Hamiltonian Integral Transformation: Insight Needed

In summary, the conversation revolves around the Hamiltonian transformation properties and the integration over a 3-surface in 4-dimensional space. The main question is whether the transformed Hamiltonian is equivalent to the time component of the transformed four-momentum. The conversation delves into more advanced notions involving 4-volume-forms and differential geometry, and suggests using an introductory textbook on calculus on manifolds for a better understanding.
  • #1
Reverend Shabazz
19
1
Hello all,
I am reading through the Jackson text as a hobby and have reached a question regarding the Hamiltonian transformation properties. I will paste the relevant section from the text below:
upload_2018-5-22_15-5-36.png


I don't understand what he's getting at in the sentence I highlighted.

To attempt to see what he is saying, I tried to show that by transforming the Hamiltonian density ##\mathcal {H} -> \mathcal {H'}## and multiplying it by the transformed 3-d volume element ##d'^3x##, the resulting Hamiltonian ##H'=\mathcal {H'}*d'^3x## is basically equivalent to the time component of the transformed four-momentum, but it did not equate (I used a simple case of constant velocity in the ##x## direction with ##E##&##B## fields in the ##y## and ##z## directions).

Could someone please provide some insight into this problem?
Thank you much
 

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  • #2
What you are integrating over is a 3-surface in 4-dimensional Minkowski space. Just as a 2-dimensional surface element ##d\vec S## in 3-dimensions is a vector quantity, you could view the 3-surface element in Minkowski space as a vector. In the case of ##d^3x## when integrating over a simultaneity, it should be pretty clear that the direction of that vector is in the time direction.

A more subtle point is the transformation of the Hamiltonian itself (not the Hamiltonian density). In general, when transforming from one inertial frame to another, it is not obvious that the total 4-momentum should satisfy the same transformation rules as the 4-momentum of a single particle. This is because the simultaneity surfaces in the new inertial frame are not the same as those in the original one and therefore the integration domain is different. Indeed, if there is any external 4-force-density acting on the field, the total 4-momenta of a time in ##S## and a time in ##S'## will generally be different (just as the total 4-momenta in either frame will generally change with time). However, in the absence of an external 4-force-density, the divergence theorem will essentially tell you that they indeed are the same 4-vector.More advanced notions (care only if you understand it): The 4-volume-form in Minkowski space is ##\eta = dx^0\wedge dx^1 \wedge dx^2 \wedge dx^3## and is a 4-form, i.e., it takes four 4-vector arguments. When you integrate over a 3-surface (e.g., a simultaneity) in Minkowski space, this 3-surface gives you three directions (given by ##x^1##, ##x^2##, and ##x^3## in the case of a simultaneity). To integrate, these three directions must be contracted with the 4-form, but since there are only three of them, the result of this is a 1-form that is orthogonal to all of these directions (since differential forms are completely anti-symmetric). The only remaining direction if you integrate over a simultaneity is the time direction. The general map from a 1-form to a 4-vector is a type (2,0) tensor and so the time-component of the result must be the 00 component of that type (2,0) tensor.
 
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  • #3
Orodruin said:
More advanced notions (care only if you understand it): The 4-volume-form in Minkowski space is ##\eta = dx^0\wedge dx^1 \wedge dx^2 \wedge dx^3## and is a 4-form, i.e., it takes four 4-vector arguments. When you integrate over a 3-surface (e.g., a simultaneity) in Minkowski space, this 3-surface gives you three directions (given by ##x^1##, ##x^2##, and ##x^3## in the case of a simultaneity). To integrate, these three directions must be contracted with the 4-form, but since there are only three of them, the result of this is a 1-form that is orthogonal to all of these directions (since differential forms are completely anti-symmetric). The only remaining direction if you integrate over a simultaneity is the time direction. The general map from a 1-form to a 4-vector is a type (2,0) tensor and so the time-component of the result must be the 00 component of that type (2,0) tensor.

Thanks for the reply. Per your last paragraph, I guess it would be nice to see what you're saying explicitly written out mathematically. That would probably help me grasp the fundamental logic of what's going on...but since I haven't studied differential geometry yet, I imagine that's too difficult to do in this forum. Perhaps if you know of any good resources on this topic, please feel free to mention them.
 
  • #4
Reverend Shabazz said:
Thanks for the reply. Per your last paragraph, I guess it would be nice to see what you're saying explicitly written out mathematically. That would probably help me grasp the fundamental logic of what's going on...but since I haven't studied differential geometry yet, I imagine that's too difficult to do in this forum.
That depends on what you mean by "too difficult". There are certainly people here with sufficient knowledge to teach differential geometry. However, a forum is no substitute for lectures or a good textbook.

