Absolute Time Clock Experiments: Einstein's Special Relativity

[JesseM]

I haven't read this whole thread, but a question for roineust: do you understand that according to relativity, if there's a rocket moving at relativistic velocity relative to us, and two clocks on board the rocket are synchronized at a single location and moved apart with slow velocities relative to the rocket, then in our frame the clocks will not remain synchronized even in the limit as their velocity relative to the rocket approaches zero? It's not too hard to prove this using the relativistic formulas for time dilation and velocity addition.

 

[Dale]

This still doesn't explain the relation between, the one way speed of light, slow transport and an experimental action.

As far as I could understand, the question wasn't about slow transport, but I am glad to discuss it now that you have clarified your intention.

Slow transport synchronization is equivalent to the Einstein synchronization convention. So, in an inertial reference frame (which by definition uses the Einstein synchronization convention) we determine that slowly-transported clocks maintain their synchronization. However, if we were to use another synchronization convention then slowly transported clocks would not remain synchronized. Here is an example:

Suppose that we have a lab with three clocks. Clock A and clock B are separated by a distance of 5 feet and are synchronized by the Einstein synchronization convention, so if A emits a pulse at t = 0 ns, then B receives the pulse at t = 5 ns, and the reflection returns back to A at t = 10 ns. The one way speed of light is c = 1 ft/ns in both directions. Clock C is initially at rest next to A and is synchronized with A. It is then slowly transported to B and found to be synchronized with B.

Now, instead suppose that A and B are synchronized by "everyday" synchronization where B simply sets its clock to match the signal that it is currently receiving from A. Using this synchronization convention if A emits a pulse at t = 0 ns, then B receives the pulse at t = 0 ns, and the reflection returns back to A at t = 10 ns. The "outward" speed of light is infinite, and the "inward" speed of light is 0.5 ft/ns. Clock C is initially at rest next to A and is synchronized with A. It is then slowly transported to B and found to be 5 ns ahead of B.

So, the synchronization convention determines the one way speed of light and the effect of slow clock transport. Slow clock transport is equivalent to the Einstein synchronization convention.

The moving frame, is a jet flying a device. The Stationary is the on-ground station with the same device.

Careful, the frame is not a physical object, it is a mathematical object. It is a mistake to identify the frame with the jet or the ground. The "moving" frame is a coordinate system where the jet is stationary, and the "stationary" frame is a coordinate system where the ground is stationary.

One theory will claim that, because of the affect of slow transport mechanism of separating the clocks, no matter how slow, the on-ground clock will measure the speed of light as X and the jet device measures the speed of light differently, say as Y.

The other theory will claim that the slow transport mechanism, doesn't change the measurements results in comparison between the two frames, exactly because the separation of the two clocks, is done slowly, and that both on-ground and jet devices, will measure the speed of light to be the same, let it be X and X or Y and Y or Z and Z.

Which two theories are you discussing here? Assuming that all the frames are inertial then they will all measure the speed of light to be c.

 

[roineust]

OK,
Here is the way I understand things, logically, up to this point.

If there is no way to indicate a logic error, regarding an SR scenario analysis, please try to explain to me why. On the other hand, if it is possible, please try to explain where, and what, is the error, in this presented situation.

I see only two possibilities, that will be described shortly. Although, it seems that most of the comments in this thread, indicate towards only one possibility, it seems, somewhat, still a fuzzy choice, and not a clear cut, so I treat the outcome of both possibilities.

This fuzziness in my understanding, of what is the true possibility, in my opinion, arises from the almost immediate operation, commonly done while analyzing SR scenarios, of 'looking at the situations from other points of views, or frames' – I am interested in this post here, to look at things, only from within the 'moving' frame.

So- either the A and B clocks, after separated by slow transport, are considered, as seen from within the moving frame, as synchronized, or they are considered, not synchronized. I don’t see any more possibilities. It seems that most of the comments, if not all of them, agree that they stay synchronized, as seen from within the moving frame, after a slow transport. But again, as explained, I will relate to both possibilities.

