# Invariance of spacetime interval

1. May 31, 2010

### motoroller

I've tried proving the invariance of the spacetime interval from Lorentz transformations 3 times now, but every time I end up with two extra terms that don't cancel! Could I have some help?

2. May 31, 2010

### haushofer

I'm trying to visualize your calculations via my paranormal abilities, but somehow I fail.

3. May 31, 2010

### motoroller

My LaTeX isn't so good, but substituting:

$$x'=\gamma(x+vt)$$
$$t'=\gamma(t+\frac{vx}{c^2})$$

into
$$(x')^2+(y')^2+(z')^2-c^2(t')^2$$

trying to get
$$x^2+y^2+z^2-c^2t^2$$

i.e. invariant interval

4. May 31, 2010

### Cyosis

Yes you told us that in your first post in words. What you need to show us is your entire calculation, otherwise we cannot possibly know where you went wrong or what these 'extra' terms are.

Hint: It will become a lot more obvious if you work in units where c=1.

Last edited: May 31, 2010
5. May 31, 2010

### haushofer

Ok, we try to show that

$$ds^2 = dx^2 - dt^2 = dx'^2 - dt'^2$$

With your transformation rules, write down

$$dt' = \frac{\partial t'}{\partial x}dx + \frac{\partial t'}{\partial t}dt$$

and a same expression for dx'. Now calculate the interval ds^2 in these primed coordinates. The cross terms cancel because of the minus sign in the interval, and you can check that

$$\gamma^2 v^2 - \gamma^2 = 1$$

This should give that

$$ds^2 = ds'^2$$

That's all there is :)

6. Jun 1, 2010

### haushofer

Did you manage to do the calculation?

7. Jun 1, 2010

### motoroller

Yes - thanks, it was the $$(1-\beta^2)$$ factor I was missing when re-arranging

8. Jun 2, 2010

### haushofer

Ok, glad we could help :)

9. Jun 2, 2010

### stevmg

Try this - straight from the Lorentz transformation equations:

x' = $$\gamma$$(x - vt) and x = $$\gamma$$(x' + vt')

x'/x = (x - vt)/(x' + vt') or x2 - xvt = x'2 - x'vt'
----------------------------------------------------------------------
t' = $$\gamma$$(t - xv/c2) and t = $$\gamma$$(t' + x'v/c2)

t'/t = (t - xv/c2)/(t' - x'v/c2) or t2 - xvt/c2 = t'2 + x'vt'/c2.

mulitply this last equation by c2: c2t2 - xvt = c2t'2 + x'vt'

now subtract this last equation from the above equation above the dotted line and you get:

x2 - c2t2 = x'2 - c2t'2

or, the other way:

c2t2 - x2 = c2t'2 - x'2

This establishes the hyperbolic relationship between time and distance. Again, this is the space interval or time interval (depending on which is positive).

This is as simple an algebraic derivation of this that you can get.

10. Jun 2, 2010

### Fredrik

Staff Emeritus
All of these things get easier when you're used to working with matrices. We want to prove that $(x-y)^T\eta(x-y)$ is invariant, i.e. that it's equal to $(\Lambda(x-y))^T\eta\Lambda(x-y)$. So we stare at it for two seconds and realize that the equality follows immediately from the definition of a Lorentz transformation and a trivial fact about the transpose of a product.

I have said before that I think you can learn SR and matrices in less time than you can learn just SR, and I still think that's right.

11. Jun 3, 2010

### stevmg

True, but this requires the acquisition of a working knowledge of matrix algebra. I did learn a lot of that years ago but have lost my working knowledge of this long ago and have no current knowledge of matrix operations. Thus, using the basic algebra is a little simpler for me. To wit, I don't even know what $(\Lambda(x-y))^T\eta\Lambda(x-y)$ means yet I can still derive the basic hyperbolic relationship between t and x without it.

Of course in more complex matters, one must use the matrix notation and operations to get somewhere.