On the idea of time in physics-relativity

I'll hold off on responding until you get around to addressing [post=3057220]post #511[/post]

 

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I haven't read the book before. Now don't go off-tangent, my point is I don't see how it relates to the train gedanken we've been discussing..

 

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The point is that a MME observed in relative motion will not be seen to have different results. If the distance of one beam was observed to travel a longer distance than the other beam then the wavelengths would not match up. They accuse me of using simple layman descriptions to say exactly what happens in reality and in turn they are doing just that. I am saying that there is supersymmetry that may even go beyond gauge symmetry, where gauge symmetry could be found even if both observers are in relative motion. I am not sure if this kind of symmetry is included in gauge symmetry.

It is like in my derivation of the proper time, I think was post #266. For a hundred years now scientist have used a derivation that doesn't even actually give an accurate equation just because it was simple to understand. My derivation is just as simple, but it actually gives the equation of the accurate equation that would be used in practice. The only thing I had to do really is assign the variables correctly. But, it has to assume that both observers actually observe the beam of light to be in the same location simultaneously. This is actually what happens in MME experiments. So then I got the correct answer, and it agrees with experiment. So then the only way both observers could measure the same speed of light for a photon that is a certain distance away from the ship simultaneously is if they measure different amounts of time on their clock. SO then different times on each others clocks can happen simultaneously. Their clocks can just say two different things while experiencing the same event at the "same" time. The difference in their clocks is more than just a consequence of light travel times. If you took two atomic clocks and sent one around the world in an airplane, the one sent around the world would still measure a different time if set back next to the one on Earth. It is just become a pain showing this since Einsteins TE has already gone viral on the internet and it doesn't even agree with the MME while seen from an outside observer in relative motion. In my derivation they both say the beam is in the same location at the same time even though both of their clocks read different times. Simultaneous events happen when different frames have clocks that read two different times.

But, even that example still wouldn't have gauge symmetry. The same amount of spacetime dilation and contraction couldn't account for a beam traveling in both directions over the same location so that they both are measured to travel at the speed of light.

 

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OK I admit it, sure you're a troll, but these comments are pretty damn funny.

This one though; I laughed out loud. Simultaneity has already gone viral on the internet! Outstanding, really, just outstanding.

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He would make a really good comedian

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Too bad comedy doesn't work as a good learning tool.

 

Comedy can work as a form of learning, the daily show for instance, but this case is just entertainment, and don't act like you don't know it.

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Let's just face it, once you see the word symmetry it just falls completely out of your league, even when trying to apply it to a topic as basic as this one. You could replace the words symmetry with utter nonsense and it will still register in your brain the same way. You truely have no idea do you? You have no basic concept of symmetry or what it even means.

 

Hey, be nice! Not everyone has your ability to learn at such amazingly accelerated rates. When this thread started you apparently weren't even very familiar with concept simultaniety and in the course of a few pages you realized that Einstein and the entire physics community in general didn't understand it, but you figured it all out! How can we mere mortals compete with that? It took you like 1 day to completely understand gauge symmetry AND super symmetry. You are clearly just and awesome guy.

I think you're a hoot, it just took me a while to appreciate you. Bringing laughter into the world is an improtant thing, so if you can troll to annoy people and be so over the top that it becomes humorous, well thats not too bad, not too bad at all.

 

Do you agree that relativity predicts the MME will be length contracted according to an outside frame in relative motion?

 

lol.

 

Yes and that it will be only in the direction of motion. If you consider c t' in the direction perpendicular to the direction of motion d' = d, but this is just for that line. This is because a light beam will follow with the object in its direction of motion. It will have a vector added to it, so that if the object in motion assumed that it was at rest, then he would only see light traveling out in a straight line. While an observer at rest will see the beam travel at an angle. So then any observer in constant motion will always notice that a beam of light will travel out in straight lines, and not at an angle because of its degree of motion.

All frames are created equal, there is no absolute frame where a degree of motion will be different than another. So then no matter how fast your going, you will notice that beams of light travel out in straight lines. But, while another observer in relative motion would notice it to travel at an angle that is a longer distance, its time is not dilated for just a straight line perpendicular to the direction of motion. Then the length contraction only makes up for the light having to travel this longer distance at an angle, if there was no angle then it wouldn't need to be contracted. It would be seen to travel the same distance in both frames, but I don't think a photon could actually follow that line in both frames.

In other words, the equation only considers one beam of light measurement. And the distance they would use in order to accurately take that measurement of the beam of light. I don't think the distance of the beam of light measurment can really be used for something other than just measuring that beam of light. The contracted distance comes from d = c t. So then if you tried to find a velocity with that distance you would be really saying that is a new velocity of the photon. I think I may have been in error before when trying to find velocity from the equation, since that distance would only apply really to a speed of light measurement. For instance, you wouldn't use the distance traveled by your car in order to figure out how fast your bike is going. But, I still think that an objects velocity would be independent of outside of observers (they wouldn't have an effect on it).

 

You can't and all you can do is try. Maybe I could give you an "A" for effort. You shouldn't let it discourage you and you should keep trying harder. There are some people with natural talent while others have to work hard at it, unfortunately.

Actually you have it all wrong, I first realized that the TE didn't exactly match up with the interpretations of the MME the first time I looked at it. That was almost 15 years ago. I read about MME about 15 times from different books, they all said the beams arrive at the same time and even arrive the same time when in another frame of reference. Then the TE flat out says that they don't arrive at the same time. I was then able to put two and two together and realize that one says they always arrive at the same time and in the other case it says they don't. You may have a lot of work ahead of you until you get to this point.

It is more like they arrive at the same time, but come from different time frames. If every clock is different and ticks at a different rate, then something would still have to be going on at the same time even though their clocks never say the same thing. The events would have to be linked somehow to reality. Things are going to happen at the same time even though there clocks read differently.

 

Fifteen years and you still don't get it. This is a record.

False. Lorentz transform for time says:

\(t'=\gamma(t-\frac{vx}{c^2})\)

In terms of time intervals, the above results into:

\(dt'=\gamma(dt-\frac{vdx}{c^2})\)

In frame F, the events being simultanneous, \(dt=0\)

In the case of:

TE) \(dx \ne 0\) so:

\(dt'=\gamma(0-\frac{vdx}{c^2})\)

MME) \(dx=0\) so:

\(dt'=0\)

Another 15 years and you might, just might, get it.

 

MME can be explained with SR. Please look at the message that i sent. It will be over by then. Very good explanation that presentation has..

 

In that case, I assume you would agree that the train is length contracted according to the platform? And that the platform is length contracted according to the train?

 

But does that length contraction have any effect on the results of simultaneity? I thought about it once before and thought the effects are cancelled out so length contraction is not important to the outcome.

 

I wasn't the one that said that they don't arrive at the same time in the TE, Einstein did genius.

 

Sure, why not?

 

Does it hurt?