# Difficulty with derivatives using the Lorentz transformations

• B
• etotheipi
Sorry.Oh, wait, never mind. Mixing up accelerations and accelerating frames. Sorry.In summary, two frames, S and S', measure the position of a particle as a function of time, with S in terms of x and t and S' in terms of x' and t'. The acceleration in frame S is given by $$\frac{d^{2}x}{dt^{2}}$$ and in frame S' by $$\frac{d^{2}x'}{dt'^{2}}$$. To write the expression for the acceleration in frame S' in terms of that measured in frame S, we can use the two coordinate transformations $$x' = \gamma(x-vt)$$ and $$t' = etotheipi Two frames measure the position of a particle as a function of time: S in terms of x and t and S', moving at constant speed v, in terms of x' and t'. The acceleration as measured in frame S is$$ \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$$is some function. I was wondering if anyone could give me some tips or guidance since I don't really know if I'm going about this the right way. Thanks a bunch. Last edited by a moderator: I'm not sure why you got stuck. You are nearly there for the so called "velocity transformation". Use equation (2) and what you have for ##dt/dt'##. Note that ##dx/dt## is the velocity of the particle in frame S. Call this ##u## say. And ##u' = dx'/dt'##. Finish this, then the acceleration transformation is just more of the same. etotheipi PeroK said: I'm not sure why you got stuck. You are nearly there for the so called "velocity transformation". Use equation (2) and what you have for ##dt/dt'##. Note that ##dx/dt## is the velocity of the particle in frame S. Call this ##u## say. And ##u' = dx'/dt'##. Finish this, then the acceleration transformation is just more of the same. Thank you, substituting in I get the relation$$u' = \frac{u - v}{1- uv}$$so I'll go with that and see if it works! PeroK Note that what you are after is$$
\frac{du'}{dt'}.
$$Just as you expressed ##u' = dx'/dt'## using the chain rule, you can do the same for ##du'/dt'##. You have already computed ##dt/dt'## so that part should be no problem. etotheipi etotheipi said: Thank you, substituting in I get the relation$$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}}$$etotheipi said: I get an answer of$$\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}}$$Are you sure you have ##a## and ##a'## the right way round? etotheipi PeroK said: Are you sure you have ##a## and ##a'## the right way round? Yeah you're right, sorry about that! It should then actually be$$\frac{d^{2}x'}{dt'^{2}} = \frac{(1-v^{2})^{\frac{3}{2}}}{(1-uv)^{3}} \frac{d^{2}x}{dt^{2}}$$PeroK etotheipi said: Two frames measure the position of a particle as a function of time: S in terms of x and t and S', moving at constant speed v, in terms of x' and t'. The acceleration as measured in frame S is$$ \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?

haushofer said:
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.

PeroK said:
That's the gamma factor between the frames.
Oh, wait, never mind. Mixing up accelerations and accelerating frames.

## 1. What are the Lorentz transformations?

The Lorentz transformations are a set of equations used in special relativity to describe how measurements of space and time are affected by the relative motion between two observers.

## 2. Why do we need to use derivatives when working with Lorentz transformations?

Derivatives are used to calculate the rate of change of a quantity, which is necessary when dealing with the time dilation and length contraction effects predicted by the Lorentz transformations.

## 3. What are some common difficulties encountered when working with derivatives in Lorentz transformations?

Some common difficulties include understanding the concept of a derivative in the context of special relativity, correctly applying the chain rule and product rule, and dealing with the non-linear nature of the Lorentz transformations.

## 4. How can I improve my understanding and proficiency in using derivatives in Lorentz transformations?

Practicing with various examples and exercises, seeking clarification from a teacher or mentor, and studying the underlying mathematics and physics principles can all help improve understanding and proficiency in using derivatives in Lorentz transformations.

## 5. Are there any alternative methods for dealing with Lorentz transformations besides using derivatives?

Yes, there are alternative methods such as using geometric interpretations, tensor calculus, or computer simulations. However, derivatives are the most commonly used and efficient method for dealing with Lorentz transformations in most cases.

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