Relativistic Force & Velocity Relation to Acceleration

[lriuui0x0]

According to this link here https://en.wikipedia.org/wiki/Relativistic_mechanics#Force , we can inverse the relation of force in terms of velocity and acceleration:

$$
\mathbf{F} = \frac{m\gamma^3}{c^2}(\mathbf{v} \cdot \mathbf{a})\mathbf{v} + m\gamma\mathbf{a}
$$

to get:

$$
\mathbf{a} = \frac{1}{m\gamma}(\mathbf{F} - \frac{(\mathbf{v}\cdot\mathbf{F})\mathbf{v}}{c^2})
$$

I'm not sure how this derivation is done, especially how to inverse the $\mathbf{v}\cdot\mathbf{a}$ term.

 

[PeroK]

Why not take $\mathbf F$ as in the first equation and using this calculate the RHS of the second equation?

 

[lriuui0x0]

Why not take $\mathbf F$ as in the first equation and using this calculate the RHS of the second equation?

Good point! But I wonder what happens if you don't already know the RHS of the second equation? So how do you derive $\mathbf{a}$ in the first place?

 

[PeroK]

Good point! But I wonder what happens if you don't already know the RHS of the second equation? So how do you derive $\mathbf{a}$ in the first place?

First, you have the inspiration to look at $\mathbf v \cdot \mathbf F$. Then you look for another good idea.

 

[Ibix]

If you want to know $\mathbf{v}\cdot\mathbf{a}$ then it's the last term that doesn't help you. Is there anything you can do to $m\gamma\mathbf{a}$ to turn it into a term like $\mathbf{v}\cdot\mathbf{a}$?

 

[lriuui0x0]

Thanks for the help. I'm able to derive the relationship betwen $\mathbf{F}\cdot\mathbf{v}$ and $\mathbf{a}\cdot\mathbf{v}$:

$$
\begin{aligned}
&\phantom{{}={}} \mathbf{F}\cdot\mathbf{v} \\
&= m\gamma(\frac{\gamma^2}{c^2}\mathbf{v}^2(\mathbf{v}\cdot\mathbf{a}) + \mathbf{v}\cdot\mathbf{a}) \\
&= m\gamma(\frac{\gamma^2}{c^2}\mathbf{v}^2 + 1)(\mathbf{v}\cdot\mathbf{a}) \\
&= m\gamma(\frac{v^2}{(1-\frac{v^2}{c^2})c^2} + 1)(\mathbf{v}\cdot\mathbf{a}) \\
&= m\gamma(\frac{c^2}{c^2-v^2})(\mathbf{v}\cdot\mathbf{a}) \\
&= m\gamma(\frac{1}{1-\frac{v^2}{c^2}})(\mathbf{v}\cdot\mathbf{a}) \\
&= m\gamma^3(\mathbf{v}\cdot\mathbf{a}) \\
\end{aligned}
$$

 

[Ibix]

Yes. Note that $v^2/c^2=1-\gamma^{-2}$, which gets you from your second line to your last somewhat more rapidly.

 

[lriuui0x0]

Yes. Note that $v^2/c^2=1-\gamma^{-2}$, which gets you from your second line to your last somewhat more rapidly.

Thanks for the note!

 

[vanhees71]

This is very complicated, because it's using the non-covariant version of the point-mechanics equation of motion. It's much more convenient to use the manifestly covariant formulation,
$$\dot{p}^{\mu}=m \ddot{x}^{\mu}=K^{\mu},$$
where the dot stands for the derivative wrt. proper time. Since $\dot{x}_{\mu} \dot{x}^{\mu}=c^2$ the Minkowski force $K^{\mu}$ must obey
$$\dot{x}^{\mu} K_{\mu}=0,$$
because
$$\dot{x}^{\mu} \dot{x}_{\mu}=c^2 \; \Rightarrow \; \ddot{x}^{\mu} \dot{x_{\mu}}=0.$$
Due to this constraint of course only 3 of the 4 equations are independent, as it should be.

