Ive been reading ON THE ELECTRODYNAMICS ...

[Galadirith]

Hi guys, I came across On the Electrodynamics of Moving Bodies while searching the net on relativity. I started reading it cause I thought It would be quite fun to do, not hoping to understand allot of it. I understand the basic (no-mathematical) concepts behind relativity, but I would like to understand the maths a bit better. What has been bugging me for ages is how is this step made:

\frac{1}{2} \left [ \ \tau (0,0,0,t) \ + \ \tau \left(0,0,0,t + \frac{x^\prime}{c - v} + \frac{x^\prime}{c+v} \right) \right ] = \tau \left(x^\prime,0,0,t + \frac{x^\prime}{c - v} \right)

Hence if x' be chosen infinitesimally small

\frac{1}{2} \left( \frac{1}{c - v} + \frac{1}{c+v}\right)\frac{\partial\tau}{\partial t} = \frac{\partial\tau}{\partial x^\prime} \ + \ \frac{1}{c-v}\frac{\partial\tau}{\partial t}

now I am still at A-level so havnt been to uni, dont know a whole lot on partial dirvatives, only what I have read in articles etc I can find online, so if peoples recomendation is leave this till I get to uni and then can do it there then thats fine, but from the looks of it this is a relatively easy step in the grand scheme of the paper, so if anyone could explain that step it would be really appreciated. Please don't flame me for asking this, I only ask because I want to understand, thanks guys :-)

[Mentz114]

I'd like to help but I don't have the text ( and you haven't given a page reference). Is x' a derivative or a transformed coordinate like x' = x + v*t ?

[Galadirith]

Hi Mentz114, thanks sorry about that I should have posted the file reference, the paper can be found here 'On the Electrodynamics of Moving Bodies' (http://www.fourmilab.ch/etexts/einstein/specrel/specrel.pdf) and the equations are on page 6, x' is a transformed coordinate, however that would have been evident from the paper if I had posted the link, so that my bad again thanks Mentz114.

[Mentz114]

Hi Galadirith,
after some head scratching I think I've got it. Start with the principle

\delta\tau = \frac{\partial\tau}{\partial x}\delta x + \frac{\partial\tau}{\partial t}\delta t.

In the left hand side of the first equation \delta x = 0 and

\delta t = \frac{1}{c - v} + \frac{1}{c + v}

This gives the LHS of the second equation. Repeating for the right hand side we get \delta x = 1 and

\delta t = \frac{1}{c - v}

which gives the RHS of the second equation.

I wonder how calculus was taught in 1905 ?

M

[Galadirith]

Thank so much Mentz114, In fact I didn't even know the first principle you stated, I know the chain rule but had no idea it could also apply to a function of several variables, I will reflex on that, clearly I do need to learn more but that will come with time, thanks so much Mentz114 thats really helpful :-)

[Mentz114]

I'm glad to be of help. I'm posting again with some advice. The early Einstein papers are important but not mathematically elegant. Hermann Minkowski read the SR papers and realised that it all becomes becomes simple if space and time are united in a special space, in which distance is measured in four dimensions according to the following rule -

(ds)^2 = -c^2(dt)^2 + (dx)^2 + (dy)^2 + (dz)^2

and clock time is given by the same formula divided by the constant c^2.

(d\tau)^2 = -(dt)^2 + \frac{1}{c^2}\left((dx)^2 + (dy)^2 + (dz)^2\right)

I find the original "On the Electrodynamics ..." fairly hard. Have fun.

M

[Galadirith]

Thanks so much Mentz114 for the post, now that looks far more intuitive, obviously this is a a small glimpse of Minkowski interpretation but to me that looks far more, well ill say it again, intuitive. Thanks so much Mentz114, i can go on and explore more :-)

[Integral]

Here is my approach to this question.

[Galadirith]

Thanks Integral, thats a big help, I am kind of surprised I couldn't find anywhere on the net that actually explained that, in fact almost every site I have come across doesn't seem to make an awful lot of reference to 'on the electrodyna...', this has been really annoying me for probably a year. Thanks so much to you Integral and Mentz114.

PS: Could I ask did you write that PDF specifically for this Integral?

[Integral]

Yes, I dwelled on that gap in AE's paper for several weeks before I came up with that simple algebraic solution. That was about 7yrs ago, I think I used a adware PDF writer to create something other then a MS doc file. I think what I have is really very similar (though less elegant) to what Mentz114 has posted.

