/sci/ - Science & Math

>>2148427
No problemo... are you the OP?

>>2148387
HA Einstein is disapoint

But we already solved for B, C, and D in terms of A, so we know them all! So finally, substituting A, ..., D into the Lorentz transformation, we get:
<span class="math">x' = (x-vt)/ \sqrt {1 - v^2/c^2}[/spoiler]
<span class="math">y' = y[/spoiler]
<span class="math">z' = z[/spoiler]
<span class="math">t' = (t - vx/c^2) / \sqrt {1 - v^2/c^2}[/spoiler].

TADA! We now have the Lorentz transformations, the backbone of Special relativity. Everything can be deduced from these equations, such as time dilation, length contraction, the structure of causality, etc. I will do this in due time, but not now.
If you're interested in learning more, keep this thread alive, and whenever I'm on /sci/ again (probably tomorrow) I'll continue where I left off. If you have questions, ask them here and I'll get back to you when I can.

Why? Well the pulse will travel in the x'-y' plane, and the distance it will travel after time = t' will be <span class="math">\sqrt{x'^2 + y'^2}[/spoiler] by the Pythagorean theorem. This equals ct'.

Note that, using C = -Av/c^2, we can rewrite the equtaion <span class="math">t' = Cx + At = A(t - vx/c^2)[/spoiler].

Plugging in shit, we get
<span class="math">x'^2 + y'^2 = (A(x-vt))^2 + y^2 = A^2 (0 - vt)^2 + (ct)^2 = c^2 t'^2 = c^2 A^2 (t - v0/c^2)^2 [/spoiler].

Solving for A, we get
<span class="math">A = 1/ \sqrt{1 - v^2/c^2}[/spoiler]

Now finally, Case 4: Light is emitted along the y-axis in the (x,y) frame at t = 0. Evidently, in the (x',y') frame the light will have components in the x' and y' axes.

We know that x = 0, and y = ct. We also know that
<span class="math"> x'^2 + y'^2 = c^2 t'^2[/spoiler].

>>2148322
1) The relative speeds wouldn't 60% or 60% + 60%. It would end up being very close to the speed of light.

2) As you said, A and B can't be at the speed of light, but let's suppose that they're close to the speed of light. So when they collide, they're relative velocities will still be smaller than the speed of light, but a bit higher than before. If you think that this "bit higher" is a negligible amount, think again. The mass of a particle is
<span class="math">m = m_0/sqrt{1 - v^2/c^2}[/spoiler]. Small changes in the speed when the particle is going close to the speed of light make enourmous changes in the mass, and hence the energy of the collision

Case 3: A light pulse is sent out from the origin at t = 0. Now, by the constancy of the speed of light, x = ct and x' = ct'. Note that
<span class="math">t' = Cx + Dt = Cx + At[/spoiler] because D = A. Now,
<span class="math">x' =A(x-v)[/spoiler]. So
<span class="math">x' = ct' = c(Cx + At) = c(Cct + At) = A(ct - vt)[/spoiler]
Canceling the Act 's from both sides, we get
<span class="math">C = -Av/c^2[/spoiler]

Plugging this into the equation
<span class="math"> x' = A(x-vt)[/spoiler], we get
<span class="math">-vt' = A(0-vt)[/spoiler]. But
<span class="math">t' = Cx + Dt = Dt[/spoiler] (because x = 0). Plugging this into the last equation, we get
<span class="math">-vDt = -Avt, D = A [/spoiler].

Note that now <span class="math">x' = Ax + Bt = Ax -Avt = A(x-vt)[/spoiler] because B = -Av

Case 2: We'll consider the situation the other way around. The observer in the (x',y') frame sees the origin of (x,y) move along the x' axis at a speed of -v. Then, using the same logic as before,
<span class="math">x=0, x'=-vt'[/spoiler]

>>2148246
Our "event" is the position of O' 's origin at time t. Obviously, x' = 0, because that's the definition of the origin. And because O' is moving at a speed of v away from O in the x direction, x = vt.

Now, we can plug this into the equation x' = Ax + Bt. Substituting x' = 0 and x = vt, we get 0 = Avt + Bt, and thus B = -Av.

I'll post the rest of the algebra so that you can read it while you're in the shower.

>>2148203
Is it the algebra or the idea?

>>2148192
Well, general relativity needs special relativity, so if GR is useful, then SR is necessarily useful.

