Do you accelerate through time when you stand still on Earth?

[Grasshopper]

TL;DR Summary

Since you don't appear to have "coordinate" acceleration, yet still feel the ground pressing up against you, are you only accelerating through time, but not space?

Disclaimer: I'm not actually sure which acceleration is proper and which is coordinate, and I can't recall the source for the half-remembered equation. I spent some time going through my bookmarks, but it was to no avail. Sorry about that.

So, general relativity removes the idea of gravity as a force, and instead, when you're standing on the ground (and therefore not in free fall), you have a force upon you (normal force), but no counter force balancing it. That must mean you are accelerating. However, you don't appear move through space. Of course you clearly are "moving" through time.

Could it be that the acceleration you experience is entirely through time?
Why do I feel this way? I saw an equation that looked something like this:

$$ \frac{dr^2}{dτ^2}= a^r - Γ^r u^2 = 0$$

I've neglected the subscripts and wrote pseudomath here, but the point is, it looks like total acceleration equals a particular type of acceleration minus Christoffel symbols times the square of a four velocity. I don't have the source at the moment, but if I recall correctly, the only component that mattered in this particular velocity was the time component. I believe the others were zero (which makes sense since you're not moving through space in your own reference frame).So, interpreting this pseudomath, it seems to say that you ARE feeling an acceleration, but you're not moving through space because your felt acceleration is canceled out by that term which includes the time component of the four-velocity squared (a spacetime curvature term, perhaps?). It would appear, then, that your motion is all tied up in motion through time, and it cancels out the acceleration, meaning you don't change spatial coordinates.
How far off is that? Please feel free to correct and instruct as much as you want. I don't mind seeing sophisticated math so long as you're not writing paragraphs in it without explanation as you go. Thanks as always!

 

[PeroK]

"Acceleration through time" has no meaning. Instead, acceleration entails changing your inertial reference frame. But does not imply motion in the direction of the acceleration. Consider, for example, uniform circular motion. The acceleration vector points to the centre of the circle, but there is no velocity component in that direction.

When accelerating in flat spacetime, you move relative to any inertial reference frame. But, in your rest frame (which is non-inertial) you do not move - by definition.

The situation on the surface of the Earth is similar. Your rest frame is an accelerating frame - but you do not move relative to it - again by definition.

Note that there is no sense of absolute space. In all these scenarios you are only moving through space (or not) relative to some reference frame. There is no such thing as abolutely moving through space. In any accelerating frame, you have a measurable proper acceleration, but there is no measureable absolute velocity - hence no concept of absolutely moving through space.

 

[Ibix]

In addition to PeroK's comment, four acceleration is always a spacelike vector. It's perpendicular to your worldline's tangent vector. When you are stationary on the Earth's surface your worldline is parallel to the timelike Killing vector, so your four acceleration is about as close to "definitely nothing to do with time" as you'll ever get in relativity.

 

[Grasshopper]

I'll have to digest these responses, but, I was wondering if anyone be willing to break down what it each part of the actual equation that I saw means. Here it is written fully as I saw it:

$$\frac{dr^2}{dτ^2}= a^r - Γ_{tt}^r (u^{t})^2 $$

It seems obvious that in flat spacetime, the last term doesn't exist. But why does u only have time components? (I'm assuming this is a special case).

I've seen a tensor before, but I'm still struggling to understand that difference between a tensor and a vector. And I've not really messed around with Einstein summation convention. However, I'd still be interested in trying to grasp the equation if someone wants to spend the time telling what the parts mean, physically speaking.

 

[PeroK]

I'll have to digest these responses, but, I was wondering if anyone be willing to break down what it each part of the actual equation that I saw means. Here it is written fully as I saw it:

$$\frac{dr^2}{dτ^2}= a^r - Γ_{tt}^r (u^{t})^2 $$

It seems obvious that in flat spacetime, the last term doesn't exist. But why does u only have time components? (I'm assuming this is a special case).

I've seen a tensor before, but I'm still struggling to understand that difference between a tensor and a vector. And I've not really messed around with Einstein summation convention. However, I'd still be interested in trying to grasp the equation if someone wants to spend the time telling what the parts mean, physically speaking.

Have you been watching this:

 

[Ibix]

But why does u only have time components?

Your four velocity is the rate of change of your coordinates with respect to your proper time. If you are on the surface of a non-rotating planet and using Schwarzschild coordinates then your $r$, $\theta$ and $\phi$ coordinates aren't changing.

 

[PeterDonis]

Here it is written fully as I saw it

Where did you see it? Please give a reference.

