Thread: relativity

I believe relativity deals with changes in space and time in certain situations. In special relativity, when you move at a high fraction of the speed of light, time will pass more quickly for you than unmoving observers and you'll look flattened in the direction you're moving in.
In general relativity, the time dilations occurs seem to occur when you're a in gravitational field. That's all I can think of for now.

I'm not implying anything like "google before you ask", but there are hours of lectures on youtube to keep you busy if you're really interested.

no, time moves more slowly when you accelerate. It seems to be a common misconception that it's the other way around.

Slash, what sort of math do you know? do you know what a dot product (also called inner product or scalar product) and vectors are? Do you know what a line integral is?

Really, all special relativity can be derived from two postulates:

1) the laws of physics look the same in all unnaccelerated reference frames
2) The speed of light will be measured the same in all reference frames.

A reference frame is just a coordinate system.

EDIT: @bonsay, yeah but explaining is good practice for those that understand. I'm just being selfish really.

GAH! Time DOESN'T exist!

The philosophy department is one door down.

Originally Posted by Hercuflea

GAH! Time DOESN'T exist!

jeezo who cares. i want to know about einstein's theory of relativity. A THEORY!

just like your THEORY

and philosopher, yes, i know about dot products etc. vectors and line integrals.

and timestopper, i know a bit about time dillation. is that part of relativity?


by reading all this that you guys are saying, that suggests my friend pays no attention in class whatsoever. he told me it was something to do with compunds forming or something, and the energy given or something.

Sounds like he was talking about fusion.

Originally Posted by Bonsay

Sounds like he was talking about fusion.

well, he diefinetally said relativity. and we dont go into much detail about fusion in our course. i dont think we do anyway. but the way he described it seemed nothing like fusion anyway.

Originally Posted by slash112

jeezo who cares. i want to know about einstein's theory of relativity. A THEORY!

just like your THEORY

It's nothing like einstein's theory. relativity makes testable predictions.

Originally Posted by slash112

and philosopher, yes, i know about dot products etc. vectors and line integrals.

Great. Do you understand the relationship between the pythagorean theorem and the dot product of two vectors? That would be a good place to start if you don't.

Originally Posted by PhilosopherStoned

It's nothing like einstein's theory. relativity makes testable predictions.



Great. Do you understand the relationship between the pythagorean theorem and the dot product of two vectors? That would be a good place to start if you don't.

well, i kinda worded that wrongly, what i meant to say was relativity is a theory, just like your thing is a theory too.

but no i dont know the relationship between them. but i might if i think about it for a minute, so hold on a sec.

yes, i think i might understand.

i havent thought about dot products in a while, so i had forgotten what it was.

is the relationship the fact that the dot product is the horizontal compnent and vertical compnent combined, just like pythagoras theorum?

Originally Posted by slash112

is the relationship the fact that the dot product is the horizontal compnent and vertical compnent combined, just like pythagoras theorum?

That's one way of putting it. Let's make it a little more precise though. Lets work in 2-space so that we only have two components. The ideas are the same in higher dimensions. Let's pretend that we're trying to figure out how to multiply vectors and get a scalar. so lets just naively multiply to vectors X = (x1 x2) and Y=(y1 y2). we get:

X*Y = g11*x1*y1 + g12*x1*y2 + g21*x2*y1 + g22*x2*y2 (eqn 1)


now, it's a theorem that we can choose the numbers {g_ij} to be anything we want and still get a sane product (meaning that it commutes and is linear in each variable) and that all sane products will be of this form. Suppose we want the product X*X to be the length of X. the if we choose g11 = g22 = 1 and g12 = g21 = 0, we get:

X*X = g11*x1*x1 + g12*x1*x2 + g21*x2*x1 + g22*x2*x2
= 1*x1*x1 + 1*x2*x2
= (x1)^2 + (x2)^2 (eqn 2)

so then we can say L(X) = sqrt(X*X). We can call that the 'euclidian' dot product because it's essentially just a new way of writing the pythagorean theorem. generally, we call the numbers {gij} the 'metric tensor' of a space because they let us define length. Does that make sense?

Originally Posted by PhilosopherStoned

Does that make sense?

no

i dont know where all those equations came from. except i recognize the end of the second one as pythagoras theorum

Originally Posted by PhilosopherStoned

Let's pretend that we're trying to figure out how to multiply vectors and get a scalar. so lets just naively multiply to vectors X = (x1 x2) and Y=(y1 y2). we get:

X*Y = g11*x1*y1 + g12*x1*y2 + g21*x2*y1 + g22*x2*y2 (eqn 1)


now, it's a theorem that we can choose the numbers {g_ij} to be anything we want and still get a sane product (meaning that it commutes and is linear in each variable) and that all sane products will be of this form.

