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Author  Topic: Geometry of Spacetime  
nwr Member Posts: 5631 From: Geneva, Illinois Joined: Member Rating: 5.0 
And the way I see it, is that in spacetime there is no such thing as "true 90 degrees" between a timelike direction and a spacelike direction. It is all relative to the observers frame. Fundamentalism  the antiAmerican, antiChristian branch of American Christianity


New Cat's Eye Inactive Member 
Sure. But if you have a particle that is at rest (in whatever reference frame), then aren't the timelike and spacelike directions at 90 degrees?


nwr Member Posts: 5631 From: Geneva, Illinois Joined: Member Rating: 5.0 
I'm not sure if that even makes sense. Take ordinary 2dimensional space. If we stretch out the xaxis, say rescale it so that what was one unit of length becomes 2 units, then angles change  assuming that we don't also stretch out the yaxis. So, in some sense, the magnitude of angles is an artifact of how we measure them. In the case of spatial directions, we normally require rotational symmetry. And if we require rotational symmetry, we cannot stretch out the xaxis without also stretching out the yaxis. As far as I know, we cannot rotate things from a spacelike direction to a timelike direction. So we don't have something like rotational symmetry to normalize our way of measuring. So I think that unavoidably leaves measurements of angles between spacelike and timelike directions to be dependent on our arbitrarily chosen standards. Fundamentalism  the antiAmerican, antiChristian branch of American Christianity


NoNukes Inactive Member 
This is true. However, there is a natural set of units that eliminates this problem. If the time and distance units are chosen so that the speed of light equals one, then we have eliminated the issue you describe above. Under a government which imprisons any unjustly, the true place for a just man is also in prison. Thoreau: Civil Disobedience (1846) I would say here something that was heard from an ecclesiastic of the most eminent degree; ‘That the intention of the Holy Ghost is to teach us how one goes to heaven, not how the heaven goes.’ Galileo Galilei 1615. If there is no struggle, there is no progress. Those who profess to favor freedom, and deprecate agitation, are men who want crops without plowing up the ground, they want rain without thunder and lightning. Frederick Douglass


Iblis Member (Idle past 2586 days) Posts: 663 Joined: 
Ok, I'm still missing something, probably many things.
I'm taking this to mean that I can keep working in terms of just two components, distance and time. They are sufficient, in the same way that the distance on the ground from me to the bottom of a flagpole, plus its height from the ground, would be sufficient to calculate the true distance from my feet to the top of the pole.
This, I'm taking to mean, I need to subtract the time instead of adding it. Here let me try it d²  t² = x² This seems to be a step in, call it the right dimension, because
but in the whole wrong direction altogether!
Look, in my original dealie, distance is 3, the square is 9; time is 4, the square is 16; 916=7, the root of minus 7 is a broken calculator. This is telling me that it is slower than light travel that is impossible / absurd !!!! my ftl version comes out a perfectly tolerable 2.64etc. Please tell I just got the terms backward or something. Please?
That first ² doesn't really belong there does it?
You sound like you might make sense. Could you give some examples or something, I want to flash on this.
I agreed with this completely when I originally posted. The legs of my triangle were my attempt to approximate the view of the "at rest" observer, who sees them at right angles to one another. The hypotenuse and its calculable components was supposed to give the pov of the traveler, who was bending them together by moving. Now I've got all this "minus" shit and I'm not sure what to do. I'm hoping I can declare something "zero" and just turn my triangle like, upsidedown or something. But I'm not holding my breath ...
Is that De Broglie? Can you explain him? Perhaps using pictures, or a word problem? If my audience understood Greek, they wouldn't be New Testament believers, would they.


NoNukes Inactive Member 
Assuming units where c=1, you can express the proper time as follows: Where distances are changing at a constant rate with respect to time, then you can use this expression: This is the form that I have seen used to show how the twin paradox works. By the way, just because an expression generates an error on your calculator does not mean that the expression is improper. Edited by NoNukes, : No reason given. Under a government which imprisons any unjustly, the true place for a just man is also in prison. Thoreau: Civil Disobedience (1846) I would say here something that was heard from an ecclesiastic of the most eminent degree; ‘That the intention of the Holy Ghost is to teach us how one goes to heaven, not how the heaven goes.’ Galileo Galilei 1615. If there is no struggle, there is no progress. Those who profess to favor freedom, and deprecate agitation, are men who want crops without plowing up the ground, they want rain without thunder and lightning. Frederick Douglass


Iblis Member (Idle past 2586 days) Posts: 663 Joined: 
Yeah yeah, this is just my expressive way of saying / showing the idea that the objection to ftl travel is supposed to be, the "imaginary" numbers involved. Ok, this looks like, a step in the actual right direction; in the sense that we now appear to be subtracting the space from the time. This will give us i's in the right place, I suspect. Why is this different from what we were seeing before, as from Son Goku for example? Also, what do those d's signify? Is there something I'm supposed to be multiplying everything by? Or do they just mean distance, and if so, why is there one on the t? And even moreso, what do those deltas signify?