Possibility no.1: A & B clocks stay synchronized.
In this case, as much as I understand, we are back to the previous spot. We are left with no more ways to explain, why the situation, as depicted in the moving frame, in diagram no.1 - is NOT a true depiction.

Possibility no.2: A & B clocks are NOT synchronized.
In this case, please look at attached diagram no.2 – If the slow transport actually does de-synchronize A & B clocks, as seen from within the moving frame, then:

The difference of time measurement, indicated by clocks A & B, in diagram no.2, can not be explained, only by the horizontal distance difference R, that light travels between clocks, but also has to be explained, as a combination of both distance difference R that light travels, and a result of a previous slow transport de-synchronization.

Conclusion:
If we started from the assumption, that it is not possible, that the clocks A & B are both synchronized and De-synchronized at the same time, as seen ONLY from within the moving frame – But that only one possibility is true, then:

Possibilities 1 & 2 can not both be right, and can not both be wrong.
Either one is wrong, while the other is right, or vise versa.


So, which possibility is correct, or where is my error here?


Thanks,
Roi.

 

[harrylin]

OK,
Here is the way I understand things, logically, up to this point.

If there is no way to indicate a logic error, regarding an SR scenario analysis, please try to explain to me why. On the other hand, if it is possible, please try to explain where, and what, is the error, in this presented situation.

I see only two possibilities, that will be described shortly. Although, it seems that most of the comments in this thread, indicate towards only one possibility, it seems, somewhat, still a fuzzy choice, and not a clear cut, so I treat the outcome of both possibilities.

This fuzziness in my understanding, of what is the true possibility, in my opinion, arises from the almost immediate operation, commonly done while analyzing SR scenarios, of 'looking at the situations from other points of views, or frames' – I am interested in this post here, to look at things, only from within the 'moving' frame.
[..]

That's ambiguous - for "only from within the 'moving' frame", the view is that everything is in rest.
If you meant that (which I doubt), for sure you already know the answer of what will be measured in rest!

Perhaps you want to look at how things will be measured with the moving frame, as determined with the rest frame? In that case you must distinguish between "are" (=according to rest frame A) and "measured as" (=according to moving frame B).

As long as it is not clear what you mean, no answer can be given to your questions.

Harald

 

[Dale]

I am interested in this post here, to look at things, only from within the 'moving' frame.

I assume by "the moving frame" you mean the frame in which the apparatus is moving. If you mean the frame in which the apparatus is stationary please let me know.

So- either the A and B clocks, after separated by slow transport, are considered, as seen from within the moving frame, as synchronized, or they are considered, not synchronized.

They are not synchronized, but technically it is impossible to do slow transport of the clocks in the frame in which the apparatus is moving. The clocks only remain synchronized in the frame in which they are moving at an infinitessimal speed (slow transport). In other frames where they are moving at finite speed (not slow transport) they do not remain synchronized. Whether or not the clocks are transported slowly is a frame-dependent concept, just like speed and synchronization.

Possibility no.2: A & B clocks are NOT synchronized.
In this case, please look at attached diagram no.2 – If the slow transport actually does de-synchronize A & B clocks, as seen from within the moving frame, then:

The difference of time measurement, indicated by clocks A & B, in diagram no.2, can not be explained, only by the horizontal distance difference R, that light travels between clocks, but also has to be explained, as a combination of both distance difference R that light travels, and a result of a previous slow transport de-synchronization.

Yes, both effects are present and exactly counteract each other. Thus, if they are detected as simultaneous in one frame they are detected as simultaneous in all frames.

At this point I think that you really need to actually work this problem out quantitatively using the Lorentz transform. You are just confusing yourself verbally and need to work through the math in order to gain some clarity.

 

[roineust]

I mean a person standing in the same frame, in which the apparatus is stationary, equipped with the undisputed knowledge that: 1. He is no longer in the same inertial frame that he was before. 2. Time dilation phenomenon exists in relation to the previous frame. 3. Devices depicted in diagram no.1 and no.2 functioned in a certain way, that was observed and recorded, by that person, before he was accelerated to the next inertial frame. The current, or call it - the last fame, I call the 'moving frame', the previous frame to that one, I call the 'stationary frame'.

