Why is the longest worldline the unaccelerated one?

[thegeniusrule]

Homework Statement

How does one prove that, of all timelike worldlines connecting two events A and B, the accelerated one is the longest.

Homework Equations

dT^2 = dt^2 - dx^2
(accelerated reference frames:
t = (sinh aT)/a
x = (cosh aT)/a
c = speed of light = 1)

The Attempt at a Solution

If there is no acceleration, then x = vt, therefore T = t(1 - v^2)^0.5
I don't know where to go from there.

Conceptually, I understand this, however, I can't seem to prove this with mathematics. Any ideas?

 

[anathema]

Thank you for your question. To prove that the accelerated worldline is the longest, we need to compare it to other timelike worldlines connecting the same two events A and B. Let's call the accelerated worldline W1 and another timelike worldline W2.

First, let's consider the equation for proper time, dT^2 = dt^2 - dx^2. This equation represents the spacetime interval between two events, and it is invariant for all observers. This means that no matter what reference frame we are in, we will always get the same value for the spacetime interval.

Now, let's look at the accelerated reference frame described in the problem: t = (sinh aT)/a and x = (cosh aT)/a. We can rewrite this as T = arcsinh(at) and x = arccosh(at). Note that this reference frame is only valid for t > 0, as we cannot have imaginary values for time.

Next, let's consider the other timelike worldline W2. We can write the equation for this worldline as x = vt, where v is the velocity of the object following this worldline. We can also rewrite this as t = x/v. Now, let's substitute this into our equation for proper time: dT^2 = dt^2 - dx^2 = (1 - v^2)dt^2 - dx^2. We can see that this equation is similar to the one for the accelerated worldline, except for the factor of (1 - v^2) in front of the dt^2 term.

To compare the proper times for these two worldlines, we can take the ratio of the two equations: dT1^2/dT2^2 = [(1 - v^2)dt^2 - dx^2]/[(1 - at)^2dt^2 - dx^2]. Simplifying this expression, we get dT1^2/dT2^2 = (1 - v^2)/(1 - at)^2. We can see that this expression is always greater than or equal to 1, as both (1 - v^2) and (1 - at)^2 are always positive. This means that the proper time for the accelerated worldline, dT1, is always greater than or equal to the proper time for the other worldline, d