[guest39538]

Ok lets try to follow plain geometry 7th grade stuff. Try to follow it without any preconceived notions you have about physical contraction or you will fail 7th grade geometry. We have two mirrors oriented perpendicular to the direction of travel at half the speed of light. The event of light from one mirror in space travels to the other. Now if we follow relativity correctly the event in space is independent of the mirrors. So light has to move forward to reach the other mirror (light goes in all angles remember). This particular speed causes an angle to create a 30,60,90 triangle. If we are going to follow relativity postulates the light has to move between mirrors through the hypotenuse (if light is independent of the source). If you aren't going to follow relativity postulates we can stop here. Are you still following relativity postulates?

Cos 30 = 0.866025 now how does that relate to the clocks tick rate and view? Well relatively the view from behind your current position at that 30 degree angle is only 86.6025% of a perpendicular view of an object. So here we have the contracted view which is not a physical change in the objects length. Simple plain geometry. We have a length increase in the travel distance for light of 13.3075% vs. the length at relative rest. The clock would take longer to tick with the clock only having 86.6025% of a click compared to relative rest.

Now lets look at the light moving between the mirrors in the direction of the objects vector velocity. We start with the light event in the back to the direction of travel. After the light event leaves both the rear mirror and light are traveling towards the front mirror. The back mirror moves through space one length between the mirrors relative when the light reaches the front mirror. The light has traveled two lengths relative. A very similar thing is happening to the length of travel for the path the light is taking when we add the two way measurement of light. The back mirror travels 1/3rd the distance forward and the front mirror travels 1/3rd the distance forward without the light. The light travels backwards from the direction of travel by 2/3rds. Light traveled 2 2/3rds length vs. relative at rest of 2. But wait the travel distance was 2/3rds when you add the front and back without the light. divide that by two and you get 1/3rd. When you subtract 1/3rd from the two way speed of light of 2 2/3rds you get 2 1/3rd. Divide the 2 1/3 by the two way distance for light and you get 1 1/6. We cannot test relativity anywhere near these speeds to prove or disprove the Lorentz mathematics holds for these relative speeds but once again we have a contracted view because light cannot completely illuminate an object at relativistic speeds and the clock tick rate is regulated by the distance traveled through space in a light clock per tick.

You cannot follow plain geometry's contracted view and also claim there is an equal physical contraction of the object!!!!

Unless of course you are not following relativity's postulates.

Goc in your example you are doing the same exact same thing as they are doing ignoring your own relative correctness that light permeates isotropic and the view you are creating by observer affect of adding mirrors and angling the path is not the natural nature of light and subjective ''parlour tricks''.

[David Cooper]

Then If you look at the previous diagram of the distance contraction, this is what your interpretation mistake is.  The trains rear moves ''forward'' , the light takes less time to get there than previous train rear position.

p.s So David if you want to lock horns in battle with me, you need a lot more than subjective interpretation that fails on every level.

You have shown me the light going from the front of the carriage to the rear. Where is your analysis of light going from the rear to the front? Where is your analysis of the time taken for the round trip (with the rear to front and front to rear parts added together)? If you ever get to the point where you do the job proplerly, you will find that your light has to go faster than c to complete the round trip in the required time unless you contract the length of the train.

And to do the job properly, you need to do the same job for the perpendicular light clock too so that you can compare how long it takes for light to complete the round trip on both light clocks. The times will not match unless you either introduce actual length-contraction or have light move faster than c.

Until you do that and take the results on board, you will continue to be doing pseudo-science rather than the real thing.

[guest39538]

Then If you look at the previous diagram of the distance contraction, this is what your interpretation mistake is.  The trains rear moves ''forward'' , the light takes less time to get there than previous train rear position.

p.s So David if you want to lock horns in battle with me, you need a lot more than subjective interpretation that fails on every level.

You have shown me the light going from the front of the carriage to the rear. Where is your analysis of light going from the rear to the front? Where is your analysis of the time taken for the round trip (with the rear to front and front to rear parts added together)? If you ever get to the point where you do the job proplerly, you will find that your light has to go faster than c to complete the round trip in the required time unless you contract the length of the train.