 

[harrylin]

I mean a person standing in the same frame, in which the apparatus is stationary, equipped with the undisputed knowledge that: 1. He is no longer in the same inertial frame that he was before. 2. Time dilation phenomenon exists in relation to the previous frame.

Dear Roi, everyone is always in all inertial frames!

So, I perhaps you mean from the perspective of someone who is at rest on a platform that he perceives as "moving" at a velocity v.
That is basically the same as the second option that I mentioned to you: it means that he takes the "rest frame" as reference frame, and not the platform on which he is standing.
He will thus account and correct for the effects of his motion with respect to the rest frame.
Do you understand that? I will continue if you do.

 

[Dale]

I mean a person standing in the same frame, in which the apparatus is stationary, equipped with the undisputed knowledge that: 1. He is no longer in the same inertial frame that he was before. 2. Time dilation phenomenon exists in relation to the previous frame.

This reply confuses me. It doesn't matter where he is standing nor wether or not he previously accelerated. What matters is the coordinate system (reference frame) that he is using to analyze the apparatus. Is he using:
a) the coordinate system where the apparatus is stationary
b) a coordinate system where the apparatus is moving

 

[roineust]

a) the coordinate system where the apparatus is stationary.

 

[Dale]

a) the coordinate system where the apparatus is stationary.

Sorry about my misunderstanding. Please disregard my post 180, it is not relevant, but I cannot edit it now.

So- either the A and B clocks, after separated by slow transport, are considered, as seen from within the moving frame, as synchronized, or they are considered, not synchronized. I don’t see any more possibilities. It seems that most of the comments, if not all of them, agree that they stay synchronized, as seen from within the moving frame, after a slow transport.

The clocks stay synchronized in the rest frame of the apparatus after slow transport. Any comments to the contrary were probably a result of misunderstanding which frame was being discussed.

Possibility no.1: A & B clocks stay synchronized.
In this case, as much as I understand, we are back to the previous spot. We are left with no more ways to explain, why the situation, as depicted in the moving frame, in diagram no.1 - is NOT a true depiction.

You do not seem to grasp the relativity of simultaneity. Clocks A and B are synchronized in the moving frame (the frame where the apparatus is stationary). Clocks A and B are not synchronized in the rest frame (the frame where the apparatus is moving). This is what the relativity of simultaneity means.

Due to the relativity of simultaneity the light will be detected as simultaneous in both frames even though it is not simultaneous in the frame where the apparatus is moving. The only thing which is wrong in the diagram is the underlined word "not" in the bottom half.

Again, I recommend that you work out the math.

 

[roineust]

Please try to explain to me, why does this not mean, that in the bottom half of diagram no.1 (attached here, and now I am referring to clock C - not to clocks A&B) - Time dilation just plainly doesn't exist. If time dilation exists, without relation to acceleration, that brought the frame to its new inertial state, how could it exist and not exist at the same time?

Taking a clock and putting it on a jet (after synchronizing it with an on-ground clock), then, at the end, getting at the conclusion that there was time dilation, also as a result of the inertial movement by itself (putting aside hight and acceleration as other causes of time dilation), then arguing that it did not happen in the jet itself? The dilated clock was only there, all along! I can't get it. What does it have to do with working out the math?

Thanks,
Roi.

 

[JesseM]

What is clock C doing? It seems to be blocking the path of the light beam, does it send a pulse out the back end as soon as it receives one from the front end? And in the "stationary diagram" detectors A and B are different distances from the sources, why do you say they detected the light beams simultaneously in this frame? Did the light sources emit the beams at different times?

 

[Dale]

Please try to explain to me, why does this not mean, that in the bottom half of diagram no.1 (attached here, and now I am referring to clock C - not to clocks A&B) - Time dilation just plainly doesn't exist.

Yes it does.

