# Transformation of an acceleration vector under a basis change

1. Oct 28, 2012

### Mentz114

This thread is spawned from an earlier one

For the stationary ( ie comoving ) frame in the Schwarzschild spacetime the co-basis of the frame field is
$$s_0= \sqrt{\frac{r-2m}{r}}dt,\ \ s_1=\sqrt{\frac{r}{r-2m}}\ dr,\ \ s_2=r\ d\theta,\ \ s_3=r\sin(\theta)\ d\phi$$
The 4-velocity of the stationary observer in the local frame is $u^\mu=\partial_t$ (or $(1,0,0,0)$).

This, $u^\nu\nabla_\mu u_\nu$, which looks like a covariant vector is
$$a_\mu= \frac{m}{{r}^{\frac{3}{2}}\,\sqrt{r-2\,m}} \ dr$$
All this is well known. Now we boost the co-basis covectors in the r-direction with velocity $-\sqrt{2m/r}$. The metric is unchanged in form by this. This gives a new basis h
$$h_0= -dt - \frac{\sqrt{2mr}}{r-2m}\ dr,\ \ h_1=\sqrt{\frac{2m}{r}}\ dt + \frac{r}{r-2m} \ dr,\ \ h_2=s_2,\ \ h_3=s_3$$
According to my calculations, in this basis the acceleration aµ is now zero. Readers are bound to be sceptical about my calculations but I'm very confident there's been no gross error. If it is an error I'll be happy to have found it.

2. Oct 28, 2012

### Staff: Mentor

I am highly skeptical of this claim. Have you tried actually working out the metric in the new coordinates?

3. Oct 28, 2012

### Muphrid

He should be right. The metric is indifferent to rotational degrees of freedom (and boosts are "rotations"). Nevertheless, it should be intuitive that a uniform boost of a vector should not change the magnitude of the vector. That said, I'm aware that intuitive results are not necessarily always right.

One thing I'm curious about: why compute the acceleration covector, rather than the acceleration vector?

4. Oct 28, 2012

### Staff: Mentor

This boost is equivalent to a transformation from Schwarzschild to Painleve coordinates. This does *not* leave the metric unchanged. [Edit: Actually you can, if you want, do the calculation in the same coordinates; but in that case the physical meaning of the boost is obviously that you are changing observers. So in any case my following comment stands:] Basically what you are doing is showing that the 4-acceleration of a Painleve observer is zero; you are *not* showing that the 4-acceleration of a static observer changes when you transform coordinates.

http://en.wikipedia.org/wiki/Frame_fields_in_general_relativity

It uses the term "Lemaitre observers" for what I'm calling Painleve observers.

The metric is invariant under *spatial* rotations. It is *not* invariant under space-time rotations, i.e., boosts.

Last edited: Oct 28, 2012
5. Oct 28, 2012

### Staff: Mentor

I'm still working through calculations, but this looks wrong; what you've written looks like the 4-acceleration vector $a^\mu$, not the 4-acceleration covector $a_\mu$. Remember that in order to compute $u^\mu \nabla_\mu$ you need the 4-velocity *vector* (which you didn't write down in your post), not the covector (which you did write down).

6. Oct 28, 2012

### Mentz114

Thanks for the replies. I'll try to answer the questions but I think now that this is correct.

@DaleSpam, you asked if I'd worked out the new metric. ?
The metric given by $g=-h_0\otimes h_0 + h_1\otimes h_1 + h_2\otimes h_2 + h_3\otimes h_3$ is the Schwarzschild metric

Last edited: Oct 28, 2012
7. Oct 28, 2012

### Mentz114

This $u^\mu \nabla_\mu$ is not the acceleration ( it's the expansion scalar ). The acceleration is $u^\nu\nabla_\nu u_\mu$ which has one free covariant index.

The velocity vector is (1,0,0,0) and the covector is (-1,0,0,0) so uμuμ = -1.

8. Oct 28, 2012

### Muphrid

I don't see why you're using the covariant velocity instead of the contravariant. Isn't it usual to use the contravariant acceleration/velocity?

I don't follow. The whole point of using a metric instead of frame fields is that it removes all six degrees of generalized rotation freedom (i.e. 3 degrees of boost freedom and 3 degrees of spatial rotation freedom).