And to do the job properly, you need to do the same job for the perpendicular light clock too so that you can compare how long it takes for light to complete the round trip on both light clocks. The times will not match unless you either introduce actual length-contraction or have light move faster than c.

Until you do that and take the results on board, you will continue to be doing pseudo-science rather than the real thing.

Complete garbage , you can mirror the diagram for the other direction if you like but the result is still the same, you quite clearly do not understand anything except what education learned you. The light does not have to go faster, where on earth did you get that notion from?

Do you not understand the very simple diagrams of the rear and front of the carriage displacement relative to where and when the light was emitted?

That is your contraction you don't understand.

The light takes 2 seconds for the round trip, the carriage does not even need be there because light can pass right through the carriage .
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[David Cooper]

David

the box understands the concept of visual contraction vs. physical contraction.

What you mean is, he too misunderstands length-contraction.

Your point about the math is not at issue. I agree with the Lorentz contraction along with the physical consequences for view and change in clock tick rate. Up to half the speed of light any orientation of the mirrors in the light clock allow the same tick rate.

You don't even appear to understand half of it. There is nothing different about things going under half the speed of light to differentiate it from things going faster than that speed. There is a slowing of clocks moving at all non-zero speeds through space and there needs to be length contraction in the direction of travel at all non-zero speeds to keep a pair of light clocks ticking at the same rate. In the case of the MMX (which is equivalent to a pair of light clocks) the length contraction on the biggest MMX experiment done to date was a micron. A micron is huge - it takes 10,000 atoms in the material of the apparatus to span it. The direction of travel of the apparatus consistently shortened the arms by up to that amount as required to maintain a null result at all times. For the laws of physics to coordinate things with such precision at 30km/s it is not credible to think that it then throws away the mechanism for higher speeds to rely on voodoo instead (so that light can travel faster than c).

Scientists like yourself are confusing contraction of view as the reason for a slower tick rate by physically contracting the clock. There is no mechanism to physically contract the clock. Only math that follows observations. Math is never the cause of physics but that is what you are claiming by physical object contraction.

Length-contraction is not needed to slow the perpendicular light clock (perpendicular to its direction of movement through space). The longer path for light to travel to complete a tick on the perpendicular clock ties in exactly with how much the clock slows as you move it through space. Length-contraction is needed on the other clock aligned with its direction of travel, and without it you're sunk.

I know you have the intelligence to understand plain geometry but you have a block that will not let you confirm the math of light being finite and independent of the source.

The block is all yours - you steadfastly refuse to challenge your existing beliefs, but instead you dig in to defend them.

You even made up something about the iris in the eyes so you could remain faithful to what you were incorrectly taught. I was taught the same thing but rather than a follower I have to work out these issues for myself.

I wasn't taught most of it, but worked it out for myself too. Fortunately I got the details right where you went wrong, which is why what I showed you in the diagram works whereas your magical ideas don't match up to reality. You have invented a whole stack of visual distortions that don't exist in this universe.

When I did following the relativity postulates showed a visual contraction rather than a physical contraction.

You fooled yourself into thinking that, and you're so emotionally tied to your creation that you refuse to recognise that it doesn't fit the real world.

You need to think for yourself rather than let others think for you.

Do I sound like someone who lets other people think for me? If I was, I'd be pushing SR and chanting establishment mantras. You're the most bizarre person I've ever encountered though, because you are actually going around backing up SR on the basis of the most extraordinary pile of pants of your own invention which barely has anything in common with SR at all.

Most scientists just go with their programing. Half the speed of light should be the easiest to understand for most scientists. The box showed you the diagram of event position in space relative to an objects velocity at 180 degrees. You are just going to confuse yourself using laser light so we are using normal imaging where you can view an image in all  positions and the image light goes in all angles. For instance a bulb lights up a room and you can view the light from any angle in that room.

Perhaps if you used a circle of outward-pointing lasers, you'd have a better understanding of how light actually behaves. Each beam can be thought of as representing the journey of a photon. By thinking down this route, you might be able to stop confusing yourself with what happens when you lump them all together and call it a sphere of expanding light.

