Mary23 Posted October 29, 2015 Share Posted October 29, 2015 Hi! I have to make a project called Minkowski space and i don't know exactly what should include in. It's very complicated what i've founded about Minkowski space on google and it's beyond my means...could you suggest me what should i refer to?? some ideas?? and i want to know if there are practical applications of Minkowski space and also why is it important.. Link to comment Share on other sites More sharing options...
studiot Posted October 29, 2015 Share Posted October 29, 2015 You don't say what sort of project. So I suggest thinking about a 'light cone' and basing your project on this. https://en.wikipedia.org/wiki/Light_cone The picture at the beginning of Wiki lends itself to A sculputure An SF short story A factual essay Drawings, sketches, discussions. Also look up Tachyons. Link to comment Share on other sites More sharing options...
ydoaPs Posted October 30, 2015 Share Posted October 30, 2015 Hi! I have to make a project called Minkowski space and i don't know exactly what should include in. It's very complicated what i've founded about Minkowski space on google and it's beyond my means...could you suggest me what should i refer to?? some ideas?? and i want to know if there are practical applications of Minkowski space and also why is it important..[/size] You can start out with its defining feature: its metric. The diagonal of the metric is 1, -1, -1, -1. This means that the pythagorian theorem doesn't hold. Instead, you get s2=(cdt)2-dx2-dy2-dz2. If you factor out the negative sign, and apply the pythagorean theorem to the spatial components, you end up with (spacetime distance)2=(temporal distance)2-(spatial distance)2. Spacetime distance is what is called "frame invariant". That means it's the same in any (inertial) frame. So, what we see is that for the same spacetime distance, the more one moves through space, the less one moves through time. Link to comment Share on other sites More sharing options...
studiot Posted October 30, 2015 Share Posted October 30, 2015 ydoaPs Spatial distance is what is called "frame invariant". That means it's the same in any (inertial) frame. So, what we see is that for the same spacetime distance, the more one moves through space, the less one moves through time. Hi ydoaPs, did you mean this? 1 Link to comment Share on other sites More sharing options...
ajb Posted October 30, 2015 Share Posted October 30, 2015 and i want to know if there are practical applications of Minkowski space and also why is it important..[/size] It is worth knowing that the notion of Minkowski space-time is vital in our understanding of high energy physics. Without any details, special relativity is needed in our theories of particle physics, and because of this it is needed in understanding the results of colliders at CERN (for example). Link to comment Share on other sites More sharing options...
ydoaPs Posted October 30, 2015 Share Posted October 30, 2015 Hi ydoaPs, did you mean this? Good catch. I meant "spacetime". Link to comment Share on other sites More sharing options...
Mary23 Posted October 31, 2015 Author Share Posted October 31, 2015 Which is the difference between Euclidian space and minkowski space?? and also, which is the connection between special relativity and minkowsky space..?? Link to comment Share on other sites More sharing options...
ajb Posted October 31, 2015 Share Posted October 31, 2015 Which is the difference between Euclidian space and minkowski space?? and also, which is the connection between special relativity and minkowsky space..?? Don't you have a thread on this already? Anyway, the difference stems from a single minus sign difference in the metric. Special relativity states that space-time (as long as we ignore gravity) is Minkowski space-time. All the causal structure and so on comes from the minus sign. I a sure Wikipedia will help you here. Link to comment Share on other sites More sharing options...
ydoaPs Posted October 31, 2015 Share Posted October 31, 2015 Don't you have a thread on this already? ! Moderator Note Yes, she does. The threads are now merged. Link to comment Share on other sites More sharing options...
ydoaPs Posted November 10, 2015 Share Posted November 10, 2015 Which is the difference between Euclidian space and minkowski space?? and also, which is the connection between special relativity and minkowsky space..??Do you have any specific questions? Special Relativity is just classical physics that uses the Minkowski space. Like I said previously, it generalizes from a space with a time parameter to a spacetime, so time is part of the 'distance'. s2=(time distance)2-(space distance)2. This s2 (the distance through spacetime) is the same between reference frames. So, if I have my own frame, and you have yours, we may disagree on how far you (or I) travel in space over a given amount of time, but we'll agree on how far you (or I) go through spacetime. Using the spacetime distance equation above, we can change co-ordinates from yours to mine and vice versa. In my frame of reference, I'm standing still and you're moving. In yours, you're standing still and I'm moving. So, let's get how much time you experienced as told by my frame of reference. (your time distance)2-(your space distance)2=(my time distance)2-(my space distance)2 Velocity (speed with the direction that you're going) is often written as ẋ. I'll use that here. The speed of light in a vacuum is written as c. I'll use that as well. When using two different co-ordinate frames, we usually give one an apostraphe (a prime) just so we can tell them apart. You'll be the primed frame. The speed of light in a vacuum is the same in all frames, so it doesn't get a prime. To get a reasonable answer, our distance through time and our distance through space have to have the same units. So, we use the speed of light as a conversion factor. The distance through time is given by c times time. So, my distance through time is ct and yours is ct'. Distance through space is given by velocity through space times time. So, mine is ẋt and yours is ẋ't'. (ct')2-(ẋ't')2=(ct)2-(ẋt)2 That's just restating what I said up there with the variables I explained below it. This is just that we both have the same spacetime distance. So, we're going to see how our times compare when you're going a certain speed relative to me. On your side of the equation, we can factor out a t'2, and on mine, we can factor out a t2. t'2(c2-ẋ'2)=t2(c2-ẋ2) To get your time by itself, we'll divide both sides by the term your time is multiplied by. t'2=t2(c2-ẋ2)/(c2-ẋ'2) Since we're looking at this from my frame, and I'm stationary in my own frame, my velocity will be zero through space. So, we can set ẋ=0 t'2=c2t2/(c2-ẋ'2) Let's go ahead and get rid of that c2 on top by cancelling it. To do that, we'll multiply ẋ'2 by 1 in the form of c2/c2, then we'll factor out a c2. t'2=(c2t2)/(c2-(c2/c2)ẋ'2) t'2=t2/(1-(ẋ2/c2)) Now to get your time, we just need to take the square root of both sides. We get [math]t'=t\frac{1}{\sqrt{1-\frac{\dot{x}'^2}{c^2}}}[/math], which is the famous time dilation equation. So, the faster you go relative to me, the longer your seconds are relative to mine. According to my frame, your time is slowed according to your speed relative to me. Now, we could have done this all the exact same way for my time according to your frame. In that case, my time is slowed according to your frame. The faster you're going relative to me, the more of my seconds each one of yours is, according to my frame. 2 Link to comment Share on other sites More sharing options...
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