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/sci/ - Science & Math

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7002241 No.7002241 [Reply] [Original]

Given the curves <span class="math">u^{\mu}=u^{\mu}(\lambda)[/spoiler] and <span class="math">a^{\mu} = \frac{d u^{\mu}}{d \lambda}[/spoiler] and the metric tensor <span class="math">g_{\mu \nu} = diag(-1,1,1,1)[/spoiler] such that:

<span class="math">g_{\mu \nu}u^{\mu}u^{\nu}=-1[\math]

And therefore,

<span class="math">g_{\mu \nu}a^{\mu}u^{\nu}=0[\math]

(i.e. u^{\mu} and a^{\nu} are orthogonal), is it possible to prove that g_{\mu \nu}a^{\mu}a^{\nu}>0[\math]? I'm trying to prove, in the context of special relativity, that 4-acceleration is spacelike without using the definition of 4-velocity.

I want to prove it only using that 4-accel and 4-vel are orthotogonals and one is the derivative of the other, but I haven't been able to do it.[/spoiler][/spoiler]

>> No.7002243

>>7002241
Fuck, here is the rest:

<span class="math">g_{\mu \nu}u^{\mu}u^{\nu}=-1[/spoiler]

And therefore,

<span class="math">g_{\mu \nu}a^{\mu}u^{\nu}=0[/spoiler]

(i.e. u^{\mu} and a^{\nu} are orthogonal), is it possible to prove that <span class="math">g_{\mu \nu}a^{\mu}a^{\nu}>0[/spoiler]? I'm trying to prove, in the context of special relativity, that 4-acceleration is spacelike without using the definition of 4-velocity.

I want to prove it only using that 4-accel and 4-vel are orthotogonals and one is the derivative of the other, but I haven't been able to do it.

>> No.7002262

How did you show they are orthogonal?

>> No.7002264

>>7002262
Just differentiate <span class="math">g_{\mu \nu}u^{\mu}u^{\nu}=-1[/spoiler] (<span class="math">\frac{d}{d \lambda}[/spoiler])

>> No.7002265
File: 27 KB, 1920x1080, earth-moon.jpg [View same] [iqdb] [saucenao] [google]
7002265

>>7002241
horrible picture op, there's like 30 earth diameters between the moon and earth

>> No.7002267

>>7002265
t-thank you

>> No.7002277

>>7002243
It's been a while but I think you can.

>> No.7002279

>>7002277
I have no idea how to do it without using <span class="math">x^{\mu}=(t,x,y,z)[/spoiler] and <span class="math">u^{\mu}= \frac{d x^{\mu}}{d \lambda}[\math].[/spoiler]

>> No.7002280

>>7002279
...and <span class="math">u^{\mu}=\frac{d x^{\mu}}{d \lambda}[/spoiler]

>> No.7002293

>>7002262
Their dot products are 0 in the OP.
When you're using tensors you can compute the dot product of two vectors using the metric tensor <span class="math"> g_{\mu \nu} [/spoiler] (parewise dot products of the covarisnt basis vectors in this case) and the vector components <span class="math"> U^{\mu} W^{\nu} [/spoiler]
in this case is using the velocity 4-vector and the acceleration 4-vector which are orthogonal to one another (standard velocity and acceleration vectors from babby physics also have the same property).

>> No.7002427

Forget u2,u3, solve for u1 in terms of u0
u1^2 = -1+u0^2 (u0^2>1 ??)

- u0^2 + u1^2 = -1
u1 = (-1 + u0^2)^(1/2)

a0 = d/dlambda u0
a1 = d/dlambda u1
= 1/2 (-1+u0^2)^(-1/2) 2 u0 d/dlambda u0
= (-1+u0^2)^(-1/2) u0 a0
Plug these in...
-(a0)^2 + ((-1+u0^2)^(-1/2) u0 a0)^2
=-a0^2 + 1/(-1+u0^2) u0^2 a0^2
=( -1 + u0^2/(-1+u0^2) ) *a0^2
=( (1-u0^2 + u0^2)/(-1+u0^2) ) *a0^2
=( 1/(-1+u0^2) ) *a0^2
= a0^2/u1^2

I really have no idea what I'm doing. Just bored while watching House Hunters.