NoNukes Inactive Member 
You can develop the equations with either negative time contributions and positive space contributions or vice versa. The "d"s in the first equation indicate that dt, dx, dy, dz are differential values. The integral is a line integral over the trajectory of a particle. In the equation with the deltas, deltas mean 'change'. The second equation can be be used where the change in time, and x, y, z coordinates in a case where those quantities each vary linearly with time from a starting point to the ending point. Your questions suggest that you've got quite a bit of studying including learning a tiny bit of math before you are going to understand special relativity. Try reading the wikipedia article I linked as a starting point. Edited by NoNukes, : No reason given. Under a government which imprisons any unjustly, the true place for a just man is also in prison. Thoreau: Civil Disobedience (1846) I would say here something that was heard from an ecclesiastic of the most eminent degree; ‘That the intention of the Holy Ghost is to teach us how one goes to heaven, not how the heaven goes.’ Galileo Galilei 1615. If there is no struggle, there is no progress. Those who profess to favor freedom, and deprecate agitation, are men who want crops without plowing up the ground, they want rain without thunder and lightning. Frederick Douglass


Iblis Member (Idle past 2586 days) Posts: 663 Joined: 
WTMFF?!? . . . Ok, is there in fact anybody anybody on this site with the insight and patience to help me understand how this minus stuff for duration works and why we are acting like this
and this are in some sense interchangeable, or descriptive of the same situation, or, whatever it is they are that allows them both to be here. . . . To reiterate, I expect imaginary numbers when my d is larger than my t, and ordinary decimals when my t is larger than my d. Yeah?


nwr Member Posts: 5631 From: Geneva, Illinois Joined: Member Rating: 5.0 
Some of this is basic calculus, and some of it comes from the mathematical model that Einstein used in special relativity.
Einstein's intuition told him that the velocity of light should be the same for all observers, and the MichelsonMorley experiment seemed to confirm that. But that was incompatible with the traditional Newtonian/Euclidean view of space, which saw time and distance as independent. So the problem was one of finding a new metric which connected time and space in such a way that the velocity of light could be the same for all observers. It worked very well. It accurately predicted motion in particle accelerators. One could deduce which seemed to accurately account for the energy seen in radioactive materials, and which was confirmed by nuclear physics. And yes, many people at that time thought it counterintuitive. But science is a pragmatic enterprise, and it is hard to beat "it works very well." Fundamentalism  the antiAmerican, antiChristian branch of American Christianity


New Cat's Eye Inactive Member 
Have you studied calculus?


NoNukes Inactive Member 
I have some patience, but to date you don't seem willing to make much of an effort on your own. Let's discuss what the equations are mean. The equation for proper time gives the time that an observer traveling a trajectory would measure. Observers not following that same trajectory could measure different times. This is explained in a fairly detailed manner with a couple of example calculations at the Wikipedia article on proper time located here. We might also discuss the "proper length" between events rather than proper time. That formulation leads to those equations with the negative signs. http://en.wikipedia.org/wiki/Proper_length [quote]In special relativity, the proper length between two spacelikeseparated events is the distance between the two events, as measured in an inertial frame of reference in which the events are simultaneous. So if the two events occur at opposite ends of an object, the proper length of the object is the length of the object as measured by an observer which is at rest relative to the object.[\quote] Under a government which imprisons any unjustly, the true place for a just man is also in prison. Thoreau: Civil Disobedience (1846) I would say here something that was heard from an ecclesiastic of the most eminent degree; ‘That the intention of the Holy Ghost is to teach us how one goes to heaven, not how the heaven goes.’ Galileo Galilei 1615. If there is no struggle, there is no progress. Those who profess to favor freedom, and deprecate agitation, are men who want crops without plowing up the ground, they want rain without thunder and lightning. Frederick Douglass


Iblis Member (Idle past 2586 days) Posts: 663 Joined: 
Sorry, had to table this question for a bit until I could get a lot more generic Excedrin supercheap.
I'm starting to grok this. In my example, 3 light years in 4 years = 75% of c, the time experienced by my astronaut is only 2.64ish years. And he can't travel faster than light because he gets the imaginary numbers only if d is bigger than t. This is good, even if it's disappointing because I don't yet have a good selfevident easytoexplain reason why I'm subtracting.
So again, I really need more of that kind of stuff.
I'm really not getting this. By the definition there, wouldn't the "proper length" in my example just be the 3? The equation is making it look like it would be 2.64etc * I, what am I missing? And how do I figure out the "improper" length, ie the distance experienced by my astronaut?
If you only knew, brother. My big triangle was a lot easier to explain. The relative distance wouldn't by any chance be 2.35ish would it? That would be Swell. Edited by Iblis, : TALKING triangles Edited by Iblis, : wtfever 

Iblis Member (Idle past 2586 days) Posts: 663 Joined: 
If the events were simultaneous, what would the t represent?


Iblis Member (Idle past 2586 days) Posts: 663 Joined: 
These guys are telling me that your equation is for calculating "proper length", which from the description I got seems like it should just be the 3 light years in my example. I'm beginning to understand how to calculate the relative time experienced by my astronaut, which I'm understanding to be the reasonable 2.64 and change I get from the square root of (4 squared minus 3 squared). But I could probably flash on this if I could calculate the relative distance.



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