If time dilation exists, without relation to acceleration, that brought the frame to its new inertial state, how could it exist and not exist at the same time?

Nonsense.

Taking a clock and putting it on a jet (after synchronizing it with an on-ground clock), then, at the end, getting at the conclusion that there was time dilation, also as a result of the inertial movement by itself (putting aside hight and acceleration as other causes of time dilation), then arguing that it did not happen in the jet itself? The dilated clock was only there, all along! I can't get it.

Also nonsense.

What does it have to do with working out the math?

You have had almost 200 replies in words and you still keep making self-contradictory statements, so there is some sort of communication barrier that prevents you from understanding in words. So let's try the math just to see if that helps you understand. Start with the top diagram (stationary apparatus) and do the following:

1) Make your apparatus design concrete by:
1a) assume that the light sources are at x=0
1b) specifying the x location of clocks A, B, and C
1c) specifying the time delay introduced by C
1d) use units of ft for distance and ns for time and use c=1 ft/ns
1e) assume that the device is small enough in y and z that there is no measurable delay between pushing the button and the lights flashing
2) Assume that the button is pushed at t=0
3) Calculate the t for light to arrive at clock B (use c and 1b)
4) Calculate the t for light to arrive at clock C (use c and 1b)
5) Calculate the t for light to leave clock C (use 4 and 1c)
6) Calculate the t for light to arrive at clock A (use c and 5 and 1b)
7) Lorentz transform your problem into the frame where the apparatus is moving at v=.6c
8) Confirm the following:
8a) The light pulses traveled at c in this frame (use distances and times from 7)
8b) Clocks A, B, and C are not synchronized (use 1b and Lorentz transform for t=0)
8c) Clocks A, B, and C read the same times as determined above (use 7, 8b, and the spacetime interval)
8d) Clocks A, B, and C are all time dilated (use 7, 8b, and 8c)

That is a lot of steps, but each one should be clear and straightforward. If you are confused on any please ask and I will be glad to help.

 

[JesseM]

Clock C sends the pulse forward, as soon as it receives one, not back.

By "back" I meant to the left, as one usually draws the horizontal x-axis with the increasing x direction going to the right.

Also, you didn't answer my question about why you think the detectors will go off simultaneously in the stationary frame, given that the light seems to travel a different distance from the bottom source to detector B than from the top source to detector A (assuming the presence of clock C makes no difference to the time). Can you give the x-coordinate of each source and detector in the stationary frame (along with clock C), along with the t-coordinate that both the top and bottom sources emit light, and the t-coordinate that you think detectors A and B receive the light simultaneously?

What is depicted in the bottom, under 'moving apparatus', represents, apparently, a false situation. The white and blue circles, are supposed to be the same as in the upper 'stationary apparatus' e.g. two blue circles, if it was to depict the apparently true situation.

Since I don't follow the stationary diagram I don't understand this one. Also, is this diagram supposed to be drawn from the perspective of the airplane's frame where the apparatus is at rest, or from the ground frame where the apparatus is moving? Whichever frame you want to use, can you specify the x-coordinate of all sources/detectors and clock C at t=0 in this frame, when the sources emit the light? (or when one of them does, if they don't emit at the same moment...in that case, please specify the delay between each one emitting light in this frame)

 

[roineust]

Clock C sends the pulse forward, as soon as it receives one, not back (or maybe it is the same, as what you call 'back-end'). What is depicted in the bottom, under 'moving apparatus', represents, apparently, a false situation. The white and blue circles, are supposed to be the same, as in the upper 'stationary apparatus' e.g. two blue circles (not one blue and one white), if it was to depict, the apparently true situation.

The device is configured in the 'stationary diagram', in order for the light beams to be detected simultaneously. And it is exactly the same (or an exact duplicate) configured device, that is sent on the 'moving diagram'. The question is, why the device functions exactly the same, also in the 'moving diagram', although we know that time dilation exists (it says, in the 'moving diagram' 'time dilation' with an arrow pointing at clock C).

