km/sec}
ball initial velocity in accel. system: v'_{b0} = 0.1 \,\text{km/sec}
ball initial velocity in inertial systme: v_{b0} = v_{r0} + v'_{b0} = 10.1 \,\text{km/sec}

For the acceleration of the rocket in the inertial frame we have:
a = F/m_r = 2 \,\text{km/sec}^2

Next we'll work out the trajectory of the ball in the inertial frame, where we are in agreement about everything.
The position of the rocket (i.e. the origin of the accel. frame) in the inertial frame is
x_r(t) = v_r t + \frac{1}{2}at^2
where v_r(t) = v_{r0}+at

The ball moves inertially once it's released, so its position in the inertial frame is simply
x_b(t) = v_{b0} t

So, what is the distance from the point of release in the rocket (i.e. the spaceman's hand) to the ball at time t? It will be:
h(t) = x_b - x_r = v_{b0} t - v_r t - \frac{1}{2}at^2 = (v_{b0}-v_r)t - \frac{1}{2}at^2

Now let's go to the accelerated frame. The ball is subject to the inertial force in the "downward" direction, so its postion will be:
x'_b(t) = v'_b t - \frac{1}{2}at^2

But we note that since the ball's (constant) velocity in the inertial frame is just
v_b = v_{b0} = v_r + v'_b
we get
v'_b = v_b - v_r
so
x'_b(t) = (v_{b0}-v_r)t - \frac{1}{2}at^2 = h(t)

In other words, by invoking Newton's Laws in his accelerated frame, the spaceman predicts exactly the same answer that we do in our inertial frame.

 

[Dale]

Hi belliott, here is my work. I have attached free-body diagrams for all four parts of this problem. All forces are indicated on the diagrams and the accelerations are worked out for the rocket and the ball. Note in particular that I have highlighted the blue 3rd law pair for the thrust and the green third law pair for the contact force. Note also that none of the inertial forces in black are part of a 3rd law pair, i.e. they do not satisfy the 3rd law.

Below are the resulting equations of motion where capital X indicates the position of the rocket, lower-case x indicates the position of the ball, subscript 1 indicates when the ball is held, subscript 2 indicates when the ball is released, unprimed is the inertial coordinate system, primed is the accelerated coordinate system with transformation equation between the two x' = x - t² and inertial forces of -2m on every mass in the accelerated frame.

The equations of motion for the inertial frame while the ball is held (upper left) are:
\begin{array}{l}<br /> X_1=t^2 \\<br /> x_1=t^2<br /> \end{array}

The equations of motion for the accelerated frame while the ball is held (lower left) are:
\begin{array}{l}<br /> X_1&#039;=0=X_1-t^2 \\<br /> x_1&#039;=0=x_1-t^2<br /> \end{array}As you can see, these equations transform correctly, indicating agreement between the frames. As you can also see, the inertial forces were required in order to achieve this agreement.

Evaluating the above expressions at t=1 for the initial conditions of the next part gives:
\begin{array}{l}<br /> X_1(1)=x_1(1)=1 \\<br /> V_1(1)=v_1(1)=2<br /> \end{array}

\begin{array}{l}<br /> X_1&#039;(1)=x_1&#039;(1)=0 \\<br /> V_1&#039;(1)=v_1&#039;(1)=0<br /> \end{array}

The equations of motion for the inertial frame while the ball is dropped (upper right) are:
\begin{array}{l}<br /> X_2=1.0001 t^2-0.0002 t+0.0001 \\<br /> x_2=2. t-1.<br /> \end{array}

The equations of motion for the accelerated frame while the ball is dropped (lower right) are:\begin{array}{l}<br /> X_2&#039;=0.0001 t^2-0.0002 t+0.0001=X_2-t^2 \\<br /> x_2&#039;=-t^2+2. t-1.=x_2-t^2<br /> \end{array}As you can see, these equations again transform correctly, indicating agreement between the frames. As you can also see, the inertial forces were again required in order to achieve this agreement.

