Sketching curves on a plane with a given metric

[cazlab]

Homework Statement

Consider a coordinate transformation from (t,x) to (u,v) given by
$$t=u\sinh v$$$$x=u\cosh v$$
Suppose (t,x) are coordinates in a 2-dimensional spacetime with metric
$$ds^2=-dt^2+dx^2$$
Sketch, on the (t,x) plane, the curves u=constant and the curves v=constant.

Homework Equations

None that I know of for this graphical part of the question.

The Attempt at a Solution

Sketching the given curves on the plane in the way I would have in earlier courses, I end up with the curves not being orthogonal. However, with the metric that is provided, it is easy to show that the two curves are orthogonal. If I do the same calculation with the metric assumed to be
$$\left[\begin{array}{cc}1&0\\0&1\end{array}\right]$$
I find that the curves are not orthogonal, which is expected given that they are not orthogonal when I sketch them. Considering that the question involves first sketching the curves and then using the metric to prove that they are orthogonal, I am assuming that the metric can be used to sketch them in such a way as for them to appear orthogonal on the plane...unless the whole point of the question is to show that they are not orthogonal with the standard metric (i.e. how I have sketched them) and then to show that they are orthogonal if the space has the metric given in the problem. I'm really not sure how to approach this.

Thanks in advance.

 

[diazona]

Regardless of what sort of help you're looking for, it'd still be better if you follow the template. It was designed that way because posts that don't follow the template generally don't give us the information we need to make a useful reply.

 

[cazlab]

Okay, thanks. I have edited my above post to provide the question statement in detail.

 

[cazlab]

weird, instead of a matrix at the top it should read t=u sinh v

No idea why it does not display properly, it's like it copied the metric that I displayed further down. It's mental. I can't even put in a line in my first post saying that the matrix is irrelevant there as it just adds more metrics and moves some of the tex around to the wrong spots.

 

[diazona]

weird, instead of a matrix at the top it should read t=u sinh v

No idea why it does not display properly, it's like it copied the metric that I displayed further down. It's mental. I can't even put in a line in my first post saying that the matrix is irrelevant there as it just adds more metrics and moves some of the tex around to the wrong spots.

Actually it did come out properly, it's just a problem with the browser caching images. If you refresh the page then hopefully it should show it the way it really appears.

Anyway, curves will only be perpendicular if they are orthogonal under a Euclidean metric. In order to make these curves look perpendicular, you would have to do a coordinate transformation that converts the metric under which they are orthogonal to a Euclidean metric. In this case I don't believe that's possible.

 

[cazlab]

Ah, thanks. Strangely it's still wrong to me even after refreshing, but as long as it looks right to you guys it's all okay.

Thanks for your response. It sounds then like the only option is to sketch it as I already have. There is more to the question (which I have mostly already done) which is why the metric is given in this part even though it may not necessarily be relevant for the graphical part of the question. I think you've solved my problem though. Thanks again.