Dynamics of the Slowly Accelerated Electron from Einstein's original SR paper

[SamRoss]

I am having difficulty working out one of the steps in Einstein's original SR paper, in the Section entitled "Dynamics of the Slowly Accelerated Electron". Einstein describes an electron in motion in an electromagnetic field. Its equation of motion in a rest frame K (not moving with the electron) is

(1) m(d^2 x/dt^2)=F

where m is the mass of the electron and F is the force on the electron in the x direction. (I am ignoring a constant e that Einstein put in). This is just f=ma.

At a given instant of time, the electron is moving with velocity v. In a frame of reference k, traveling at the same velocity, the equation of motion looks like this...

(2) m(d^2 x'/dt'^2)=F'

Einstein then lists the previously obtained Lorentz transformations...

(3) x'=B(x-vt) and t'=B(t-vx/c^2) where B=(1-v^2/c^2)^-1/2

as well as the transformation for the force on the electron, which turns out to simply be...

(4) F=F' (for this post I am leaving out the y and z directions).

Now, using these transformations, Einstein transforms equation (2) from system k to system K. The result is

(5) mB^3(d^2 x/dt^2)=F

In case this is hard to read the way I wrote it, it turns out to be equation (1) with the left side multiplied by B^3.

My question is, how did Einstein get from (2) to (5)? I'm pretty sure the chain rule is involved but I keep getting the wrong answer. If anyone can take me step by step from (2) to (5) I would greatly appreciate it. Thanks!

 

[bcrowell]

Hi, SamRoss,

Welcome to PF!

You will be more likely to get good responses if you make your posts easy to read by marking them up in LaTeX. For example, here's your equation (5): [itex]m\beta^3(d^2 x/dt^2)=F [/itex]. To see how I did that, hit the QUOTE button at the bottom of my post. When you preview your own posts with math in them, the math will often be displayed incorrectly, showing up as some other, unrelated equation. Don't worry about this. After you submit your post, it will be shown correctly. This is a known bug.

Although Einstein is talking about electromagnetism and Newton's second law, I believe this is simply a matter of relativistic kinematics. If I see someone else moving, and I see him measure an acceleration, I consider his result to be off by a factor of [itex]\gamma^3[/itex] ([itex]\beta^3[/itex] in Einstein's notation). His clocks are slow by a factor of γ, which leads to a factor γ² in the acceleration. His measuring rods are also too short, so he judges distances to be too big by a factor of γ. Putting this all together, you get γ³.

IMO part I of Einstein's 1905 paper is a good thing for a modern reader to study carefully, but wading through the archaic notation of part II is a bad idea unless you're a historian of science.

-Ben

 

[jtbell]

To see how I did that, hit the QUOTE button at the bottom of my post.

Or simply click on the equation.

 

[SamRoss]

His clocks are slow by a factor of γ, which leads to a factor γ² in the acceleration. His measuring rods are also too short, so he judges distances to be too big by a factor of γ. Putting this all together, you get γ³.

-Ben

Thanks for the quick reply Ben. I'm still a bit confused though. His clock being off by a factor of gamma is clear, but I don't see how this immediately leads to a factor of gamma squared in the acceleration. More explicitly, what equation are we taking the second derivative of that leaves us with gamma squared? Also, how do we "put together" distance being off by a factor of gamma and the second time derivative being off by gamma squared?

Thanks again for all your help! I'm definitely going to use this site more often

 

[bcrowell]

Thanks for the quick reply Ben. I'm still a bit confused though. His clock being off by a factor of gamma is clear, but I don't see how this immediately leads to a factor of gamma squared in the acceleration. More explicitly, what equation are we taking the second derivative of that leaves us with gamma squared? Also, how do we "put together" distance being off by a factor of gamma and the second time derivative being off by gamma squared?

It's just a generic argument about scaling. When you take a second derivative d²y/dx², that's the effect you get from a rescaling of y and x. It's independent of the actual functional dependence of y on x.

 

[SamRoss]

It's just a generic argument about scaling. When you take a second derivative d²y/dx², that's the effect you get from a rescaling of y and x. It's independent of the actual functional dependence of y on x.

I'm sure you're right and I'm just not clicking on something, but using the method you described, couldn't I say that the formula for transformation of velocities is just v'=v times gamma squared (sorry, I still haven't gotten LaTeX yet, but I will) to account for the length contraction together with the time dilation. That is obviously not the known result.

 

[bcrowell]

I'm sure you're right and I'm just not clicking on something, but using the method you described, couldn't I say that the formula for transformation of velocities is just v'=v times gamma squared (sorry, I still haven't gotten LaTeX yet, but I will) to account for the length contraction together with the time dilation. That is obviously not the known result.

The argument I gave in #2 is just a plausibility argument, not a rigorous derivation. It only takes into account the scaling factors that go into the Lorentz transformation, not the additive terms. Einstein is making some of those additive terms go away by making the simplifying assumption that the electron is at rest at t=0 in one of those frames. If you don't make that assumption, then the transformation of the acceleration is not just a factor of [itex]\gamma^3[/itex], it's
[tex]a'=a\frac{(1-v^2)^{3/2}}{(1-uv)^3}[/tex],
where c=1, u is the velocity in the unprimed frame, and v is the velocity of one frame relative to the other. Because he assumes u=0, he gets the simpler form.

 

[SamRoss]

Ok, it's clearer now. Thanks for all your help, Ben. I really appreciate it.