Could you please expand your calculations JohnDMalcolm because I'm struggling to get from the first expression to last. Thank You

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- Aug 9th 2012, 05:32 AMAA23Re: Timelike, Spacelike and Null Seperation
Could you please expand your calculations JohnDMalcolm because I'm struggling to get from the first expression to last. Thank You

- Aug 9th 2012, 06:24 AMJohnDMalcolmRe: Timelike, Spacelike and Null Seperation
Sorry, I've noticed my mistake. We are working in units where the speed of light is given the value $\displaystyle c = 1 $. This makes time measured in metres. One metre of time is defined as the amount of time it takes light to travel one metre of distance. With this, the Lorentz transformation is

$\displaystyle t' = \gamma \left( t - vx \right) $

$\displaystyle x' = \gamma \left( x - vt \right) $

$\displaystyle \gamma = \frac{1}{\sqrt{1 - v^{2}}} $

$\displaystyle ds^{2} = dt^{2} - dx^{2} $

If we were working in SI units, where $\displaystyle c = 3 \times 10^{8} \frac{m}{s} $, then we use the other transformations we were trying before AND the spacetime interval is

$\displaystyle ds^{2} = c^{2}dt^{2} - x^{2} $

So to solve your problem of $\displaystyle t^{2} - x^{2} $, not $\displaystyle c^{2}t^{2} - x^{2} $, use the set of equations above where $\displaystyle c = 1 $. Sorry for the earlier confusion. - Aug 9th 2012, 06:38 AMAA23Re: Timelike, Spacelike and Null Seperation
Thank You again JohnDMalcolm, I feel a lot more confident answering these types of questions now (Happy)

I do have one final question which I have attached that has been causing me a lot of problems. Now I can solve similar questions easily when given the relationship betweem (x,y) and (p,q) but am unsure how to establish one?

Do you have any suggestions or advice? Thank You once again - Aug 9th 2012, 06:52 AMHallsofIvyRe: Timelike, Spacelike and Null Seperation
- Aug 9th 2012, 06:53 AMAA23Re: Timelike, Spacelike and Null Seperation
My mistake, thank you HallsofIvy

- Aug 9th 2012, 07:27 AMJohnDMalcolmRe: Timelike, Spacelike and Null Seperation
Hey, call me John. I'm happy to help out. Helping someone through these problems helps to solidify my own understanding of the topic.

For the perpendicular case I think it's easiest to derive the metric using vectors. The unit vectors in the new coordinate system are

$\displaystyle \hat{p} = \hat{x} $

$\displaystyle \hat{q} = \cos{\theta} \hat{x} + \sin{\theta} \hat{y} $

The position of A can be written as the vector

$\displaystyle \vec{A} = x_{A} \hat{x} + y_{A} \hat{y} $

The (p,q) coordinates would be the projection of this vector onto those coordinate axes.

$\displaystyle p_{A} = \vec{A} \cdot \hat{p} $

$\displaystyle q_{A} = \vec{A} \cdot \hat{q} $

Since the coordinates of A are general, these dot products can be used to construct the metric. - Aug 9th 2012, 07:31 AMJohnDMalcolmRe: Timelike, Spacelike and Null Seperation
As for the parallel case, I find it easiest to use trig geometry. Draw a line from A that is parallel to the p-axis and intersects the q-axis. Try to find the length from the origin to this intersection (ie, the coordinate q) in terms of y and theta. Similarly, draw a line from A parallel to the q-axis that intersects the p-axis. The length from the origin to this intersection will be a linear combination of x and y involving theta.

- Aug 9th 2012, 07:43 AMAA23Re: Timelike, Spacelike and Null Seperation
Hi John, so going from what you have said I have started my calculations (see attachment). I believe the next step is to establish the differential coefficients and use them to obtain the metric tensor. Is this correct??

- Aug 9th 2012, 08:15 AMJohnDMalcolmRe: Timelike, Spacelike and Null Seperation
Your transformations are linear so we can do this by inspection. So the transformation in going from (x,y) to (p,q) is the 2x2 matrix below.

$\displaystyle \left( \begin{array}{c} p \\ q \end{array} \right) = \left( \begin{array}{cc} E&F \\ G&H \end{array} \right) \left( \begin{array}{c} x \\ y \end{array} \right) $

This gives the two equations

$\displaystyle p = Ex + Fy $

$\displaystyle q = Gx + Hy $

So, by comparing to the expressions you derived, what are E, F, G, and H? These are the differential coefficients you're looking for. With what the problem is asking you to do, you didn't need to express x and y in terms of p and q. Those expressions would allow you to find the transformation from (p,q) to (x,y). That could be useful, but, like I said, the problem doesn't require it.

Let's define the transformation matrix as

$\displaystyle \Lambda = \left( \begin{array}{cc} E&F \\ G&H \end{array} \right) $

The metric in the (x,y) coordinates is just the 2x2 identity matrix.

$\displaystyle g = \left( \begin{array}{cc} 1&0 \\ 0&1 \end{array} \right) $

The metric in the (p,q) coordinates is

$\displaystyle g' = \Lambda^{T} g \Lambda $ - Aug 9th 2012, 08:23 AMAA23Re: Timelike, Spacelike and Null Seperation
Would:

E= 1 , F=0 , G = cos(theta) and H = sin(theta)

Is this what you were asking for? - Aug 9th 2012, 10:23 AMJohnDMalcolmRe: Timelike, Spacelike and Null Seperation
Yes, for the perpendicular case. Were you able to find the metric from those values?

- Aug 9th 2012, 10:57 AMAA23Re: Timelike, Spacelike and Null Seperation
Brilliant (Clapping)

I've done most of the calculation but as you can see from the attachment, I am unsure on one bit - Aug 9th 2012, 12:03 PMJohnDMalcolmRe: Timelike, Spacelike and Null Seperation
That T superscript means transpose. The transpose of a matrix (doesn't have to be square) is where you take the rows and write them as columns.

$\displaystyle A = \left( \begin{array}{ccc} a&b&c \\ d&e&f \\ g&h&i \end{array} \right) $

$\displaystyle A^{T} = \left( \begin{array}{ccc} a&d&g \\ b&e&h \\ c&f&i \end{array} \right) $ - Aug 9th 2012, 12:36 PMAA23Re: Timelike, Spacelike and Null Seperation
OK I understand so my final answer would be (see attachment)

- Aug 9th 2012, 01:26 PMJohnDMalcolmRe: Timelike, Spacelike and Null Seperation
Remember that the commutative property is not true for multiplication of matrices in general. You have

$\displaystyle g' = \Lambda \Lambda^{T} \neq \Lambda^{T} \Lambda $