At this point the question is even more defined: If it is wrong to say that time dilation occurred, while the dilated clock was in the 'moving frame' - then how come time dilation can be detected, from the beginning? How can it be there, but still, yet, not be there. 'There' doesn't exist for the time dilation phenomenon? But the dilated clock was all the experiment time only 'there' and nowhere else...

Dalespam brings forward a set of calculations, that are supposed to clear things up, but they include a Lorentz transformation. What I am trying to explain is, that all measurements in this experiment, are done only inside the moving frame. So why use the transformation, in this case? Using the transformation means that time dilation, experimentally, never happens in the 'moving frame'. But we know it was detected experimentally in the past, so how come, when inside the 'moving diagram', it can't be detected? Again, the clock was all along the past experiment, that did prove time dilation, only there and nowhere else...

 

[Dale]

all measurements in this experiment, are done only inside the moving frame. So why use the transformation, in this case?

You are clearly analyzing it from two different frames. Stop making excuses, just do the math.

 

[roineust]

Using the transformation means that time dilation, experimentally, never happens in the 'moving frame' (transformation means you 'jump' 'out' of the frame, in order, to calculate what is, 'inside' the frame). But we know it was already detected and verified, experimentally, when a clock was in the 'moving' frame, so how come, when inside the 'moving diagram' or let it be, 'moving frame', it can't be detected? Again, the clock was all along the past experiment, that did prove time dilation, only in the 'moving frame' of that experiment, and nowhere else...

 

[Dale]

transformation means you 'jump' 'out' of the frame, in order, to calculate what is, 'inside' the frame

Have you or have you not drawn your diagrams in two different frames? If you have then you MUST use the Lorentz transformation to go from one to the other.

All that is required is basic algebra. If you are incapable of doing algebra, then you have no business doing classical Newtonian physics, let alone relativity. If you are capable of doing algebra then do so and maybe you will learn something.

 

[harrylin]

Using the transformation means that time dilation, experimentally, never happens in the 'moving frame' (transformation means you 'jump' 'out' of the frame, in order, to calculate what is, 'inside' the frame). But we know it was already detected and verified, experimentally, when a clock was in the 'moving' frame, so how come, when inside the 'moving diagram' or let it be, 'moving frame', it can't be detected? Again, the clock was all along the past experiment, that did prove time dilation, only in the 'moving frame' of that experiment, and nowhere else...

First of all, and I already told you twice (or thrice?!), transformation does not mean you 'jump' 'out' of the frame, in order, to calculate what is, 'inside' the frame.

Transformation means that you interpret what is measured (or can be measured) from the perspective of a different reference system.
And you don't calculate what is, but only what will be measured ("observed"). What "is", is a matter of perspective!

In special relativity you can only move or be at rest in a frame; it's impossible to "jump out"! Perhaps that misunderstanding of concepts caused confusions? Or perhaps it's even the main cause?!

Now I have no time, maybe later more - as a last effort

Note that as far as I can see, I already answered all your questions at least once, and others did too...

 

[roineust]

This question is addressed to anyone, who thinks that Hafele–Keating experiment (1971), was not a: superfluous experiment, or just plainly not an accurate experiment.

Where there ever any attempts to build, an experiments that verifies, not by observing or measuring particles, the postulate: "The laws of Nature, the ways in which things behave, are the same in all inertial systems regardless of their speeds." ?

Thanks,
Roi.

 

[Dale]

Where there ever any attempts to build, an experiments that verifies, not by observing or measuring particles, the postulate: "The laws of Nature, the ways in which things behave, are the same in all inertial systems regardless of their speeds."

I don't know what you have against particles, but there are many such experiments. See http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html especially sections 3.5 and 3.6.

Then do the math.

 

[harrylin]

OK,
Here is the way I understand things, logically, up to this point.

If there is no way to indicate a logic error, regarding an SR scenario analysis, please try to explain to me why. On the other hand, if it is possible, please try to explain where, and what, is the error, in this presented situation. [..]

