This discussion centers on the implications of relativity and the behavior of clocks in different frames of reference, particularly in accelerated frames. A blind man's experience with synchronized clocks on a train illustrates how perception of time can differ based on motion and sound. Participants clarify that while Newton's laws apply universally, their interpretation changes in non-inertial frames, such as a rocket in acceleration. The conversation emphasizes that no fundamental laws are violated; rather, the laws must be adapted to account for acceleration and the effects of forces like centrifugal and Coriolis forces.
PREREQUISITESPhysics students, educators, and anyone interested in the nuances of relativity and the application of Newton's laws in various frames of reference.
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.DaleSpam said:If it is not necessary for a force to have an equal and opposite pair then Newton's 3rd law is not a law..
I am really sorry,i was not able to communicate the problem properly.YellowTaxi said: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.
Which is exactly why the laws of motion are violated in an accelerated frame.newTonn said: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.
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 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.
If you agree my statement,Action reaction pair is spacecraft and thrust.Ball is at uniform motion.Then how comes the law is violated?DaleSpam said:Which is exactly why the laws of motion are violated in an accelerated frame.
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?newTonn said:If you agree my statement,Action reaction pair is spacecraft and thrust.Ball is at uniform motion.Then how comes the law is violated?
Thanks for your response on the problem.DaleSpam said: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.
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.newTonn said: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.
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.DaleSpam said: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.
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).DaleSpam said: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 said: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
I am talking about the vacuum inside the container(clock) not the surrounding vaccum.I never mentioned that universe is enclosed.DaleSpam said: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?
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 said:I am talking about the vacuum inside the container(clock) not the surrounding vaccum.I never mentioned that universe is enclosed.
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.DaleSpam said: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?
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.newTonn said:Dalespam as far as analogy of sound clock to light clockis considered,we must not be bothered about the velocity of surrounding.
Sure, you have a right triangle with two legs of length:newTonn said:Anyhow l am attaching some of my doubts and reasoning about the moving clock.
Can you please clear the doubts?
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.DaleSpam said: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.
That is incorrect:newTonn said: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.
But L=ct ; so the triangle will remains open always.DaleSpam said: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.
No.newTonn said:But L=ct ; so the triangle will remains open always.
Does it means there is a length expansion in the direction perpendicular to the motion when v>0.?DaleSpam said:No.
L^2 + (vt)^2 = (ct)^2
So L=ct iff v=0.
DaleSpam said:...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.DaleSpam said: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.
newTonn said: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.
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.newTonn said: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 .
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.DaleSpam said:.
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.
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'..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