Bottom line: in the accelerating frame the inertial forces do not follow Newton's 3rd law, as shown in the free-body diagrams, but are required in order to satisfy Newton's 1st and 2nd laws and to obtain the correct equations for motion. In order to satisfy Newton's 1st and 2nd laws the 3rd law is violated. I didn't look at your posts in detail, but from my cursory reading nothing even appeared to address the 3rd law nor rebut my claims that it is violated.

 

[Dale]

Sorry, I do think you're missing some forces here, ... The ball: ... the spaceman is holding it in his hand, exerting an equal and opposite force of +2 N.

You are correct, I neglected that force (+0.2 N contact force on the ball) in the initial description. I have included it and it's 3rd law pair (-0.2 N contact force on the rocket/spaceman) in my analysis above.

Okay, let's do this more carefully, then. First, let's look at it in inertial coordinates, where we know we have no disagreement. Before the spaceman releases the ball, the rocket thrust is accelerating the entire system of rocket + ball, so the acceleration is a_1=F/(M+m), where, as before, F is the thrust of the motor, m is the mass of the ball (I think it was a rock last time ...), and M is the remaining accelerated mass. After he releases the ball, the thrust is applied only to the remaining mass, so the acceleration is now a_2 = F/M. After the ball hits the floor and comes to rest, the acceleration will return to its initial value.

What is the difference in the acceleration: it's simply
\delta a = a_2 - a_1 =\frac{F}{M} - \frac{F}{(M+m)} = \frac{F}{(M+m)} * \frac{m}{M}

But since the original acceleration is a_1=\frac{F}{(M+m)} we have,
\delta a = a_1 * \frac{m}{M}

Yes.

Now, what does the spaceman see in the accelerated frame? He feels the ball with its "weight" of ma, and when he releases it, he sees it accelerate "downward" with an acceleration of a. He also feels the floor accelerate upward, however, due to the equal and opposite force exerted by the ball on the floor. What acceleration does he detect? Well, the force is given by ma, so that is the force "upward" on the floor. That force must accelerate the entire rocket (minus ball), so it produces an acceleration
a = \frac{F}{M} = a*\frac{m}{M}
which is the same acceleration we saw from the inertial point of view. Therefore by assuming a force equal and opposite to the inertial force he felt on the ball, he predicts exactly the same resulting acceleration as we did in the inertial frame.

OK, there are several errors in this analysis. First and most importantly, you left out the thrust force (2000 N) acting on the rocket. Second, you left out the inertial force of the rocket (-1999.8 N). These two forces are sufficient to account for the acceleration and obtain agreement between the two frames. By arbitrarily adding another 0.2 N force on the rocket you wind up with disagreement between the frames since the delta acceleration in the non-inertial frame would be twice the delta in the inertial frame.

 

[Dale]

This is the remainder of my response, which the Forum wouldn't let me post in one piece:

I think there must be a limit on the number of LaTeX objects in a post because I was getting database errors replying too.

Okay, let's go through this. I'm not sure I really see the point, since we're just applying the 1st and 2nd laws, which you've already agreed work fine in the accelerated system, but since if you believe the error in my reasoning will be revealed, then on we go ...

I have consistently agreed that, by including the inertial forces, the 1st and 2nd laws work. I have consistently stated that the inertial forces required for the 1st and 2nd law violate the 3rd law.

Let's define some variables so that the equations aren't just a mash-up of numbers:
...

We're using primed coordinates, velocities, and accelerations for the accelerating system, unprimed for the inertial system.
I'm going to override your specification that the ball is released at t=1 s ... I'll just say that t=0 at the moment of release ...

I'll also say that the origins of both coordinate systems are aligned with the point of release of the ball at t=0, so that point is x = x' = 0 as well.

Then we have initial velocities, all at time t = 0:
...

For the acceleration of the rocket in the inertial frame we have:
a = F/m_r = 2 \,\text{km/sec}^2

This is incorrect. It should be
a = F/(m_r - m_b) = 2.0002 \,\text{m/s}^2
for t>0

Next we'll work out the trajectory of the ball in the inertial frame, where we are in agreement about everything.
The position of the rocket (i.e. the origin of the accel. frame) in the inertial frame is
x_r(t) = v_r t + \frac{1}{2}at^2
where v_r(t) = v_{r0}+at

I think you mean x_r(t) = v_{r0} t + \frac{1}{2}at^2. I corrected the following equations too.

