Spacetime Inverval Invariance using Lorentz Transformations

In summary, the task is to prove that the spacetime interval -(ct)^2 + x^2 + y^2 + z^2 is invariant, using Lorentz transformations. By simplifying the given formula and collecting terms, it can be shown that the coefficient of x^2 is \gamma^2(1-u^2/c^2), which can be further simplified using the definition of \gamma.
  • #1
maccyjj
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Homework Statement


Prove that the spacetime interval
-(ct)[itex]^{2}[/itex] + x[itex]^{2}[/itex] + y[itex]^{2}[/itex] + z[itex]^{2}[/itex]
is invariant.

[/itex][itex]

Homework Equations


Lorentz transformations
[itex]\Delta[/itex][itex]x' = \gamma(\Delta[/itex][itex]x-u\Delta[/itex][itex]t)[/itex]
[itex]\Delta[/itex][itex]y' = \Delta[/itex][itex]y[/itex]
[itex]\Delta[/itex][itex]z' = \Delta[/itex][itex]z[/itex]
[itex]\Delta[/itex][itex]t' = \gamma(\Delta[/itex][itex]t-u\Delta[/itex][itex]x/c^{2})[/itex]


The Attempt at a Solution


I have tried to prove that [itex]\Delta S = \Delta S'[/itex]
So first I said that [itex]\Delta S' = - \Delta (ct')^{2} + \Delta (x')^{2} + \Delta (y')^{2} + \Delta (z')^{2}[/itex]

And inserted all the Lorentz Transformations above into the above formula.

I end up simplyfying it to get

[itex]\gamma^{2} (x^{2} + u^{2}t^{2} - c^{2}t^{2} - \frac{u^{2}x^{2}}{c^{2}}) + y^{2} + z^{2}[/itex]

How does this equal [itex]S = - \Delta (ct)^{2} \Delta (x)^{2} + \Delta (y)^{2} + \Delta (z)^{2}[/itex] ? I can't see a way to get rid of the extra terms to get this simple function.

Any help would be really really great!
 
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  • #2
If you collect the terms, you'll see that the coefficient of x2 is [itex]\gamma^2(1-u^2/c^2)[/itex]. Use the definition of [itex]\gamma[/itex] to simplify that.
 
  • #3
Oh of course! How did I miss that?

Thank you so much I got it out now!
 

FAQ: Spacetime Inverval Invariance using Lorentz Transformations

What is Spacetime Interval Invariance?

Spacetime Interval Invariance is a fundamental concept in special relativity that states that the measurement of the interval between two events in spacetime is the same for all observers, regardless of their relative motion.

How is Lorentz Transformation related to Spacetime Interval Invariance?

Lorentz Transformations are mathematical equations that describe how measurements of space and time change between two reference frames that are moving at a constant velocity relative to each other. These transformations are necessary to maintain the principle of Spacetime Interval Invariance.

3. Why is Spacetime Interval Invariance important in special relativity?

Spacetime Interval Invariance is important because it is a fundamental principle that helps reconcile the discrepancies between measurements of space and time made by different observers in different reference frames. It also forms the basis for many of the predictions and observations in special relativity, such as time dilation and length contraction.

4. Can Spacetime Interval Invariance be violated?

No, Spacetime Interval Invariance cannot be violated. It is a fundamental principle of special relativity that has been experimentally tested and confirmed. Any observations or measurements that appear to violate this principle are due to incorrect interpretations or assumptions.

5. How is Spacetime Interval Invariance used in practical applications?

Spacetime Interval Invariance is used in many practical applications, such as GPS navigation and particle accelerators. In GPS, it is necessary to account for time dilation effects due to the high speeds of satellites in orbit. In particle accelerators, it helps calculate the energy and momentum of particles moving at near-light speeds.

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