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Author  Topic: Geometry of Spacetime  
nwr Member Posts: 6421 From: Geneva, Illinois Joined: Member Rating: 4.1 
The way I understand it, the only way for the time component of your path to be at a true 90 degrees would be if you were not moving.
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 
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. 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: 6421 From: Geneva, Illinois Joined: Member Rating: 4.1 
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? 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 
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. 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 3981 days) Posts: 663 Joined: 
Ok, I'm still missing something, probably many things.
Son Goku writes: Minkowski space, as a plane, just like the normal plane of Euclidean geometry. 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.
nwr writes: This, I'm taking to mean, I need to subtract the time instead of adding it. Here let me try it
d  t = x This seems to be a step in, call it the right dimension, because
this retard writes: I expect this problem somewhere to make me subtract but in the whole wrong direction altogether!
subtract the square of distance directly from the square of time. 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?
Son Goku writes: That first doesn't really belong there does it?
Catholic Scientist writes: It "takes" time to gain distance. You sound like you might make sense. Could you give some examples or something, I want to flash on this.
Catholic Scientist writes: The time coordinate is at a right angle, but the time direction of your path is only at a right angle to your distance when you're at rest. As you increase your velocity, the angle of the time direction becomes more acute. When you start approaching the speed of light, the time direction approaches being parallel to your space direction, and that's how you get length contraction. 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 ...
nwr writes: 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 
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. 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 3981 days) Posts: 663 Joined: 
NoNukes writes: just because an expression generates an error on your calculator 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 
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? 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 3981 days) Posts: 663 Joined: 
You can develop the equations with either negative time contributions and positive space contributions or vice versa. 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
nwr writes: 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: 6421 From: Geneva, Illinois Joined: Member Rating: 4.1 
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 
why we are acting like 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. 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.
Proper length  Wikipedia
[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 3981 days) Posts: 663 Joined: 
Sorry, had to table this question for a bit until I could get a lot more generic Excedrin supercheap.
NN writes: The equation for proper time gives the time that an observer traveling a trajectory would measure. 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.
CS writes: It "takes" time to gain distance. So again, I really need more of that kind of stuff.
NN writes: "proper length" 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?
NN writes: you don't seem willing 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 3981 days) Posts: 663 Joined: 
in which the events are simultaneous. If the events were simultaneous, what would the t represent?


Iblis Member (Idle past 3981 days) Posts: 663 Joined: 
Those paths can be drawn on both Euclidean space and Minkowski space. 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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