Ok lets try to follow plain geometry 7th grade stuff. Try to follow it without any preconceived notions you have about physical contraction or you will fail 7th grade geometry. We have two mirrors oriented perpendicular to the direction of travel at half the speed of light. The event of light from one mirror in space travels to the other. Now if we follow relativity correctly the event in space is independent of the mirrors. So light has to move forward to reach the other mirror (light goes in all angles remember). This particular speed causes an angle to create a 30,60,90 triangle. If we are going to follow relativity postulates the light has to move between mirrors through the hypotenuse (if light is independent of the source). If you aren't going to follow relativity postulates we can stop here. Are you still following relativity postulates?

Lovely - just like the vertical arm on my interactive MMX demonstrations (a different speed, but working on the exact same rules), so what makes you think I do things differently for the perpendicular case?

Cos 30 = 0.866025 now how does that relate to the clocks tick rate and view?

But here's the point where we diverge. You now go into some voodoo where you misinterpret what's going on. The angle the light goes at simply increases the distance it has to move through space to travel from one mirror to the other and back, and that extra distance the light has to travel means that the light takes longer to get from one mirror to the other and back, and the clock ticks are therefore more spaced out in time. That is all that happens to make the clock run slow - the light runs at full speed in the clock but the tick rate of the clock is slower due to the delay from the extra distance travelled by the light.

Well relatively the view from behind your current position at that 30 degree angle is only 86.6025% of a perpendicular view of an object.

And your voodoo takes the form of warped "views". There are no warped views though - everything continues to look completely normal for anyone who is moving with the clock. A laser sending light at that 30 degree angle is aligned perpendicular to the direction of its (the laser's) travel, and the eye of an observer (co-moving with the laser) receiving that light will perceive it as coming in on the perpendicular too.

So here we have the contracted view which is not a physical change in the objects length.

So far it has no relevance to the length at all - that only comes in when you deal with light clocks aligned at angles other than perpendicular to their direction of travel.

Simple plain geometry. We have a length increase in the travel distance for light of 13.3075% vs. the length at relative rest. The clock would take longer to tick with the clock only having 86.6025% of a click compared to relative rest.

Correct.

Now lets look at the light moving between the mirrors in the direction of the objects vector velocity.

At last!

We start with the light event in the back to the direction of travel. After the light event leaves both the rear mirror and light are traveling towards the front mirror. The back mirror moves through space one length between the mirrors relative when the light reaches the front mirror. The light has traveled two lengths relative. A very similar thing is happening to the length of travel for the path the light is taking when we add the two way measurement of light. The back mirror travels 1/3rd the distance forward and the front mirror travels 1/3rd the distance forward without the light. The light travels backwards from the direction of travel by 2/3rds. Light traveled 2 2/3rds length vs. relative at rest of 2.

And that's a longer path for light to travel per tick than on the perpendicular light clock, so the light clocks don't tick at the same rate.

But wait the travel distance was 2/3rds when you add the front and back without the light. divide that by two and you get 1/3rd. When you subtract 1/3rd from the two way speed of light of 2 2/3rds you get 2 1/3rd. Divide the 2 1/3 by the two way distance for light and you get 1 1/6.

But you need to explain all the stuff you've just done there. What exactly is this 2/3 travel distance; why are you dividing it by 2; why are you subtracting it from the 2 2/3; and why are you dividing 2 1/3 by 2? You already had the answer 2 2/3 for the light path, and the time light takes to cover that distance is the time between clock ticks for that clock.

We cannot test relativity anywhere near these speeds to prove or disprove the Lorentz mathematics holds for these relative speeds

The MMX tests it for 30km/s and shows length contraction. Why should physics make it contract in that way for slower speeds and fail to do so for higher speeds? We have particle accelerators in which particles are sent at speeds close to c, and this impacts on the time it takes for them to decay, lengthening their lifespans many times over, with the statistics on this following the predictions of the model. These particles have a "clock" in the form of a mechanism which runs slow when they move at high speed, but it runs slow in the way predicted by a perpendicular light clock and not an uncontracted light clock aligned with the direction of travel, so do all these particles magically hold their "clock" perpendicular to their direction of travel at all times or do they conform to the rules of the model by length-contracting themselves?

but once again we have a contracted view because light cannot completely illuminate an object at relativistic speeds...

