B Difficulty with derivatives using the Lorentz transformations

  • Thread starter etotheipi
  • Start date
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

Science Advisor
Homework Helper
Insights Author
Gold Member
2018 Award
9,510
3,492
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.
 
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

Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
2018 Award
15,759
5,758
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

Science Advisor
Homework Helper
Insights Author
Gold Member
2018 Award
9,510
3,492
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.
 
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

Science Advisor
Homework Helper
Insights Author
Gold Member
2018 Award
9,510
3,492
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?
 
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

Science Advisor
Insights Author
2,190
541
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

Science Advisor
Homework Helper
Insights Author
Gold Member
2018 Award
9,510
3,492
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.
 

Want to reply to this thread?

"Difficulty with derivatives using the Lorentz transformations" You must log in or register to reply here.

Related Threads for: Difficulty with derivatives using the Lorentz transformations

Replies
2
Views
3K
Replies
27
Views
3K
Replies
4
Views
457
Replies
1
Views
6K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads

Top