# Ive been reading ON THE ELECTRODYNAMICS

1. Sep 21, 2008

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 :-)

2. Sep 21, 2008

### 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 ?

3. Sep 21, 2008

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' 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.

4. Sep 21, 2008

### Mentz114

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

5. Sep 21, 2008

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 :-)

6. Sep 21, 2008

### 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

7. Sep 21, 2008

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 :-)

8. Sep 21, 2008

### Integral

Staff Emeritus
Here is my approach to this question.

#### Attached Files:

• ###### SRdevelopment3.pdf
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58.4 KB
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9. Sep 21, 2008

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?

10. Sep 21, 2008

### Integral

Staff Emeritus
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.

11. Sep 22, 2008

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 :-)

12. Sep 22, 2008

### Staff: Mentor

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.

13. Sep 22, 2008

### Integral

Staff Emeritus
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.

14. Sep 30, 2008

### 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 (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):

Last edited: Sep 30, 2008
15. Sep 30, 2008

### DrGreg

For what it's worth, see also this thread, 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.