# Are worldlines/geodesics coordinate/frame independent

My professor states in the lecture that geodesics are frame independent.

But to me geodesics are just special type of wordlines and worldlines aren't coordinate or frame independent. In the rest frame of the particle the worldline on the Minkowski diagram would be along the time axis but in a moving frame they wouldn't be.

Even a geodesic of a free particle would be a straight line in inertial (cartesian or polar coordinates) but in a rotating or accelerating frame they wouldn't be straight lines.

So why did my professor state that geodesics are frame/coordinate independent.

Will the geodesics/worldlines be (look like ) the same in all (inertial or accelerating or whatever) frames (coordinate system).

Is my understanding wrong. Is what I have said above not correct or does the professor mean something else.

• A geodesic is always a straight line locally. How you plot it in your coordinates is a different matter. I think that is what your professor is trying to tell you. – Apoorv Dec 31 '20 at 21:31

It depends on what you mean here. I'm fairly certain that what your instructor meant was that whether a worldline is or is not a geodesic is a frame-independent notion. This is because geodesics $$\gamma$$ extremize the path length functional: $$S[\gamma] = \int d\lambda \sqrt{ g(\dot \gamma,\dot \gamma)}$$ Nothing in the definition of $$S$$ requires a choice of coordinates, so whether $$\gamma$$ extremizes $$S$$ or not doesn't depend on your choice of coordinates.

A different way to see this is to note$$^\dagger$$ that geodesics are solutions to the autoparallel equation $$\nabla_{\dot \gamma} \dot \gamma = 0$$ which is a tensor equation, so once again coordinate-independent.

Of course, the coordinates of the points along that worldline will clearly depend on which coordinate chart you use.

$$^\dagger$$This assumes that you have chosen the Levi-Civita connection, which we typically do in GR.

• So a curved/straight worldline (in SR or GR) in one frame/coordinates would be curved/straight (drawn) differently in another coordinate/frame. And this holds for geodesics too. The same solutions of geodesic equation when drawn using one coordinate/frame would be (say a straight line in this coordinate/frame) would be **drawn differently if drawn using another coordinate/frame ( maybe a curved line in this coordinate/frame). Is this understanding correct. Please let me know this. And what I understood that the lecturer meant lthat a geodesic would be a geodesic in any frame. – Shashaank Jan 1 at 6:11
• While letting me know what I have said above is correct or not please read "SEEN as or drawn as" where ever I have just written "drawn as". – Shashaank Jan 1 at 6:21
• The descriptor "straight" in this context means "autoparallel" - that is, $\nabla_{\dot \gamma} \dot \gamma =0 \implies \ddot \gamma+ \Gamma^\mu_{\nu\rho} \dot \gamma^\mu \dot \gamma^\rho = 0$. If that is what you mean, then yes, your understanding is correct. – J. Murray Jan 1 at 7:44
• Maybe, I can be more precise. Suppose, in GR, I solve the geodesic equation in some frame. I get a solution, say a parabola(so in those coordinates, the plot would be a parabola/ see the particle moving in parabola). No I try to solve the geodesic equation in another frame(an accelerated one). Then in this new frame will I get a parabola again or some other curve( so plotting the solution in this new frame will be a parabola or some other curve). Please let me know this. I am thinking it should be another curve because worldlines depend on the coordinates you plot them in. Is that correct. – Shashaank Jan 1 at 9:34
• cont.. -worldline of a particle at rest would be parallel to the time axis but the worldline of the same particle in another moving frame wouldn't be parallel to the time axis. It would have changed. And geodesics are just a special kind of a worldline. I am thinking along these lines. So I said the geodesic which is a parabola in one frame shouldn't be a parabola in another frame/coordinates. Is this wrong... – Shashaank Jan 1 at 9:48

The statement you have attributed to your professor is not very rigorous on its own. It is trying to address a common source of confusion amongst students.

Imagine a world without coordinates first. Say you have a ball moving in absence of any forces. GR tells us that it will move along a straight line through spacetime. In the frame attached to the ball there is no motion in space. Time passes uniformly as $$\tau$$ (proper time). The change in its four velocity ($$\mathbf{u}$$) with respect to proper time is $$\mathbf{0}$$ as this ball is a free particle. The following is thus its geodesic equation: $$\frac{\mathrm{d}\mathbf{u}}{\mathrm{d}\tau} = \mathbf{0}$$

Now you sitting somewhere else record this ball's position in space as a function of time. You are therefore introducing a coordinate system. The ball is still moving in a straight line but your coordinate system could be flat or curved depending upon the energy-momentum distribution.