Second, typically elementary particles travel fast enough for relativistic effects become important. For instance, in gold atoms electrons travel at something like half the speed of light, maybe faster. So our understanding of matter requires SR. Molecular electronics, I think, is sensitive for SR to become important, and Molecular electronics will be the source of modern technology in the future.

But anyway, even if it wasn't useful, why does that make it not worth studying? Does something have to have applications to merit attention? Relativity is a beautiful theory, and changes our concepts of space and time...doesn't that make it worth learning?

>>2148174
Recommended music for relativity: http://www.youtube.com/watch?v=VqANh4QnNxA

So, does >>2148093 make sense? I'm trying to, by considering a few cases where we know the relationships between x' and x etc., to solve for the values of A, B, etc.

>>2148149
So, the question remains: what ARE the correct equations to replace the Galilean transformations?

We can, as Einstein did, solve for them algebraically. However, we will first assume that the equations adopt the form written above, with all the A, B, C's etc. THESE ARE UNKOWNS. We are trying to find out what their values are. Originally, A = 1, B = v, C = 0, and D =1. We will now solve for the correct values. Does that make sense?

>>2148132
Okay, do you see why the regular transformations
<span class="math">x' = x-vt[/spoiler] etc. don't work?

>>2148103
Can you be more specific about what you don't understand? Is it the A, B, C, and D? Or the Lorentz transformations in general?

>>2148081
Does >>2148093 make you understand? If not, stop me know and I'll explain.

>>2148072
Our goal is to solve for A, B, C, and D. If you're wondering why the Lorentz transformations take this form, I can justify it later...let me first solve for A, B, C, D.

We will do this by considering four separate cases where we know how the transformations will behave.

Case 1: Observer in (x,y) sees origin of (x',y') moving along the x-axis at speed v. We know that
<span class="math">x = vt[/spoiler], <span class="math"> x' = 0[/spoiler]. Therefore
<span class="math">0 = Avt + Bt, B = -vt [/spoiler].

Since the Galilean transformations do not obey the law that the speed of light is a universal constant, Einstein derived a new prescription for describing the same event in different (non-accelerating) frames of reference. Because Einstein derived them, we call them the Lorentz transformations...yeah.

We will assume that the Lorentz transformations take the form
<span class="math">x' = Ax + Bt[/spoiler]
<span class="math">y' = y[/spoiler]
<span class="math">z' = z[/spoiler]
<span class="math">t' = Cx + Dt[/spoiler].

>>2148011
In that case the light isn't actually slowed down. If light travels inside a material, like glass, the photons will travel at the speed of light until they hit an atom. When they hit the atom, the photon gets absorbed, and released a bit later. The overall effect is for light to travel more slowly, but at any point it's traveling at the speed of light.

Now, the equations given about are called the Galilean transformations. But they're wrong! If they were right, then necessarily observers moving relative to one another would measure light to travel at different speeds. Why?

Let x be the position of a light beam at time t. We have
<span class="math">x' = x - vt[/spoiler]
<span class="math">\Delta x' = \Delta x - v \Delta t[/spoiler]
<span class="math">\Delta x' / \Delta t' = \Delta x' / \Delta t = \Delta x / \Delta t - v[/spoiler]
But <span class="math">\Delta x / \Delta t[/spoiler] is the speed of light as measured by O, and <span class="math">\Delta x' / \Delta t'[/spoiler] is the speed of light as measured by O'. And these are not the same! Ergo, the Galilean transformations are incorrect.

We will now derive the correct transformation laws.

>>2147990
Should be
<span class="math"> x' = x - vt[/spoiler]
<span class="math">y' = y[/spoiler]
<span class="math">z' = z[/spoiler]
<span class="math">t' = t[/spoiler]

>>2147988
You were trolled I think. Light always travels at c...although there have been weird experiments where light has travelled much faster than c, and where light has stood still. I don't know what's up with that, but it apparently doesn't affect relativity.

Well, this should be simple enough. Looking at the diagram, it appears that
<span class="math">x' = x - vt
y' = y
z' = z
t' = t[/spoiler]
The last step is justified on the grounds that both observers should measure time the same way.

>>2147953
If I say that something happens 5m from me, someone else might say that it happened 3m from her, because she's closer. x would be 5m, and x' (the position from the other observer's vantage point) would be 3m. Does that make sense?