This fuzziness in my understanding, of what is the true possibility, in my opinion, arises from the almost immediate operation, commonly done while analyzing SR scenarios, of 'looking at the situations from other points of views, or frames' – I am interested in this post here, to look at things, only from within the 'moving' frame.

So- either the A and B clocks, after separated by slow transport, are considered, as seen from within the moving frame, as synchronized, or they are considered, not synchronized. I don’t see any more possibilities. It seems that most of the comments, if not all of them, agree that they stay synchronized, as seen from within the moving frame, after a slow transport. But again, as explained, I will relate to both possibilities.

Possibility no.1: A & B clocks stay synchronized.
In this case, as much as I understand, we are back to the previous spot. We are left with no more ways to explain, why the situation, as depicted in the moving frame, in diagram no.1 - is NOT a true depiction.

Possibility no.2: A & B clocks are NOT synchronized.
In this case, please look at attached diagram no.2 – If the slow transport actually does de-synchronize A & B clocks, as seen from within the moving frame, then: [..]

So, which possibility is correct, or where is my error here?

Thanks,
Roi.

OK then, here's my last try!

Transformation means that you interpret what is measured (or can be measured) from the perspective of a different reference system.
And when people say "as seen from within the moving frame", they commonly mean "as measured by someone who considers "the moving frame" to be not moving but in rest" - so that the "rest frame" is moving. Apparently that is also what you mean.

I will now recycle an old answer here and add some elaboration.

Slow transport is only slow relative to the platform on which the clocks are transported.

Clocks that are slowly moved over a not too far distance will remain approximately synchronized with other clocks in that system - according to the platform's synchronization with light rays!

As I formulated it, the transport may be fast: both clocks will always remain exactly synchronized with each other according to measurements on the moving platform. For your example it doesn't matter if they are both behind, as long as they are equally behind.

However, the effect of speed on clock rate is not linear.

In approximation, for not too high speeds, the clock rate decrease is proportional to the square of the speed (by a factor 1/2 v²/c²). Thus the difference of clock rate depends very much on your assumption to be moving fast or to be in rest.

Just consider the difference in effect of clock transport on clock rate at 1 m/s:
a. From "rest"= 0 m/s: 1x1 - 0x0 = 1
b. From "moving" at 1000 m/s: 1001x1001 - 1000x1000= 2001

As a result, the clock retardation effect of "slow clock transport" is roughly 2000 times bigger if you assume the platform to be moving at 1000 m/s than if you assume it to be in rest.

Try to calculate this for yourself, to improve your intuition!

According to measurements with a stationary frame of the moving platform, the clock that is transported in the direction of motion will be ticking slower, during that transport, than the clock that is transported against the direction of motion.

The result is as follows for motion to the right, first purely in theory and then from measurements (which are not free from theory):

1a. An observer on the moving platform moves the clocks apart. He assumes to be in rest, so that the effect on clock rate is very small and anyway, the effect on each is equal. Thus for him they should still be in tune with each other.

1b. According to interpretation from the stationary frame, those clocks will not be in tune with each other, as I explained above.

2a. The observer on the moving platform checks with light signals if his clocks are synchronized:
- He finds that light from the light sources arrive at each clock at the same clock time.
- He also measures the same time for light from A to B as from B to A.
Thus the clocks are synchronized to the moving platform according to the synchronization convention.

2b. According to measurements with the stationary frame, the extra delay from C makes that light takes longer to reach clock A than if the system were in rest. But there are a lot other effects, compensating each other!
The light and clock A are moving towards each other, which reduces the time. Clock A is ticking slower and it was also delayed more due to transport in the direction of motion, while the platform is contracted in length. And clock B is also ticking slower but was delayed less.

It is partly the other way round for motion in the opposite direction; and again more complex in 3D.

The combined effect is always that the clocks indicate the same arrival time of light from the light sources; and they indicate the same time as when the platform was in rest. Therefore the observations are the same as if the platform is in rest.

Let's hope that this helped...
Harald

 

[roineust]

Thanks Harald...
You did not relate to what really bothers me (posts #186, #190, #192), but your clarification was, non the less, enlightening.

