Transformation of an acceleration vector under a basis change

[PeterDonis]

4-acceleration is a contravariant vector, regardless of the fact it can be called a rank 1-tensor.

I'm confused; in the other thread you said 4-acceleration was a tensor, not a contravariant vector:

Have you tried reading Carroll notes for instance? Then you should realize that in GR the expression with christoffel corrections transforms as a tensor, not as a contravariant vector.

The "expression with Christoffel corrections" is the 4-acceleration.

 

[Mentz114]

I will try to do so, but it will take a while. However, if you are working in local Minkowski coordinates then the form of the metric cannot be Schwarzschild. I think my skepticism is justified.

It only takes 5 minutes. Yes, the metric is Schwarzschild but we have a local Minkowski metric. What I've done is standard frame field stuff.

 

[Mentz114]

In which case the metric is not Minkowski; the cobasis vectors you wrote down are not those of the Minkowski metric.

If the "local frame" is supposed to be Minkowski then the connection coefficients are zero.

The tetrad gives expressions for unit vectors in the four basis directions, in terms of the global (Schwarzschild) metric, yes. That isn't the same as actually using the local Minkowski metric.

But those don't match the metric expression you wrote down. They aren't unit vectors unless the line element you are using is the Minkowski line element, $ds^2 = - dt^2 + dr^2 + r^2 d\theta^2 + r^2 sin^2 \theta d\phi^2$.

These remarks don't help because what I've done is standard frame field stuff which I suspect you are not familiar with.

 

[Mentz114]

You guys seem to want to confuse things deliberately.
Doesn't anybody here see the difference between the transformation properties of tensors versus vectors? Or following Dalespam policy since all of them can be called n-tensors they all behave equally?
4-acceleration is a contravariant vector, regardless of the fact it can be called a rank 1-tensor.

Please stop polluting my thread with this stuff. Start your own thread if you want to argue about whether 4-vectors are tensors or whatever.

 

[Mentz114]

Well, it's possible I've made an error in believing that the acceleration
...
...

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.

This is interesting. The connections are different between the frames, which is why they have different accelerations. I'll study what you've written later when I have more time.

 

[PeterDonis]

These remarks don't help because what I've done is standard frame field stuff which I suspect you are not familiar with.

I have a basic understanding of what frame fields are. I don't understand which coordinate chart you are using; you seem to be mixing together expressions written in the global Schwarzschild chart and expressions written in a local Minkowski frame.

I agree with what you said in this quote about the actual physics involved:

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.

But I don't see how your calculations correspond to this physics.

 

[Mentz114]

...
I agree with what you said in this quote about the actual physics involved:

But I don't see how your calculations correspond to this physics.

I don't think I can make it any clearer. There are two frame fields - one is stationary wrt the source, the other is free-falling. They are represented by the basis covectors I gave. They have different acceleration vectors - no problem.

 

[Muphrid]

I'm curious: you weren't suggesting that we look at purely temporal (not sure how else to say this...purely t-direction?) velocities for each of the two different frame fields and see that one had a nonzero acceleration while the other did not, were you?

 

[PeterDonis]

I don't think I can make it any clearer. There are two frame fields - one is stationary wrt the source, the other is free-falling. They are represented by the basis covectors I gave. They have different acceleration vectors - no problem.

Yes, I understand that part, at least the physics of it. But there's a lot more that has been said in this thread than just the above. For example: my understanding is that, by the definition of a frame field, the timelike vector (or covector) of the basis *is* the 4-velocity of the observer whose frame field it is. For example, the 4-velocity of a static observer *is* the vector $s_0$ of the first frame field you wrote down. But you say the 4-velocity of that observer is (1, 0, 0, 0).

 

[Mentz114]

I'm curious: you weren't suggesting that we look at purely temporal (not sure how else to say this...purely t-direction?) velocities for each of the two different frame fields and see that one had a nonzero acceleration while the other did not, were you?

