Am I taking crazy pills or is Einstein missing a factor of gamma?

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SamRoss
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Main Question or Discussion Point

This has been bothering me for a while. In Einstein's original derivation of the Lorentz transformations, he finds...

[tex]\epsilon[/tex]=[tex]\frac{c^2}{c^2-v^2}[/tex]x'

Here, [tex]\epsilon[/tex] is what we would normally call x', and x'=x-vt (sorry if that's a bit confusing). He then says, "Substituting for x' its value, we obtain"...

[tex]\epsilon[/tex]=[tex]\gamma[/tex](x-vt)

Now, am I taking crazy pills, or should that be gamma squared instead of gamma? I'm getting a similar result for the time transformation.
 
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  • #2
SamRoss
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I'm not very good with Latex yet. That should be c^2/(c^2-v^2)
 
  • #3
JesseM
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That should be [tex]\xi[/tex] (xi), not [tex]\epsilon[/tex] (epsilon). Also you're missing the factor of a. And after "Substituting for x' its value", he doesn't go directly to the equation [tex]\xi = \gamma*(x - vt)[/tex] as you suggest, instead he writes [tex]\xi = \phi(v)*\beta*(x - vt)[/tex], using the implicit substitution [tex]\phi(v) = \frac{ac}{\sqrt{c^2 - v^2}}[/tex] (a was an undetermined
function of v in the first place, so he just writes [tex]\phi(v)[/tex]) and [tex]\beta = \frac{c}{\sqrt{c^2 - v^2}} = \frac{1}{\sqrt{1 - v^2/c^2}}[/tex]. Then he goes on to give some arguments to demonstrate that [tex]\phi(v) = 1[/tex].

In an old thread I ran through Einstein's derivation and tried to explain everything in my own words since there were a few places where it wasn't immediately obvious how he got from one step to another, see [post=720738]here[/post] if you're interested (also see this thread, especially DrGreg's post, for an alternate explanation of one of the steps that's probably better than my original explanation).
 
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  • #4
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I'm not very good with Latex yet. That should be c^2/(c^2-v^2)
If the edit button is still available on your initial post, then change the tex code to this:

<tex>
\xi = x \prime \frac{c^2}{c^2-v^2}
</tex>

but with [] brackets rather than <> brackets and it should display OK.

To see the code used in the equations posted by other users, hover your mouse over their equations or click on the equations to see a pop up window with the code. You can also hit the quote button as if you are replying to the post and see the unparsed tex code of their equations. Hope that helps :smile:
 
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  • #5
SamRoss
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using the implicit substitution [tex]\phi(v) = \frac{ac}{\sqrt{c^2 - v^2}}[/tex]
Where are you getting [tex]\phi(v) = \frac{ac}{\sqrt{c^2 - v^2}}[/tex] ?

When defining a and phi in his paper, Einstein says "where a is a function phi(v) at present unknown..." Doesn't that mean they are one and the same? He's not saying one is a function of the other. I left the factor a out of my original paraphrasing of Einstein's paper, as you correctly noted, because Einstein later finds that phi(v)=1 and it was (and still is, although I'm hoping that you can show me the error of my ways) my understanding that phi(v)=a.
 
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  • #6
JesseM
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Where are you getting [tex]\phi(v) = \frac{ac}{\sqrt{c^2 - v^2}}[/tex] ?
Again, the substitution is implicit, you have to deduce it from his equations. He had written the equation:

[tex]\xi = a \frac{c^2}{c^2 - v^2} x'[/tex]

Then he says "Substituting for x' its value" (with x' having been earlier defined as x' = x - vt) and writes:

[tex]\xi = \phi(v) \beta (x - vt)[/tex]

This sequence of equations only makes sense if [tex]\phi(v) \beta = a \frac{c^2}{c^2 - v^2}[/tex]. And he says that he's defining [tex]\beta[/tex] by [tex]\beta = \frac{1}{\sqrt{1 - v^2/c^2}}[/tex], which can be rearranged as [tex]\beta = \frac{c}{\sqrt{c^2 - v^2}}[/tex], so to make things work out this means we must have [tex]\phi(v) = a \frac{c}{\sqrt{c^2 - v^2}}[/tex].
SamRoss said:
When defining a and phi in his paper, Einstein says "where a is a function phi(v) at present unknown..." Doesn't that mean they are one and the same?
He's speaking a bit sloppily (could just be the translation), but they are both just different symbols for unknown functions of v, to be determined later. Note that after giving the equation [tex]\xi = \phi(v) \beta (x - vt)[/tex] and defining the meaning of [tex]\beta[/tex], he just says "and [tex]\phi[/tex] is an as yet unknown function of v." Again, you can deduce the exact relation that must hold between the two unknown functions of v from the other equations I mention above.
 
  • #7
SamRoss
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Then he says "Substituting for x' its value" (with x' having been earlier defined as x' = x - vt) and writes:

[tex]\xi = \phi(v) \beta (x - vt)[/tex]

I think I see now. After Einstein says, "Substituting for x' its value", he's not deriving the value of gamma (his beta), he's choosing it. He has the freedom to do so because there is still a factor of phi(v) whose value is not yet determined. So he chooses what he knows will be the most convenient form and then goes on to derive the value of phi(v) (which he does by considering a third system of coordinates which we have so far not been talking about in these posts). Would you say this is correct?
 
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  • #8
JesseM
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I think I see now. After Einstein says, "Substituting for x' its value", he's not deriving the value of gamma (his beta), he's choosing it. He has the freedom to do so because there is still a factor of phi(v) whose value is not yet determined. So he chooses what he knows will be the most convenient form and then goes on to derive the value of phi(v) (which he does by considering a third system of coordinates which we have so far not been talking about in these posts). Would you say this is correct?
Yes, I'd say that's a good description.
 
  • #9
SamRoss
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Cool, thanks for all your help.
 

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