Understanding the Invariancy of Space-Time Interval Equations with Mathematical Proof!!!

 

In the last article, I explained about Lorentz Transformation in a very simple way and in this article, we are going use that beast.

In this article, I going to explain the invariancy of Spacetime interval equations in a mathematical way.

But before you continue, you have to read my articles on Introduction to Space-Time and Lorentz Transformation to understand this article in a better way.

Let’s get started….

As I have explained in my article on Introduction to Space-time, as it sounds, made up of two critical words, space and time. Space, as per the traditional physics, represents the 3-dimensional coordinate system of our universe. Scientists before Einstein believed that our universe is limited to these 3 coordinates (x, y and z-axis) and the universe according to this convention is called Newtonian. But, in 1905, Einstein showed in his Special Theory of relativity that time is also an integral factor. Well, it was not before 1908, when one of the Einstein’s own professor and renowned physicist Herman Minkowski proposed the concept of the union of space and time as a way to reformulate Special Relativity and by this way, the concept of space-time was born.

To derive the equations to prove its invariancy, we need to understand the above picture. Here in this picture, there are two coordinate systems, S and S’. Space and time coordinate of S are x, y, z and t, and for S’ they are x’, y’, z’ and t’.

Now, suppose, in the course of motion of S’ relative to S in the x-direction with a speed v, a flash of light is emitted at the origins of S and S’ when they were just coincident. Let’s identify that instant as t=0=t’.

Using this picture, I derived the Lorentz transformation equations in my Introduction to Lorentz Transformation article and in this article, we are going to find the space-time interval equation using it.

Proof of Invariancy

As we all know,

Time = Distance / Velocity

t = √ (x² + y² + z²) / c

c² t² = x² + y² + z²    (1)

Distance = √ (x² + y² + z²), because it has 3 axes.

Now, to prove invariancy, we need to show,

x² + y² + z² – c² t² = x’² + y’²+ z’² – c² t’²    (2)

Why??? Because invariancy means in every condition, the space-time interval equation in both S and S’ will be equal. Which means that we can write the space-time interval equation for S and S’ plane and equate them as we did in equation 2.

Now, proving LHS = RHS,

Where, LHS = x² + y² + z² – c² t²     (space-time interval equation in S plane)

And, RHS =  x’² + y’² + z’² – c² t’²   (space-time interval equation in S' plane)

Now, using Lorentz Transformations in RHS,

 

[x – vt / (√ 1 – v²/c²)]² + y² + z² – c² [(t – vx/c²) / (√ 1 – v²/c²)]²     (3)

Since,

x’ = x – vt / (√ 1 – v²/c²)

y’ = y   

z’ = z      

t’ = (t – vx/c²) / (√ 1 – v²/c²)

Now if you solve equation 3, you will get,

x² + y² + z² – c² t²

which is our LHS.

Thus, LHS = RHS and invariancy of the space-time interval is proved.

Same can also be proved by inverse Lorentz Transforms and putting those in LHS and then deriving RHS from it. The choice is yours always!!!

So, that’s its guys…that’s all about Space-time intervals and its invariancy.

In my upcoming articles, I will write about 4-vectors, tensors etc. are very important to understand relativity and cosmology in general. So, I will be writing about these gems, very soon…

Once these articles will be available links will be provided here...

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                                                                 -Ratnadeep Das Choudhury

                                                  Founder and writer of The Dynamic Frequency