No, I wasn't suggesting that.

Can I make it clear that as far as I'm concerned my original problem was because I was wrong in my expectation. The physics is clear to me now.

There's no point in pursuing this further.

 

[PeterDonis]

For example, the 4-velocity of a static observer *is* the vector $s_0$ of the first frame field you wrote down. But you say the 4-velocity of that observer is (1, 0, 0, 0).

Just to elaborate a bit on this: as I understand it, the 4-velocity *covector* of the static observer in the s frame field should be $s_0$, i.e.,

$$u_\mu = \sqrt{\frac{r - 2m}{r}} dt$$

The 4-velocity *vector* is obtained by raising an index, thus:

$$u^\mu = g^{\mu \nu} u_{\nu} = g^{tt} u_t \partial_t = \sqrt{\frac{r}{r - 2m}} \partial_t$$

This, of course, is just the timelike vector of the *vector* frame field corresponding to the covector frame field you wrote down, i.e., it is the vector $s^0$ of the vector frame field $s^0$, $s^1$, $s^2$, $s^3$; the other three vectors would be obtained by raising an index on the other three covectors, i.e.:

$$s^1 = \sqrt{\frac{r - 2m}{r}} \partial_r$$

$$s^2 = \frac{1}{r} \partial_\theta$$

$$s^3 = \frac{1}{r sin \theta} \partial_\phi$$

[Edit: I understand we're in agreement about the physics, this is more for my understanding of the frame field mathematical machinery.]

 

[Mentz114]

Just to elaborate a bit on this: as I understand it, the 4-velocity *covector* of the static observer in the s frame field should be $s_0$, i.e.,

$$u_\mu = \sqrt{\frac{r - 2m}{r}} dt$$[snip]

Those vectors look right. If you do the tensor product you should get the metric.

 

[PeterDonis]

Those vectors look right.

Ok, good.

If you do the tensor product you should get the metric.

The tensor product $g=-s_0\otimes s_0 + s_1\otimes s_1 + s_2\otimes s_2 + s_3\otimes s_3$ gives the Schwarzschild metric. Is that what you mean?

 

[Mentz114]

Ok, good.

The tensor product $g=-s_0\otimes s_0 + s_1\otimes s_1 + s_2\otimes s_2 + s_3\otimes s_3$ gives the Schwarzschild metric. Is that what you mean?

Yes. Where are these questions leading ? I think there's an error the h₀ in my OP. Should be 1 not -1 I think.

 

[PeterDonis]

Yes. Where are these questions leading ?

I'm trying to do the calculations myself, for my own edification.

I think there's an error the h₀ in my OP. Should be 1 not -1 I think.

Meaning, +dt instead of -dt? I think I agree; I'm trying to relate the covectors you wrote to the frame field vectors for the infalling observer that are given (in terms of the Schwarzschild chart) on the Wiki page on frame fields:

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

 

[PeterDonis]

Meaning, +dt instead of -dt? I think I agree; I'm trying to relate the covectors you wrote to the frame field vectors for the infalling observer that are given (in terms of the Schwarzschild chart) on the Wiki page on frame fields:

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

Hmm. When I lower indexes on the components of the frame field vectors for Lemaitre observers, I get the same signs that are in the OP; $h_0$ has a minus sign and $h_1$ has a plus sign for the $dt$ term.

 

[Mentz114]

Hmm. When I lower indexes on the components of the frame field vectors for Lemaitre observers, I get the same signs that are in the OP; $h_0$ has a minus sign and $h_1$ has a plus sign for the $dt$ term.

The h cobasis is correct. It's h⁰ that has the positive time component. I don't why I suddenly thought it was wrong. If you did confirm my result it would be a bonus although it's unlikely to be wrong because it gives the expected accelerations.

 

[PeterDonis]

The h cobasis is correct. It's h⁰ that has the positive time component.

Yes, that's what I get as well.