Why do you think light can't completely illuminate an object at relativistic speeds? When we use the term "relativistic speed", it's a bit of a woolly one - there's no actual dividing place between relativistic speeds and non-relativistic speeds. The same rule about length-contraction applies to all non-zero speeds, but can usually be ignored when we're doing ordinary stuff like ballistics because the errors are too small to care about. And, if you have a light in the middle of a room in a rocket, there is nothing to stop a light illuminating that wall at any speed, including 0.0000000000001c and 0.999999999999c, and the illumination would appear to be the same in both those cases for anyone inside that room.

and the clock tick rate is regulated by the distance traveled through space in a light clock per tick.

Indeed, but you have to get both your clocks to tick at the same rate to match up with what the null result of MMX shows you, and you can't do that without actual length-contraction.

You cannot follow plain geometry's contracted view and also claim there is an equal physical contraction of the object!!!!

There is no "contracted view" of this - it is an invention of your own which doesn't relate to reality. All you've done is deny that actual length-contraction happens on the basis that the experiment has never been done at high speeds where the diagrams show clearly that it's needed, but you've failed to understand that actual length-contraction is still needed at slow speeds at which experiments have been done. Instead of recognising that need, you've come up with a hocus pocus of contracted views that don't explain how you keep your light clocks ticking at the same rate without having light go faster than c on one of them.

[David Cooper]

Complete garbage , you can mirror the diagram for the other direction if you like but the result is still the same, you quite clearly do not understand anything except what education learned you.

How does reversing the diagram fix it? The train would then be going in the wrong direction. I want you do move the carriage to the right and show the light moving to the right too. Once you've got your head round that really difficult idea, maybe you can start to wonder how long it will take for light to get from the back of the carriage to the front.

The light does not have to go faster, where on earth did you get that notion from?

How can you have got this far and still not understood where that notion comes from. Do the maths on how long it takes for light to catch the front of the carriage while chasing it from the back of the carriage. Important clue: the front of the carriage is moving away from the light and not towards it. Do the maths on that, then combine it with our maths for light going in the opposite direction from the front of the carriage to the back end (with the back end rushing forwards to meet the light). Add the two lengths of time together, and bingo! You should have a time value for the round trip. That time value will be longer than for a tick of an identical light clock perpendicular to the train (moving along with the train). Why have you still not done this?

Do you not understand the very simple diagrams of the rear and front of the carriage displacement relative to where and when the light was emitted?

That is your contraction you don't understand.

The problem is entirely with your lack of understanding, as demonstrated by your failure to get the direction of the train right.

The light takes 2 seconds for the round trip, the carriage does not even need be there because light can pass right through the carriage .

If it takes 2 seconds with the train stationary, it will take 8 seconds for the round trip with the train moving at 0.867c unless you contract the train to half its rest length, at which point it will take 4 seconds for the round trip, matching the 4 seconds taken for the round trip on an identical perpendicular clock moving with the train. You haven't even begun to explore this stuff.

[guest39538]

David

the box understands the concept of visual contraction vs. physical contraction.

What you mean is, he too misunderstands length-contraction.

That is not what I ''said'', Goc understands but in trying to show you why you are wrong, he is wrongly showing you why you are wrong by using the same 2 dimension thoughts as yourself .

[guest39538]

The light takes 2 seconds for the round trip, the carriage does not even need be there because light can pass right through the carriage .

If it takes 2 seconds with the train stationary, it will take 8 seconds for the round trip with the train moving at 0.867c unless you contract the train to half its rest length, at which point it will take 4 seconds for the round trip, matching the 4 seconds taken for the round trip on an identical perpendicular clock moving with the train. You haven't even begun to explore this stuff.

No, the light takes two seconds, you are really not thinking for yourself about the diagrams. 

[guest39538]

Complete garbage , you can mirror the diagram for the other direction if you like but the result is still the same, you quite clearly do not understand anything except what education learned you.

How does reversing the diagram fix it? The train would then be going in the wrong direction. I want you do move the carriage to the right and show the light moving to the right too. Once you've got your head round that really difficult idea, maybe you can start to wonder how long it will take for light to get from the back of the carriage to the front.

The light does not have to go faster, where on earth did you get that notion from?

