Acceleration is perpendicular to velocity

[Naty1]

Somebody posted a reference to Fermi Walker transport, maybe Ben Crowell, anyway,
The first section here:

http://en.wikipedia.org/wiki/Fermi-Walker_transport

says:

we can regard acceleration in spacetime as simply a rotation of the 4-velocity.

That's a new perspective for me. Can anyone help provide an intuitive explanation? I see the lower diagram on the left, two velocity vectors with an acceleration vector between the arrowheads, but I'm having trouble trying to form mental picture of the physical.

I can relate to an acceleration in space forming a curve in spacetime: for example a particle accelerating in two dimensions traces out a curve in three dimensional spacetime. And a circular rotation in two space dimensions traces out a corkscrew in three dimensionalspacetime...that's a clear example of the acceleration being perpendicular to the velocity...

As I think about this for the first time, It seems to relate to a description Doc Greg helped me with sometime ago...that acceleration results in a TYPE of spacetime curvature, but not the same type as gravitational curvature...Is THAT what's going on?? Or do i have unrelated concepts muddled??

..I copied it for my own notes...maybe I can actually find it and post it...

 

[Naty1]

Amazing..I found it: (from Dr. Greg) (my boldface)

"... let's restrict our attention to 2D spacetime, i.e. 1 space dimension and 1 time dimension, i.e. motion along a straight line. …
In the absence of gravitation, an inertial frame corresponds to a flat sheet of graph paper with a square grid. If we switch to a different inertial frame we "rotate" to a different square grid, but it is the same flat sheet of paper. (The words "rotation" and "square" here are relative to the Minkowski geometry of spacetime, which doesn't look quite like rotation to our Euclidean eyes, but nevertheless it preserves the Minkowski equivalents of "length" (spacetime interval) and "angle" (rapidity).)

If we switch to a non-inertial frame ([an accelerated observer] but still in the absence of gravitation), we are now drawing a curved grid, but still on the same flat sheet of paper. Thus, relative to a non-inertial observer, an inertial object seems to follow a curved trajectory through spacetime, but this is due to the curvature of the grid lines, not the curvature of the paper which is still flat.

When we introduce gravitation, the paper itself becomes curved. (I am talking now of the sort of curvature that cannot be "flattened" without distortion. The curvature of a cylinder or cone doesn't count as "curvature" in this sense.) Now we find that it is impossible to draw a square grid to cover the whole of the curved surface. The best we can do is draw a grid that is approximately square over a small region, but which is forced to either curve or stretch or squash at larger distances. This grid defines a local inertial frame, where it is square, but that same frame cannot be inertial across the whole of spacetime.

So, to summarize, "spacetime curvature" refers to the curvature of the graph paper, regardless of observer, whereas visible curvature in space is related to the distorted, non-square grid lines drawn on the graph paper, and depends on the choice of observer...."

 

[DrGreg]

When we talk of curvature in spacetime (either curvature of a worldline, or curvature of spacetime itself) we don't mean the kind of curves that result from using a non-inertial coordinate system, i.e, non-square graph paper in my analogy.

In Minkowski coordinates in Special Relativity, 4-acceleration is just the coordinate derivative of 4-velocity with respect to arc-length (proper time), and the 4-velocity is the unit tangent vector of worldline. As the 4-velocity has a constant length its derivative must be orthogonal to it. The 4-acceleration is the curvature vector; orthogonal to the worldline and its length is the reciprocal of the worldline's "radius of curvature".

In non-inertial coordinates in GR, the 4-acceleration is defined as a covariant derivative. This takes into account (and removes) any curvature of spacetime or "apparent curvature" due to using a "non-square grid", and leaves us with curvature that is a property of the worldline itself, not the spacetime or the choice of coordinates. Then everything else I said in the last paragraph is still true in a coordinate-independent sense.

A geodesic has zero 4-acceleration and zero curvature.

 

[Phrak]

In non-inertial coordinates in GR, the 4-acceleration is defined as a covariant derivative. This takes into account (and removes) any curvature of spacetime or "apparent curvature" due to using a "non-square grid", and leaves us with curvature that is a property of the worldline itself, not the spacetime or the choice of coordinates. Then everything else I said in the last paragraph is still true in a coordinate-independent sense.

Can you explain what you mean by removing curvature or apparent curvature?

I see that A^u is a four vector, but as the covariant derivative of the 4-velocity, then it is not itself generally covariant. Though I don't see how
[tex]\frac{dU^\mu}{d\tau}[/tex]
is found to be a covariant derivative of U.

I find that [itex]\partial_\mu U_\nu - \partial_\nu U_\mu[/itex] is a quantity that is covariant, but perhaps not as useful a definition of 4-acceleration.

 

[Naty1]

Hello, Dr Greg, and thanks again for a response...

