Infinitesimal form of the Lorentz Transformation

In summary: Simply apply ##[W^i,W^j]=W^iW^j-W^jW^i## to a function ##f(x,y,z)##.There's a typo in the solution you attached. Is that what's confusing you?Any tips appreciated.ta.
  • #1
binbagsss
1,254
11

Homework Statement



attached:
generatorq.jpg

Homework Equations



where ##J_{yz} ## is
jyz.jpg


The Attempt at a Solution


[/B]
In a previous question have exponentiated the generator ##J_{yz}## to show it is the generator of rotation around the ##x## axis via trig expansions

so ##\Phi(t,x,y,z) \to \Phi(t,x,y cos \alpha - z sin \alpha, y sin \alpha + z cos \alpha ) ## and so via small angle expansions have:

##y \to ( y(1-\frac{\alpha^2}{2}+O(\alpha^4))-z(\alpha-\frac{\alpha^3}{3!}+O(\alpha^5))) ##
and
## z\to ( ( z(1-\frac{\alpha^2}{2}+O(\alpha^4))+y(\alpha-\frac{\alpha^3}{3!}+O(\alpha^5)))##I am unsure now how expand. I thought perhaps a taylor expansion in multivariables - y and z- was the idea, but I can't see how you would arrive at the answer attached with this:
generatorsol.jpg


Any tips appreciated.ta.
 

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  • #2
binbagsss said:

Homework Statement



attached:View attachment 227270

Homework Equations



where ##J_{yz} ## is View attachment 227273

The Attempt at a Solution


[/B]
In a previous question have exponentiated the generator ##J_{yz}## to show it is the generator of rotation around the ##x## axis via trig expansions

so ##\Phi(t,x,y,z) \to \Phi(t,x,y cos \alpha - z sin \alpha, y sin \alpha + z cos \alpha ) ## and so via small angle expansions have:

##y \to ( y(1-\frac{\alpha^2}{2}+O(\alpha^4))-z(\alpha-\frac{\alpha^3}{3!}+O(\alpha^5))) ##
and
## z\to ( ( z(1-\frac{\alpha^2}{2}+O(\alpha^4))+y(\alpha-\frac{\alpha^3}{3!}+O(\alpha^5)))##I am unsure now how expand. I thought perhaps a taylor expansion in multivariables - y and z- was the idea, but I can't see how you would arrive at the answer attached with this:View attachment 227272

Any tips appreciated.ta.
It is far easier, although not really short. Simply apply ##[W^i,W^j]=W^iW^j-W^jW^i## to a function ##f(x,y,z)##.
 
  • #3
There's a typo in the solution you attached. Is that what's confusing you?
 
  • #4
binbagsss said:
Any tips appreciated.ta.
I'm not sure whether this is giving too much of a hint on the first round, but here it is...

For a 1-parameter coordinate transformation ##x^\mu \to x'^\mu(x,\alpha)##, where ##\alpha## is the parameter, and ##\alpha=0## corresponds to the identity transformation, the generator ##G## can be computed from the following formula:$$G ~=~ \left. \frac{\partial x'^\mu}{\partial\alpha} \right|_{\alpha=0} \, \frac{\partial}{\partial x^\mu} ~~,$$(with implicit summation over the index ##\mu##).

HTH.
 
  • #5
strangerep said:
I'm not sure whether this is giving too much of a hint on the first round, but here it is...

For a 1-parameter coordinate transformation ##x^\mu \to x'^\mu(x,\alpha)##, where ##\alpha## is the parameter, and ##\alpha=0## corresponds to the identity transformation, the generator ##G## can be computed from the following formula:$$G ~=~ \left. \frac{\partial x'^\mu}{\partial\alpha} \right|_{\alpha=0} \, \frac{\partial}{\partial x^\mu} ~~,$$(with implicit summation over the index ##\mu##).

HTH.

okay thanks
ive never come across this before, could someone give me a brief summary of some background or point me to a link?(also may i ask who takes credit for the last quote in your sig :P ? )
 
  • #6
binbagsss said:
[...] could someone give me a brief summary of some background [...]
You just have to fill in a couple of steps in the solution you posted (which, as vela said, does indeed contain a typo).

