Differentials? Can you elaborate?

In summary: So the limit above equals (u_x - v)/(1-(u_x v/c^2)) = u_x' = \gamma (u_x - v)\\u_x' = \gamma (1-\gamma^2) u_x + \gamma^2 v In summary, the velocity transformation between reference frames in the x direction is done by differentiating the spacetime Lorentz transforms for the space-time coordinates. This method is applicable in situations where the Lorentz transformation equations are satisfied and has implications in understanding the relationship between variables in different reference frames.
  • #1
BiGyElLoWhAt
Gold Member
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Disclaimer: This isn't a homework assignment, so maybe it shouldn't be in the homework forums. If you feel it should be located elsewhere, feel free to move it, but the template doesn't really apply to this question so...

* * * * * * * *

So I'm reading Fundamentals of Modern Physics by Eisberg, and I just got through all the Lorentz transforms for space-time coordinates.

In order to do a Lorentz transform on the velocity between reference frames, he differientated (somehow) the spacetime lorentz transforms for the ST coords.

The velocity is presumed to be exclusively in the x direction between both sets of coordinates.

He started with

##x'(x,t) = \frac{1}{\sqrt{1-\beta ^2}}(x-vt)##
and differentiated to get
##dx'(x,t)=\frac{1}{\sqrt{1-\beta ^2}}(dx-vdt)##

I have a couple questions about the legitimacy of this.

1) I think this is called the total differential (?) is this correct? It appears to take the form ##\frac{\partial}{\partial x}x' *dx + \frac{\partial}{\partial t}x' *dt##

2) Assuming this is 100% legitimate (I don't necessarily doubt that it is), when is this method applicable? What are the implications of this method?

I understand that x' is a function of x and t, (it's actually denoted simply x', not x'(x,t) in the book), so the velocity should be dependent on both differentials, but I'm not really seeing the logic behind differentiating this way.

Any help would be greatly appreciated.
Thanks
 
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  • #2
If ##f:\mathbb R^2\to\mathbb R##, then you can define ##df:\mathbb R^4\to\mathbb R## by
$$df(x,y,h,k)=D_1f(x,y) h+D_2 f(x,y)k,$$ for all x,y,h,k. If you denote the numbers h and k by dx and dy respectively, and simplify the notation for the left-hand side to just df, the above can be written
$$df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy.$$ You're dealing with four variables x,t,x',t', that satisfy two constraints, the Lorentz transformation equations. These equations ensure that the values of the primed variables are fixed by a choice of values for the unprimed variables. In particular, we can define a function f by
$$f(x,t)=\gamma(x-vt)$$ for all x,t. It's confusing but common to use the notation x' instead of f. So x' now denotes two different things, both a real number and a function.
$$dx'=df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial t}dt =\frac{\partial x'}{\partial x}dx+\frac{\partial x'}{\partial t}dt.$$ Why do it this way? I have no idea.
 
  • #3
Fredrik said:
Why do it this way? I have no idea.
Ok, thanks for the reply.
I'm just hoping to get a little more insight into the why.
 
  • #4
BiGyElLoWhAt said:
Disclaimer: This isn't a homework assignment, so maybe it shouldn't be in the homework forums. If you feel it should be located elsewhere, feel free to move it, but the template doesn't really apply to this question so...

* * * * * * * *

So I'm reading Fundamentals of Modern Physics by Eisberg, and I just got through all the Lorentz transforms for space-time coordinates.

In order to do a Lorentz transform on the velocity between reference frames, he differientated (somehow) the spacetime lorentz transforms for the ST coords.

The velocity is presumed to be exclusively in the x direction between both sets of coordinates.

He started with

##x'(x,t) = \frac{1}{\sqrt{1-\beta ^2}}(x-vt)##
and differentiated to get
##dx'(x,t)=\frac{1}{\sqrt{1-\beta ^2}}(dx-vdt)##

I have a couple questions about the legitimacy of this.

1) I think this is called the total differential (?) is this correct? It appears to take the form ##\frac{\partial}{\partial x}x' *dx + \frac{\partial}{\partial t}x' *dt##

2) Assuming this is 100% legitimate (I don't necessarily doubt that it is), when is this method applicable? What are the implications of this method?

I understand that x' is a function of x and t, (it's actually denoted simply x', not x'(x,t) in the book), so the velocity should be dependent on both differentials, but I'm not really seeing the logic behind differentiating this way.

Any help would be greatly appreciated.
Thanks

You can avoid "differentials", by using finite differences and the basic definition of derivative. For a space-time increment ##(\Delta x, \Delta t)## in the first frame, you get an increment ##(\Delta x', \Delta t')## in the second frame, and these are related by
[tex] \Delta x' = \gamma\,(\Delta x - v \Delta t)\\
\Delta t' = \gamma \, (\Delta t - \frac{v}{c^2} \Delta x ) [/tex]
Here, ##x = x(t)## is on some trajectory. Now use the definitions
[tex] u_x = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t}\\
u_x' = \lim_{\Delta t' \to 0} \frac{\Delta x'}{\Delta t'} = \lim_{\Delta t' \to 0} \frac{\Delta x - v \Delta t}{\Delta t - (v/c^2) \Delta x}\\
= \lim_{\Delta t' \to 0} \frac{(u_x-v) \Delta t}{(1-(u_x v/c^2)) \Delta t} [/tex]
and note that ##\Delta t' \to 0 ## if and only if ##\Delta t \to 0##.
 

1. What is a differential and how does it work?

A differential is a mechanical device that allows two axles to turn at different speeds while receiving power from the same source. It consists of a set of gears that help distribute torque evenly among the wheels, allowing for smooth turning and preventing wheel slippage.

2. Why do some vehicles have limited slip differentials?

Limited slip differentials are used to improve traction and handling in vehicles. They distribute torque more evenly among the wheels, providing better grip and preventing wheel spin in slippery conditions.

3. What is the difference between an open and a locked differential?

An open differential allows the wheels on one axle to spin at different speeds, while a locked differential locks the wheels on the same axle to spin at the same speed. This is useful for off-roading and other situations where all wheels need to spin at the same speed for better traction.

4. How often should I change the differential fluid in my car?

It's recommended to change the differential fluid every 30,000 to 50,000 miles, or according to your vehicle's manufacturer's guidelines. This helps maintain proper lubrication and prolong the life of your differential.

5. What are the signs of a failing differential?

Some common signs of a failing differential include leaking fluid, strange noises (such as whining or clunking), difficulty turning or handling, and uneven tire wear. If you notice any of these symptoms, it's important to have your differential checked and repaired by a professional.

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