# On the idea of time in physics-relativity

Discussion in 'Physics & Math' started by ash64449, Mar 15, 2013.

1. ### Prof.Laymantotally internally reflectedRegistered Senior Member

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I have no idea what your talking about or where this is all coming from. Getting to hard to have two conversations at once. I think that there is length contraction in the other frame so that the beams of light in the MME will reach the detector at the same time. Then they will measure the same speed of light in both directions, the added distance it has to travel in the direction of motion is shortened just enough so that it is seen as being the same distance from another frame. Same speed, less distance traveled, with less time then counteracts the longer distance the beam is seen to travel from another frame that would think it would have to travel a longer distance since it is in relative motion.

3. ### Prof.Laymantotally internally reflectedRegistered Senior Member

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Look at them, the flashes arrive at the front and back of the train at different times. They travel the same distance that is the distance of the train. It then takes different amounts of time for them to do this. If they are traveling the same distance and taking a different amount of time to get there, then you can draw only one conclusion, they are traveling at different relative speeds to the train. All I can see is flashes of light traveling the same distance in different amounts of time, except the blue lines in the right diagram actually travel the same distance in the same amount of time, so then they would have the same velocity.

5. ### PeteIt's not rocket surgeryRegistered Senior Member

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You are mixing platform frame times with train frame distances.

At different times according to platform clocks (clocks at rest with respect to the platform).
At the same time according to train clocks (clocks at rest with respect to the train).

The same distance measured by train rulers (rulers at rest in the train frame).
Different distances measured by platform rulers (rulers at rest in the platform frame).

Different amounts of time according to platform clocks.
Same amounts of time according to train clocks.

In both diagrams, distance is on the vertical axis, time is on the horizontal axis.
The speed of an object is the slope of the line representing it.
All light flashes have the same slope (some upward, some downward).
All light flashes have the same speed.

7. ### Prof.Laymantotally internally reflectedRegistered Senior Member

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Then how do you suppose how they would then travel at the same speed, but then arrive at the front and back of the train at different times from the opposite end of the train?

Just consider the points the lines intersect and the points they come from. If you calculated the velocity from starting point to the time it came to the ending point, they wouldn't have the same velocity. A point represents a certain location at a certain time. You would have said that it goes from this point to another that is the same real distance in two different amounts of times.

Say in the first diagram, the top yellow flash then takes forever to get to the other end of the train, while the bottom red flash takes no time at all to get to the bottom line or the other end of the train. The yellow line is trying to catch up to the train and the red line is able to get there sooner because of the motion of the train. Then in the diagram both flashes are sent at the same time. Then take different amounts of time to reach the black lines or opposite ends of the train.

8. ### PeteIt's not rocket surgeryRegistered Senior Member

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Which lines and which points? In which diagram?
Have you tried these calculations yourself?

The red and yellow flashes have the same slope, and therefore the same speed, in both diagrams.

In the platform rest frame:
The yellow flash begins at t = 0, x = -10, and reaches the front of the train at t = 50, x = 40.
It travels 50 units distance in 50 units time. Its speed is 1.
The red flash begins at t = 0, x = 10, and reaches the back of the train at t = 12.5, x=-2.5.
It travels -12.5 units distance in 12.5 units time. Its speed is -1.

In the train rest frame:
The yellow flash begins at t'=0, x'=0, and reaches the front of the train at t' = 12.5, x' = 12.5.
It travels 12.5 units distance in 12.5 units time. Its speed is 1.
The red flash begins at t'=0, x'=0, and reaches the back of the train at t' = 12.5, x' = -12.5.
It travels -12.5 units distance in 12.5 units time. Its speed is -1.

9. ### Prof.Laymantotally internally reflectedRegistered Senior Member

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The red and yellow lines are two flashes of light that begin on opposite ends inside of the train. The black lines are then the end of the train, (IDK, why I am explaining this to you in your own diagrams). The distance between the black lines would then be like the length of the train. From the time the flashes are sent from each end, they then intersect the black lines ( each end of the train ) at different times. So then it has taken a different amount of time for each flash to move from black line to black line, or in other words to the opposite end of the train. I didn't do any calculations, I can tell this just from looking at them. The time it takes for the flashes to go from black line to black line is different, so then according to the diagram they would be traveling at different relative speeds as compared to the train.

10. ### Aqueous Idflat Earth skepticValued Senior Member

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6,152
Because time and space are relative, which means that they warp (rotate) as per the Lorentz transform. This has been the same answer given since the outset of this thread, the answer you keep ignoring.

False.

False.

"Real"is meaningless. The operative term is "relative".

Endless meaningless BS.

Meaningless endless BS.

No, no, no, no, no.

No.

No.

Just deal with it. Time and space are relative. Spacetime rotates when the train leaves the inertial reference frame of the platform, accelerates to cruise speed and levels off. When it passes the platform, there are two copies of space time superimposed on one another, but they are no longer identical, since the train's reference frame has gone into rotation. Until you begin to understand and accept this basic principle of nature, all of your evasive maneuvers are just repetitive nonsense.

There is no way around it. You have to comply with the laws of nature. They will not defer to your personal preferences about how you think they should work. It really doesn't get any simpler than that. You are wrong, dead wrong, and the more you kick and scream about it, the more paint yourself the fool.

11. ### PeteIt's not rocket surgeryRegistered Senior Member

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10,166
I think that you're making a mistake, because you're not thinking it through rigorously, and you are mixing the train rulers with platform clocks.