How can you have got this far and still not understood where that notion comes from. Do the maths on how long it takes for light to catch the front of the carriage while chasing it from the back of the carriage. Important clue: the front of the carriage is moving away from the light and not towards it. Do the maths on that, then combine it with our maths for light going in the opposite direction from the front of the carriage to the back end (with the back end rushing forwards to meet the light). Add the two lengths of time together, and bingo! You should have a time value for the round trip. That time value will be longer than for a tick of an identical light clock perpendicular to the train (moving along with the train). Why have you still not done this?

Do you not understand the very simple diagrams of the rear and front of the carriage displacement relative to where and when the light was emitted?

That is your contraction you don't understand.

The problem is entirely with your lack of understanding, as demonstrated by your failure to get the direction of the train right.

The light takes 2 seconds for the round trip, the carriage does not even need be there because light can pass right through the carriage .

If it takes 2 seconds with the train stationary, it will take 8 seconds for the round trip with the train moving at 0.867c unless you contract the train to half its rest length, at which point it will take 4 seconds for the round trip, matching the 4 seconds taken for the round trip on an identical perpendicular clock moving with the train. You haven't even begun to explore this stuff.

You are not understanding, I think it may be beyond you, no insult intended.

[guest39538]

How can you have got this far and still not understood where that notion comes from. Do the maths on how long it takes for light to catch the front of the carriage while chasing it from the back of the carriage. Important clue: the front of the carriage is moving away from the light and not towards it. Do the maths on that, then combine it with our maths for light going in the opposite direction from the front of the carriage to the back end (with the back end rushing forwards to meet the light). Add the two lengths of time together, and bingo! You should have a time value for the round trip. That time value will be longer than for a tick of an identical light clock perpendicular to the train (moving along with the train). Why have you still not done this?

2 seconds round trip, you are doing it wrongly

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The drawing is not your scenario, it is to get you to understand how wrong you are.


Relative correctness( I have drawn you a diagram, study it , understand where you are going wrong.


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[David Cooper]

Here's the root of your misunderstanding. Look at your diagram again - I've replaced parts of it with my own lines and named them A, B and C (in blue text). The length of A added to the length of B comes to twice the length of C. That's your 1.2 + 0.8 = 2.

However, what you've still failed to grasp after all this time is that if the light is taking 1.2 seconds to make the trip from the rear to the front of the carriage, the carriage will move further during that part of the trip for the light than it does on the return journey from the front to the rear where the light makes that trip in only 0.8 seconds, so the carriage moves less far during that part of the trip. You have it moving the same distance for both parts of the light's journey, which means you aren't moving the train at a constant speed. The maths of this is a teeny weeny bit more complicated than you realise.

[guest39538]

Here's the root of your misunderstanding. Look at your diagram again - I've replaced parts of it with my own lines and named them A, B and C (in blue text). The length of A added to the length of B comes to twice the length of C. That's your 1.2 + 0.8 = 2.

However, what you've still failed to grasp after all this time is that if the light is taking 1.2 seconds to make the trip from the rear to the front of the carriage, the carriage will move further during that part of the trip for the light than it does on the return journey from the front to the rear where the light makes that trip in only 0.8 seconds, so the carriage moves less far during that part of the trip. You have it moving the same distance for both parts of the light's journey, which means you aren't moving the train at a constant speed. The maths of this is a teeny weeny bit more complicated than you realise.

Quite clearly you have failed to understand  the diagram which is very correct and did not need no edit from yourself.
let us remove the train carriage so David can understand.

A:_______________________________________________________

B:_______________________________________________________

C:_______________________________________________________

there you go can you understand now?

David ignores that light can pass through things. David ignores that the surface of the train wall does not even reflect light and we would have to add mirrors and a medium.

[guest39538]

I am still waiting for you to explain what suppose to be the problem?  It is relative straight forward.   Light is emitted from one point of the train, (the rear or front).

The train is in motion relative to the lights motion, the light travels less distance and more distance in a round trip because the rear is moving forward contracting the distance the light has to travel, the front recedes away from the light so it has to travel further.

Nothing more to it.

Event 1:light emitted, travelling left to right

Event 2:train moving relative forward

Event 3: spacial distance contracting between the light and train rear

Event 4: reflection and return trip

Event 5: The front of the train recedes away from the chasing light

Event 6: the light eventually catches up with the front

Event 7: -t/dx+t/dx=t/dx

[David Cooper]

Box,

The reason I reworked your diagram a little was to put names on some of the lines to make it easier to refer to them. I've added a new line called D to the latest version of the diagram for the same reason.

