That's a good thing (if so). Those polygraph tests are retarded and proven to be ineffective. Can't believe any stupid intelligence agency would ever use them
If a ship passes me at .995c, and, at the instant of passing, has an event that's 1 light second in its reference frame, then, in my reference frame, the time and distance of that event is 9.96 seconds after time of passing and 10.01 light seconds from my frame's origin.
@NeuroFuzzy I have a feeling my reading course this summer with my advisor will be Guillemin & Pollack, Milnor and then Hawking-Ellis to cover some of the more topological aspects of GR :D
And I know Beem et al. has Morse theoretic aspects of GR.
And maybe I'll finally understand the topological conditions required for a Lorentz metric to exist...(I know what the conditions are, but I can't prove it)
@ACuriousMind Surprisingly not. But I now know by heart that there are 15 tensors of the form $g_{ab,c}g_{de,f}$ with exactly two contractions. And lots of other algebraic things :(
@barrycarter Because "when" is subjective. That's kinda the entire point of relativity. You can't say the event occurs "when" the ship is passing you, that's only true from the ship's point if view
I don't get how this is so confusing. I'm saying that by choosing two pairs of indices to contract there are fifteen distinct tensors you can generate.
@ACuriousMind My expectation would be: the event occurs when the other ship passes me. It's 1 light second away from him, so, my Lorentz contraction, it should appear 1/10 light second away from the ship for me. Thus, I expect to see it at time 0, at distance 1/10 light second. But that's NOT what happens?
@0celo7 well $A$ is symmetric in the first two indices. if the two distinct indices are $a$ and $b$ you'd have $A_{acb}A_{def} g^{cd}g^{ef}$. Distinct from $A_{bca}A_{def} g^{cd}g^{ef}$. Distinct from $A_{acd}A_{bef} g^{cd}g^{ef}$. etc.
@0celo7 Lol alright. Actually Penrose notation or something similar is a really nice way to solve the problem. You'd hate the thing trying to be proved because it's an abuse of notation
@barrycarter Nope, SR is indeed a bit more than just time dilation and length contraction. Inertial frames are related by Lorentz transformations, which can turn a pure temporal or pure spatial distance into a mixture of both
@barrycarter But this isn't really different from the classical Galilean world: If you move at a constant velocity w.r.t. an observer, all things happening along your trajectory have no spatial separation for you, but they surely have for the observer
@ACuriousMind So, if I know the time and distance between two events in one frame, the Lorentz contraction and time dilation taken individually can tell me the time and distance between those two events in another frame... BUT... can't tell me when and where the events will occur in the other frame?
@barrycarter Not really - time dilation tells you what the temporal component for a pure temporal distance in one frame will be in another, and length contraction tells you what the spatial component of a pure spatial distance in one frame will be in another.
It's really easier to just apply the correct Lorentz transformation for general cases instead of trying to do something with time dilation and length contraction
@ACuriousMind Yes, using the matrix seems easier, but I'm trying to understand how it ties in to the Lorentz contractoin and time dilation. So the main point here is: just because two things occur at the same time in one frame, they don't occur at the same time in another frame?
@ACuriousMind By pure temporal distance, do you mean two events occurring at the same x coordinate but at different times?
@ACuriousMind Oh! So it's not a per-coordinate transformation. A time distance of 0 doesn't get multiplied by time dilation. A time distance of 0 just means the distance factor gets multiplied.
@ACuriousMind Bloody hell, that's ugly.
@ACuriousMind Oh, and all of this has nothing to do with light travel time, correct?
@ACuriousMind If I say "frame x's origin is 7 light seconds away at time t=5", I actually don't know that until t=12, right? In other words I'm not saying "I see x's origin 7 light seconds away".
@barrycarter Indeed, all those statements are purely about coordinates in your frame, not about perception. For instance, you don't actually see a contracted, but a rotated object.
@ACuriousMind OK, so if two events occur at the same time for me but at different places, I can tell you the distance at which you'll see the two events, but you won't see them at the same time.
@ACuriousMind And if I see two events occur at the same place in my frame but at different times, I can tell you how long you'll have to wait between the events, but they won't occur at the same place for you?
@ACuriousMind BUT... this is the Gallilean/Newtonian part: the distance apart they occur from each other won't simply be because I'm traveling at a given velocity?
@ACuriousMind Example for low speeds: I'm traveling at 20mph compared to Bob. At noon, an event occurs at my origin. At 1pm, a different event occurs at my origin. I compute the distance as 0, and the time as 1 hour. Now Bob will obviously see those two events 20 miles apart, but, if these speeds were relativistic, that wouldn't be the case?
@ACuriousMind Oh, so if I look at Bob's clock after an hour, I might say "only 59 minutes have ticked off on his clock (time dilation)", so he's still seeing me at 59/60 of 20 miles, not the whole 20 miles. Is that what you meant?
@barrycarter von Neumann once said "You don't understand things in mathematics, you just get used to them" While meant to be tongue-in-cheek, some parts of physics are the same ;)
@ACuriousMind Something I've never understood: If $L$ is the length functional of a curve and $E$ is its energy functional, then $L^2\le(b-a)E$ (I understand this part)
but why does this imply that minima of $L$ are also minima of $E$
I am a novice at this, and I believe I have only asked a handful of questions in total. The first were all rapidly tagged as "opinion-based", and I can see why. I don't see how to ask for suggestions in any context if that sort of thing is declared to be off limits.
Today for the first time in...
First, let me state the form of Lagrangian for YM and GR
\begin{align}
L_{YM} = \alpha \textrm{tr}(F^2), \qquad L_{GR} = \beta R
\end{align}
I heard, YM is a gauge theory but GR isn't a really gauge theory. Due to the fact of its Cartan-Killing form of YM is positive, but GR is negative.
Is i...
@0celo7 am just 13 so i don't know any complex physics then how can i say something useful or constructive
And i have apologized for my non sense questions many times.
we are saying particles are created by field disturbance. on the other, we say fields are excited by the presence particle. so it’s a bit chicken and egg. which comes to being first then?
I am a novice at this, and I believe I have only asked a handful of questions in total. The first were all rapidly tagged as "opinion-based", and I can see why. I don't see how to ask for suggestions in any context if that sort of thing is declared to be off limits.
Today for the first time in...
First, let me state the form of Lagrangian for YM and GR
\begin{align}
L_{YM} = \alpha \textrm{tr}(F^2), \qquad L_{GR} = \beta R
\end{align}
I heard, YM is a gauge theory but GR isn't a really gauge theory. Due to the fact of its Cartan-Killing form of YM is positive, but GR is negative.
Is i...
