Space-time curvature is incorrect

Discussion in 'Physics & Math' started by Frencheneesz, Aug 26, 2002.

1. (Q)Encephaloid MartiniValued Senior Member

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Rav

As I mentioned in a previous post, I am no mathematician.

Sorry, I must have missed that.

If you feel inclined, I'd like it if you could try to communicate your ideas in view of that.

Hmmm... I'm don't think I could do that without regurgitating Greene. I'd have to give it more thought.

3. ProsoothusRegistered Senior Member

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Q,

Maybe you believe Crisp's and James' conclusion that there can be multiple time dilations in a single frame of reference, but I don't. Especially since atomic clocks show that there is only ONE unique time in a single frame of reference. To tell you the truth, I'm surprised that Crisp and James even suggested it (I guess since they found it in some physics books, it must be true

)

It seems that some people will do anything to preserve special relativity, even if it defies logic. Very unscientific, isn't it???

By the way, give me multiple time dilations and I'll prove that I can piss at the same speed in any frame of reference.

Tom

5. FrencheneeszAmazing MemberRegistered Senior Member

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I for one see no reason why rav's conclusion is so heatedly debunked. It makes perfect sence, conceptually. To prove it wrong would involve math, no doubt. But it would explain time dialations very well if it were the case.

I do have a question for rav though. If you have a triple axis grid and you go the speed of light diagonal to all the axes, you would have to calculate that the coordinance (x,y,z) added up (z+x+y) should equal the distance of the line. This is clearly not the case mathmatically. Another example would be pythagoreans theorem in a two dimentional grid. How do you explain this?

7. FrencheneeszAmazing MemberRegistered Senior Member

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James R:
you said: "Instead, you're plugging in numbers which represent the distance between each observer and the light pulse at some arbitrary time. "

I fully realize what i am doing mathmatically. This is because to calcualte it any other way would be not what objects I want to calculate with.

You want me to calculate the speed that the distance between the light and the source, is increasing. I want to calculate relative speed of the light and the observers. I can only calculate speed taking into account ONLY 2 observers. By calculating it using the source I would have to remove the observers.

The entire idea of relativity is that different objects are different distances from eachother. Thus distance is relative. If distance is relative then speed must also be relative because V=D/T.

Do you see that the speed of light compared to an observer takes in to account no other information than the distance between the observer and the light, and time?

8. James RJust this guy, you know?Staff Member

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<b>Rav</b>

The 4-position vector of an object is (ct,x,y,z).
The 4-velocity is the derivative with respect to the proper time:

v = d/d(tau) (ct,x,y,z)

Now, my problem is what Greene means when he talks about velocity through spacetime. I think he means that the 4-scalar made from the 4-velocity is constant. We get something like:

(dt/d(tau))<sup>2</sup> - (dx/d(tau))<sup>2</sup> - (dy/d(tau))<sup>2</sup> - (dz/d(tau))<sup>2</sup> = K

But this thing K is the square of the derivative of the spacetime interval with respect to the proper time. I'm not sure if that is a constant or not. That's the problem I have with Greene. He does not specify clearly what he is talking about. It is certainly not true to say that the 4-velocity vector is always equal to the speed of light, because it is a 4 component object whose components can vary. It <b>might</b> be true to say that some 4-scalar formed from the 4-velocity is a constant related to the speed of light, but I need more information to confirm that.

Greene is doing some serious hand-waving here.

<b>Tom</b>:

I'm starting to get sick of your misrepresentations.

Neither Crisp nor I have ever said there can be two time dilations in a single reference frame. That is entirely your fantasy. In fact, we have taken pains to point out to you that that is not the case. The fact that you continue to ignore explanations which do not suit and twist the truth when it does suit you puts you in the crackpot basket.

<b>Frencheneesz</b>:

<i>I fully realize what i am doing mathmatically. This is because to calcualte it any other way would be not what objects I want to calculate with.</i>

I don't think you know what you're doing mathematically. If you think you can explain yourself clearly, please give it another try.