[Galadirith]

Thanks so much Integral, yeh I kind of cant see why he left the maths out of that step I mean its not like its and entirely logical step as you put its 'obvious'. I ask my Maths teacher and he suggested that Einstein left that gap in the maths so that people could do it for them selves; he intentionally left that gap in the maths. I suppose that does sound like something he might do, but I say that from a very limited knowledge of what Einstein was like. Anyway many thanks, that will put my mind at rest ... at least until I find another gap in maths somewhere :-)

[jtbell]

I kind of cant see why he left the maths out of that step I mean its not like its and entirely logical step as you put its 'obvious'.

When people write formal papers for publication, the "audience" that they have in mind is their peers, i.e. mostly Ph.D.'s, or at least with a similar level of training. They assume that their readers know enough math to fill in missing steps.

[Integral]

When people write formal papers for publication, the "audience" that they have in mind is their peers, i.e. mostly Ph.D.'s, or at least with a similar level of training. They assume that their readers know enough math to fill in missing steps.

The steps left out by AE are EXACTLY the type you would expect to be left out. He gives the starting point and the result. The reader is expected to be able to bridge the gap. I think that there are several other ways of filling in this gap, mine is one which requires little mathematical sophistication. I have no idea how AE acutally did the steps in his development of the paper.

[JesseM]

A while ago I made my own attempt to follow along with Einstein's derivation of the Lorentz transform which I posted on this thread (http://www.physicsforums.com/showthread.php?t=85796) (post #4). Here's what I wrote for the step you were asking about (no real calculus rules are needed aside from the fact that the curved surface you get when you plot a function of two variables in 3D space will look like a tilted 2D plane when you zoom in on an infinitesimally small region of it):
1/2[tau(0, 0, 0, t0') + tau(0, 0, 0, t0' + xm'/(c-v) + xm'/(c+v))] =
tau(xm', 0, 0, t0' + xm'/(c-v))


Then he goes from this to the equation 1/2(1/(c-v) + 1/(c+v))*(dtau/dt')
= dtau/dx' + (1/c-v)*(dtau/dt'), which also confused me for a little
while because I didn't know what calculus rule he was using to go from
the last equation to this one. But then I realized that if you just
ignore the y' and z' coordinates and look at tau(x',t'), then since he
says "if x' is chosen infinitesimally small", you can just assume tau is
a slanted plane in the 3D space with x',t' as the horizontal axes and
tau as the vertical axes. The general equation for a slanted plane in
these coordinates which goes through some point xp', tp', and taup would be:


tau(x',t') = Sx'*(x' - xp') + St'*(t' - tp') + taup


Where Sx' is the slope of the plane along the x' axis and St' is the
slope of the plane along the t' axis. If we say this plane must go
through the three points tau0, tau1 and tau2 earlier, then we can use
tau0's coordinates for xp', tp' and taup, giving:


tau(x',t') = Sx'*x' + St'*t' + tau0


So, plugging in tau1 = tau(xm', t0' + xm'/(c-v)) gives


tau1 = Sx'*xm' + St'*(t0' + xm'/(c-v)) + tau0


And plugging in tau2 = tau(0, t0' + xm'/(c-v) + xm'/(c+v)) gives:


tau2 = St'*(t0' + xm'/(c-v) + xm'/(c+v)) + tau0


So plugging these into 1/2(tau0 + tau2) = tau1 gives:


1/2(tau0 + St'*(t0' + xm'/(c-v) + xm'/(c+v)) + tau0 ) = Sx'*xm' +
St'*(t0' + xm'/(c-v)) + tau0


With a little algebra, this reduces to:


(1/2)*St'*(1/(c-v) + 1/(c+v)) = Sx' + St'*(1/(c-v))


And since tau(x',t') is just a plane, of course St' = dtau/dt' and Sx' =
dtau/dx', so this gives the equation 1/2(1/(c-v) + 1/(c+v))*(dtau/dt') =
dtau/dx' + (1/c-v)*(dtau/dt') which Einstein got.

[DrGreg]

For what it's worth, see also this thread (http://www.physicsforums.com/showthread.php?t=95942), and, in particular, my post #6. However my reply assumes knowledge of the chain rule for partial derivatives so may mystify you. But it's essentially the same argument as Mentz114's in slightly different language.