The ball moves inertially once it's released, so its position in the inertial frame is simply
x_b(t) = v_{b0} t

So, what is the distance from the point of release in the rocket (i.e. the spaceman's hand) to the ball at time t? It will be:
h(t) = x_b - x_r = v_{b0} t - v_{r0} t - \frac{1}{2}at^2 = (v_{b0}-v_{r0})t - \frac{1}{2}at^2

OK

Now let's go to the accelerated frame. The ball is subject to the inertial force in the "downward" direction, so its postion will be:
x&#039;_b(t) = v&#039;_{b0} t - \frac{1}{2}at^2

But we note that since the ball's (constant) velocity in the inertial frame is just
v_{b0} = v_{r0} + v&#039;_{b0}
we get
v&#039;_{b0} = v_{b0} - v_{r0}
so
x&#039;_b(t) = (v_{b0}-v_{r0})t - \frac{1}{2}at^2 = h(t)

In other words, by invoking Newton's Laws in his accelerated frame, the spaceman predicts exactly the same answer that we do in our inertial frame.

This is fine, your frame is accelerating at a different rate than mine, but everything works out OK.

However, you never addressed the 3rd law question here. Where is the 3rd law pair for the -0.20002 N "inertial force in the 'downward' direction" on the ball? You have an inertial force on the rocket of -2000 N which exactly balances the thrust of 2000 N, but that force is neither equal nor opposite. Again, from your own work, it is clear that in order to get Newton's 1st and 2nd laws to work in the accelerated frame you must postulate inertial forces which violate the 3rd law.

 

[newTonn]

This is fine, your frame is accelerating at a different rate than mine, but everything works out OK.

However, you never addressed the 3rd law question here. Where is the 3rd law pair for the -0.20002 N "inertial force in the 'downward' direction" on the ball? You have an inertial force on the rocket of -2000 N which exactly balances the thrust of 2000 N, but that force is neither equal nor opposite. Again, from your own work, it is clear that in order to get Newton's 1st and 2nd laws to work in the accelerated frame you must postulate inertial forces which violate the 3rd law.

Let us analys the problem logically.I think there is no need to eloborate the problem with putting y,z co-ordinate(we can take x co-ordinate which is the only relevant one in this example) .
Initally the relative velocity will be zero and hence there will be no relative change in position between the ball and spacecraft floor because ball is physically connected to the spacecraft by the person who holds it.
A thrust,applied to the spacecraft will accelerate both spacecraft and ball.F= (M+m)a will be the equation of motion.But again the relative velocity and relative motion will be zero because both the bodies are accelerated at same rate.
At the instance the ball is released,it will have a velocity which will be the velocity of spacecraft and ball at that instance as calculated from the above equation.the ball will continue it's state of motion with that uniform velocity.The position of ball after 't' seconds will be decided by this velocity.
Once the ball is released,the equation of motion will be F=Ma'.the position of spacecraft floor after 't' seconds can be calculated from a'.
In both these None of the laws are violated.
Now,if we don't have a clear picture as above.ie. we don't have a third reference which is absolute with respect to both object,still we can calculate the position of ball after 't' seconds by introducing a fictious force.
Since we know this is a fictious force,it is not necessary for it to have an equal and opposite pair.
Still if it is so necessary,i can say there is a fictious equal and opposite pair for it ,by using the same logic.

 

[Dale]

Let us analys the problem logically.I think there is no need to eloborate the problem with putting y,z co-ordinate(we can take x co-ordinate which is the only relevant one in this example) .
Initally the relative velocity will be zero and hence there will be no relative change in position between the ball and spacecraft floor because ball is physically connected to the spacecraft by the person who holds it.
A thrust,applied to the spacecraft will accelerate both spacecraft and ball.F= (M+m)a will be the equation of motion.But again the relative velocity and relative motion will be zero because both the bodies are accelerated at same rate.
At the instance the ball is released,it will have a velocity which will be the velocity of spacecraft and ball at that instance as calculated from the above equation.the ball will continue it's state of motion with that uniform velocity.The position of ball after 't' seconds will be decided by this velocity.
Once the ball is released,the equation of motion will be F=Ma'.the position of spacecraft floor after 't' seconds can be calculated from a'.
In both these None of the laws are violated.