Last edited: Oct 28, 2012
9. Oct 28, 2012

### Staff: Mentor

No, it isn't. It's the directional derivative operator along the 4-velocity $u^\mu$. It's a scalar, yes, since it contracts the 4-velocity, a vector, with the covariant derivative operator, a covector. The point I was making is that it's the 4-velocity *vector* that appears in it; you only wrote down the *covector*, so I'm not sure what 4-velocity vector components you are using to compute the operator $u^\mu \nabla_\mu$.

No, they aren't. I didn't notice this before, but in looking through the OP, I see that you made the same mistake there. The 4-velocity is a unit vector: (1, 0, 0, 0) is not a unit vector unless r -> infinity (i.e,. unless the metric is Minkowski). For the original coframe s that you wrote down, $s_0$ is the 4-velocity covector (and its corresponding frame vector $s^0$ is the 4-velocity vector). For the boosted coframe you wrote down, you have to rewrite $s_0$ in that frame if you want to express the 4-velocity of a static observer in that coframe (and similarly you have to express $s^0$ in the corresponding frame). What you actually did was (more or less) to compute the 4-acceleration of an observer whose 4-velocity covector is $h_0$, which is a *different* 4-velocity, that of a Painleve observer.

10. Oct 28, 2012

### Staff: Mentor

I think you're misunderstanding what metrics and frame fields are. They both express the same physics and the same "degrees of freedom", just in different forms. A metric (by which I think you actually mean a "line element") expresses the physics in terms of a coordinate chart and the basis vectors (or covectors) of that chart. A frame field expresses the physics in terms of the basis vectors (or covectors) carried by a particular family of observers as they travel along their worldlines. There is the same amount of freedom in either form.

Also, whether or not a metric is invariant under a particular set of transformations depends on the specific spacetime geometry under discussion. Schwarzschild spacetime is spherically symmetric, so the metric is invariant under spatial rotations. There are other spacetimes that are not spherically symmetric (such as Kerr spacetime), and their metrics are not invariant under spatial rotations.

11. Oct 28, 2012

### Mentz114

Peter,
the metric is Minkowski. We are working in the local frame. I never calculate the scalar $u^\mu \nabla_\mu$. I'm calculating the covariant derivative of the (co)vector and projecting in the direction of the vector. I used to write it as $u_{\mu;\nu}u^\nu$.

Does this clear up some of your questions ?

It occurs to me that tensors transform like tensors under holonomic transformations, ie we can write
$$dx'^\mu = \frac{\partial x'^\mu}{\partial x^\mu}dx$$
and going from the coordinate basis to a frame basis is not holonomic.

Last edited: Oct 28, 2012
12. Oct 28, 2012

### PAllen

This way of doing things is all too modern for me. For me, 4 acceleration is absolute derivative of U along a path. For a constant r,theta,phi world line in SC coordinates this works out to be zero except for the r component which is:

(m/r^2 ) (1-2m/r)^-1

with a norm of: (m/r^2 )(1-2m/r)^(-3/2) [if i haven't made a stupid mistake somewhere].

So, what I thought was claimed was that there was some magic coordinate transform which would make this vector go to zero and its norm vanish, which is clearly impossible.

13. Oct 28, 2012

### Staff: Mentor

Doesn't matter; in order to calculate the 4-acceleration of a static observer (or indeed of any observer), you have to know the connection coefficients, which means you have to know the actual metric (Schwarzschild), not just the approximate metric in the local frame. If the metric were really Minkowski, you could use the partial derivative operator $\partial_\mu$ instead of the covariant derivative operator $\nabla_\mu$ and simplify your formulas considerably.

Also, if the metric were Minkowski, the first two components of the coframe would not be what you wrote down; they would be (-dt, dr). The fact that you wrote down a different coframe means that you implicitly agree that the spacetime curvature affects the calculation.

That's fine, but you still have to know the vector components (not just the covector components) to do the projection. What vector components are you using to compute that?

14. Oct 28, 2012

### Staff: Mentor

That's true, but you're not going from a coordinate basis to a frame basis. You're going from one frame basis to another frame basis. The first frame basis happens to have vectors pointing in the same direction as the coordinate basis vectors, but that doesn't make it a coordinate basis.