In a little more detail, express ##\Phi'\Big(x^\mu(\alpha)\Big)## as a Taylor series in ##\alpha## (to 1st order is enough). Then use the chain rule to re-express ##d\Phi'/d\alpha## in terms of ##\partial\Phi'/\partial x^\mu## (since ##\Phi'## only depends on ##\alpha## indirectly, via the dependence of ##x'^\mu## on ##\alpha##).

(also may i ask who takes credit for the last quote in your sig :P ? )
That's mine. But "credit" is probably the wrong word.
 
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  • #7
I don't think that a Taylor series is needed. Just calculate things like ##\frac{\partial }{\partial x_\mu}(x_v f_\nu(x))## etc.
 
  • #8
fresh_42 said:
I don't think that a Taylor series is needed. Just calculate things like ##\frac{\partial }{\partial x_\mu}(x_v f_\nu(x))## etc.
Huh? :oldconfused:
 
  • #9
strangerep said:
Huh? :oldconfused:
I mean, we have differential operators and the multiplication times ##x_\nu##. As we are only interested in the commutator relations of them, we can simply take any twice differentiable function in three variables and calculate what these operators do. No need for any power series, the manifold itself, or a parameter. Simply differentiate: (I set ##\partial_x := \dfrac{\partial}{\partial x}\; , \;etc.)##
$$
[W^x,W^y](f)=(y\partial_z - z \partial_y)(z\partial_x - x\partial_z)(f) - (z\partial_x - x\partial_z)(y\partial_z - z \partial_y)(f) = \ldots
$$
I would further set ##f_x:=\partial_x (f)\, , \,etc.## to keep the workload low, but there is nothing else needed than good old Leibniz rule. It's trivial.
 
  • #10
fresh_42 said:
No need for any power series,
Oh, I see. You're answering a different part of the question.

I was explaining how to compute the generator corresponding to a particular (finite) coordinate transformation. You're explaining how to calculate commutator(s) between generators.
 
Last edited:
  • #11
strangerep said:
Oh, I see. You're answering a different part of the question.

I was explaining how to compute the generator(s) corresponding to a particular (finite) coordinate transformation. You're explaining how to calculate commutator(s) between generators.
Uh, o.k., I only concentrated on the image, and then I didn't understand where the entire rest was necessary for. How to get the tangents versus how do they behave or vice versa.
 

1. What is the infinitesimal form of the Lorentz Transformation?

The infinitesimal form of the Lorentz Transformation is a mathematical representation of the relationship between space and time in Einstein's theory of special relativity. It describes how measurements of space and time change for an observer moving at a constant velocity relative to another observer.

2. How is the infinitesimal form of the Lorentz Transformation derived?

The infinitesimal form of the Lorentz Transformation is derived from the Lorentz Transformation, which is a set of equations that describe the relationship between space and time in special relativity. The infinitesimal form is obtained by taking the limit as the velocity approaches zero, resulting in a simpler form of the equations.

3. What are the applications of the infinitesimal form of the Lorentz Transformation?

The infinitesimal form of the Lorentz Transformation is used in many areas of physics, including particle physics, electromagnetism, and cosmology. It is also used in engineering and technology, such as in GPS systems and particle accelerators.

4. What is the significance of the infinitesimal form of the Lorentz Transformation?

The infinitesimal form of the Lorentz Transformation is significant because it allows us to understand the effects of special relativity in situations where the velocity is small. It also provides a mathematical framework for understanding the relationship between space and time in the presence of high velocities.

5. How does the infinitesimal form of the Lorentz Transformation differ from the Galilean Transformation?

The infinitesimal form of the Lorentz Transformation differs from the Galilean Transformation in that it takes into account the effects of special relativity, such as time dilation and length contraction, which are not accounted for in the Galilean Transformation. It also includes a term for the speed of light, which is constant in all frames of reference in special relativity.

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