Relativistic Velocity Transformation

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  • #1
Daniel5423
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I'm reading modern physics, Tipler 5th edition, pages 21 and 22, and I'm not understanding how the differentiation was done from the position to find the velocity.
Equation for position: x'= y(x - vt)
y is the gamma constant.

Then in the first step to find the velocity, a derivative was done:
dx'= y(dx - vdt)


From the above equation, it was used to find the relativistic equation for velocity, and I understand that part. but what I don't understand is what the differential operator was is in the derivative, such as something like d/dt. If I take d/dt to both sides of the equation, then I do not see how the above equation was found. Could someone please explain to me what the differential operator was used to find the above equation? Thanks.
 

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  • #2
Meir Achuz
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Use [itex]dt'=\gamma(t-vx/c^2)[/itex], and divide dx' by dt' to get dx'/dt'=v'.
 
  • #3
Ibix
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Take two nearby events, (x,t) and (x+dx,t+dt). In the primed frame the coordinates of these two events are (x',t') and (x'+dx',t'+dt'). The Lorentz transforms give you a relationship between x' and x and t, and between x'+dx' and x+dx and t+dt. Then you should be able to get the relationship you want.

Formally, you should probably start with x+Δx (etc) and then take the limit as Δ becomes infinitesimal. Generally physicists just assume that would work (because it always does in physically realistic situations), but that attitude tends to drive mathematicians to drink.
 
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  • #4
Ibix
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[itex]dt'=\gamma(t-vx/c^2)[/itex]
I don't think this formula is correct. The RHS should have dt and dx or the LHS should not have dt'.
 
  • #5
SiennaTheGr8
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I'm reading modern physics, Tipler 5th edition, pages 21 and 22, and I'm not understanding how the differentiation was done from the position to find the velocity.
Equation for position: x'= y(x - vt)
y is the gamma constant.

Then in the first step to find the velocity, a derivative was done:
dx'= y(dx - vdt)


From the above equation, it was used to find the relativistic equation for velocity, and I understand that part. but what I don't understand is what the differential operator was is in the derivative, such as something like d/dt. If I take d/dt to both sides of the equation, then I do not see how the above equation was found. Could someone please explain to me what the differential operator was used to find the above equation? Thanks.

Sounds like you're confusing the infinitesimal with the derivative.

First, the Lorentz transformation (using ##c=1##):

##\Delta x^\prime = \gamma (\Delta x - v \Delta t)##
##\Delta t^\prime = \gamma (\Delta t - v \Delta x)##.

Now, to find ##dx^\prime / dt^\prime##, there's a more "rigorous" way and there's a shortcut. The "rigorous" way is:

##\dfrac{dx^\prime}{dt^\prime} = \lim_{\Delta t^\prime \rightarrow 0} \dfrac{\Delta x^\prime}{\Delta t^\prime} = \lim_{\Delta t \rightarrow 0} \dfrac{\gamma (\Delta x - v \Delta t)}{\gamma (\Delta t - v \Delta x)} = \dfrac{\lim_{\Delta t \rightarrow 0} \left( \frac{\Delta x}{\Delta t} - v \right)}{\lim_{\Delta t \rightarrow 0} \left( 1 - v \frac{\Delta x}{\Delta t} \right)} = \dfrac{\frac{dx}{dt} - v}{1 - v \frac{dx}{dt}}##.

The shortcut is to regard ##dx^\prime## and ##dt^\prime## as infinitesimals in the first place—i.e., infinitesimally small "versions" of the finite changes ##\Delta x^\prime## and ##\Delta t^\prime##—and just divide them. So use ##d##'s instead of ##\Delta##'s in the first equations above:

##d x^\prime = \gamma (d x - v \, d t)##
##d t^\prime = \gamma (d t - v \, d x)##

and divide:

##\dfrac{dx^\prime}{dt^\prime} = \dfrac{ \gamma (d x - v \, d t)}{\gamma (d t - v \, d x)} = \dfrac{\frac{dx}{dt} - v}{1 - v \frac{dx}{dt}}##.
 
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  • #6
Daniel5423
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Sounds like you're confusing the infinitesimal with the derivative.

First, the Lorentz transformation (using ##c=1##):

##\Delta x^\prime = \gamma (\Delta x - v \Delta t)##
##\Delta t^\prime = \gamma (\Delta t - v \Delta x)##.

Now, to find ##dx^\prime / dt^\prime##, there's a more "rigorous" way and there's a shortcut. The "rigorous" way is:

##\dfrac{dx^\prime}{dt^\prime} = \lim_{\Delta t^\prime \rightarrow 0} \dfrac{\Delta x^\prime}{\Delta t^\prime} = \lim_{\Delta t \rightarrow 0} \dfrac{\gamma (\Delta x - v \Delta t)}{\gamma (\Delta t - v \Delta x)} = \dfrac{\lim_{\Delta t \rightarrow 0} \left( \frac{\Delta x}{\Delta t} - v \right)}{\lim_{\Delta t \rightarrow 0} \left( 1 - v \frac{\Delta x}{\Delta t} \right)} = \dfrac{\frac{dx}{dt} - v}{1 - v \frac{dx}{dt}}##.

The shortcut is to regard ##dx^\prime## and ##dt^\prime## as infinitesimals in the first place—i.e., infinitesimally small "versions" of the finite changes ##\Delta x^\prime## and ##\Delta t^\prime##—and just divide them. So use ##d##'s instead of ##\Delta##'s in the first equations above:

##d x^\prime = \gamma (d x - v \, d t)##
##d t^\prime = \gamma (d t - v \, d x)##

and divide:

##\dfrac{dx^\prime}{dt^\prime} = \dfrac{ \gamma (d x - v \, d t)}{\gamma (d t - v \, d x)} = \dfrac{\frac{dx}{dt} - v}{1 - v \frac{dx}{dt}}##.

Thanks! That helped me understand it better.
 

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