Correct, this is all done in the inertial reference frame.

Now,if we don't have a clear picture as above.ie. we don't have a third reference which is absolute with respect to both object,still we can calculate the position of ball after 't' seconds by introducing a fictious force.
Since we know this is a fictious force,it is not necessary for it to have an equal and opposite pair.

If it is not necessary for a force to have an equal and opposite pair then Newton's 3rd law is not a law.

Still if it is so necessary,i can say there is a fictious equal and opposite pair for it ,by using the same logic.

No, you cannot, as I have demonstrated twice now. Inertial forces violate Newton's 3rd law. You are welcome to try to demonstrate otherwise, but handwaving like this is insufficient.

 

[newTonn]

If it is not necessary for a force to have an equal and opposite pair then Newton's 3rd law is not a law..

In this case ,when we introduce a fictious force,we misinterpret that the change in relative position of floor and ball is due to the acceleration of ball(ball is accused of having a force),instead the actual relative motion is because of the acceleration of floor.

 

[newTonn]

In all honesty I only looked at this thread because I was wondering if there were any startling similarities between how things might 'look' for a blind man, and how they look to people who have eyes to perceive the transmission of light. Actually I'm rather disappointed that the subject being discussed hasn''t really addressed that question, and has veered off into a different direction altogether to what I expected. But maybe I misunderstood what the OP was getting at. I didn't follow all that he said.

I am really sorry,i was not able to communicate the problem properly.
I was keeping quite because the discussion was going in another direction.
Now i will come back to my point with another example.
Consider a clock similar to the clock we are using to explain the time dilation.The difference in this clock,is instead of light,we are using sound as signal and instead of mirrors we are using a sound sensor.
When stationary,a tick sound is produced from the speaker attatched to the first sensor.when this sound reach the sensor on the opposite side(say ,L meter apart),it will produce a tick.consider this interval is set as one second.

Now if the clock is set in a moving object,we can hear that the clock starts slowing down.As the velocity of clock is increased,it become more slower.And if the speed exeeds the speed of sound you will not hear the ticks any more.Time will be stopped? or in other words,relative velocity is restricted to the speed of sound.

All the transformation equations will work in this case also,by replacing 'c' with the velocity of sound.

So,if i claim that you will be aged lesser or even stop ageing,if you are traveling together with this clock,Do anybody agree with me?.
If no,why?and why this reasoning is not applicable in the case of light.

 

[Dale]

In this case ,when we introduce a fictious force,we misinterpret that the change in relative position of floor and ball is due to the acceleration of ball(ball is accused of having a force),instead the actual relative motion is because of the acceleration of floor.

Which is exactly why the laws of motion are violated in an accelerated frame.

 

[Dale]

I am really sorry,i was not able to communicate the problem properly.
I was keeping quite because the discussion was going in another direction.
Now i will come back to my point with another example.
Consider a clock similar to the clock we are using to explain the time dilation.The difference in this clock,is instead of light,we are using sound as signal and instead of mirrors we are using a sound sensor.
When stationary,a tick sound is produced from the speaker attatched to the first sensor.when this sound reach the sensor on the opposite side(say ,L meter apart),it will produce a tick.consider this interval is set as one second.

Now if the clock is set in a moving object,we can hear that the clock starts slowing down.As the velocity of clock is increased,it become more slower.And if the speed exeeds the speed of sound you will not hear the ticks any more.Time will be stopped? or in other words,relative velocity is restricted to the speed of sound.

All the transformation equations will work in this case also,by replacing 'c' with the velocity of sound.

So,if i claim that you will be aged lesser or even stop ageing,if you are traveling together with this clock,Do anybody agree with me?.
If no,why?and why this reasoning is not applicable in the case of light.

