# Derivation of Lorentz Transformations

## Main Question or Discussion Point

Could anyone provide a derivation of the Lorentz transformations for me? And if the Lorentz transformations existed before Einstein came up with special relativity, then why wasn't the Lorentz guy able to come up with special relativity? It seems to me that he did all the work in showing that space contracts and time slows down as you approach the speed of light.

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Ed Quanta said:
Could anyone provide a derivation of the Lorentz transformations for me? And if the Lorentz transformations existed before Einstein came up with special relativity, then why wasn't the Lorentz guy able to come up with special relativity? It seems to me that he did all the work in showing that space contracts and time slows down as you approach the speed of light.
I will let somebody else do the derivation, or point you to a link for it, but I can tell you why Lorentz didn't discover relativity. He wouldn't give up his traditional intuitions. He left the status of his transformations ambiguous in his own mind. They were useful, ant surely true, in some Maxwellian sense, but he could not see to revolutionize not only physics but all thinking about space and time.

HallsofIvy
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Here's a derivation that has nothing to do with relativity!

Imagine a person who can swim in still water with speed c but who is swimming in a river with a downstream current of v. If he swims a distance downstream a distance l, his speed, relative to the bank (oops! a little relativity there!) is c+ v and so he take time $\frac{l}{c+v}$.
He now turns and swims the same distance upstream: now his speed, relative to the bank is c- v (of course, we must have v< c for that to be possible) and so the time required is $\frac{l}{c-v}$. The time required for the "round trip" is the sum of those: $\frac{l}{c+v}+ \frac{l}{c-v}= \frac{lc-lv+lc+lv}{c^2- v^2}= \frac{2lc}{c^2- v^2}$.

Now, he swims a distance l' ACROSS the river. Of course, in order to go directly across the river, he must swim slightly up stream. Drawing a triangle, it is easy to see that his "speed made good" is $\sqrt{c^2- v^2}$ and so the time to go l' and back is $\frac{2l'}{sqrt{c^2- v^2}}$.

Suppose he finds that the two times are exactly the same? What must the relationship be between l and l'? Setting the times equal,
$\frac{2lc}{c^2- v^2}= \frac{2l'}{sqrt{c^2- v^2}}$ so that
$l'= \frac{cl}{\sqrt{c^2- v^2}}= \frac{l}{\sqrt{1-\frac{v}{c}^2}}$.

Do you see the connection? The Michaelson- Morely experiment which was supposed to measure the effect of the earth's movement through the "ether" gave a null result. Although the two arms of the "supposed" to be of the same length, Lorentz calculated how much the one pointing "upstream" must have contracted in order to give the same time for the path of the light.

Actually, Lorentz did come up with a theory to explain that: What if, as a result of the fact that magnetic force is dependent upon speed (which is what really opened that whole can of worms to start with!) caused electrons to exert more force in the direction of movement so that the arm facing in the direction of movement contracted exactly enough to give a null experiment! That's lovely! That's exactly how a lot of things happen in science! Unfortunately (for Lorentz anyway) a variation on the Michaelson-Morley experiment (called, I think, the "Kennedy" experiment) showed that that was not the case, leading Einstein to conjecture that it was space itself (the space between the electrons themselves) that contracted.

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HallsofIvy said:
Now, he swims a distance l' ACROSS the river. Of course, in order to go directly across the river, he must swim slightly up stream. Drawing a triangle, it is easy to see that his "speed made good" is $\sqrt{c^2- v^2}$ and so the time to go l' and back is [\itex]\frac{2l'}{sqrt{c^2- v^2}}[/itex].
How did you get the value as you called it "speed made good"?

pervect
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Code:
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\       /\
\     |
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I'll jump in here uninvited

The horizontal line (vector) above at the top of the ascii diagram is the "speed made good"

The slanting line is the path that the swimmer swims

The vertical line is the current of the river, which is flowing up the page.

The vector sum of the path that the swimmer swims plus the current is equal to the "speed made good"

It is assumed that the swimmer swims in a direction so that the "speed made good" is strictly horizontal. This is not the path of least effort for the swimmer, but it's something a swimmer could do if he had a reason (perhaps there's no safe landing further down the stream, so he can't let himself be swept downstream).

HallsofIvy
Homework Helper
See the very nice picture by pervect! Except that the vertical arrow (representing the current) should be downward (downstream).

The vertical line is the current: v
The hypotenuse is the swimmers speed (relative to the water): c
The horizontal line is the swimmers speed (relative to the shore): u

By the Pythagorean theorem c2= u2+ v2 so u2= c2- v2.

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Regarding Lorentz and why he was never credited for any work on SR (not really credited, but no one said he started it first) was because he came up with the transformations but failed to provide a suitable physical description for it.

Lorentz got a lot of credit for his work - they are still called Lorentz transforms - but Lorentz's derivation was based upon motion wrt to an ether - moreover he did not interpret the time factor as a reality - Einstein's contribution was based upon the counterintuitive proposition that time is not universal.

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HallsofIvy said:
Although the two arms of the "supposed" to be of the same length, Lorentz calculated how much the one pointing "upstream" must have contracted in order to give the same time for the path of the light.
I have been thinking for this for a few days, and here is my problem. I understand mathematically how you got what you got which is indeed the Lorentz transformation for length. But why are the two arms supposed to be of the same length? If the velocity going across the stream is different than the velocity going up and down the stream and we know the time is equal, then wouldn't we expect different lengths?

If V does not equal V', and t=t, then V/t=l does not equal V'/t=l'. It seems like that is what is going on here. I don't know. Tell me where or how I am seeing this wrong.

Ed - the MMx experiment was conducted using arms having the same length - Lorentz and others proferred the idea that measurements contracted along the line of motion - so this could explain the apparent constancy of light as long as arms of the same lenght were used. Later in the KennedyThorndike experiments, the arms were of different length but the result (null) was the same. But this required a modification of Lorentz's theory - in addition to length contraction, the observed time rate of clocks had to be different in the relatively moving frames. So the original ether theory became the modified Lorentz ether theory (MLET) - the modified theory could still explain the null result, but by then SR had been generally accepted and it provided a simplier solution which did not require that rulers and other distances be physically contracted.

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Ok fair enough, but in our mathematical derivation (not for instance. We set the times to be equal for our two terms, but we have no reason to believe the velocity across the stream is the same as the velocity up and down the stream, and thus no reason to believe l=l'.