Take a non GR example before we attempt to formalize this further. Imagine an airplane flying the shortest distance from A to B. If you zoom in sufficiently (in other words, just look locally and in formal terms look in the infinitesimal subspace around the plane's position at any point on its trajectory) the velocity vector is parallel to the trajectory. Which implies no change in trajectory. If you now plot this over a globe you get a great circle. If you plot this on Mercator projection you get a strange curve.

Formally, if your coordinate basis is given by $$\mathbf{e}_\alpha$$ and $$\Gamma^\alpha_{\beta\gamma}$$ are the Christoffel symbols for your coordinate system then you can simplify the geodesic equation using the chain rule as follows:

$$\frac{\mathrm{d}\mathbf{u}}{\mathrm{d}\tau} = \frac{\mathrm{d}\mathbf{u}^\alpha \mathbf{e}_\alpha}{\mathrm{d}\tau} = \mathbf{u}^\alpha \frac{\mathrm{d}\mathbf{e}_\alpha}{\mathrm{d}\tau} + \mathbf{e}_\alpha \frac{\mathrm{d}\mathbf{u}^\alpha}{\mathrm{d}\tau} = \mathbf{u}^\alpha \Gamma^\gamma_{\alpha \beta} \mathbf{u}^\beta \mathbf{e}_\gamma + \mathbf{e}_\alpha \frac{\mathrm{d}\mathbf{u}^\alpha}{\mathrm{d}\tau} = \mathbf{u}^\alpha \Gamma^\gamma_{\alpha \beta} \mathbf{u}^\beta \mathbf{e}_\gamma + \mathbf{e}_\gamma\frac{\mathrm{d}\mathbf{u}^\gamma}{\mathrm{d}\tau} = \mathbf{0}$$

Rearranging you get the geodesic equation in the coordinate form.

$$\frac{\mathrm{d}\mathbf{u}^\gamma}{\mathrm{d}\tau} = - \Gamma^\gamma_{\alpha \beta} \mathbf{u}^\alpha \mathbf{u}^\beta$$

Quite clearly, this is a frame dependent equation. But on its own the geodesic is still unchanged. How you plot it has changed.

• This a very nice answer, thus the upvote. So if a geodesic in one frame is (drawn as or seen as) one curve (say a straight line or any other curve) then if it is drawn or seen in another coordinate/frame, then it will a different curve. This statement is correct. Let me know this, please. I understood that a geodesic would be a geodesic in any frame just that it will be drawn or seen differently in different frames. Is this understanding correct. Please let me know this. – Shashaank Jan 1 at 6:18
• When you say " The ball still moves in a straight line.... " . But worldlines for the same particle are different in different frames. In a rest frame the worldlines would be parallel to time axis but in a moving inertial frame it won't be. – Shashaank Jan 1 at 9:39
• @Shashaank you are right in your first comment in that if a particle is moving on a geodesic (formally, its trajectory satisfies the geodesic equation) in one frame then it will satisfy the geodesic equation in all other frames. The shape of the curve will of course depend on the observer. When I say the ball moves in straight line, I mean straight as defined for that coordinate system and when viewed locally (i.e., in the infinitesimal subspace at any point in spacetime). There is no requirement for unaccelerated particles to move in straight lines in GR when you zoom out. – Apoorv Jan 1 at 10:23
• So that means if I solve (GR) the geodesic equation in one frame , I will get a curve, say a parabola. That is I will see that particle move in a parabola ( I will plot a parabola). Now if another observer, in another frame (say an accelerated one etc) observes that particle then will not see the particle as moving in a parabola, but he will plot another curve for the particle which will nit be a parabola. Is that right. The solution to the geodesic eqn was a parabola in one frame and will something different. If that is correct why do we say geodesics on a sphere are circles. – Shashaank Jan 1 at 11:04
• They will be circle in one frame and another curve in another coordinate – Shashaank Jan 1 at 11:05

The way physicists generally talk is that things like scalars, vectors, and tensors are classified according to their transformation properties. However, it's understood by the people talking this way that there is an equally valid point of view in which something like a momentum vector is an invariant quantity, but its components have transformation properties that depend on what coordinate system you pick.

Even a geodesic of a free particle would be a straight line in inertial (cartesian or polar coordinates) but in a rotating or accelerating frame they wouldn't be straight lines.

Changing the coordinates doesn't change whether it's curved. A solution to the geodesic equation is still a solution regardless of the coordinates you use.