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## Trajectory of Sun's motion

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Hungry4info
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### Re: Trajectory of Sun's motion

I believe the ' is a reference to the derivative of a function.

d/dx f[x] = f'(x).
Where d/dx (x^n) = nx^(n-1).
Thus, if f(x)=x^3, f'(x)=3x^2.
And if f(x)=2x^2, f'(x)=4x.

Underscores denote subscripts, I believe.

If any of that's wrong, please, somebody tell me.
Current Setup:
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t00fri
Developer
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### Re: Trajectory of Sun's motion

Epimetheus, Hungry4info,

Epimetheus wrote:
t00fri wrote:
Consider the components of the vector X in 2 such frames, this means

X.X=X^2 = (c*t)^2 - x^2 -y^2 -z^2 = (c*t')^2 - x'^2 -y'^2 - z'^2

Forgive my budding mathematical knowledge ( intermediate algebra and intro to discrete math), Dr. Shrempp, but in the above equation are we using ' as logical not operator? The right-hand side is the inverse? ...
Epimetheus

Sorry if this prime notation was unclear. It is very simple. The primed quantities are merely labelling the components of the 4-vector in another relativistic reference frame.

In the original, unprimed frame, the position 4-vector had the form

X = (c*t, x,y,z)

after a Lorentz transformation to another reference frame (the "primed" one), all components of our vector change into

X' = (c*t', x',y',z')

Note the most fundamental feature that is at the basis of all this: the speed of light c stays the same in all possible frames (it gets NO prime)!!

Despite all components of the 4-vector change, the length of that 4-vector must remain the same, hence we get the equation that I wrote down above:

X.X=X^2 = (c*t)^2 - x^2 -y^2 -z^2 = (c*t')^2 - x'^2 -y'^2 - z'^2 = X'^2

Got it?

F.

Epimetheus
Posts: 42
Joined: 30.03.2008
With us: 12 years 4 months

### Re: Trajectory of Sun's motion

t00fri wrote:Note the most fundamental feature that is at the basis of all this: the speed of light c stays the same in all possible frames (it gets NO prime)!!

Despite all components of the 4-vector change, the length of that 4-vector must remain the same, hence we get the equation that I wrote down above:

X.X=X^2 = (c*t)^2 - x^2 -y^2 -z^2 = (c*t')^2 - x'^2 -y'^2 - z'^2 = X'^2

Got it?
F.

With all due respect, sir, I can't even remember where I left my keys! I bet they're following one of those worldlines and I'm on the wrong plane of reference!

BTW, doc, does X.X represent a multi-dimensional matrix? The velocity of light remains constant in ALL reference frames, regardless of an observer at rest at any cartesian coordinate within any hyper-polytropal dimension, correct?
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Celestia 1.5.1.4342 Qt4 Experimental

Hungry4info
Posts: 1133
Joined: 11.09.2005
With us: 14 years 10 months
Location: Indiana, United States

### Re: Trajectory of Sun's motion

Current Setup:
Windows 7 64 bit. Celestia 1.6.0.
AMD Athlon Processor, 1.6 Ghz, 3 Gb RAM

Epimetheus
Posts: 42
Joined: 30.03.2008
With us: 12 years 4 months

### Re: Trajectory of Sun's motion

viewtopic.php?f=4&t=11959 Computer Info:

CPU: Intel Core 2 6700 @ 2.66GHz
RAM: Ocz 2GB DDR2 800MHz
HDD: Seagate 400GB SATA
VD: Nvidia GeForce 7950 GT OC 512MB
OS: MS XP Pro SP2
Celestia 1.5.1.4342 Qt4 Experimental

t00fri
Developer
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Location: Hamburg, Germany

### Re: Trajectory of Sun's motion

Epimetheus wrote:
t00fri wrote:Note the most fundamental feature that is at the basis of all this: the speed of light c stays the same in all possible frames (it gets NO prime)!!

Despite all components of the 4-vector change, the length of that 4-vector must remain the same, hence we get the equation that I wrote down above:

X.X=X^2 = (c*t)^2 - x^2 -y^2 -z^2 = (c*t')^2 - x'^2 -y'^2 - z'^2 = X'^2

Got it?
F.

With all due respect, sir, I can't even remember where I left my keys! I bet they're following one of those worldlines and I'm on the wrong plane of reference!

BTW, doc, does X.X represent a multi-dimensional matrix? The velocity of light remains constant in ALL reference frames, regardless of an observer at rest at any cartesian coordinate within any hyper-polytropal dimension, correct?

Well, this forum is not well suited for math notation, so we should really activate the LaTex feature ASAP!