 

[Grasshopper]

Have you been watching this:

Yes that is the one I got the equation from. Been watching a ton of videos on it lately.

Where did you see it? Please give a reference.

It's the video PeroK posted. I had watched like 15 on the topic yesterday and forgot to bookmark, but I did find it again after digging through history.

 

[PeroK]

Yes that is the one I got the equation from. Been watching a ton of videos on it lately.

Are you not impressed that I figured that out?

 

[Grasshopper]

Are you not impressed that I figured that out?

I am quite impressed. You found it easier than I did when I had already watched it earlier in the day. ;)

EDIT- for the record I didn't click on your vid (which was my original source) until after I already posted. :D

 

[Grasshopper]

Your four velocity is the rate of change of your coordinates with respect to your proper time. If you are on the surface of a non-rotating planet and using Schwarzschild coordinates then your $r$, $\theta$ and $\phi$ coordinates aren't changing.

In retrospect this seems kind of obvious, but just the concept of objects feeling an acceleration opposite of each other yet not moving through space away from each other is really weird to me (objects on opposite sides of the world, for example).

 

[PeroK]

I'll have to digest these responses, but, I was wondering if anyone be willing to break down what it each part of the actual equation that I saw means. Here it is written fully as I saw it:

$$\frac{dr^2}{dτ^2}= a^r - Γ_{tt}^r (u^{t})^2 $$

This comes from the Geodesic Equation specifically for the curved spacetime near a spherical mass in Schwarzschild coordinates. And, for an observer falling radially inward. One reason not to read too much physical significance into the fact that it is the time parameter of the four-velocity $u^t$ is that this equation would look different in different coordinates. Although it would still say the same physical thing.

In particular, in any spacetime the choice of the time coordinate is arbitrary. There is no such thing as a unique absolute time coordinate. For example, if you use these Schwarzschild coodinates to study infalling all the way into a black hole, then there is a coordinate singularity at the event horizon. You are forced, therefore, to switch to an alternative coordinate system if you want to analyse things below the event horizon. E.g. the wonderfully named Eddington-Finkelstein coordinates.

The above equation would look very different in these coordinates and any talk of "accelerating through time" would be seen to be coordinate dependent and hence physically meaningless.

Note that you mustn't confuse the coordinate singularity at the event horizon with the physical singuarity at the centre of a black hole. That one doesn't go away no matter what coordinates you choose.

 

[A.T.]

...the concept of objects feeling an acceleration opposite of each other yet not moving through space away from each other is really weird to me (objects on opposite sides of the world, for example).

Note, that even in Newtonian mechanics you can have objects with proper acceleration towards each other, yet not moving towards each other (objects on opposite sides of a turn table).

General Relativity allows the same (without rotation) in the opposite direction (away from each other). This is only possible if spacetime is curved between the objects.

This is what is looks like locally in a frame at the surface (note that there is no intrinsic spacetime curvature yet):

And this is the global picture that connects the local geometries, and has intrinsic spacetime curvature:

Note that the free falling wordline always deviates towards the "more stretched" proper time, which corresponds to greater gravitational time dilation (along its gradient). Gravitational time dilation has an extreme point at the center of the mass (gradient is zero), so there is no gravity there (but the maximal gravitational time dilation).

 

[cianfa72]

Your four velocity is the rate of change of your coordinates with respect to your proper time. If you are on the surface of a non-rotating planet and using Schwarzschild coordinates then your $r$, $\theta$ and $\phi$ coordinates aren't changing.

Does that definition of 4-velocity apply regardless the chosen coordinate system ?

 

[Ibix]

Does that definition of 4-velocity apply regardless the chosen coordinate system ?

The first sentence you quoted, yes: Your four velocity is the rate of change of your coordinates with respect to your proper time, although it does assume that your basis vectors are a coordinate basis. I don't think you can have a coordinate system where you can't define a coordinate basis, although you may choose not to use it.

The second sentence depends on your coordinate system.

 

[cianfa72]

although it does assume that your basis vectors are a coordinate basis. I don't think you can have a coordinate system where you can't define a coordinate basis

If you have a coordinate system then you also have a coordinate basis, don't you ?

The second sentence depends on your coordinate system.

Do you mean the following ?

If you are on the surface of a non-rotating planet and using Schwarzschild coordinates then your $r$, $\theta$ and $\phi$ coordinates aren't changing.

 

[Ibix]

If you have a coordinate system then you also have a coordinate basis, don't you ?

One can be defined in all cases as far as I'm aware. You aren't obligated to use it.

Do you mean the following ?