<i>If distance is relative then speed must also be relative because V=D/T.</i>

Time is relative too, remember. If both D and T change, V can remain constant. One example of that type of thing was shown when we solved Tom's problem in the "Does light have a mass?" thread.

<i>Do you see that the speed of light compared to an observer takes in to account no other information than the distance between the observer and the light, and time?</i>

Yes, that's fine. It still works out that all observers measure the same speed for light in a vacuum.

9. FrencheneeszAmazing MemberRegistered Senior Member

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James R:

I think you are a bit full of yourself in this discussion. I am trying to make a point and all you can say to me is that I am stupid and don't understand what i am talking about, or you say that I can't accept what you have said before. It is a bit tiring to hear you just call me an idiot the whole time. I even used YOUR math when you gave it to me and you just ignored my conclusion.

I think it would be wise to keep an open mind, even if you do think you know everything.

"Time is relative too, remember. If both D and T change, V can remain constant."

Ok, I see this. But after considering that in the example I find that both of our math calculations were wrong because they didn't take this into account.
SO your saying that the moving observer's clock is going slower than the stationary observer? Thus the equations must be altered to find light-speed constant?

Ok, that works but we must then talk about something else: the direction that the objects are traveling. According to relativity, the moving observer would also see the stationary observer's time run slower.

How can the moving observer see the stationary observer's clock run slower and still be able to say his time was going slower than that of the stationary observer's clock?

10. (Q)Encephaloid MartiniValued Senior Member

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French

How can the moving observer see the stationary observer's clock run slower and still be able to say his time was going slower than that of the stationary observer's clock?

That's not quite right. Consider that you and I are the moving observer and the stationary observer. You will view your own clock running normally and I will also view my own clock running normally. We will view each others clock running slower.

11. FrencheneeszAmazing MemberRegistered Senior Member

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Q:
"You will view your own clock running normally and I will also view my own clock running normally."

Right, I probably just didn't make that clear. Let me rephrase the question.

If the moving observer sees the stationary observer moving slower, wouldn't he conclude that that observer would see the light moving faster?

12. James RJust this guy, you know?Staff Member

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Frencheneesz:

<i>I think you are a bit full of yourself in this discussion.</i>

You keep demonstrating that you don't understand relativity. I keep trying to explain it to you. You keep trying to avoid facing the implications of the explanations you are given. Who, then, is full of himself here? I'm not sure. BTW, I've never said you're stupid; I've said your ideas are wrong. There is a difference.

<i>I think it would be wise to keep an open mind, even if you do think you know everything.</i>

I don't think I know everything, but I'm fairly confident I know more about relativity than you, and that's what's relevant here. If that sounds arrogant, I apologise.

<i>Ok, I see this. But after considering that in the example I find that both of our math calculations were wrong because they didn't take this into account.</i>

Possibly. I will need to think about that.

<i>How can the moving observer see the stationary observer's clock run slower and still be able to say his time was going slower than that of the stationary observer's clock?</i>

It's not a problem. Each can predict what the other will see. There's no conflict whilst the two observers are in different frames of reference. If they ever come back into the same reference frame, the acceleration involved in doing so will sort out discrepancies so that they end up agreeing with each other on how much time has passed for each.

13. FrencheneeszAmazing MemberRegistered Senior Member

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"If that sounds arrogant, I apologise."

Quite frankly, it does. You do seem to take critisism very rationaly, which is more than I can say for myself.

"You keep trying to avoid facing the implications of the explanations you are given. "

I find your explanations hard to understand sometimes, what can I say. I'm just trying to make sence of some of these relativity ideas that i have never heard of before, like the idea that acceleration effects time dialation.

"There's no conflict whilst the two observers are in different frames of reference. "

It seems that the only way to make the equations show light to go the same speed is to slow down the moving observer's time. If you slow down the moving observer's time, obviously his time would be going slower than the stationary observer, am i correct?

