Jump to content

B F Schutz and proof of invariance


JohnFromAus

Recommended Posts

Have moved on from SR and started Schutz book "A first course in general relativity". I do not understand some of the symbols used in the proof of the invariance of the interval. I am referring to equ 1.2 to equ 1.3. What I am missing is what is meant by M with 2 subscripts followed by 2 quantities in brackets.

 

M[math]_{ab}[/math] (dx[math]^{a}[/math]) (dx[math]^{b}[/math])

 

Umm have not quite worked out how to show equations yet either! The a and b are superscripts to the dx.

 

Any help appreciated

John

Link to comment
Share on other sites

I don't have the book so not sure what I can do.

 

[math]M_{ab}\{dx^{a}\}\{dx^{b}\}[/math]

 

Is this some strange notation for the tensor or wedge product? It is not standard notation in geometry. The book such explain it somewhere.

Link to comment
Share on other sites

Thanks ajb I am still working on it - think the Ms may be functions of the various dxs involving v as well.

 

Have figured out the maths equation formating - the terms I am interested in are

 

[math]M_{\alpha\beta}(\Delta x^{\alpha})(\Delta x^{\beta})[/math]

 

These are summed over alpha and beta ranging from 0 - 3 for t,x,y,z giving 16 terms in all.

 

John

Link to comment
Share on other sites

Given it's supposedly about invariance, I suspect M to be a rank-2 tensor and the dx to be infinitesimal shifts in coordinates that can be considered Lorentz-vectors in this case. The result of the product then is a scalar, i.e. a number that has the same value in all coordinate systems. I assume what Schutz is about to do is showing that if the objects dx and M are expressed in a different coordinate system (where they could be written as [math]M'_{ab}, dx'^a [/math]) the value of the product remains the same i.e. s'=s.

Link to comment
Share on other sites

Thanks ajb I am still working on it - think the Ms may be functions of the various dxs involving v as well.

 

Have figured out the maths equation formating - the terms I am interested in are

 

[math]M_{\alpha\beta}(\Delta x^{\alpha})(\Delta x^{\beta})[/math]

 

These are summed over alpha and beta ranging from 0 - 3 for t,x,y,z giving 16 terms in all.

 

John

 

As Atheist says, you should think of the [math] \Delta x^{\alpha}[/math] as a "small change in [math]x^{\alpha}[/math]". i.e. [math] \overline{x}^{\alpha} = x^{\alpha} + \Delta x^{\alpha}[/math]. The expression you give should be interpreted as a scalar. I don't think this is the best way to understand tensors, but ok.

 

[math]M_{\alpha \beta}= M_{\alpha \beta}(x)[/math], i.e. it is a (collection of ) functions of [math]x[/math].

 

The important thing is how these objects change under changes of coordinates.

Link to comment
Share on other sites

OK - I have a good maths background and can handle all the algebra ok except for the use of M which I assumed had some special meaning. Turns out here it is the coefficient of the quadratics dt.dt, dt.dx, dt.dy, dt.dz.....12 more which may be a function of v, the relative frame velocity.

 

This is in chap 1 of the book - chap 2 deals with vectors and chap 3 with tensors - so expect some more posts in a few weeks!

 

I still have some questions re the interval but cannot yet be specific enough to ask anything sensible!

 

Thanks again

John

Link to comment
Share on other sites

My advice is to get to grips with some elementary differential geometry and in particular tensors and tensor like objects.

 

They are (usually) defined in terms of how the components and basis change under (passive) diffeomorphisms i.e. changing the coordinates at a point. (We also have active ones which are also important in special relativity).

 

The lecture notes by Carroll are very good. (The notes discuss general relativity, but also explain special relativity very well, they would compliment any other text).

 

Once you understand this, most of special relativity is quite straight forward.

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.