Yes. That isn't part of the definition of a four velocity, but you quoted it so I was making clear that it was coordinate dependent.

 

[cianfa72]

One can be defined in all cases as far as I'm aware. You aren't obligated to use it.

Not sure to grasp it. We defined four velocity as the rate of change of your coordinates with respect to your proper time.
4-velocity -- being a vector belonging to the tangent space at a given point (event) -- has components in every chosen basis of the tangent (vector) space. We can just pick the coordinate basis in the given coordinate chart.

 

[Ibix]

A coordinate-free definition of four-velocity is a unit vector tangential to your worldline.

If you define a coordinate system and a matching coordinate basis, the components of the four-velocity are the rate of change of your coordinates with respect to your proper time. If you define a coordinate system and a non-coordinate basis, though, that statement is not true. You may always choose to use the coordinate basis, but you do not have to do so.

For example, on a flat sheet of paper I can draw a Cartesian x-y grid. At any point I can draw an arrow. It is simple to use the x- and y- axis directions as the basis vectors to describe that vector, but nothing requires me to do so. I can use radial and tangential vectors as the basis, for example. It would be peculiar to do so in this case, but it's perfectly well defined and unambiguous - it just needs to be used with care.

Edit: A better example is on the surface of the Earth. Typically you express location using longitude and latitude. However, the coordinate basis is orthogonal but not orthonormal, because one degree change of longitude is a smaller distance the further from the equator you are but one degree change of latitude is the same distance everywhere. So, for example, the vector (1,1) in the coordinate basis means different angles away from North at different latitudes. You might prefer that (1,1) always means 45° clockwise from North, in which case you are adopting a local orthonormal basis, which is a non-coordinate basis.

 

[cianfa72]

A coordinate-free definition of four-velocity is a unit vector tangential to your worldline.

If you define a coordinate system and a matching coordinate basis, the components of the four-velocity are the rate of change of your coordinates with respect to your proper time. If you define a coordinate system and a non-coordinate basis, though, that statement is not true. You may always choose to use the coordinate basis, but you do not have to do so.

Thus, even in the context of GR (curved spacetime), starting from the very definition of 4-velocity in a coordinate chart as $$u^ {\alpha} = \frac {dx^{\alpha}} {d \tau} $$ the following $g_ {\alpha\beta}$ metric tensor contraction should results in: $$g_ {\mu \nu}u^ {\mu}u^ {\nu} = 1$$ Is that correct ?

 

[PeroK]

Thus, even in the context of GR (curved spacetime), starting from the very definition of 4-velocity in a coordinate chart as $$u^ {\alpha} = \frac {dx^{\alpha}} {d \tau} $$ the following $g_ {\alpha\beta}$ metric tensor contraction should results in: $$g_ {\mu \nu}u^ {\mu}u^ {\nu} = 1$$ Is that correct ?

Yes.

 

[cianfa72]

Yes.

Anyway that should be true just for massive objects, I believe. For light (massless particles) my understanding is that its 4-velocity is orthogonal to itself thus resulting in $g_ {\mu \nu}u^ {\mu}u^ {\nu} = 0$

 

[PeterDonis]

Is that correct ?

Assuming you are using a $+ - - -$ metric signature convention, yes. Many textbooks and papers use a $- + + +$ signature convention instead; with that convention you will have $g_{\mu \nu} u^\mu u^\nu = - 1$.

 

[PeterDonis]

For light (massless particles) my understanding is that its 4-velocity is orthogonal to itself thus resulting in $g_ {\mu \nu}u^ {\mu}u^ {\nu} = 0$

$g_ {\mu \nu}u^ {\mu}u^ {\nu} = 0$ is the definition of a null vector, yes. Calling it a "4-velocity" is not really justified, since that term implies a unit vector and of course no null vector can be a unit vector. Calling it the "tangent vector" to the null worldline is fine.

Also, $g_ {\mu \nu}u^ {\mu}u^ {\nu} = 0$ for null vectors is not derived from "orthogonal to itself"; it's the other way around, you first define null vectors as $g_ {\mu \nu}u^ {\mu}u^ {\nu} = 0$ and then show (which is obvious) that that implies that any such vector is orthogonal to itself.

 

[cianfa72]

About the relative velocity between an observer $O$ having 4-velocity $V$ and the Null tangent vector $N$, on math grouds the value of the 'scalar' product $g_ {\mu \nu}v^ {\mu}n^ {\nu}$ should be related to the (local) 3-velocity $c$ of the light.

However, from a physical point of view, which is the operational procedure the observer $O$ should employ to measure the instantaneous (local) speed of the light ?