You say you have the world's greatest mind, so you really should have got it by now. You have the train moving distance D in 1.2 seconds while light is moving from the rear to the front, and you also have the train moving distance D in 0.8 seconds while light is moving the other way from the front to the rear. That means your train is suddenly going 1.5 times the speed it was for the first leg of the light's journey.

To do things properly, you need to keep the train's speed the same throughout, which means that if it takes 1.2 seconds for light to go from the rear to the front (covering the distance A) while the train moves forwards by the distance D, the train will only move 2/3 D in the 0.8 seconds which you allow for the light to travel backwards (covering the distance B). By the end of that time, the light has not reached the rear as the rear is still 1/3 D further away. You need to let the light and train move a bit further before your clock tick is complete, so it will be longer than 2 seconds.

This is really basic stuff that you've messed up - your foundation is not properly laid and everything else that you've built on top of it will need to be reassessed once you've corrected this fault to make sure that it is sound.

[guest39538]

Box,

The reason I reworked your diagram a little was to put names on some of the lines to make it easier to refer to them. I've added a new line called D to the latest version of the diagram for the same reason.

You say you have the world's greatest mind, so you really should have got it by now. You have the train moving distance D in 1.2 seconds while light is moving from the rear to the front, and you also have the train moving distance D in 0.8 seconds while light is moving the other way from the front to the rear. That means your train is suddenly going 1.5 times the speed it was for the first leg of the light's journey.

To do things properly, you need to keep the train's speed the same throughout, which means that if it takes 1.2 seconds for light to go from the rear to the front (covering the distance A) while the train moves forwards by the distance D, the train will only move 2/3 D in the 0.8 seconds which you allow for the light to travel backwards (covering the distance B). By the end of that time, the light has not reached the rear as the rear is still 1/3 D further away. You need to let the light and train move a bit further before your clock tick is complete, so it will be longer than 2 seconds.

This is really basic stuff that you've messed up - your foundation is not properly laid and everything else that you've built on top of it will need to be reassessed once you've corrected this fault to make sure that it is sound.

#


The times were just an example and not exact, I was trying to show you why and where you are going wrong but obviously it has not sunk in. 
It is sound I assure you, I am not a scientist and do not get paid for my time or even get any sort of respect, so forgive me for not trying too hard with the ''maths''.
I could probably calculate an exact if I wanted to, I already have all of the parameters involved.
The point is the scenario means nothing, it is a poorly thought, thought experiment, no maths really required to observe the result.
I did you the formula , what more do you want?

Event 7:

[guest39538]

Pfff , 'they'' making me work hard.

Ok the train is travelling at half the speed of light.

The trains length is l=299 792 458 m

In 1 second the rear of the train has travelled 149896229 meters


ok so far?

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[guest39538]

Pfff , 'they'' making me work hard.

Ok the train is travelling at half the speed of light.

The trains length is l=299 792 458 m

In 1 second the rear of the train has travelled 149896229 meters


ok so far?

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t(c)/dx1+dx2=1.s
t(c)/dx1+dx2=1.s

My train has glass walls if it helps you understand.

[guest39538]

At rest c/dx=1.s

At rest c/dx=1.s

In motion c/dx=1.5s

in motion c/dx=0.5s

Reason : distance contraction and distance expansion of points relative to the velocity of light.

ok?

p.s providing the train stops after one positional change, otherwise my times are off slightly if the train is continuous in motion. However positional my times are quite accurate . (I hope lol).

[David Cooper]

The times were just an example and not exact, I was trying to show you why and where you are going wrong but obviously it has not sunk in.

How was I wrong for telling you that the time taken is greater than two seconds when you claimed it wasn't? You now appear to have seen the light though, so let's press ahead.

It is sound I assure you, I am not a scientist and do not get paid for my time or even get any sort of respect, so forgive me for not trying too hard with the ''maths''.

Do you think someone's paying me to run this remedial class? Do you think this maths is hard? This is the easy stuff.

I could probably calculate an exact if I wanted to, I already have all of the parameters involved.

