Special Relativity Spacetime Intervals

In summary, the spacetime interval between two events is given by Δs^2 = -c^2(Δt^2) + Δx^2 + Δy^2 + Δz^2. The time component of the spacetime interval is negative because the first one is frame invariant and the second one is not.
  • #1
swashtecher
3
0
Just a quick question on spacetime intervals in spacetime. Why is the spacetime interval between two events given by Δs^2 = -c^2(Δt^2) + Δx^2 + Δy^2 + Δz^2, rather than c^2(Δt^2) + Δx^2 + Δy^2 + Δz^2 (as if it were the distance between two points in four spatial dimensions)? Or more succinctly, why is the time component of the spacetime interval negative?
 
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  • #2
Because the first one is frame invariant (it takes the same value in every frame), whereas the second one is not.

Take the Lorentz Transformation:

[itex]\Delta x'=\gamma (\Delta x-v\Delta t)[/itex]
[itex]\Delta t'=\gamma (\Delta t-v\Delta x/c^2)[/itex]

Now try calculating [itex](\Delta x')^2-(c\Delta t')^2[/itex]. You'll find that [itex](\Delta x')^2-(c\Delta t')^2=(\Delta x)^2-(c\Delta t)^2[/itex].

If you try doing the same thing with [itex](\Delta x')^2+(c\Delta t')^2[/itex] you'll see that [itex](\Delta x')^2+(c\Delta t')^2 \neq (\Delta x)^2+(c\Delta t)^2[/itex]
 
  • #3
The whole point of the spacetime interval is to arrive at something analogous to a distance in space and/or a period of time that comes out the same no matter what frame you do the calculation in. If you use the second formula, it doesn't come out the same in all frames whereas the first one does.
 
  • #4
swashtecher said:
Just a quick question on spacetime intervals in spacetime. Why is the spacetime interval between two events given by Δs^2 = -c^2(Δt^2) + Δx^2 + Δy^2 + Δz^2, rather than c^2(Δt^2) + Δx^2 + Δy^2 + Δz^2 (as if it were the distance between two points in four spatial dimensions)? Or more succinctly, why is the time component of the spacetime interval negative?

elfmotat and ghwellsjr summed it up pretty well. Here's basically the same point they've made using some graphics.
4_D_Metric_SpaceE.jpg

From this point of view you can see why the Pythagorean theorem doesn't work for the blue guy when he tries to compute an invariant value for a red interval.

In a way, the root cause of this situation (from the standpoint of the space-time diagram) has to do with the unique orientation of the blue and red X1 and X4 coordinates. For example, the direction the blue guy moves in the 4-dimensional universe determines the slope of his world line (his X4 coordinate slope with respect to the black rest system).

Now, the mysterious aspect of special relativity is that blue's X1 coordinate (his 3-D cross-section view of the 4-D universe) rotates such that the 45-degree photon world line always bisects the angle between X4 and X1. This is a perfectly general phenomenon, regardless of the observer's speed, holding for any observer moving relative to the black rest system. Accordingly, this explains why the speed of light is always c, for any frame of reference.

So, in a way, you could ask of nature: Why is it that the observer's X1 axis always rotates in that special manner when the X4 rotates? One possible answer that has been suggested is that consciousness can only pick out a cross-section view of the 4-D universe that is intelligible. And The 4-dimensional objects are arranged geometrically so as to present symmetric relationships among objects that are consistent with the laws of physics. Some would argue that this underlies the discovery of laws of physics. And this is consistent with the fact that the laws of physics are the same in all Lorentz frames.
 
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Related to Special Relativity Spacetime Intervals

1. What is the concept of spacetime intervals in special relativity?

Spacetime intervals refer to the distance between two events in both space and time. In special relativity, this distance is a fundamental concept and is used to understand the relationship between space and time.

2. How is the spacetime interval calculated?

The spacetime interval is calculated using the Minkowski metric, which takes into account the differences in space and time coordinates. It is given by the formula: Δs^2 = c^2Δt^2 - Δx^2 - Δy^2 - Δz^2, where c is the speed of light and Δt, Δx, Δy, and Δz are the differences in time and space coordinates between two events.

3. What is the significance of spacetime intervals in special relativity?

Spacetime intervals are significant because they are invariant. This means that they have the same value for all observers, regardless of their relative motion. This is a fundamental principle in special relativity and leads to the concepts of time dilation and length contraction.

4. What is the difference between spacetime intervals in special relativity and classical mechanics?

In classical mechanics, the concept of distance and time are independent of each other. However, in special relativity, the distance and time components are combined into a single spacetime interval. This is due to the fact that space and time are relative and depend on the observer's frame of reference in special relativity.

5. How does the concept of spacetime intervals relate to Einstein's theory of relativity?

Einstein's theory of relativity is based on the idea that the laws of physics are the same for all observers in uniform motion. Spacetime intervals play a crucial role in this theory by providing a way to measure and compare distances and times between events for different observers. This allows for the prediction and understanding of phenomena such as time dilation and the twin paradox.

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