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Comments:
<0> http://google.com/patents?q=person Patent search?
<1> hehe http://www.google.com/patents?q=linus+torvalds
<2> Hello. I am working feverishly on a paper o relativity and I am looking at the Lorentz-transformation written as t=\gamma[t'+\beta x'/c]. Trying to figure out how to apply this to an example, I am completely at loss.
<2> Could someone help me out?
<2> Basically what my ***ignment tells me to do is "apply" this to an arbitrary example, but I can't quite figure out what I should set up as parameters.
<3> Saraphim: well its a transformation, so transform a frame to another frame, thats really all it does
<2> JabberWalkie: The problem is, I guess, that (I'm in over my head, and) I simply can't conjure up an example. Whenever I start, I seem to think I am violating the very consequences of SR..
<3> just do time dialation....
<2> I tried something along these lines: A car moving at speed B=1/100 (a very fast car) in frame S' that has relative speed to S=v - and then I'm stuck.
<2> Because I needed to have the times be the same somehow, but how can I ever synchronize the times t'=0=t when these frames are moving relatively to each other?
<2> For the record, I'm studying at a high school equivalent.
<3> Saraphim: just do this, in the unprimed frame we shall have an observer at t=0, x=0, he is watching a car p*** by by him at t=0,x=0, then you want to know what things are like for the car
<3> so to get the time in the cars frame you calculate t'
<3> and so on...
<2> That's all?
<2> I think I have somehow managed to misplace the forest among all the trees.
<3> yeah, thats not a very interesting example though
<2> No, but at least I understand it. O:-)
<3> it gets better when you start comparing how each fames measure time
<3> but what the lorentz transfroms give you is this, you have a set of coordinates in one frame, x,y,z,t.....you want to know what they are in another frame, say x
<3> x',y',z',t'
<3> you use the lorentz transformations to find primed coordinates
<2> Yes, I have the formulas for that written down, it's their application that's giving me a headache.
<3> yep, you can play around with them and get some interesting things
<2> So, to take the example you proposed, I would simply insert t=0 and x=0 into the formula for t' and I'd have at least SOME kind of example?
<3> yeah, you would get t'=0
<2> Fascinating......
<4> a lorentz transformation is simply a hyperbolic rotation
<3> like i said, not very interesting :)
<2> *grins* Indeed.
<2> If I were to compare velocity (speed), then, that would provide some more interesting figures, wouldn't it?
<3> try taking the derivative with respect to t and see what happens...
<3> set acceleration at zero
<2> Hmm. I'm going to try figuring that out.
<2> Well if I can calculate x' easily and t' easily, would dx'/dt' not still be velocity?
<2> And in that case, would I not get some sort of relativistic effect when comparing dx/dt to dx'/dt'?
<2> (Choosing to write about this subject has made me feel unreasonably dense)
<3> Saraphim: no no, find a formula for dt/dt'
<3> you can do that by taking the derivative of that thing you had there
<2> Alright.
<3> make sure its the derivative with respect to time..
<2> Aye, I am trying. As I said, my level of understanding of even mathematics is fairly basic - we have not had calculus yet.. If that is what you call it over there. Differentiation and integration. So doing these things take a little time.
<2> I appreciate your help, by the way.
<3> np
<0> Saraphim: The basic concepts of differentiation and integration are the simplest things you may ever encounter, problems usually arise from algebraic properties and rules and whatnot; the idea of finding the slope of a secant line as the secant line progressively becomes a tangent line is largely a matter of algebra and limits more than anything else. The differentiation/integration is just nifty notation :)
<2> kanzure: Yes I realize that, but it takes some practice to get used to it none the less. :)
<0> Saraphim, Yep. It also helps to hear such explanations rather than having a teacher just go through many example problems without actually explaining what's going on. I found the secant-to-tangent-line transformation spectacular and an excellent method of what's actually going on :)
<0> *method of explaining what's
<2> I have seen that example and I found it very explanatory too. Also, every time I talk about it or get told about it from anyone, it helps my understanding along a little bit.
<4> i happened upon the transformation by thinking of velocity as a constant that could be varied between space and time
<4> so i derived a sin/cos transformation
<4> which was incorrect
<4> now i realized that because the time dimension is imaginary, you should rotate via an imaginary angle in the sin/cos
<4> making it a hyperbolic rotation instead
<2> I suppose that is what I will see when I plot the velocity function.
<2> (?)
<4> basically, speed = cosh theta, theta = angle between t-axis and velocity
<4> time dilation = sinh theta
<0> Saraphim: Yes, talking is always very, very helpful for me as well. The problem is that not many people know how to talk about these sorts of mathematical topics! Kind of bizarre if you ask me.. but nontheless interesting to note.