Why have you never done it? Why wait till now? I've been setting examples in front of you in which the extra distance light has to travel in an uncontracted carriage would lead to a light clock ticking four times less often than a stationary clock due to the light path being four times as long and you've told me that that can't happen. For example,

I said,

If it takes 2 seconds with the train stationary, it will take 8 seconds for the round trip with the train moving at 0.867c unless you contract the train to half its rest length, at which point it will take 4 seconds for the round trip, matching the 4 seconds taken for the round trip on an identical perpendicular clock moving with the train. You haven't even begun to explore this stuff.

and you replied,

No, the light takes two seconds, you are really not thinking for yourself about the diagrams.

Even with the much more modest speeds of travel in other examples, length-contraction has a crucial role in reducing the length of the path light has to follow from rear to front and back again in order to keep the light clock in sync with a perpendicular light clock (which itself runs slower than a stationary clock). This is all necessary to account for the null result of MMX, but you've been writing it all off as nonsense while claiming neither light clock is slowed.

The point is the scenario means nothing, it is a poorly thought, thought experiment, no maths really required to observe the result.

It's a well thought out experiment which directly illustrates how lengths of light paths are increased by movement of clocks. How the blazes do you imagine it can be explored otherwise?

I did you the formula , what more do you want?

I don't want anything from you at all. It's entirely up to you how much you want to understand and how much you are happy to go on misunderstanding. I'm simply offering you help with getting your head around it if you're prepared to put in the necessary effort (which isn't greatly taxing at this stage). If you want to understand length contraction, you need to work through the numbers by looking at a light clock aligned with a moving vehicle. If you want to understand the slowing of apparent time, you need to do the same kine of work with a perpendicular light clock to find out how much extra distance light has to travel on that if the vehicle is moving.

Here are my numbers for a vehicle moving at 0.5c:-

Length of vehicle = d

Time for light to travel distance d = t

Time for light to make round trip lengthways when vehicle at rest = 2t

Time for light to make first part of trip when vehicle moving at 0.5c = 2t
(Front of vehicle was ahead of light by d and moving at 0.5c while light is moving at c, so light is gaining on front of vehicle at 0.5c and will take 2t to catch it.)

Distance vehicle has moved by this point = d
(The light moved 2d and the vehicle moved half that.)

Distance light has moved by this point = 2d

Time for light to make second part of trip = 2/3t
(This time we add the speeds together instead of subtracting, so it's a "closing speed" of 1.5c to cover distance d.)

Distance vehicle has moved during the time the light was coming back = 1/3d

Distance light has moved during second part of trip = 2/3d

We now have a round trip for the light completed in 2 2/3t. The light has moved 2 2/3d through space. The vehicle has moved a total of 1 1/3d, which is half the distance the light travelled, and that's no surprise as the light was moving twice as fast as the vehicle.

Do your numbers match mine? If not, why not? Let's see if we can get agreement on this before we go on to look at the perpendicular clock.

[guest39538]

The times were just an example and not exact, I was trying to show you why and where you are going wrong but obviously it has not sunk in.

How was I wrong for telling you that the time taken is greater than two seconds when you claimed it wasn't? You now appear to have seen the light though, so let's press ahead.

It is sound I assure you, I am not a scientist and do not get paid for my time or even get any sort of respect, so forgive me for not trying too hard with the ''maths''.

Do you think someone's paying me to run this remedial class? Do you think this maths is hard? This is the easy stuff.

I could probably calculate an exact if I wanted to, I already have all of the parameters involved.

Why have you never done it? Why wait till now? I've been setting examples in front of you in which the extra distance light has to travel in an uncontracted carriage would lead to a light clock ticking four times less often than a stationary clock due to the light path being four times as long and you've told me that that can't happen. For example,

I said,

If it takes 2 seconds with the train stationary, it will take 8 seconds for the round trip with the train moving at 0.867c unless you contract the train to half its rest length, at which point it will take 4 seconds for the round trip, matching the 4 seconds taken for the round trip on an identical perpendicular clock moving with the train. You haven't even begun to explore this stuff.

and you replied,

No, the light takes two seconds, you are really not thinking for yourself about the diagrams.

