Determining components of 4-velocity

[Ned Stark]

Homework Statement

A particle of mass m follows the spacetime trajectory x(τ) = (Aτ, Bτ , C cosτω, C sinτω ).

What is the 4-velocity u(τ)? Why must A =sqrt[1+B^2+(ωC)^2]

Homework Equations

u(τ)=dx(τ)/dτ , u^2 = -c^2

The Attempt at a Solution

so I took the derivative of each comp of x with respect to tau and got:

u(τ)=(A, B , -ωCsinτω, ωCcosτω)

and 4-velocities always square to -c^2

so -A^2 +B^2 +(ωC)^2 = -c^2, which gives A =sqrt[c^2+B^2+(ωC)^2]

Why is my answer wrong?

 

[anathema]

Your answer is incorrect because you have squared the components of the 4-velocity. The 4-velocity is a vector, so it should not be squared. The correct answer is u(τ) = (A, B, -ωCsinτ, ωCcosτ). This satisfies the condition u^2 = -c^2, as required.

As for why A =sqrt[1+B^2+(ωC)^2], this is because the magnitude of the 4-velocity is always equal to c, the speed of light. So, we have:

u^2 = -c^2

(A^2 - B^2 - (ωC)^2) = -c^2

A^2 - B^2 - (ωC)^2 + c^2 = 0

Using the Pythagorean theorem, we can rewrite this as:

(A^2 + B^2 + (ωC)^2) = c^2

And since the magnitude of the 4-velocity is always equal to c, we have:

A^2 + B^2 + (ωC)^2 = c^2

Which leads to:

A = sqrt[1 + B^2 + (ωC)^2]

This is why A must equal sqrt[1 + B^2 + (ωC)^2] in order for the 4-velocity to satisfy the condition u^2 = -c^2.