Okay, say we have a basis for our vector space, e1 and e2, so that

X = (x1 x2) = x1*e1 + x2*e2 and Y = (y1 y2) = y1*e1 + y2*e2

for example, e1 = (1 0) and e2 = (0 1) Then (2 3) = 2*(1 0) + 3*(0 1). If we naively multiply X*Y we get:

X*Y = (x1*e1 + x2*e2) * (y1*e1 + y2*e2)

we can just multiply this out like binomials (we call this the FOIL rule in the states) and we get (eqn 1) from my last post if we specify that ei * ej = gij.

You should multiply out the above equation to see what I'm talking about.

Where all this is going is that we use the 'minkowski' metric tensor in relativity which is given by g11 = 1, g22 = g33 = g44 = -1 with all the other gij = 0 if i =/= j.

oh right, that makes sense now, we call it FOIL aswell. but isnt that like the noob way of doing it

yeah, it's sort of the noob way of doing it but it was a point of reference. strictly speaking though, it's the only way of multiplying binomials. It's only the noob way because it doesn't work for polynomials or vectors with more than two terms.

okay, the importance of the metric tensor is that it tells us how to form the dot product and that allows us to measure the distance between points in space. if we have a point X = (1, 2) and Y = (3, 4) then to measure the square of the distance between them we just do:

(X - Y)^2 = (X - Y) * (X - Y) = (-2, -2) * (-2, -2) = 4 + 4 = 8

so the distance between them is sqrt(8). remember, this is with g11 = g22 = 1 and g12 = g21 = 0 in (eqn 1) above. Again, the importance of understanding that the formula for the distance between two points comes from the metric tensor is that we change it in relativity.

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Really, a form relativity is already in newtonian mechanics. It's called Galilean relativity and has to do with changing from one frame of reference to another that is moving with constant velocity in relation to the first. First, we say that a frame of reference is an inertial frame if newtons first law holds in it. Essentially, this means that the frame isn't accellerating. So if you are in a car that's accelerating towards a street sign, that sign does not obey newtons first law because it is accelerating towards you even though there is no force acting on it. So we restrict ourselves to inertial frames. Even though it's not necessary, we will also restrict ourselves to frames of reference that aren't rotated with respect to each other so that all the coordinate axes are parallel. Otherwise, we have to bring matrices into it.

If an observer measures an event, E, with coordinates (x1, x2, x3) in one frame, lets call it O, and a second observer measures it in another frame that's at rest with respect to the first, let's call it O', then the we can determine the coordinates (x1', x2', x3') that the second observer assigns to E as follows:

We measure the distance from O to O' and get a vector D=(d1, d2, d3) so that the coordinates of O' in O are just D. Then the coordinates of E in the O' frame of reference are just (x1 - d1, x2 - d2, x3 - d3).

Does that make sense? We can get to the physics next post if so.

Cross product is way bettar than dot product. :3

I don't know what a line integral is though... probably will in a couple of months.

Originally Posted by Xei

Cross product is way bettar than dot product. :3

Nah.... Dot product owns.

a) the dot product is simpler
b) the dot product can be used to define the cross product
c) the cross product doesn't generalize to higher dimensions
d) the dot product completely defines the geometry of a space

dot product ftw.

Originally Posted by Xei

I don't know what a line integral is though... probably will in a couple of months.

isnt a line integral just like a line that has been integrated, or something?

wait a minute, a 2 dimentional line that has been integrated gives a single point.

ok i dunno, maybe it isnt that.

Try doing general problems involving lines and planes just via the dot product... >:V

Cross product is heaven.

isnt a line integral just like a line that has been integrated, or something?

wait a minute, a 2 dimentional line that has been integrated gives a single point.

ok i dunno, maybe it isnt that.

It's to do with vectors and things combined with calculus... I really don't know.

Originally Posted by Xei

It's to do with vectors and things combined with calculus... I really don't know.

yes... a line is a vector, and integration is calculus

Originally Posted by Xei

Try doing general problems involving lines and planes just via the dot product... >:V

Give me one. But yes, the vector product is useful in three dimensions.

Equation of the line that intersects with and is mutually parallel to x - 2 = y/5 + 3 = z + 4 and x - 1 = y + 5 = z/3.

:V

(It's a few months since I've done any maths so I could be talking nonsense bollocks).

yes... a line is a vector, and integration is calculus

Not really, the lines you look at in calculus are hardly ever 'vectors'... i.e. straight. They're wobbly, that's the whole point. And they're infinite.

Originally Posted by Xei

Not really, the lines you look at in calculus are hardly ever 'vectors'... i.e. straight. They're wobbly, that's the whole point. And they're infinite.

no, the stuff i did in calculus was intergration of quadratics etc. which as you said arent straight. but it is true infact that you can intergrate a straight line. which is still calculus. and a line is a vector so...