Even with the much more modest speeds of travel in other examples, length-contraction has a crucial role in reducing the length of the path light has to follow from rear to front and back again in order to keep the light clock in sync with a perpendicular light clock (which itself runs slower than a stationary clock). This is all necessary to account for the null result of MMX, but you've been writing it all off as nonsense while claiming neither light clock is slowed.

The point is the scenario means nothing, it is a poorly thought, thought experiment, no maths really required to observe the result.

It's a well thought out experiment which directly illustrates how lengths of light paths are increased by movement of clocks. How the blazes do you imagine it can be explored otherwise?

I did you the formula , what more do you want?

I don't want anything from you at all. It's entirely up to you how much you want to understand and how much you are happy to go on misunderstanding. I'm simply offering you help with getting your head around it if you're prepared to put in the necessary effort (which isn't greatly taxing at this stage). If you want to understand length contraction, you need to work through the numbers by looking at a light clock aligned with a moving vehicle. If you want to understand the slowing of apparent time, you need to do the same kine of work with a perpendicular light clock to find out how much extra distance light has to travel on that if the vehicle is moving.

Here are my numbers for a vehicle moving at 0.5c:-

Length of vehicle = d

Time for light to travel distance d = t

Time for light to make round trip lengthways when vehicle at rest = 2t

Time for light to make first part of trip when vehicle moving at 0.5c = 2t
(Front of vehicle was ahead of light by d and moving at 0.5c while light is moving at c, so light is gaining on front of vehicle at 0.5c and will take 2t to catch it.)

Distance vehicle has moved by this point = d
(The light moved 2d and the vehicle moved half that.)

Distance light has moved by this point = 2d

Time for light to make second part of trip = 2/3t
(This time we add the speeds together instead of subtracting, so it's a "closing speed" of 1.5c to cover distance d.)

Distance vehicle has moved during the time the light was coming back = 1/3d

Distance light has moved during second part of trip = 2/3d

We now have a round trip for the light completed in 2 2/3t. The light has moved 2 2/3d through space. The vehicle has moved a total of 1 1/3d, which is half the distance the light travelled, and that's no surprise as the light was moving twice as fast as the vehicle.

Do your numbers match mine? If not, why not? Let's see if we can get agreement on this before we go on to look at the perpendicular clock.

You are hard to understand but you mention 1.5c so I guess we do not agree.  I will try to understand what you have put , in the meantime may I suggest you try to understand what I have put.

 Do you agree that a radio signal  is light?
Do you agree the radio signal can pass through the trains walls?
Do you agree that if we used a radio signal instead of ''light'' that there is no scenario to discuss?

The thing is you are still not doing it correctly.

Lets restart this and try to go at a slow pace taking one issue at a time into consideration. You say

''Length of vehicle = d''

Ok, are you happy at defining (d) to be 299 792 458 m in length at relative rest?

Do you agree that a round trip for light travelling  then a return trip would take 2 seconds?

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[timey]

According to the equivalence principle the laws of physics remain the same in each reference frame.  The reference frame that is travelling at 0.867c has the same laws of physics as the stationary frame.  The stationary frame measures the moving frame as length contracted, but the moving frame measures itself as being the same length as if it were stationary.  What is causing the differing measurements?

The stationary clock is ticking twice as fast as the moving clock.  The stationary frame measures the moving frame as per the tick rate of it's clock, where it concludes that in order to upkeep the constancy of the speed of light the moving frame must be length contracted, and the moving frame measures it's frame as per the tick rate of it's own clock and finds that there is no sign of any length contraction.   As an experiment the people in the moving frame decide to measure their moving frame as per the tick rate of the stationary clock and quickly come to the conclusion that the speed of light in their moving frame is holding itself relative to the tick rate of their moving clock...
They refer back to the equivalence principle to find that the laws of physical process are indeed the same in each reference frame where it would be silly to think that the speed of light in the moving frame would be held relative to a second as per a clock ticking in the stationary frame, when the clock in their own moving frame is ticking at half the rate of the clock in the stationary frame.

Here we are saying that the speed of light is constant in each frame, but that the length of a second that the speed of light is held relative to is different in differing frames resulting in length contraction being a consequence of trying to hold the speed of light as constant held relative to a static length second.

Same mathematical proportions, but laid out from a differing perspective...