So if the moving observer's time IS actually slower, then why do both observers see eachother's time to be going slower?

"If they ever come back into the same reference frame, the acceleration involved in doing so will sort out discrepancies"

As food for thought: If the moving observer accelerates back to the stationary observers speed, isn't the stationary observer also accelerating to the moving observer's speed?

14. c'est moiall is energy and entropyRegistered Senior Member

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could somebody here explain to me how this would be done PRACTICALLY

how on earth, could I "view" a clock running slower when I pass it with a speed of 0.9/ v of light?

same for the other one in the other FOR

further, this has never been done, so what makes you all thick to be talking about this in a way that it is somehow a fact (and I recall, the experiment with the planes was one-way as all the reports and articles about it have revealed to me -unless you can show me that this isn't so)

15. ProsoothusRegistered Senior Member

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James,

Let me remind you in case you forgot: You and Crisp claimed that two events that take the same time to occure in one frame of reference can take two DIFFERENT times to occure in another frame of reference. Let me repeat that I used the term "occure" and not "observed". The ONLY way to obtain the two unique times is to use two unique TIME DILATIONS in the same frame of reference. There is no other way. If you take a look at the Lorentz' boost transformations and think about it a little, you will see what I mean.

Tom

16. (Q)Encephaloid MartiniValued Senior Member

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cest moi

how on earth, could I "view" a clock running slower when I pass it with a speed of 0.9/ v of light?

same for the other one in the other FOR

Neither the moving observer nor the stationary observer know absolutely which of them is in motion relative to the other. In other words, you may use either frame of reference as the stationary frame or the moving frame. In both cases, the stationary frame will view the moving frame's clock as running slower.

17. ProsoothusRegistered Senior Member

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c'est moi,

You can't. It's illogical. But I guess since it's relativity it doesn't have to follow logic, right James and Crisp??

The fact is that according to the "twin paradox", one clock runs slower than the other. That means that the other clock runs faster. They both can't run slower than each other because it's impossible.

This is just another one of those illogical ideas physicists "hold on" to in order to preserve SR. It appears that relativity, to some people, is a religion, and not science.

Tom

18. c'est moiall is energy and entropyRegistered Senior Member

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common man, you're not even trying to answer my question

and this of course, "In both cases, the stationary frame will view the moving frame's clock as running slower", does not need comfirmation by experiment

19. (Q)Encephaloid MartiniValued Senior Member

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cest moi

common man, you're not even trying to answer my question

Or more precisely, it is not what you wanted to hear.

20. FrencheneeszAmazing MemberRegistered Senior Member

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"how on earth, could I "view" a clock running slower when I pass it with a speed of 0.9/ v of light? "

Lets say you had a guy in a glass bubble in space, he waved his hands at a uniform rate. Then you have a spaceship that has a window in the back; it speeds away from the bubble (lets say the bubble is not destroyed). When the guy in the spaceship looks back with his telescope he should see the guy behind him waving his hands slower than he was before.

This is how, if you actually look at them they will be moving slower.

"this has never been done"

As far as I know, thats true. Also as far as I know we have never built or seen anything that we could use for measurement, that has been traveling at fast enough speeds to test this. I am still trying to figure out certain paradoxes of this, I can't just say it is incorrect until I know what they are thinking.

Could someone try to answer my questions now?....

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22. c'est moiall is energy and entropyRegistered Senior Member

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Did anyone travel with a meson to confirm that the same thing happens for him in our FOR?

23. James RJust this guy, you know?Staff Member

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c'est moi:

<i>Did anyone travel with a meson to confirm that the same thing happens for him in our FOR?</i>

Did anybody see any living dinosaurs? No? Then how can we know anything about them?

Did anybody witness continental drift directly? No? Then how do we know it happens?

Has anybody ever seen a lithium atom? No? Then how do we know lithium exists?

Has anybody ever been inside the sun? No? Then how can we say what processes happen there?