15. Oct 28, 2012

### Mentz114

There's nothing approximate here. These are exact calculations. We have the (inverse) metric. It is encoded in the coframe cobasis vectors ( see the tensor product expression I give earlier ).

In the local frame the connection coefficients are Ricci rotation coefficients and they are used in calculating covariant derivs instead of the CS2's.

The global metric is in the coframe, as I said. The tetrad formed from the coframe basis projects the metric to the Minkowski (local) metric.

I stated both, uμ=(1,0,0,0) and uμ=(1,0,0,0).

16. Oct 28, 2012

### Mentz114

Yes, the boost is also non-holonomic. I'm just thinking out loud.

17. Oct 28, 2012

### Mentz114

Well, I can make it vanish by boosting a frame field. This effectively changes the local frame from an accelerating one to a geodesic one. Whatever is going on, I'm not making the components of the same tensor vanish - as you say, that's impossible.

18. Oct 28, 2012

### PAllen

Can you express this as a coordinate transform matrix? Such that I can apply it to the 4-acceleration I computed?

[edit: wait, I see you agree that can't be done. So I bow out because I have not studied frame fields.]

Last edited: Oct 28, 2012
19. Oct 28, 2012

### Staff: Mentor

Hmm, I don't find that very helpful. I cannot even tell what coordinates you are using after your radial boost, and this expression doesn't make it clear to me at all that the form of the metric is unchanged in those coordinates.

Btw, do you intend to have a different boost at each r?

20. Oct 28, 2012

### Staff: Mentor

I think I am with you here.

21. Oct 28, 2012

### Mentz114

I calculated that tensor product and the metric is unchanged. What more can I say. If you don't believe me, re-calculate it. It's a frame field so I'm working in local Minkowski coords when the acceleration is calculated.

Yes, as I stated in the OP the boost is $\sqrt{2m/r}$

Having thought a bit more about this, I don't think there's anything amiss. The static frame field is accelerating and will always be accelerating from any viewpoint. The boosted static frame is now moving and so is a different physical setup with a different acceleration vector which happens to be zero for the boost I chose. All observers will agree on this too. It just looks as if the same tensor is being considered because the math and the vectors look the same in the local frame. For me the mists have cleared.

Thanks for the responses.

Last edited: Oct 28, 2012
22. Oct 28, 2012

### Muphrid

Let me try to cut through some of the confusion. Let there be a fundamental flat spacetime with basis vectors $e_t, e_r, e_\phi, e_\theta$. These basis vectors need not be unit--in this case, $e_\theta \cdot e_\theta = r^2$ and $e_\phi \cdot e_\phi = r^2 \sin^2 \theta$. Define a linear operator $\underline h$ such that

$$\underline h^{-1}(e_t) = (1 - 2M/r)^{1/2} e_t \equiv g_t\\ \underline h^{-1}(e_r) = (1- 2M/r)^{-1/2} e_r \equiv g_r\\ \underline h^{-1}(e_\theta) = e_\theta \equiv g_\theta\\ \underline h^{-1}(e_\phi) = e_\phi \equiv g_\phi$$

This linear operator goes by many names--frame field, displacement gauge field, etc. Let's just call it the h-field. The vectors $g_\alpha$ obey the relationship $g_\alpha \cdot g_\beta = g_{\alpha \beta}$, so they are the actual coordinate basis vectors, and this gives us the metric in terms of the h-field. $g_{\alpha \beta} = \underline g(e_\alpha) \cdot e_\beta = \underline h^{-1}(e_\alpha) \cdot \underline h^{-1}(e_\beta)$. Simplified, the metric is $\underline g(s) = \overline h^{-1} \underline h^{-1}(s)$, for any vector $s$.

So far so good? So far so good. Now, let us apply an operator $\underline L$ such that its adjoint is equal to its inverse--$\overline L = \underline L^{-1}$--to the h-field. This equally describes spatial rotations as well as Lorentz boosts. This operator may be position dependent. If $\underline h \mapsto \underline L \underline h$, then the metric goes to

$$\underline g(s) = \underline h^{-1} \underline L^{-1} \overline L^{-1} \overline h^{-1}(s)$$

But again, the adjoint of $\underline L$ is equal to the inverse. The L's cancel, and you recover the original metric. This is what I meant when I said the metric is invariant under rotations and boosts. No rotation or boost to the h-field affects the metric.