 

[vanhees71]

I don't understand what you mean. If $O$ has a four-velocity $V$, $V \cdot V=1$, and since this is by definition also a tangent vector at any point if his world-line, there cannot be a null-tangent vector.

 

[Ibix]

About the relative velocity between an observer $O$ having 4-velocity $V$ and the Null tangent vector $N$, on math grouds the value of the 'scalar' product $g_ {\mu \nu}v^ {\mu}n^ {\nu}$ should be related to the (local) 3-velocity $c$ of the light.

Assuming that $n^\nu$ is the tangent vector to a light pulse worldline, $g_{\mu\nu}v^\mu n^\nu$ has nothing to do with the speed of light. You can always pick a coordinate system where, locally, $g_{\mu\nu}=\eta_{\mu\nu}$ and the observer is at rest so $v^t=1$ and the other components are zero. Thus $g_{\mu\nu}v^\mu n^\nu$ is just the $t$ component of $n^\nu$ in these coordinates. That will be the energy of the light pulse as measured by your observer, with units depending on how you normalised $n^\nu$. (Edit: and modulo your metric sign convention.)

However, from a physical point of view, which is the operational procedure the observer $O$ should employ to measure the instantaneous (local) speed of the light ?

Conceptually, get a laser, a mirror, a ruler, and a stopwatch. Bounce a laser pulse off the mirror back to the laser. Measure distance and time. Fizeau's toothed wheel experiment is a neat way of doing it practically, and can be found online.

 

[cianfa72]

Assuming that is the tangent vector to a light pulse worldline, has nothing to do with the speed of light. You can always pick a coordinate system where, locally, and the observer is at rest so and the other components are zero. Thus is just the component of in these coordinates. That will be the energy of the light pulse as measured by your observer, with units depending on how you normalised

Maybe I am wrong, but the scalar product between two 4-velocity tangent vectors (even if in this case one of them is a light path -- null tangent vector) shouldn't be related in some way to the 3-velocity relative speed between them ?

Conceptually, get a laser, a mirror, a ruler, and a stopwatch. Bounce a laser pulse off the mirror back to the laser. Measure distance and time. Fizeau's toothed wheel experiment is a neat way of doing it practically, and can be found online.

Sure in principle that's a round-trip speed measurement.

 

[Ibix]

Maybe I am wrong, but the scalar product between two 4-velocity tangent vectors (even if in this case one of them is a light path -- null tangent vector) shouldn't be related in some way to the 3-velocity relative speed between them ?

Four velocities are timelike. You can normalise them to unit length and their inner product is related to their relative speed - it's the Lorentz gamma factor in fact. Null tangent vectors behave differently. They can't be of unit length, for a start, so you shouldn't really think of them as four velocities. As I showed above, the inner product of a four-velocity with a null vector is the energy of the light as measured by an observer with that four velocity.

 

[vanhees71]

Just to sort this a bit. For a plane em. wave you have the wave-four-vector $(k^{\mu})=(\omega/c,\vec{k})$ which is a null-vector, i.e., $g_{\mu \nu} k^{\mu} k^{\nu}=0$. If then $(u^{\mu})$ is the normalized four-velocity vector of an observer, i.e., $u_{\mu} u^{\mu}=g_{\mu \nu} u^{\mu} u^{\nu}=1$ then $\omega_{\text{obs}}/c=g_{\mu \nu} k^{\mu} u^{\nu}$ is the frequency of the em. wave as observed by the observer (using the west-coast convention, where $g_{\mu \nu}$ has the signature $(1,3)$).

 

[PeterDonis]

the scalar product between two 4-velocity tangent vectors

The tangent vector to a null worldline is not properly referred to as a "4-velocity". The term "4-velocity" implies a unit vector.

shouldn't be related in some way to the 3-velocity relative speed between them ?

If both vectors are timelike, the scalar product is related to the 3-velocity, yes, since the scalar product is the relative $\gamma$ factor between them, and $\gamma = 1 / \sqrt{1 - v^2 / c^2}$.

If one vector is null and the other is timelike, the scalar product is, as @vanhees71 has said, the frequency of the light ray (null vector) as measured in the rest frame of the timelike vector. So no, unfortunately, it is not related to the speed of the light ray. If you think about it, you will see that it can't possibly be related to the speed, since the speed of the light ray is $c$ regardless of which timelike vector you pick, but the scalar product is different for different timelike vectors. This is related to the fact that Lorentz transformations act differently on timelike vectors than on null vectors; they rotate timelike vectors in spacetime, but they dilate null vectors.