Invariance of spacetime interval

In summary: However, for the simpler cases such as this, the algebra will suffice.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.
  • #1
motoroller
29
0
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?
 
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  • #2
I'm trying to visualize your calculations via my paranormal abilities, but somehow I fail.
 
  • #3
haushofer said:
I'm trying to visualize your calculations via my paranormal abilities, but somehow I fail.


My LaTeX isn't so good, but substituting:

[tex]x'=\gamma(x+vt)[/tex]
[tex]t'=\gamma(t+\frac{vx}{c^2})[/tex]

into
[tex](x')^2+(y')^2+(z')^2-c^2(t')^2[/tex]

trying to get
[tex]x^2+y^2+z^2-c^2t^2[/tex]

i.e. invariant interval
 
  • #4
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:
  • #5
motoroller said:
My LaTeX isn't so good, but substituting:

[tex]x'=\gamma(x+vt)[/tex]
[tex]t'=\gamma(t+\frac{vx}{c^2})[/tex]

into
[tex](x')^2+(y')^2+(z')^2-c^2(t')^2[/tex]

trying to get
[tex]x^2+y^2+z^2-c^2t^2[/tex]

i.e. invariant interval

Ok, we try to show that

[tex]
ds^2 = dx^2 - dt^2 = dx'^2 - dt'^2
[/tex]

With your transformation rules, write down

[tex]
dt' = \frac{\partial t'}{\partial x}dx + \frac{\partial t'}{\partial t}dt
[/tex]

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

[tex]
\gamma^2 v^2 - \gamma^2 = 1
[/tex]

This should give that

[tex]
ds^2 = ds'^2
[/tex]


That's all there is :)
 
  • #7
haushofer said:
Did you manage to do the calculation?

Yes - thanks, it was the [tex](1-\beta^2)[/tex] factor I was missing when re-arranging
 
  • #9
Try this - straight from the Lorentz transformation equations:

x' = [tex]\gamma[/tex](x - vt) and x = [tex]\gamma[/tex](x' + vt')

x'/x = (x - vt)/(x' + vt') or x2 - xvt = x'2 - x'vt'
----------------------------------------------------------------------
t' = [tex]\gamma[/tex](t - xv/c2) and t = [tex]\gamma[/tex](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
All of these things get easier when you're used to working with matrices. We want to prove that [itex](x-y)^T\eta(x-y)[/itex] is invariant, i.e. that it's equal to [itex](\Lambda(x-y))^T\eta\Lambda(x-y)[/itex]. 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
Fredrik said:
All of these things get easier when you're used to working with matrices. We want to prove that [itex](x-y)^T\eta(x-y)[/itex] is invariant, i.e. that it's equal to [itex](\Lambda(x-y))^T\eta\Lambda(x-y)[/itex]. 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.

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 [itex](\Lambda(x-y))^T\eta\Lambda(x-y)[/itex] 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.
 

1. What is the concept of invariance of spacetime interval?

The invariance of spacetime interval refers to the idea that the measurement of distance and time between two events in the universe is the same for all observers, regardless of their relative motion. This concept is a fundamental principle in the theory of special relativity.

2. Why is the invariance of spacetime interval important?

The invariance of spacetime interval is important because it is a key principle in understanding the nature of space and time in the universe. It helps us to reconcile the differences in measurements of time and distance between different observers and allows for the development of the theory of special relativity.

3. How is the invariance of spacetime interval mathematically expressed?

The invariance of spacetime interval is expressed mathematically using the equation: Δs² = Δx² + Δy² + Δz² - (cΔt)², where Δs is the spacetime interval, Δx, Δy, and Δz are the spatial distances, c is the speed of light, and Δt is the difference in time between two events.

4. Does the invariance of spacetime interval only apply to objects moving at the speed of light?

No, the invariance of spacetime interval applies to all objects regardless of their speed. However, it becomes more significant and apparent for objects moving at speeds close to the speed of light. This is because at these speeds, the differences in measurements of time and distance become more noticeable.

5. How does the concept of invariance of spacetime interval challenge our intuitive understanding of time and space?

The concept of invariance of spacetime interval challenges our intuitive understanding of time and space because it suggests that our perception of time and space is relative and depends on the observer's frame of reference. It also blurs the distinction between space and time, as they are both considered part of a unified concept of spacetime.

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