Roi.

 

[harrylin]

Thanks Harald...
You did not relate to what really bothers me (posts #186, #190, #192), but your clarification was, non the less, enlightening.

Roi.

You're welcome Roi!

Every day there is something else that "really bothers you" but it's all very much related. I replied to your post #178 of a few days ago.

#186: "If time dilation exists, without relation to acceleration, that brought the frame to its new inertial state, how could it exist and not exist at the same time?"

I also gave the answer to that question in my last post!
Time dilation is measured wrt what one assumes to be "rest".

It's the same as with kinetic energy: there is said to be none when we say that it is in rest, but there is said to be some when we say that it is in motion. That is Newtonian mechanics - for sure you know it!
How can kinetic energy exist and not exist at the same time? And the example can be pushed further, for it costs energy to bring something into motion and a moving charge also has a magnetic field. How could it have no kinetic energy and no magnetic field?

#190: "The question is, why the device functions exactly the same, also in the 'moving diagram', although we know that time dilation exists (it says, in the 'moving diagram' 'time dilation' with an arrow pointing at clock C)."

I also gave the answer to that question in my last post!
Without time dilation the device would not function exactly the same in motion as in rest.

I have the impression that you confuse the different points of view - a bit like asking how, if kinetic energy does not exist, it can exist. That way you can never understand it.

As measured with the moving frame, the clocks have zero kinetic energy and zero time dilation and the speed of light is c relative to that frame. For that you need no transformation.

#192: "But we know [time dilation] was already detected and verified, experimentally, when a clock was in the 'moving' frame, so how come, when inside the 'moving diagram' or let it be, 'moving frame', it can't be detected?"

That is wrongly formulated as I already answered in #194. As you don't use calculations, wrong formulations are disastrous. Everything is always in all frames; and I explained in my last post how it works!

Regards,
Harald

 

[roineust]

Hearld,
Sorry, but if you open with a complain, that the rate of me having questions, that 'bother me' is a problem - then it makes it harder for me, to read throughly what you write (maybe this is what you are wishing for?). So, maybe when I will be ready to read berating explanations, I will get back to you.

Meanwhile, maybe I will find someone, who really has an unheard of, novel way, to explain SR, or maybe corrections to SR will take place in the future.

Please - replies only relating to my questions!

 

[Dale]

I don't think you need more explanations. I think you need to work some problems. There is a reason that homework problems are an essential part of any physics course. In the process of actually calculating the answers you learn things that are not easy to explain verbally. Please work the problem I suggested in 188, I think it will help your understanding.

 

[harrylin]

Hearld,
Sorry, but if you open with a complain, [..]
Please - replies only relating to my questions!

Sorry if I sounded like complaining, that was not intended! I tried to make clear that already the answers had been there, in our earlier replies.
But this time I did even better, for just after answering your post #178 (which clarified a lot for you), I now answered in one reply:

- your post #186,
- your post #190,
- your post #192

So, I answered to all your remaining questions in my last reply!

Did you really read it and think it over? Next you could try to get a better feeling for it by doing a few calculations, as Dalespam and I recommended.

Success!
Harald

 

[ghwellsjr]

Roi, are you trying to understand how in one frame time dilation can exist for a particular clock but in another frame, there is no time dilation for the very same clock?

 

[harrylin]

Roi, are you trying to understand how in one frame time dilation can exist for a particular clock but in another frame, there is no time dilation for the very same clock?

I'm pretty sure of that - that's why I gave him the kinetic energy example!

 

[PhilDSP]

Hi Roi,

I'm wondering if it might be easiest to put everything together if one first understands why the concept of length contraction became the primary aspect of SR and LET in the first place. To understand that we really don't need to refer to the MMX and neither do we need to know any terminology or relationships which are now part of SR or LET.

To begin, we can describe how the frequency of sound waves are experienced when a person moves with respect to the source of the sound waves. Do you understand the Doppler effect and can you describe that verbally or with math?