# Difficulty with derivatives using the Lorentz transformations

• B
Gold Member
2019 Award
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.

Last edited:

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PeroK
Homework Helper
Gold Member
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
Gold Member
2019 Award
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!

PeroK
Orodruin
Staff Emeritus
Homework Helper
Gold Member
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.

etotheipi
PeroK
Homework Helper
Gold Member
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
Gold Member
2019 Award

$$\frac{d^{2}x}{dt^{2}} = \frac{1}{\gamma}\frac{1-v^{2}}{(1-uv)^{3}} \frac{d^{2}x'}{dt'^{2}}$$
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
Homework Helper
Gold Member

$$\frac{d^{2}x}{dt^{2}} = \frac{1}{\gamma}\frac{1-v^{2}}{(1-uv)^{3}} \frac{d^{2}x'}{dt'^{2}}$$
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
Gold Member
2019 Award
Are you sure you have ##a## and ##a'## the right way round?
Yeah you're right, sorry about that! It should then actually be

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

PeroK
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
Homework Helper
Gold Member
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