My Own website regarding physics and math.

[misogynisticfeminist]

I've set up my own website,

http://www.freewebs.com/mouldy-fart/

Please critique the articles on math and physics, especially the more mathematical ones. I fear that there could be problems in my presentation or derivations of certain formulas (in physics) or that my understanding of the concept is just wrong !

I have a little question as well:

In deriving the lorentz transform for the x-coordinate, $$x'=\gamma(x-vt)$$ as compared to the galilean transform $$x'=x-vt$$. By taking into account length contraction $$x'=\frac{1}{\gamma} x$$, shouldn't the lorentz transform be $$x=\gamma(x-vt)$$ instead of $$x'=\gamma(x-vt)$$, since if we substitute the length contraction formula, x' is already replaced by x divided by one over gamma?

 

[dextercioby]

Please,set up the problem.Which system is moving wrt which,in what direction,...blah,blah,...

Daniel.

 

[misogynisticfeminist]

^hmmm, apologies if i wasn't specific enough.

Frame S' is moving with velocity v relative to a rest frame S in the positive x direction (i.e. towards the right).

 

[dextercioby]

In deriving the lorentz transform for the x-coordinate, $$x'=\gamma(x-vt)$$ as compared to the galilean transform $$x'=x-vt$$. By taking into account length contraction $$x'=\frac{1}{\gamma} x$$, shouldn't the lorentz transform be $$x=\gamma(x-vt)$$ instead of $$x'=\gamma(x-vt)$$, since if we substitute the length contraction formula, x' is already replaced by x divided by one over gamma?

I'm afraid the bolded part is incorrect.It should be [itex]x'=\gamma (x-vt) [/tex] or else $$x=\gamma (x'+vt')$$.You can find the rigurous proof (which is pretty digestable) in any SR (electrodynamics) book.

Daniel.

 

[misogynisticfeminist]

^ hmm ok, understood, I'll dig up the modern physics textbook again...thanks for the help...

: )

 

[pervect]

I've set up my own website,

http://www.freewebs.com/mouldy-fart/


I have a little question as well:

In deriving the lorentz transform for the x-coordinate, $$x'=\gamma(x-vt)$$ as compared to the galilean transform $$x'=x-vt$$. By taking into account length contraction $$x'=\frac{1}{\gamma} x$$, shouldn't the lorentz transform be $$x=\gamma(x-vt)$$ instead of $$x'=\gamma(x-vt)$$, since if we substitute the length contraction formula, x' is already replaced by x divided by one over gamma?

Then inverse of the Lorentz transform

x' = gamma*(x-v*t)
t' = gamma*(t-v*x)

is just

x = gamma*(x'+v*t')
t = gamma*(t' + v*x')

You should be able to simply solve the linear equations to verify this - also note that the inverse of the Lorentz transform is a Lorentz transform with v=-v, which should be obvious if you think about it for a bit.