<2> NoorulIslaam: I think I will try and show myself the graphs for galilean and lorentz for comparison.
<4> Saraphim hmm not sure how much good that will be
<2> kanzure: hehe, well I guess it's a question of peers often. I find that every time I have asked a question here, even though I might not reach understanding immediately, it will help me understand a plethora of other things.
<2> NoorulIslaam: But if I plotted the galilean velocity transform surely I would find a straight line, while for the lorentz I would find a hyperbola?
<4> saraphim the galilean transform for velocity changes the angle and magnitude of the velocity vector
<2> (surely being a rather optimistic word in my present state of understanding)
<4> saraphim in a lorentz transform, the graph is squeezed diagonally
<4> Saraphim collapsing to two diagonals at lightspeed
<2> Ah
<4> however, think about it in this manner...
<4> 1. you that everything travels at 'c' in space-time
<4> naturally, you might think to draw a space-time diagram, and circle with radius 'c'. then any point on the circumference of this circle represents a valid velocity
<4> that make sense to you?
<2> NoorulIslaam: I'm afraid the concept of space-time eludes me and will do so for a few more days into my project.
<4> Saraphim err right
<4> you know how to draw a two-axis graph, right? :)
<2> Haha, yes.
<4> ok imagine you just did that
<2> Right.
<4> now draw a circle at the origin
<4> done?
<2> Mhm.
<4> imagine the radius of this circle is 'c'
<4> k?
<2> Yep.
<4> now label the vertical axis 't' and the horizontal 'x'
<2> Yes.
<4> now draw an arrow from the origin to the circumference of the circle at the vertical
<4> meaning, an arrow along t-axis until it touches the circle
<2> Yes.
<2> (I was just about to get my English dictionary)
<4> take this to represent the 4-velocity of a particle at rest
<4> 4-velocity is the space-time analog of regular velocity
<2> By for velocity, you mean x,y,z,t?
<2> four*
<4> we ignore y and z here
<2> Ah.
<4> just concentrate on x and t
<2> Right.
<4> as you can see, the vector you just drew has no x component
<4> so it's at rest
<4> right?
<2> Yes.
<4> now draw a horizontal vector
<4> along the x-axis
<2> Yep.
<4> this has no component along t, but full across x (space)
<2> Yes.
<4> you can imagine that you are able to basically rotate this vector anywhere along the circle
<4> you can immediately see that the velocity can never exceed the speed of light
<4> you with me so far?
<2> Yes.
<2> That is a rather clever representation.
<4> but now here's another interesting bit
<4> the amount of the vector along the t-axis represents your "time dilation", as it is called
<4> we'll do the stuff in natural units, so take c = 1, to make things simple
<4> you see then that for a particle at rest, the t-component is 1.0
<4> for a something at lightspeed, the t-component is 0.0
<2> Aah.
<4> if we define the angle between the vector and the positive t-axis to be theta
<4> this relationship is: v = c sin theta, and relativistic beta = cos theta
<4> v is the velocity of the particle in space
<4> get it?
<2> Yes.
<2> I was just drawing it
<2> That is really practical.
<4> good
<4> but wait...
<4> there's more
<4> draw the vector so it points along the negative t-axis
<4> now.. it's at rest... but you get a negative beta!
<4> guess how you can physically interpret this?
<2> Uuhh. Into the past?
<2> >_> That doesn't make sense
<4> (btw, i'm making you follow my line of thought that i went through to arrive at my unified physic model)
<4> right, until you realize that the beta factor refers to how "quickly" something can interact, atleast as far as electromagnetism goes
<2> Haha, you lost me there.
<2> But I don't expect to understand that. O:-)
<4> an electron at rest is easy to interact with... an eletric field deflects it easily, but an electon moving at 99.999% of the speed of light is much more difficult to move about
<2> Sounds like momentum. Which I'm not sure of the status of in SR.
<4> in SR its known as the relativistic m*** increase
<2> Right.
<4> m_relativistic = m_rest * gamma
<4> so do you understand how you can take beta (beta = 1/gamma) to be a factor representing how quickly something interacts?
<3> beta != 1/gamma
<4> beta = 1 at rest, meaning 100% interacting, beta = 0.0000000000001 at 99.9999999% of c, meaning very little interaction
<4> JabberWalkie oh sorry
<4> hmm
<4> 1/gamma is 1/gamma :P
<4> Saraphim heh r u following?
<2> Sort of. It's a little much to take in all at once.
<4> Saraphim i see. well now getting back to that vector drawn along the negative t-axis
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