Now then, we want to consider a four-velocity that is (edit) $u = e_t$--this is the gauge-invariant velocity. (Edit: I'm actually a bit puzzled at this point. This implies the time component of the four-velocity in the coordinate basis is no longer unit.) We need the connection, which is $\underline \omega(e_t) = \frac{M}{r^2} (1-2M/r)^{-1/2} e_{rt}$, so the acceleration is

$$a= u \cdot D u = u \cdot \overline h(\nabla) u + \underline \omega(u) \cdot u$$

I assume $u$ has no dependence on $\tau$, so that the first term vanishes. Note that $e_{rt} \cdot e_t = e_r$ under this sign convention, so the result is

$$a = \frac{M}{r^2} (1-2M/r)^{-1/2}e_r$$

(Edit: stopping here, since I'm no longer sure of the conclusion that followed.)

Last edited: Oct 28, 2012
23. Oct 28, 2012

### Mentz114

Muphrid, thanks, that is most informative. Although I'm new to this notation I can clearly follow what you're doing.

You're not doing what I was but you have shown that the acceleration for a certain frame field is a tensor and cannot be made to vanish under a boost. But I was calculating the acceleration for a certain h-field (h1 say). Then I boosted the h1-field to get h2 and - lo, I get a zero acceleration for h2. No problem because they are different h fields representing different physical setups.

Last edited: Oct 28, 2012
24. Oct 28, 2012

### Muphrid

Well, it's possible I've made an error in believing that the acceleration has the same transformation law as the inverse h-field. The most rigorous way would be to recalculate the connection $\underline \omega$ for the new h-field and see how that changes the acceleration.

So, to verify things, I would want to figure out the transformation of the connection $\underline \omega$ under this boost. I'm poring over my resources--I think, because this boost is not position dependent, the $\underline \omega$ field transforms by either a simple boost or perhaps by a boost of itself and a boost of its argument. I'm gonna have to wade through the material a bit more.

But as far as the different h-fields representing different setups, I disagree in one respect. No matter what you do--apply boosts/rotations or various coordinate system changes--the h-field either can or cannot be transformed to the identity. If it can't, I think there is intrinsic information about the system contained in the h-field, information that doesn't go away based on coordinate-system transformations or boosts and rotations. But this is a point of interpretation, not really relevant to the question at hand.

Once we know how the connection transforms, it will be clear what the acceleration should be.

Regardless, though, the benefit of working with these $e_\alpha$ running around is that the acceleration and velocity should be invariants regardless of h-field. It is the components, when evaluated with respect to different h-fields (different coordinate basis vectors $g_\alpha$), that should change.

Now, that said, it's clear that you weren't looking at the gauge invariant acceleration (which is what my $a$ is), but the components according to a set of coordinate basis vectors.

So if we take our gauge-invariant acceleration $a$ and extract components $a \cdot g^\mu$ for some choice of $g^\mu$ (some choice of h-field), can we make the radial acceleration vanish? Just kinda doing the h-field matrix inversion in my head here, let's say the coordinate-basis covectors are $g^t = K \gamma (e^t - \beta e^r)$ and $g^r = K^{-1} \gamma (e^r - \beta e^t)$ for $K = \sqrt{1-2M/r}$ (it might be 1/that, but it gets the point across).

The components in the coordinate basis are then

\begin{align*} a^t &= -\frac{M}{r^2} K \gamma \beta\\ a^r &= \frac{M}{r^2} K^{-1} \gamma \end{align*}

Hm, that would seem to indicate it's not possible to reduce these four-accleration components to zero. It's possible I'm just screwing something up royally, considering the time of night. The only way I can see, right now, that the components of acceleration could vanish in some coordinate frame is if the gauge-invariant acceleration $a$ had two terms, one in $e_r$ and the other in $e_t$. Then the required cancellation could appear. This might be the case if I improperly calculated $a$ earlier.

Edit: I think I found my error. I constructed the gauge-invariant velocity incorrectly. Fixed some things, but I'll have to look at this with fresh eyes in the morning.

Last edited: Oct 28, 2012
25. Oct 28, 2012

### PAllen

I did make a stupid mistake above. Fixing algebra:

r component of 4-acceleration is simply: m/r^2.
proper acceleration (norm) is simply: (m/r^2) (1-2m/r)^(-1/2)