X.X is to indicate the scalar product of the two vectors. To be precise,
if X is a row vector, like so, X=(c*t, x,y,z) then the other X should be a transposed, i.e. /column/ vector. But columns are hard to draw here . There is another notation in terms of the vector components. But it is also hard to display here. It uses greek indices and the usual convention that repeating indices are to be summed over.

Something like

X_mu * X^mu; mu = 0,1,2,3

So X.X is a product that contracts a vector and it's transpose into a scalar (hence scalar product) , i.e. it is just one dimensionful number, the length^2 of our vector.

I wasn't explicit here since you mentioned somewhere that you were taking math classes at University, these days?

F.

Epimetheus
Posts: 42
Joined: 30.03.2008
With us: 12 years 4 months

### Re: Trajectory of Sun's motion

t00fri wrote:Well, this forum is not well suited for math notation, so we should really activate the LaTex feature ASAP!

X.X is to indicate the scalar product of the two vectors. To be precise,
if X is a row vector, like so, X=(c*t, x,y,z) then the other X should be a transposed, i.e. /column/ vector. But columns are hard to draw here . There is another notation in terms of the vector components. But it is also hard to display here. It uses greek indices and the usual convention that repeating indices are to be summed over.

Something like

X_mu * X^mu; mu = 0,1,2,3

So X.X is a product that contracts a vector and it's transpose into a scalar (hence scalar product) , i.e. it is just one dimensionful number, the length^2 of our vector.

I wasn't explicit here since you mentioned somewhere that you were taking math classes at University, these days?

F.

Yes sir! I believe I know what you're speaking of - Big O, theta, sigma and the like? The type of symbols you use for proofs by induction, say the giant S looking symbol that would state n=0 as n tends to infinity = {0(n+1) + 1(n+2) ... }
Representing the summation of a series, or pi to represent the multiplicative of a series, and so on. I was just introduced to proof by induction, propositional logic, infinities and greater infinities, number theory, Euler and Hamilton Paths and the like in Intro to Discrete Math and struggled with the concepts a great deal. Not a natural math head, just a motivated student. We'll see where I am in a year or two with all this! p.s. yes, a LateX or mathmatica port would be nice for this subject matter. Hmmmm! Computer Info:

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RAM: Ocz 2GB DDR2 800MHz
HDD: Seagate 400GB SATA
VD: Nvidia GeForce 7950 GT OC 512MB
OS: MS XP Pro SP2
Celestia 1.5.1.4342 Qt4 Experimental

Epimetheus
Posts: 42
Joined: 30.03.2008
With us: 12 years 4 months

### Re: Trajectory of Sun's motion

t00fri wrote:Epimetheus, Hungry4info,

Epimetheus wrote:
t00fri wrote:
Consider the components of the vector X in 2 such frames, this means

X.X=X^2 = (c*t)^2 - x^2 -y^2 -z^2 = (c*t')^2 - x'^2 -y'^2 - z'^2

Forgive my budding mathematical knowledge ( intermediate algebra and intro to discrete math), Dr. Shrempp, but in the above equation are we using ' as logical not operator? The right-hand side is the inverse? ...
Epimetheus

Sorry if this prime notation was unclear. It is very simple. The primed quantities are merely labelling the components of the 4-vector in another relativistic reference frame.

In the original, unprimed frame, the position 4-vector had the form

X = (c*t, x,y,z)

after a Lorentz transformation to another reference frame (the "primed" one), all components of our vector change into

X' = (c*t', x',y',z')

Note the most fundamental feature that is at the basis of all this: the speed of light c stays the same in all possible frames (it gets NO prime)!!

Despite all components of the 4-vector change, the length of that 4-vector must remain the same, hence we get the equation that I wrote down above:

X.X=X^2 = (c*t)^2 - x^2 -y^2 -z^2 = (c*t')^2 - x'^2 -y'^2 - z'^2 = X'^2

Got it?

F.

It's starting to sink in. So the primed part happens after the event! ( e.g. The event at (x, y, z, t) in frame S has coordinates (x?, y?, z?, t?) in frame S?. ) How would one apply the Lorentz Transformation to a 4-D vector? If you want to model this behavior in virtual space would you use the Matrix Form and convert to cartesian coords?

It's interesting how this understanding of hyperspace is starting to show up in game programming. For example, the use of quarternions and slerp for rotations and camera angles, which is much smoother and no "gimbal lock," utilizes the same 4-D vectors for camera views (reference frames ) that you've laid out here for understanding spacetime and worldlines and the like. Fascinating stuff!
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