- #1
SiennaTheGr8
- 491
- 193
An operation is frame-invariant if it maintains the Lorentz-transformability of an input, yes? So for example, the coordinate ##x## transforms according to ##x^\prime = \gamma(x-vt)##, and multiplying the unprimed space and time coordinates by ##c## would give ##cx^\prime = c\gamma(x-vt)##. In other words, we multiplied the unprimed coordinates by ##c##, and we can still convert them into their primed-frame equivalents using the Lorentz transformation.
But we could multiply both sides of the equation by any old number, couldn't we? Doesn't have to be an invariant. If it weren't an invariant, though, then the inverse transformation (to go from primed frame to unprimed) would have to be multiplied by a different number, I think.
So would it be accurate to say that the key to an invariant operation is that it leaves the Lorentz transformation and its inverse "reversible" or "symmetrical"? (Word choice, sorry, but is it clear what I'm getting at? That, if the operation is multiplication, the multiplier should be the same in the two equations?)
I have another question, which is related. I'm trying to understand what happens when we take the ##d\tau##-derivative of ##c\, dt## and ##dx##. I know we end up with ##\gamma c## and ##\gamma v_x##, and I understand that these quantities are Lorentz-transformable, but I'm not sure that I understand why. Is it just that ##d\tau## is invariant, and so the transformation and inverse transformations remain "reversible" in the sense I described above? So for ##dx##:
##dx^\prime = \gamma(dx-v \, dt) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, ⇔ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, dx = \gamma(dx^\prime+v \, dt^\prime)##
##(\gamma v_x)^\prime = \dfrac{d}{d\tau}\gamma(dx-v \, dt) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,⇔\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \gamma v_x = \dfrac{d}{d\tau}\gamma(dx^\prime+v \, dt^\prime)##
(where the ⇔ indicates the "reversibility" I was speaking of).
Hope that isn't complete gibberish.
But we could multiply both sides of the equation by any old number, couldn't we? Doesn't have to be an invariant. If it weren't an invariant, though, then the inverse transformation (to go from primed frame to unprimed) would have to be multiplied by a different number, I think.
So would it be accurate to say that the key to an invariant operation is that it leaves the Lorentz transformation and its inverse "reversible" or "symmetrical"? (Word choice, sorry, but is it clear what I'm getting at? That, if the operation is multiplication, the multiplier should be the same in the two equations?)
I have another question, which is related. I'm trying to understand what happens when we take the ##d\tau##-derivative of ##c\, dt## and ##dx##. I know we end up with ##\gamma c## and ##\gamma v_x##, and I understand that these quantities are Lorentz-transformable, but I'm not sure that I understand why. Is it just that ##d\tau## is invariant, and so the transformation and inverse transformations remain "reversible" in the sense I described above? So for ##dx##:
##dx^\prime = \gamma(dx-v \, dt) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, ⇔ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, dx = \gamma(dx^\prime+v \, dt^\prime)##
##(\gamma v_x)^\prime = \dfrac{d}{d\tau}\gamma(dx-v \, dt) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,⇔\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \gamma v_x = \dfrac{d}{d\tau}\gamma(dx^\prime+v \, dt^\prime)##
(where the ⇔ indicates the "reversibility" I was speaking of).
Hope that isn't complete gibberish.