The subtlies of coordinate, covarient and contravarient derivaties may be a bit too much for me to interpret on my own right now. How is coordinate different from "ordinary" derivative...from covarient...I know the contravarient (index upstairs) derivative involves a Christoffel term...in GR...

I'm listening to Leonard Susskind's General Relativity youtube videos and he introduces the latter two in lecture #5. I'm moving ahead to watch the remainder of the lectures to get an overview and with some luck ...and ambition...will go back and take notes for mathematical details. But I suspect Susskind's very basic discussion may not get me where I need to be to really understand your comments.

You posted:

"When we talk of curvature in spacetime (either curvature of a worldline, or curvature of spacetime itself) we don't mean the kind of curves that result from using a non-inertial coordinate system, i.e, non-square graph paper in my analogy."

yes, I do get that thanks to your prior explanations...I posted your prior desciption here to distinguish for others the types of curvature since I have never seen others here in the forums provide those distinctions and those were REALLY helpful to me in gaining some physical insights.

You posted:

"In Minkowski coordinates in Special Relativity, 4-acceleration is just the coordinate derivative of 4-velocity with respect to arc-length (proper time), and the 4-velocity is the unit tangent vector of worldline. As the 4-velocity has a constant length its derivative must be orthogonal to it. The 4-acceleration is the curvature vector; orthogonal to the worldline and its length is the reciprocal of the worldline's "radius of curvature"."

This reminded me that in 2+1 dimensional (2 space and time) constant velocity is a straight line, linear acceleration is a curve and rotational acceleration a corkscrew...so I think that's the physical sense in which acceleration in Minkowski space causes a rotation? and why acceleration must be perpendicular to velocity...strange as that seems at first...


If so I now have a physical picture for:

...we can regard acceleration in spacetime as simply a rotation of the 4-velocity

and which reminded me rotation in Euclidean SPACE and Minkowski SPACETIME don't look the same...

Also I did not get at first:

As the 4-velocity has a constant length its derivative must be orthogonal to it.

until I read this:

The length of the four-velocity (in the sense of the metric used in special relativity) is always equal to c (it is a normalized vector).

here http://en.wikipedia.org/wiki/Four-velocity

That's what you meant?

I have to do more studying before tackling your last paragraph and will be interested if you and Phrak reach any conclsuions together...

 

[DrGreg]

Can you explain what you mean by removing curvature or apparent curvature?

I see that A^u is a four vector, but as the covariant derivative of the 4-velocity, then it is not itself generally covariant. Though I don't see how
[tex]\frac{dU^\mu}{d\tau}[/tex]
is found to be a covariant derivative of U.

I find that [itex]\partial_\mu U_\nu - \partial_\nu U_\mu[/itex] is a quantity that is covariant, but perhaps not as useful a definition of 4-acceleration.

Well, actually I didn't say that [itex]dU^\mu / d\tau[/itex] was a covariant derivative (I never gave a definition). And in fact the name "covariant derivative" is possibly confusing or incorrect. I find the terms "absolute derivative" or "intrinsic derivative" or "covariant derivative along a worldline" used in the literature. The definition is

[tex]
\frac{DU^\mu}{D\tau}
= U^\alpha \nabla_\alpha U^\mu
= U^\alpha \left( \frac{\partial U^\mu}{\partial X^\alpha} + \Gamma^\mu_{\alpha\beta}U^\beta\right)
= \frac{dU^\mu}{d\tau} + \frac{1}{2} g^{\mu \nu} \left( \frac{\partial g_{\nu \beta}}{\partial X^\alpha} + \frac{\partial g_{\alpha\nu}}{\partial X^\beta} - \frac{\partial g_{\alpha \beta}}{\partial X^\nu} \right) U^\alpha U^\beta
[/tex]​

In coordinates that are "locally Minkowski" all the terms vanish except [itex]dU^\mu / d\tau[/itex], so that is the sense in which I meant 'removing curvature of spacetime or "apparent curvature" due to using a "non-square grid" '.

Note that in non-Minkowski cooordinates, [itex]dU^\mu / d\tau[/itex] is not a tensor, but [itex]DU^\mu / D\tau[/itex] is.

See covariant derivative: derivative along curve for details.

 

[DrGreg]

This reminded me that in 2+1 dimensional (2 space and time) constant velocity is a straight line, linear acceleration is a curve and rotational acceleration a corkscrew...so I think that's the physical sense in which acceleration in Minkowski space causes a rotation? and why acceleration must be perpendicular to velocity...strange as that seems at first...

Remember we are talking about 4-velocity and 4-acceleration, which aren't the same as 3-velocity and 3-acceleration, so that's why it might seem strange at first. The 4-velocity does indeed have a constant length of c (or 1 if you take c=1), which explains the orthogonality of the 4-acceleration.

