Return to Answer

It is possible, but not in special relativity (because of the answer given by Youloush). Use a 2-dimensional space-time diffeomorphic to $\mathbb{R}^2$ and define a pseudo-riemannian metric (using coordinates for the tangent space from the flat case):

$g_{t,x}((a,b), (c,d))=(1+x^4)\cdot a\cdot c-b\cdot d$

The coordinate $t$ is identical to the proper time of an observer resting at point $0$. Now consider the proper time of an observer moving with constant velocity (with respect to our coordinates) from $(0,0)$ to $(2,2)$ and back to $(4,2)$ (he starts at the speed of light, but we do not have to worry, we would get the same result when we would start a little bit slower). The proper time elapsed is:

$2\cdot\int_0^2 \sqrt{(1+t^4)-1}\mathrm{d}t=2\cdot\int_0^2 t^2\mathrm{d}t=\frac{16}{3}$

This number is larger than $4$ (the proper time elapsed for the resting observer).

However, you can generalize the result from special relativity: Geodesics are those paths extremizing the elapsed proper time. In the most typical cases you actually maximise the elapsed proper time (for example in flat space-time, where the geodesics are given by movements at constant velocity, look at the example above: you move far away (to 2), where the metric is very different, to get the effect, if you would stay within $(-1,1)$, the effect would not happen and it would be like in the flat case: dilatation). Then on every non-geodesic path you get a smaller proper time. However, in general it is not the case.

It is possible, but not in special relativity (because of the answer given by Youloush). Use a 2-dimensional space-time diffeomorphic to $\mathbb{R}^2$ and define a pseudo-riemannian metric (using coordinates for the tangent space from the flat case):

$g_{t,x}((a,b), (c,d))=(1+x^4)\cdot a\cdot c-b\cdot d$

The coordinate $t$ is identical to the proper time of an observer resting at point $0$. Now consider the proper time of an observer moving with constant velocity (with respect to our coordinates) from $(0,0)$ to $(2,2)$ and back to $(4,2)$ (he starts at the speed of light, but we do not have to worry, we would get the same result when we would start a little bit slower). The proper time elapsed is:

$2\cdot\int_0^2 \sqrt{(1+t^4)-1}\mathrm{d}t=2\cdot\int_0^2 t^2\mathrm{d}t=\frac{16}{3}$

This number is larger than $4$ (the proper time elapsed for the resting observer).

However, you can generalize the result from special relativity: Geodesics are those paths (locally) extremising the elapsed proper time. In the most typical cases you actually maximise the elapsed proper time (for example in flat space-time, where the geodesics are given by movements at constant velocity, look at the example above: you move far away (to 2), where the metric is very different, to get the effect, if you would stay within $(-1,1)$, the effect would not happen and it would be like in the flat case: dilatation). Then on every non-geodesic path you get a smaller proper time. However, in general it is not the case.

It is possible, but not in special relativity (because of the answer given by Youloush). Use a 2-dimensional space-time diffeomorphic to $\mathbb{R}^2$ and define a pseudo-riemannian metric (using coordinates for the tangent space from the flat case):

$g_{t,x}((a,b), (c,d))=(1+x^4)\cdot a\cdot c-b\cdot d$

The coordinate $t$ is identical to the proper time of an observer resting at point $0$. Now consider the proper time of an observer moving with constant velocity (with respect to our coordinates) from $(0,0)$ to $(2,2)$ and back to $(4,2)$ (he starts at the speed of light, but we do not have to worry, we would get the same result when we would start a little bit slower). The proper time elapsed is:

$2\cdot\int_0^2 \sqrt{(1+t^4)-1}\mathrm{d}t=2\cdot\int_0^2 t^2\mathrm{d}t=\frac{16}{3}$

This number is larger than $4$ (the proper time elapsed for the resting observer).

However, you can generalize the result from special relativity: Geodesics are those paths extremizing the elapsed proper time. In the most typical cases you actually maximise the elapsed proper time (for example in flat space-time, where the geodesics are given by movements at constant velocity). Then on every non-geodesic path you get a smaller proper time. However, in general it is not the case.

It is possible, but not in special relativity (because of the answer given by Youloush). Use a 2-dimensional space-time diffeomorphic to $\mathbb{R}^2$ and define a pseudo-riemannian metric (using coordinates for the tangent space from the flat case):

$g_{t,x}((a,b), (c,d))=(1+x^4)\cdot a\cdot c-b\cdot d$

The coordinate $t$ is identical to the proper time of an observer resting at point $0$. Now consider the proper time of an observer moving with constant velocity (with respect to our coordinates) from $(0,0)$ to $(2,2)$ and back to $(4,2)$ (he starts at the speed of light, but we do not have to worry, we would get the same result when we would start a little bit slower). The proper time elapsed is:

$2\cdot\int_0^2 \sqrt{(1+t^4)-1}\mathrm{d}t=2\cdot\int_0^2 t^2\mathrm{d}t=\frac{16}{3}$

This number is larger than $4$ (the proper time elapsed for the resting observer).

However, you can generalize the result from special relativity: Geodesics are those paths extremizing the elapsed proper time. In the most typical cases you actually maximise the elapsed proper time (for example in flat space-time, where the geodesics are given by movements at constant velocity, look at the example above: you move far away (to 2), where the metric is very different, to get the effect, if you would stay within $(-1,1)$, the effect would not happen and it would be like in the flat case: dilatation). Then on every non-geodesic path you get a smaller proper time. However, in general it is not the case.

It is possible, but not in special relativity (because of the answer given by Youloush). Use a 2-dimensional space-time diffeomorphic to $\mathbb{R}^2$ and define a pseudo-riemannian metric (using coordinates for the tangent space from the flat case):

$g_{t,x}((a,b), (c,d))=(1+x^4)\cdot a\cdot c-b\cdot d$

The coordinate $t$ is identical to the proper time of an observer resting at point $0$. Now consider the proper time of an observer moving with constant velocity (with respect to our coordinates) from $(0,0)$ to $(2,2)$ and back to $(4,2)$ (he starts at the speed of light, but we do not have to worry, we would get the same result when we would start a little bit slower). The proper time elapsed is:

$2\cdot\int_0^2 \sqrt{(1+t^4)-1}\mathrm{d}t=2\cdot\int_0^2 t^2\mathrm{d}t=\frac{16}{3}$

This number is larger than $4$ (the proper time elapsed for the resting observer).

It is possible, but not in special relativity (because of the answer given by Youloush). Use a 2-dimensional space-time diffeomorphic to $\mathbb{R}^2$ and define a pseudo-riemannian metric (using coordinates for the tangent space from the flat case):

$g_{t,x}((a,b), (c,d))=(1+x^4)\cdot a\cdot c-b\cdot d$

The coordinate $t$ is identical to the proper time of an observer resting at point $0$. Now consider the proper time of an observer moving with constant velocity (with respect to our coordinates) from $(0,0)$ to $(2,2)$ and back to $(4,2)$ (he starts at the speed of light, but we do not have to worry, we would get the same result when we would start a little bit slower). The proper time elapsed is:

$2\cdot\int_0^2 \sqrt{(1+t^4)-1}\mathrm{d}t=2\cdot\int_0^2 t^2\mathrm{d}t=\frac{16}{3}$

This number is larger than $4$ (the proper time elapsed for the resting observer).

However, you can generalize the result from special relativity: Geodesics are those paths extremizing the elapsed proper time. In the most typical cases you actually maximise the elapsed proper time (for example in flat space-time, where the geodesics are given by movements at constant velocity). Then on every non-geodesic path you get a smaller proper time. However, in general it is not the case.