Deriving length contraction using Lorentz invariants

[azure kitsune]

Homework Statement

Derive length contraction using Lorentz invariants.

Homework Equations

$$ds^2 = dx^2 +dy^2 + dz^2 - c^2 dt^2$$

The Attempt at a Solution

Consider a rod parallel to the x-axis and moving with velocity v in the x-direction. We can measure the length of the rod of this frame by measuring the location of the two endpoints simultaneously. Call this frame S'. Then dx' is the length of the rod in frame S', and we define

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

which has the same value in all reference frames. For frame S', since the measurements are simultaneous, we have dt = 0. It follows that

$$ds^2 = (dx')^2$$

Now we switch to frame S in which the rod is at rest. The rod's length here is dx, and we have:

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

where dt is the time between the two measurements in this frame. Then:

$$(dx')^2 = dx^2-c^2dt^2$$

and now we need to get dt out the equation somehow. I tried using time dilation, (dt = γ dτ) followed by the relation ds = -c^2 dτ, followed by ds = dx'. This brings me to the incorrect equation:

$$(1-\gamma^2)(dx')^2 = dx^2$$

That's probably incorrect because I was blindly applying formulas without understanding what I was doing. I believe everything up to

$$(dx')^2 = dx^2-c^2dt^2$$

is correct. Am I right? And if that is correct, how do I continue from there? Thanks!

 

[anathema]

Yes, you are correct up to the equation (dx')^2 = dx^2 - c^2dt^2. From here, we can use the Lorentz transformation equations to relate the coordinates in frame S' to those in frame S. The Lorentz transformation for the x-coordinate is given by:

x = γ(x' + vt')

where γ is the Lorentz factor and t' is the time in frame S'. Plugging this into our equation, we get:

(dx')^2 = (γ(x' + vt'))^2 - c^2dt^2

Expanding and rearranging, we get:

(dx')^2 = γ^2(x')^2 + γ^2v^2t'^2 + 2γ^2vx't' - c^2dt^2

Since we know that dt = γdτ, we can substitute this in and get:

(dx')^2 = γ^2(x')^2 + γ^2v^2(γdτ)^2 + 2γ^2vx'(γdτ) - c^2(γdτ)^2

Simplifying, we get:

(dx')^2 = γ^2[(x')^2 - c^2dτ^2]

But we also know that ds^2 = (dx')^2 - c^2(dt')^2, so we can substitute this in and get:

ds^2 = γ^2[(x')^2 - c^2dτ^2] - c^2(dt')^2

Since the length of the rod in frame S is dx, we can rewrite the equation as:

(ds')^2 = γ^2[(dx')^2 - c^2(dt')^2] = γ^2[(dx)^2 - c^2(dt)^2] = ds^2

Therefore, the length of the rod in frame S' (dx') is related to the length in frame S (dx) by the Lorentz factor, γ. This is the length contraction formula, and it shows that as an object moves with velocity v, its length in the direction of motion shrinks by a factor of γ. I hope this helps!