In order to define the acceleration of a body in its own frame, we need to first prove that the acceleration is a four-vector so that its dot product with itself can then be labeled as acceleration squared in the rest frame. For velocity and displacement vectors, we can show that they have a constant dot product. But how do we prove that for acceleration?
$\begingroup$
$\endgroup$
7
-
$\begingroup$ Normally it's defined as a 4 vector. What alternative definition are you using? $\endgroup$– WillOCommented Oct 27, 2022 at 13:34
-
1$\begingroup$ Are position and velocity 4 vectors in SR? $\endgroup$– Jon CusterCommented Oct 27, 2022 at 13:36
-
$\begingroup$ @WillO thank you for your comment. I meant that, for velocity four vectors, I can show that their dot product is a constant, but we don’t do the same for acceleration, so do we just assume that it satisfies as well?. $\endgroup$– ShaashaankCommented Oct 27, 2022 at 14:58
-
$\begingroup$ First, we must distinguish between what is a constant and what is an invariant. A scalar quantity having a specific value in all space points of an (inertial here) reference frame is a constant. A scalar quantity having the same value in all (inertial here) reference frames is an invariant. So you have proved that $\:(\mathbf{v\cdot v})\:$ is an invariant, more exactly it's $\:c^2\:$ in all inertial frames. $\endgroup$– VoulkosCommented Oct 27, 2022 at 15:24
-
1$\begingroup$ If the value of $a \cdot a$ is different some other frame, then it's not an invariant. An invariant is the same in all frames by its very definition. $\endgroup$– Michael SeifertCommented Oct 28, 2022 at 11:37
|
Show 2 more comments
3 Answers
$\begingroup$
$\endgroup$
5
Is it not so by definition? $$ {\bf a}= \frac {d{\bf v}}{d\tau} $$ where $$ {\bf v}= \frac{d{\bf x}}{d \tau} $$ is a 4-vector and $\tau$ is a scalar.
answered Oct 27, 2022 at 13:36
-
$\begingroup$ Thank you for answering, for velocity, I can prove that the dot product of velocity with itself is a constant, how do we show that for acceleration?. $\endgroup$ Commented Oct 27, 2022 at 14:44
-
1$\begingroup$ @Shaashaank : First, we must distinguish between what is a constant and what is an invariant. A scalar quantity having a specific value in all space points of an (inertial here) reference frame is a constant. A scalar quantity having the same value in all (inertial here) reference frames is an invariant. So you have proved that $(\mathbf{v\cdot v})$ is an invariant, more exactly it's $\:c^2\:$ in all inertial frames. $\endgroup$– VoulkosCommented Oct 27, 2022 at 15:27
-
1$\begingroup$ @Shaashaank : But if you have proved this it's not difficult to prove that $(\mathbf{a\cdot a})$ is an invariant following the same footing (replace $\mathbf x,\mathbf v$ with $\mathbf v,\mathbf a$ respectively). $\endgroup$– VoulkosCommented Oct 27, 2022 at 15:35
-
1$\begingroup$ @Shaashaank : As a hint $(\mathbf{a\cdot a})=-\Vert \boldsymbol\alpha_0\Vert^2=\text{invariant}$, where $\:\boldsymbol\alpha_0\: $ is the acceleration 3-vector with respect to the rest frame of a particle. $\endgroup$– VoulkosCommented Oct 27, 2022 at 15:46
-
$\begingroup$ @Shaashaank : Because of your low reputation you could not vote any post, answer or question. But you can accept an answer to your question as best giving +15 reputation to the answerer. Try to learn how. And by the way, Welcome to PSE. $\endgroup$– VoulkosCommented Oct 27, 2022 at 15:52
$\begingroup$
$\endgroup$
Since you accept that four-velocity is a four-vector, this is an argument that four-acceleration is a four-vector:
$$a^{\mu}=\lim _{h\rightarrow 0} \frac{v^{\mu}(\tau+h)-v^{\mu}(\tau)}{h}$$.
The path is parametrized by $\tau$, the proper time, which is a scalar because it's equal to the spacetime interval (upto maybe a sign)
You can imagine carrying out this limit calculation in two different frames. If you're doing a numerical calculation, you will take $h$ to a small finite number.
The numerator will be a difference of four-vectors, hence it is a four-vector. The denominator is a scalar. Hence, the fraction is a four-vector.
This isn't a proof of course. A proof will try to argue that the limit of the sequence four-vectors, as $h\rightarrow 0$, will also be a four-vector (which makes sense sort-of)
EDIT Ok so, let's say $v^{\mu}(h)$ is a sequence of four-vectors paramerized by a real parameter $h$. Let $v^{\nu} (h)=\Lambda v^{\mu}(h)$, $\Lambda$ is a Lorentz transform. Then,
$$\lim_{h\rightarrow 0}v^{\nu} (h)= \lim_{h\rightarrow 0}\Lambda v^{\mu}(h)$$
$$=\Lambda \lim_{h\rightarrow 0} v^{\mu}(h)$$
So you see, the limit in one frame is the Lorentz transform of the limit in another frame (We could pull $\Lambda$ out of the limit because it's a constant matrix)
$\begingroup$
$\endgroup$
8
In physics, we prove things with experiments. Four-vectors are components of a mathematical model. Does that model pass experimental test? Yes, in many cases. We use it because it passes those tests, not because of any mathematical proof.
-
1$\begingroup$ The 4-accelaration is defined and by its definition is a priori 4-Lorentz vector, see @mike stone's answer, not proved by experimental tests. $\endgroup$– VoulkosCommented Oct 27, 2022 at 14:11
-
$\begingroup$ @Frobenius But if that definition did not lead to results confirmed by experiment, we would reject it as irrelevant to physics. $\endgroup$ Commented Oct 27, 2022 at 14:22
-
1$\begingroup$ In mechanics I define the following named "stupid 3-dimensional momentium" $$\mathbf{s3m}=m^2\upsilon\,\mathbf n$$ What is the meaning of the "experimental test" of my definition ??? Note that I don't mention for what purpose I will use this 3-dimensional quantity. $\endgroup$– VoulkosCommented Oct 27, 2022 at 14:43
-
$\begingroup$ @Frobenius thank you for replying, I meant that, for four velocity, we can show that it’s dot product is a constant, but for acceleration do we just assume it?. $\endgroup$ Commented Oct 27, 2022 at 14:47
-
1$\begingroup$ What do you mean by "constant", as @Frobeneius says, ${\bf a}\cdot {\bf a}$ is an "invariant" , meaning independent of reference frame, but is not a "constant" as (unlike ${\bf v}\cdot {\bf v}=-c^2$) it can vary with with $\tau$. $\endgroup$ Commented Oct 27, 2022 at 15:33