 

[jason12345]

When we talk of curvature in spacetime (either curvature of a worldline, or curvature of spacetime itself) we don't mean the kind of curves that result from using a non-inertial coordinate system, i.e, non-square graph paper in my analogy.

In Minkowski coordinates in Special Relativity, 4-acceleration is just the coordinate derivative of 4-velocity with respect to arc-length (proper time), and the 4-velocity is the unit tangent vector of worldline. As the 4-velocity has a constant length its derivative must be orthogonal to it. The 4-acceleration is the curvature vector; orthogonal to the worldline and its length is the reciprocal of the worldline's "radius of curvature".

In non-inertial coordinates in GR, the 4-acceleration is defined as a covariant derivative. This takes into account (and removes) any curvature of spacetime or "apparent curvature" due to using a "non-square grid", and leaves us with curvature that is a property of the worldline itself, not the spacetime or the choice of coordinates. Then everything else I said in the last paragraph is still true in a coordinate-independent sense.

A geodesic has zero 4-acceleration and zero curvature.

Can you recommend a book which goes into detail on all this stuff?

I've got a feeling you're going to say Gravitation by Wheeler et all

 

[Phrak]

Thanks, Dr. Greg.

 

[EWH]

"'we can regard acceleration in spacetime as simply a rotation of the 4-velocity.'

That's a new perspective for me. Can anyone help provide an intuitive explanation?"

In geometric algebra, in 4-D Minkowski signature there are:
1 scalar (numbers)
4 vectors (3 space vectors with (-) square (xyz), 1 time vector with (+) square (t))
6 bivectors (planes of rotation - 3 spatial (xy xz yz) and 3 velocities (xt yt zt)),
4 trivectors (angular velocities - (xyt xzt yzt), plus the spatial volume (xyz) and
1 pseudoscalar (xyzt).
(Component orders not guaranteed to be correct. Reordering components may cause a sign change. Usual notation is e_0, e_1, e_2,e_3; signature (+,-,-,-))
These form all the possible combinations of the 4 components, with the number of "blades" of each type specified by binomial[4,n] with n from 0 to 4.

As you can see linear velocities are essentially the same as spatial rotations, both are bivectors (oriented areas of rotation), the difference being that spatial rotations are with both components the same sign (of their squares), hence circular rotation (e.g. x^2 + y^2) whereas velocities have components of different signs (e.g. x^2 - t^2), thus a hyperbolic rotation. (The usual signature would actually have negative x^2, y^2 and positive t^2, but the principle is the same.) Thus a spatial rotation is a circular rotation, while a linear velocity is a hyperbolic rotation. A change in linear velocity (acceleration) is a change in rotation, and thus is a rotation of a rotation. Geometric algebra makes it easy to work with rotations or nested sequences of rotations in any dimension.

 

[pervect]

I'm sort of jumping in late, but there's a very easy way to show that the 4-acceleration is perpendicular to the 4-velocity, i.e that a dot v = 0, where a is the 4-acceleration vector and v is the 4-velocity.

It also helps to clarify what these terms are.

Consider the local rest frame of the object, so that the objects instantaneous velocity is zero, and the metric coefficients are of unit magnitude except for sign.

Then the 4-velocity in that frame is (dt/dtau) = 1, with only time components.
The 4-acceleration can be shown to only have spatial components, we just calculate

d(dt/dtau)/dtau = d/dtau (1-v^2)^(-1/2) = v/(1-v^2)^(-3/2). But v=0. So this expression is zero.

Therefore the 4-velocity is perpendicular to the 4-acceleration in the rest frame of the object, the dot product of a and v is zero.

The dot product of two vectors is a tensor expression, so since it's true in this simple coordinate system, it's true in all coordinate systems that a dot v = 0.

 

[robphy]

"'we can regard acceleration in spacetime as simply a rotation of the 4-velocity.'

That's a new perspective for me. Can anyone help provide an intuitive explanation?"

[note the thread was idle for a few months]

An analogy with circular motion may help.

Since acceleration is the time-derivative of velocity,
then infinitesimal changes to the velocity vector are parallel to the acceleration vector.
The component of a along v "lengthens" [or "shrinks"] of v, whereas the remaining component of a [which is perpendicular to v] "rotates" v.
If the length of v is constant (as in "uniform circular motion"), then acceleration is perpendicular to the velocity and can be thought of as the instruction to "rotate" (or, equivalently, "turn") the instantaneous velocity vector.

So [as others have said], since the 4-velocity has constant length [(v dot v)=1],
the 4-acceleration (dv/ds) is orthogonal to the 4-velocity: 0=(d/ds)(v dot v)=2 v dot (dv/ds).
Thus, with the analogous interpretation, the 4-acceleration "rotates" (or "turns") the 4-velocity.