Minkowski metric tensor computation

In summary: What is the difference between summation convention and coordinate change?Summation convention means to sum over the repeated indices, while coordinate change means changing the coordinate system.
  • #1
Advent
30
0
Hi, I'm having problem with understanding tensors and the Einsteins summation convention, so I decided to start doing explicit calculations, and I'm doing it in the wrong way. Hope someone could help me to clarify the concepts.

In flat spacetime we have [tex]\eta[/tex] with the signature (-+++). Under some coordinate change, say [tex]x_{\mu} \rightarrow x_{\overline{\mu}}[/tex], then the metric changes as [tex]g_{ \overline{\mu} \overline{\nu}}=\frac{ \partial x^{\rho}}{ \partial x_{\overline{\mu}}} \frac{ \partial x^{\sigma}}{ \partial x_{\overline{\nu}}}g_{ \overline{\rho} \overline{\sigma}}[/tex]. So, If I change the coordinate system from Cartesian [tex](t,x,y,z)[/tex] to spherical [tex](t,r, \theta, \varphi)[/tex] with the following equations

[tex] x = r \cos(\varphi) \cos (\theta)[/tex], [tex] y = r \cos(\varphi) \sin (\theta)[/tex], [tex] z = r \sin(\varphi) [/tex], [tex] t = t [/tex]. The four non-zero componentes of the metric [tex]\eta[/tex] in spherical coordinates should be:

[tex]g_{11} = (\frac{ \partial t }{ \partial t })^2 g_{1'1'}=-1[/tex]


[tex]g_{22} =(\frac{ \partial x }{ \partial r })^2 g_{2'2'}=\cos^2(\varphi) \cos^2 (\theta)[/tex]


[tex]g_{33} =(\frac{ \partial y }{ \partial \theta })^2 g_{3'3'}=r^2 \cos^2(\varphi) \cos^2 (\theta)[/tex]


[tex]g_{44} =(\frac{ \partial z }{ \partial \varphi })^2 g_{4'4'}=r^2 \cos^2(\varphi)[/tex]

And finally, the line element [tex]ds^2=g_{\mu \nu}dx^{\mu}dx^{\nu}=dt^2+\cos^2(\varphi) \cos^2 dr^2+ r^2 \cos^2(\varphi) \cos^2 d^2 \theta + r^2 \cos^2(\varphi) d^2 \varphi[/tex] which is incorrect.

Thanks for your time, any help will be appreciated.
 
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  • #2
Did you sum over the repeated indices in the change of coordinates formula (summation convention)?
 
  • #3
atyy said:
Did you sum over the repeated indices in the change of coordinates formula (summation convention)?

I think I did, in fact it explains the square of the derivatives.
 
  • #4
I'm not sure about this, but I think in grr you should have terms like (dt/dr)^2.gtt+(dx/dr)^2.gxx+(dy/dr)^2.gyy+(dz/dr)^2+(dt/dr)(dx/dr)gtr + ...

where in your formula I've taken u=r,v=r and rho and sigma must be summed over all combinations of rho=t,x,y,x and sigma=t,x,y,z
 
  • #5
atyy said:
I'm not sure about this, but I think in grr you should have terms like (dt/dr)^2.gtt+(dx/dr)^2.gxx+(dy/dr)^2.gyy+(dz/dr)^2+(dt/dr)(dx/dr)gtr + ...

Thanks for your posts.

I think now I understand your point. Do you mean that, in the equation of the line element should be 4*4=16 summands instead of only four, by varying the ro and sigma indices between all their range?
 
  • #6
Ok, just answered too fast!

Then this was my error, thank you so much!
 
  • #7
Advent said:
Ok, just answered too fast!

Then this was my error, thank you so much!

Wow, you sure compute fast! Good to know - I wasn't sure about this.
 

Related to Minkowski metric tensor computation

1. What is the Minkowski metric tensor?

The Minkowski metric tensor is a mathematical tool used in the study of special relativity. It is a 4x4 matrix that describes the geometry and distance measurements in spacetime. It allows for the calculation of distances and intervals between events in space and time.

2. How is the Minkowski metric tensor computed?

The Minkowski metric tensor is computed using the Pythagorean theorem in four-dimensional spacetime. It takes into account the time component, as well as the three spatial dimensions, to calculate the proper distance between two events. It can also be computed using the Lorentz transformation equations.

3. What is the significance of the Minkowski metric tensor?

The Minkowski metric tensor is significant because it is the mathematical foundation for special relativity. It allows for the precise measurement of distances and intervals in spacetime, which is necessary for understanding the effects of time dilation and length contraction.

4. Can the Minkowski metric tensor be visualized?

Yes, the Minkowski metric tensor can be visualized using a spacetime diagram. In this diagram, time is represented on the vertical axis and space on the horizontal axis. The Minkowski metric tensor can also be represented as a grid of lines that intersect at right angles, known as a light cone.

5. Are there alternative metrics to the Minkowski metric tensor?

Yes, there are alternative metrics to the Minkowski metric tensor, such as the Riemannian metric tensor used in general relativity. These metrics take into account the curvature of spacetime and are more complex to compute. However, in the special case of flat spacetime, the Minkowski metric tensor is the most commonly used metric.

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