# 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