I'm interested in what you get when you do the calculations that you suggested.

12. ### Prof.Laymantotally internally reflectedRegistered Senior Member

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Looks like your the one doing all the kicking and screaming. Would you like 50% more cash?

13. ### Prof.Laymantotally internally reflectedRegistered Senior Member

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982
I think the problem is that you just put in the equations in the diagram that are already the relations between the two frames, so then the equations graphed with each other do not give the same relations that the equations themselves have found.

14. ### PeteIt's not rocket surgeryRegistered Senior Member

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10,166
I don't understand what you mean. I don't think there's a problem at all. I don't know what equations you're referring to.

And I'm still interested in what you get for the calculations you suggested.

15. ### Aqueous Idflat Earth skepticValued Senior Member

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6,152
Slamming your BS down the drain is merely a mop-up operation. It's quite the opposite of your infantile banter.

Until you come to the outright admission of your guilt, that you are an anti-science denialist crank, you will continue to pollute these threads with nonsense, invented fantasy, BS and cynical whining.

Of course you could just man up and admit that you have no idea what spacetime rotation really means, and simply ask any of the folks here who bothered to get an education to help you. And this, since you obviously need someone to lead you to the information you're continuing to spit out as it has been spoonfed to you.

Grow up. Stop trolling these threads and get with the program. Get up, get and pencil in hand and start working problems. Until then you're just the intellectual equivalent of a couch potato.

16. ### Prof.Laymantotally internally reflectedRegistered Senior Member

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Both flashes start at time zero, then the yellow flash intersects the top black line at about 50. Then the red flash intersects the bottom black line at about 12. If the top black line is the front of the train and the bottom black line is the back of the train, then a flash from the front will arrive to the back in 50 (secounds?) and a flash coming from the back of the train will reach the front in about 12 (secounds?). The train is the same length for both flashes. It then takes one 50 to reach the other side and the other 12 to reach the other side. Say d = 20, that is the distance of the train front to back, v = 20/50 and v = 20/12. They clearly wouldn't have the same velocity for any value of "x" the size of the train.

17. ### Prof.Laymantotally internally reflectedRegistered Senior Member

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All your doing now is just becoming a nuisance. Your not hard, you don't scare me, I am not even phased by you one bit.

18. ### PeteIt's not rocket surgeryRegistered Senior Member

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10,166
I put some numbers in a few posts up.
In the platform rest frame:

The yellow flash begins at t = 0, x = -10, and reaches the front of the train at t = 50, x = 40.
It travels 50 units distance in 50 units time. Its speed is 1.
The red flash begins at t = 0, x = 10, and reaches the back of the train at t = 12.5, x=-2.5.
It travels -12.5 units distance in 12.5 units time. Its speed is -1.

The difference in our calculations is that you are assessing the distance the flashes travel from the length of the train.

Perhaps you're correct... let's see.
Are you calculating the speed of the flashes as measured by the train[ observer, or by the platform observer?

19. ### Prof.Laymantotally internally reflectedRegistered Senior Member

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From the internet observer looking at the diagram. The train is the same length through the whole diagram, by 20. The green train and the black train are 20 across the whole thing. It doesn't consider length contraction as seen from the rest frame.

20. ### PeteIt's not rocket surgeryRegistered Senior Member

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10,166
The internet observer looking at the flashes in the diagram should not be confused by the train.
Try it like this:

These are the same flashes as before.
What is their speed?

Which rest frame do you mean? The rest frame of the train, or the rest frame of the platform?

21. ### Aqueous Idflat Earth skepticValued Senior Member

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6,152
That's the way trolls are. Nothing matters to them except their insane self-absorption.

You cannot escape the grip of relativity no matter what you think or say. It rules over you, not you over it. Until you man up to this fact, you're just posting meaningless BS.

Until you come to grips with spacetime rotation, nothing else you post here could possibly be relevant. But that requires you to man up, which is the big obstacle here, isn't it? You simply can't handle it, can you? Not only does it challenge your intuition, it requires you to understand 8th grade math, which was your stumbling block, wasn't it?

Well get over it. No one cares. What bugs folks who come here for information is the endless BS from trolls who can't get over their low self-esteem, their lies, and their BS. It's been evident to them since your first post that you never made it into high school math and science. And many of them have generously coddled your tantrums and spoonfed you. But still you just keep kicking and screaming.

So grow up. Get with the program. Get out a pencil and figure out how to do a simple rotation on one axis. It's child's play for most of the members here, so how about catching up instead of playing games and pretending to have some magical powers, powers you gave up when you turned your back on our education.

22. ### Aqueous Idflat Earth skepticValued Senior Member

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6,152
...on your education, I meant to say.

23. ### Prof.Laymantotally internally reflectedRegistered Senior Member

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982
If you take the train out of it, then they are traveling at the same speed, they just where not traveling at the same speed relative to the train itself. The trains velocity was being added by comparing them in the diagram just like this whole ordeal started.

So then changing how the diagrams are used this way doesn't fix the problem. They only become the speed of light in all frames when you remove the train. They travel at the speed of light to the coordinate plane, but then if you put a train in there then they don't travel the speed of light relative to the train. There would be no way really to show how it is still the speed of light to the train when the trains motion on the coordinate plane will cause the difference of speed of the train and the beams to become more Newtonian.