Reverend Shabazz said:
Perhaps if you know of any good resources on this topic, please feel free to mention them.
Any introductory textbook that covers calculus on manifolds should do. I would suggest chapters 2 and 9 (and to some extent 8 and 10 with regards to the basics of classical field theory) in the one in my avatar, but I am obviously biased there and it covers many other topics as well. :rolleyes:
 
  • #5
A commercial in disguise... :-D

Orodruin said:
That depends on what you mean by "too difficult". There are certainly people here with sufficient knowledge to teach differential geometry. However, a forum is no substitute for lectures or a good textbook.Any introductory textbook that covers calculus on manifolds should do. I would suggest chapters 2 and 9 (and to some extent 8 and 10 with regards to the basics of classical field theory) in the one in my avatar, but I am obviously biased there and it covers many other topics as well. :rolleyes:
 
  • #6
MathematicalPhysicist said:
A commercial in disguise... :-D
I don't know about the disguise part ... I clearly stated my bias. :biggrin:
 
  • #7
Orodruin said:
I don't know about the disguise part ... I clearly stated my bias. :biggrin:
Explicit disguise...
 
  • #8
Orodruin said:
That depends on what you mean by "too difficult". There are certainly people here with sufficient knowledge to teach differential geometry. However, a forum is no substitute for lectures or a good textbook.

Indeed. I guess what I was looking for with respect to my original question was a mathematical demonstration of what Jackson is trying to encapsulate in that paragraph using the regular "vanilla" calculus that he's exclusively worked with thus far in the book. I was hoping that was possible since (to my knowledge) he doesn't assume the machinery/nomenclature of differential geometry. Perhaps in this instance, though, my assumption is not correct..

Orodruin said:
Any introductory textbook that covers calculus on manifolds should do. I would suggest chapters 2 and 9 (and to some extent 8 and 10 with regards to the basics of classical field theory) in the one in my avatar, but I am obviously biased there and it covers many other topics as well. :rolleyes:

Hehe, maybe I will check that out at some point :wink:. Since I'm pretty much learning E&M as a hobby, I have to determine what fields I wish/need to dive into. I imagine if I wanted to formally study it at the graduate level, it would behoove me to learn differential geometry, but at a basic hobbyist level maybe I can overlook it, at least temporarily.
 
  • #9
Reverend Shabazz said:
Indeed. I guess what I was looking for with respect to my original question was a mathematical demonstration of what Jackson is trying to encapsulate in that paragraph using the regular "vanilla" calculus that he's exclusively worked with thus far in the book. I was hoping that was possible since (to my knowledge) he doesn't assume the machinery/nomenclature of differential geometry. Perhaps in this instance, though, my assumption is not correct..
The argument is essentially the argument I gave first in #2. Of course you could perhaps flesh it out by writing some additional formulas, but the basic idea remains the same - a map from a vector to a vector is a rank 2 tensor, the original vector is in the pure time-direction and you are interested in the time component of the resulting vector so what you are interested in is the time-time component of the rank 2 tensor.

Reverend Shabazz said:
Hehe, maybe I will check that out at some point :wink:. Since I'm pretty much learning E&M as a hobby, I have to determine what fields I wish/need to dive into. I imagine if I wanted to formally study it at the graduate level, it would behoove me to learn differential geometry, but at a basic hobbyist level maybe I can overlook it, at least temporarily.
You really can do most of SR without knowing any differential geometry. GR is another matter ...
 

FAQ: Hamiltonian Integral Transformation: Insight Needed

What is a Hamiltonian Integral Transformation?

A Hamiltonian Integral Transformation is a mathematical tool used in quantum mechanics to transform the equations of motion of a system into a new set of equations, which are easier to solve. It involves expressing the Hamiltonian operator, which represents the total energy of a system, in terms of new variables, known as canonical variables.

What is the purpose of a Hamiltonian Integral Transformation?

The purpose of a Hamiltonian Integral Transformation is to simplify the equations of motion of a system, making it easier to analyze and solve. It also allows for the use of different coordinate systems, which can provide insight into the dynamics of a system and reveal symmetries that may not be apparent in the original coordinates.

How is a Hamiltonian Integral Transformation performed?

A Hamiltonian Integral Transformation is performed by finding a suitable set of canonical variables, which are related to the original variables through a transformation known as a canonical transformation. This transformation can be derived using the Hamilton-Jacobi equation or other mathematical methods.

What is the difference between a Hamiltonian Integral Transformation and a Lagrangian Integral Transformation?

A Hamiltonian Integral Transformation and a Lagrangian Integral Transformation are both used to simplify the equations of motion of a system. However, a Hamiltonian transformation involves transforming the Hamiltonian, which represents the total energy, while a Lagrangian transformation involves transforming the Lagrangian, which represents the kinetic and potential energies of a system.

Can a Hamiltonian Integral Transformation be applied to any system?

Yes, a Hamiltonian Integral Transformation can be applied to any system described by a Hamiltonian, regardless of its complexity. However, finding a suitable set of canonical variables and performing the transformation may be more difficult for some systems than others.

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