This is actually more along the line of where I thought you were going originally. I very much understand your point, as I had the same idea for a period of about 5 years. I will respond in more depth later, since I think I have some thoughts on the subject that may be useful to you also.

 

[newTonn]

Which is exactly why the laws of motion are violated in an accelerated frame.

If you agree my statement,Action reaction pair is spacecraft and thrust.Ball is at uniform motion.Then how comes the law is violated?

 

[Dale]

If you agree my statement,Action reaction pair is spacecraft and thrust.Ball is at uniform motion.Then how comes the law is violated?

I refer you back to https://www.physicsforums.com/showpost.php?p=1674418&postcount=56", and in particular the free-body diagrams. What part did you not understand?

Regarding your "sound clock". The difference between sound waves and light waves is that light waves propagate at the speed of light in any reference frame and sound waves only propagate at the speed of sound in the rest frame of the medium. Basically, this can be understood in the fact that sound propagates through a medium and light propagates through empty space. When you are moving through the air you can perform a number of experiments to determine the velocity of the air, but when you are moving through empty space there is no experiment you can do to determine the "velocity" of empty space.

So, to go back to your sound clock. Let's suppose that your sound clock consisted of an enclosed cylinder of a length such that the sound would go from the speaker, echo off the far side, and return to a microphone mounted next to the speaker over a period of 1 second. Now, let's mount this clock on a supersonic jet traveling at mach 2 wrt the air. The clock will still tick at a rate of 1 second because the sound travels at mach 1 wrt the air inside the clock, which is traveling at mach 2 wrt the ground. So, in the ground frame, the sound in the clock is traveling at 3 times the speed of sound. This is the difference between sound and light.

 

[newTonn]

Regarding your "sound clock". The difference between sound waves and light waves is that light waves propagate at the speed of light in any reference frame and sound waves only propagate at the speed of sound in the rest frame of the medium. Basically, this can be understood in the fact that sound propagates through a medium and light propagates through empty space. When you are moving through the air you can perform a number of experiments to determine the velocity of the air, but when you are moving through empty space there is no experiment you can do to determine the "velocity" of empty space.

So, to go back to your sound clock. Let's suppose that your sound clock consisted of an enclosed cylinder of a length such that the sound would go from the speaker, echo off the far side, and return to a microphone mounted next to the speaker over a period of 1 second. Now, let's mount this clock on a supersonic jet traveling at mach 2 wrt the air. The clock will still tick at a rate of 1 second because the sound travels at mach 1 wrt the air inside the clock, which is traveling at mach 2 wrt the ground. So, in the ground frame, the sound in the clock is traveling at 3 times the speed of sound. This is the difference between sound and light.

Thanks for your response on the problem.
As mentioned in your explanation,if a similar enclosed cylinder is used for the light clock,please note that the enclosed vaccum is a separate entity and will have a velocity equal to that of the apparatus(clock) and at least in the case of relative motion it will be similar to that of the air inside the sound clock.
So all the equations will works similar as in the case of sound clock. The relative position of the observer inside spacecraft with the enclosed vacuum will remain same.So,Of course in the ground frame,the light has to travel with a speed greater than 'c',to satisfy this.(if the ship is traveling near to 'c'.)

 

[Dale]

As mentioned in your explanation,if a similar enclosed cylinder is used for the light clock,please note that the enclosed vaccum is a separate entity and will have a velocity equal to that of the apparatus(clock) and at least in the case of relative motion it will be similar to that of the air inside the sound clock.

What do you mean? In what way is the empty space inside something an "entity", and how would you propose to measure the velocity of the vacuum? I can think of a couple of very easy ways to measure the velocity of air, but not vacuum.

I cannot think of any physical meaning to empty space having a velocity.

 

[newTonn]

What do you mean? In what way is the empty space inside something an "entity", and how would you propose to measure the velocity of the vacuum? I can think of a couple of very easy ways to measure the velocity of air, but not vacuum.

I cannot think of any physical meaning to empty space having a velocity.

please try to shift a vacuum chamber from one place to another.You can see the vacuum inside the chamber shifting from one place to another.

 

[Dale]

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.

 

[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} &gt; 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} &gt; 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(<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.