Since the coordinate time $t$ is related uniquely to $\tau$, because $\dot{t}=\gamma>0$ you can of course use the spatial part to got back to the non-covariant (1+3)-formulation:
$$p^{\mu}=m \dot{x}^{\mu}=m \gamma \mathrm{d}_t x^{\mu} \; \rightarrow \; \vec{p}=m \gamma \mathrm{d}_t \vec{x}.$$
Then the spatial components of the covariant EoM read
$$\dot{\vec{p}}=\gamma \mathrm{d}_t \vec{p}=m \gamma \mathrm{d}_t (\gamma \vec{x})=\vec{K}$$
or
$$\mathrm{d}_t \vec{p}=m \mathrm{d}_t (\gamma \mathrm{d}_t \vec{x})=\frac{1}{\gamma} \vec{K}:=\vec{F}.$$
The equation of the time component reads
$$\dot{p}^0=\gamma \mathrm{d}_t p^0=K^0.$$
Since
$$K^{\mu} \dot{x}_{\mu} = \gamma (c K^0-\vec{v} \cdot \vec{K})=0 \; \Rightarrow\; K^0=\vec{\beta} \cdot \vec{K}=\gamma \vec{\beta} \cdot \vec{F}$$
we get
$$\mathrm{d}_t p^0=\vec{\beta} \cdot \vec{F}$$
or since $E=c p^0$
$$\mathrm{d}_t E=\vec{v} \cdot \vec{F}.$$
So the time component is simply the work-energy theorem for relativistic mechanics.

 

[pervect]

Conceptually, it's probably easier to start with

$$ F = dp/ dt \quad p= \gamma m v $$

where $\gamma = 1 / \sqrt{1-v^2/c^2}$.

Then the chain rule tells you that

$$F =m\,v\,(d/dt) \gamma + \gamma m (d/dt) v$$

Much of the mathematical complexity can be minimized by replacing the 3-momentum p with the 4-momentum, and replacing the coordinate time t with the proper time $\tau$. This is formalized in the 4-vector approach. However, it's probably going to be too confusing to actually grasp 4-vectors from a short post, though vanhees71 has outlined the approach. Wiki has a treatment, but it's also rather terse to learn from, a book such as Taylor & Wheeler's "Space-time Physics" would be a more realistic approach.

Note that in general, regarding vectors as arrows with magnitude and direction, p and v point in the same direction, while F and v do not point in the same direction.

 

[vanhees71]

For a more detailed treatment of special relativity (and also point-particle mechanics) in terms of four-vectors:

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

I find the non-covariant coordinate-time formalism utmost confusing in comparison to the four-vector formalism, although the latter has the complication of using a redundant degree of freedom ($x^0$), which is compensated by a constraint like $\dot{x}_{\mu} \dot{x}^{\mu}=c^2$, defining the world-line parameter to be the proper time of the particle (which is the natural coordinate-independent world-line parameter with a concrete physical meaning namely the time measured by a clock comoving with the partice).

 

[lriuui0x0]

For a more detailed treatment of special relativity (and also point-particle mechanics) in terms of four-vectors:

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

I find the non-covariant coordinate-time formalism utmost confusing in comparison to the four-vector formalism, although the latter has the complication of using a redundant degree of freedom ($x^0$), which is compensated by a constraint like $\dot{x}_{\mu} \dot{x}^{\mu}=c^2$, defining the world-line parameter to be the proper time of the particle (which is the natural coordinate-independent world-line parameter with a concrete physical meaning namely the time measured by a clock comoving with the partice).

Thanks for the resource!

 

[Grasshopper]

Yes. Note that $v^2/c^2=1-\gamma^{-2}$, which gets you from your second line to your last somewhat more rapidly.

That is so sneaky. I like it.

