B Difficulty with derivatives using the Lorentz transformations

[etotheipi]

Two frames measure the position of a particle as a function of time: S in terms of x and t and S', moving at constant speed v, in terms of x' and t'. The acceleration as measured in frame S is $$ \frac{d^{2}x}{dt^{2}} $$ and that measured in frame S' is $$ \frac{d^{2}x'}{dt'^{2}} $$My question is how can we write the expression for the acceleration in frame S' in terms of that measured in frame S, noting the two coordinate transformations $$x' = \gamma(x-vt)$$ and $$t' = \gamma(t-vx)$$ I have had a go at the first derivative

\begin{align}
\frac{dx'}{dt'} &= \gamma \frac{dx}{dt'} - \gamma v \frac{dt}{dt'} \\ &= \gamma \frac{dx}{dt}\frac{dt}{dt'} - \gamma v \frac{dt}{dt'}
\end{align}
I tried deriving the time transformation with respect to t
\begin{align}
t' &= \gamma t - \gamma xv
\\ \frac{dt'}{dt} &= \gamma - \gamma v \frac{dx}{dt}
\end{align}

so that $$\frac{dt}{dt'} = \frac{1}{\gamma - \gamma v \frac{dx}{dt}}$$ but substituting this in doesn't really help and even then I'm not sure how to handle the second derivative. The aim would be to have some relationship like $$ \frac{d^{2}x}{dt^{2}} = f(\frac{d^{2}x'}{dt'^{2}})$$ where $$f$$ is some function. I was wondering if anyone could give me some tips or guidance since I don't really know if I'm going about this the right way. Thanks a bunch.

 

[PeroK]

I'm not sure why you got stuck. You are nearly there for the so called "velocity transformation".

Use equation (2) and what you have for $dt/dt'$.

Note that $dx/dt$ is the velocity of the particle in frame S. Call this $u$ say. And $u' = dx'/dt'$.

Finish this, then the acceleration transformation is just more of the same.

 

[etotheipi]

I'm not sure why you got stuck. You are nearly there for the so called "velocity transformation".

Use equation (2) and what you have for $dt/dt'$.

Note that $dx/dt$ is the velocity of the particle in frame S. Call this $u$ say. And $u' = dx'/dt'$.

Finish this, then the acceleration transformation is just more of the same.

Thank you, substituting in I get the relation

$$u' = \frac{u - v}{1- uv}$$ so I'll go with that and see if it works!

 

[Orodruin]

Note that what you are after is
$$
\frac{du'}{dt'}.
$$
Just as you expressed $u' = dx'/dt'$ using the chain rule, you can do the same for $du'/dt'$. You have already computed $dt/dt'$ so that part should be no problem.

 

[PeroK]

Thank you, substituting in I get the relation

$$u' = \frac{u - v}{1- uv}$$ so I'll go with that and see if it works!

A good test is to try $u = 1$ and check that $u' =1$, for any $v$.

This of course represents the invariance of the speed of light.

 

[etotheipi]

I get an answer of

$$\frac{d^{2}x}{dt^{2}} = \frac{1}{\gamma}\frac{1-v^{2}}{(1-uv)^{3}} \frac{d^{2}x'}{dt'^{2}}$$
Would it be ok to rewrite
$$\frac{1}{\gamma} = (1-v^{2})^{\frac{1}{2}}$$ so that this becomes
$$\frac{d^{2}x}{dt^{2}} = \frac{(1-v^{2})^{\frac{3}{2}}}{(1-uv)^{3}} \frac{d^{2}x'}{dt'^{2}}$$

 

[PeroK]

I get an answer of

$$\frac{d^{2}x}{dt^{2}} = \frac{1}{\gamma}\frac{1-v^{2}}{(1-uv)^{3}} \frac{d^{2}x'}{dt'^{2}}$$
Would it be ok to rewrite
$$\frac{1}{\gamma} = (1-v^{2})^{\frac{1}{2}}$$ so that this becomes
$$\frac{d^{2}x}{dt^{2}} = \frac{(1-v^{2})^{\frac{3}{2}}}{(1-uv)^{3}} \frac{d^{2}x'}{dt'^{2}}$$

Are you sure you have $a$ and $a'$ the right way round?

 

[etotheipi]

Are you sure you have $a$ and $a'$ the right way round?

Yeah you're right, sorry about that! It should then actually be

$$\frac{d^{2}x'}{dt'^{2}} = \frac{(1-v^{2})^{\frac{3}{2}}}{(1-uv)^{3}} \frac{d^{2}x}{dt^{2}}$$

 

[haushofer]

Two frames measure the position of a particle as a function of time: S in terms of x and t and S', moving at constant speed v, in terms of x' and t'. The acceleration as measured in frame S is $$ \frac{d^{2}x}{dt^{2}} $$ and that measured in frame S' is $$ \frac{d^{2}x'}{dt'^{2}} $$My question is how can we write the expression for the acceleration in frame S' in terms of that measured in frame S, noting the two coordinate transformations $$x' = \gamma(x-vt)$$ and $$t' = \gamma(t-vx)$$ I have had a go at the first derivative

\begin{align}
\frac{dx'}{dt'} &= \gamma \frac{dx}{dt'} - \gamma v \frac{dt}{dt'} \\ &= \gamma \frac{dx}{dt}\frac{dt}{dt'} - \gamma v \frac{dt}{dt'}
\end{align}
I tried deriving the time transformation with respect to t
\begin{align}
t' &= \gamma t - \gamma xv
\\ \frac{dt'}{dt} &= \gamma - \gamma v \frac{dx}{dt}
\end{align}

so that

$$\frac{dt}{dt'} = \frac{1}{\gamma - \gamma v \frac{dx}{dt}}$$

but substituting this in doesn't really help and even then I'm not sure how to handle the second derivative. The aim would be to have some relationship like

$$ \frac{d^{2}x}{dt^{2}} = f(\frac{d^{2}x'}{dt'^{2}})$$

where $$f$$ is some function. I was wondering if anyone could give me some tips or guidance since I don't really know if I'm going about this the right way. Thanks a bunch.

Why is the derivate of gamma w.r.t. the time coordinate t zero? We're talking accelerations here. Or am I missing something?

 

[PeroK]

Why is the derivate of gamma w.r.t. the time coordinate t zero? We're talking accelerations here. Or am I missing something?

That's the gamma factor between the frames.

 

[haushofer]

That's the gamma factor between the frames.

Oh, wait, never mind. Mixing up accelerations and accelerating frames.