1. Jan 22, 2010

### snoopies622

What physical meaning can be ascribed to the four-acceleration vector?

For example, for an object moving on the positive x axis with speed 0.6c and accelerating in this direction at rate a, the approximate components of its four-acceleration vector - at least according to my math - are

< 1.465a , 2.44a , 0, 0 >

What does this mean? How is this useful?

2. Jan 22, 2010

### George Jones

Staff Emeritus
I get the same numbers as you.

Let B be an inertial observer, and let A be an accelerated observer who moves along B's x-axis, and who, at some instant, has speed $dx/dt = 3/5$ and coordinate acceleration $d^2 x/dt^2 = a$.

Then, in B's frame, the components of A's 4-acceleration are

$$\left< \frac{375}{256}a, \frac{625}{256}a, 0, 0\right>$$.

The frame-invariant magnitude of this instantaneous 4-acceleration is

$$a' = \sqrt{-\left(\frac{375}{256}a\right)^2 + \left(\frac{625}{256}a\right)^2 } = \frac{125}{64}a.$$

Suppose A is standing on a bathroom scale such that A's body is in the direction of the acceleration, and that there is thrust acting under the scale that causes the acceleration. Then, the bathroom scale gives A's weight as $ma'$.

Suppose C is an inertial observer who moves along B's x-axis with speed 3/5. Then, A and C, are momentarily comoving (zeo relative speed), but A has coordinate acceleration

Then, at this instant, in C's frame, the components of A's 4-acceleration are

$$\left< 0, a', 0, 0\right>$$.

3. Jan 22, 2010

### Staff: Mentor

The norm of the 4-acceleration is the magnitude of the proper acceleration. It is also related to the curvature of a particle's worldline.

4. Jan 22, 2010

### snoopies622

So the four-acceleration helps one find the relationships between how different reference frames measure acceleration. OK, I understand. Thanks!

5. Jan 23, 2010

### snoopies622

Oh, a follow-up:

More than a year ago I asked this forum a question just like this one but about the four-velocity vector, and what I remember learning is that by multiplying it by the rest mass one obtains a four-vector (the four-momentum) that is conserved in interactions. This seems to me to be a very useful thing to know.

Is there a similar conservation law that involves four-acceleration?

6. Jan 23, 2010

### Altabeh

Did you ignore gravity and the gravitational acceleration felt in B's frame? The bathroom scale must show A's weight as $$ma'+mg$$ and remember that we don't use curvilinear coordinates so the gravitational affects on the frame are ruled out and thus we are still in SR.

How do you guys calculate the zeroth component of four-acceleration?? My calculation shows something else!

No because four-acceleration is not a conserved quantitiy.

AB

7. Jan 24, 2010

### George Jones

Staff Emeritus
And in epecial relativity there is no gravity. Even, in general relativity, the scale reads $ma'$. For example, the magnitude of of the 4-acceleration of someone standing on the Earth's surface is $a' = g$.

The relationship between A's coordinate acceleration $a = d^2 x / dt^2$ in B's frame and the magnitude of A's 4-acceleration $a'$ is

$$a' = \gamma^3 a.$$

(See page 35 of of Gron and Hervik,

Let C's inertial frame be denoted by primes.

Then, $a' = d^2 x' / dt'^2$. Then, C's primed components of A's 4-acceleration are

$$\left( a^0', a^1' \left) = \left( 0, a' \left) = \left( 0, \gamma^3 a \left),$$

and a Lorentz boost gives B's unprimed coordinates of A's 4-acceleration. For example,

$$a^0 = \gamma \left( a^0' +va^1' \right) = v \gamma^4 a = \frac{3}{5}\frac{a}{\left(1 - \left( \frac{3}{5} \right)^2 \right)^2}.$$

8. Jan 24, 2010

### DrGreg

Another way of doing this is to note that the 4-velocity is $(\gamma, \gamma v)$, so the 4-acceleration, which has to be orthogonal to this, must be a multiple of $(\gamma v, \gamma)$ and have a magnitude of $\gamma ^3 a$, as stated by George.

9. Jan 24, 2010

### Altabeh

Thank you both!

But I strongly disagree with George on claiming that one has to ignore the weight caused by gravity just because we are in SR. George you say:

And in special relativity there is no gravity. Even, in general relativity, the scale reads ma'. For example, the magnitude of of the 4-acceleration of someone standing on the Earth's surface is a'=g.

As I said before, if the frame itself is not affected by gravity just because we are in SR does not mean we can neglect the gravitational force of Earth exerted on A's body so the bathroom scale must read $$mg+sin(\theta)ma'$$ where $$\theta$$ is the angle between x-axis and the direction of A's motion. And of course in the present case, we get $$W=mg$$ which is the weight measured when being at rest on the surface of Earth. Now I can see that I was wrong too, because I did work ma' into the whole equation rather than paying attention to the fact that the vertical force (supposed to be directed toward the center of Earth) caused by ma' is accountable for a likely extra weight besides the weight mg.

And when I wake up every day, I weight myself on my bathroom scale which agrees with you on your last claim about the equivalnce of g and a'. But I think we've forgot about something so simple-minded: Here x-axis in not directed toward the center of Earth so ma' must not be taken as the weight measured by the scales, rather it has to come with a $$sin(\theta)$$ which is by definition zero.

AB

10. Jan 24, 2010

### snoopies622

For the record, I was only thinking about a flat space-time situation.

Regarding a useful law involving four-acceleration, I have that read F = ma where F and a are both four-vectors, but this seems to me to be more like a definition than a law.

But the fact that a and u are Minkowski-orthogonal to each other (if that's the right way to put it) in every frame of reference, and that the magnitude of a is always equal to that of the "proper acceleration", both seem like very helpful things to know.