 

[robphy]

In terms of rapidity, where $(v/c)=\tanh\theta$ and $\gamma=\cosh\theta$ (and $k=\exp \theta$),
the expression $v^2/c^2=1-\gamma^{-2}$
corresponds to $\tanh^2\theta=1-\mbox{sech}^2\theta$,
which is equivalent to $\frac{\sinh^2\theta}{\cosh^2\theta}=1-\frac{1}{\cosh^2\theta}$ or, by rearrangement, $\cosh^2\theta-\sinh^2\theta=1$.

In

$$
\begin{aligned}
&\phantom{{}={}} \mathbf{F}\cdot\mathbf{v} \\
&\stackrel{\vdots}{=}\\
&\stackrel{\vdots}{=} m\gamma(\frac{\gamma^2}{c^2}\mathbf{v}^2 + 1)(\mathbf{v}\cdot\mathbf{a}) \\
&\stackrel{\vdots}{=} m\gamma^3(\mathbf{v}\cdot\mathbf{a}) \\
\end{aligned}
$$

note that $(\frac{\gamma^2}{c^2}\mathbf{v}^2 + 1)$ is $(\sinh^2\theta+1)$,
but this is equal to $(\cosh^2\theta)$.

 

[lriuui0x0]

That is so sneaky. I like it.

Yeah, a easier to remember form is $\beta^2 + \gamma^{-2} = 1$ :)

 

[Grasshopper]

Yeah, a easier to remember form is $\beta^2 + \gamma^{-2} = 1$ :)

But that surely means you can easily change those to trig functions. I am now intrigued. That negative power raises a small problem, but surely there is a simple work around I’m missing. An i or something.

 

[DrGreg]

But that surely means you can easily change those to trig functions. I am now intrigued. That negative power raises a small problem, but surely there is a simple work around I’m missing. An i or something.

$$\tanh^2 \theta + \operatorname{sech}^2 \theta= 1$$which follows from$$\sinh^2 \theta+ 1= \cosh^2 \theta$$

 

[vanhees71]

But that surely means you can easily change those to trig functions. I am now intrigued. That negative power raises a small problem, but surely there is a simple work around I’m missing. An i or something.

I guess you mean hyperbolic functions. You have
$$\beta=\tanh \eta, \quad \gamma=\cosh \eta$$
Then, of course,
$$\beta^2+\gamma^{-2}=\tanh^2 \eta+\frac{1}{\cosh^2 \eta}=\frac{\sinh^2 \eta+1}{\cosh^2 \eta}=1.$$

 

[Grasshopper]

I guess you mean hyperbolic functions. You have
$$\beta=\tanh \eta, \quad \gamma=\cosh \eta$$
Then, of course,
$$\beta^2+\gamma^{-2}=\tanh^2 \eta+\frac{1}{\cosh^2 \eta}=\frac{\sinh^2 \eta+1}{\cosh^2 \eta}=1.$$

Yes. But presumably you could use the simple Pythagorean trig identity by replacing $γ^{-1}$ with $\frac{1}{γ}$ could you not? (I wonder if that would be of use at all) That is, $γ^{-2} = γ^{(^-1)(2)} = (\frac{1}{γ})^2$

$\beta^2+(\frac{1}{γ})^2 = 1$

looks the same to me.

But even if that's true, everything I've seen in SR screams hyperbolic.

 

[vanhees71]

Are you referring to Loedel diagrams? I find them utmost unintuitive and besides the point.

The important point of SR is that you do NOT have a Euclidean space-time, because such a spacetime would not admit a causality structure, but a pseudo-Euclidean spacetime with the signature (+---) or (-+++) depending on your choice of convention.

 

[Grasshopper]

Are you referring to Loedel diagrams? I find them utmost unintuitive and besides the point.

The important point of SR is that you do NOT have a Euclidean space-time, because such a spacetime would not admit a causality structure, but a pseudo-Euclidean spacetime with the signature (+---) or (-+++) depending on your choice of convention.

Not with knowledge of the term but I suppose yes. But yeah, as a student I will agree that standard Minkowski diagrams are more intuitive. Loedel diagrams may have some usefulness, but not at the expense of opening the door to more confusion. There’s enough unintuitive stuff to worry about without adding in a different kind of spacetime diagram.