Meaning of trajectory in four-electricity in Landau's book

In summary, the conversation discusses the concept of velocity in the context of charge density and its variations with position and time. The equation ##j^i=\rho{\frac{dx^i}{dt}}## is introduced, but the issue of using the inverse ##\frac{dt}{dx^i}## when ##dx^i## is equal to zero is raised. The speaker is still confused about the definition of velocity in this scenario and is seeking clarification.
  • #1
zhouhao
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Homework Statement


I think the in equation ##(28.2)##,##x^i## in ##\frac{dx^i}{dt}## and the ##x^i## decides ##\rho## is not the same,if they are equaivalent,##\rho## can not vary with position changing and time fixed, because ##\frac{dx^i}{dt}## indicate the ##x^i(t)## which means if position changed the time would also be changed.

##\frac{dx^i}{dt}## could define the velocity through a points' trajectory,##\rho## is not point charge,what's the meaning of the "velocity"?

Homework Equations


The four electricity vector is defined as ##j^i=\rho\frac{dx^i}{dt}##,##\rho## depend on ##(t,x^1,x^2,x^3)##

The Attempt at a Solution


I want to find some clues about the "velocity" in contents after the definition and only find LLs(Landau and Lifshitz) regarding the ##x^i## in ##\frac{dx^i}{dt}## as the same as the ##x^i## which decides ##\rho##-------In equation ##(28.5)## ##dS^i## was gotten through dividing ##dV## with ##dx^i##.
 
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  • #2
When we define ##j^k(x^i) =\rho(x^i) \frac{ dx^k}{dt}##, I don't see any need to worry about whether the ##x^k## in ##\frac{ dx^k}{dt}## is the same as the ##x^i## in ##\rho(x^i)##.

##\rho(t, x^1, x^2, x^3)## is the charge density at some arbitrary point ##(x^1, x^2, x^3)## at time ##t##. At that point and time, the charge density has a velocity ##\vec{v} = \left( \frac{dx^1}{dt},\frac{dx^2}{dt},\frac{dx^3}{dt} \right)## which is well-defined. Thus, we can define ##\frac{dx^i}{dt} =\left( c, \vec{v} \right)= \left( c, \frac{dx^1}{dt},\frac{dx^2}{dt},\frac{dx^3}{dt} \right)##.

Then ##j^k = \rho(x^i) \frac{dx^k}{dt}## is well defined. This could also be written as ##j^k(x^i) = \rho(x^i) \frac{dx^k}{dt}(x^i)## in order to show the explicit functional dependence of the factors on the choice of the spacetime point ##x^i = (x^0, x^1, x^2, x^3)##.

In the expression ##j^k(x^i) = \rho(x^i) \frac{dx^k}{dt}(x^i)## , we can imagine varying the spatial point ##(x^1, x^2, x^3)## while keeping ##t## fixed.
 
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  • #3
TSny said:
When we define ##j^k(x^i) =\rho(x^i) \frac{ dx^k}{dt}##, I don't see any need to worry about whether the ##x^k## in ##\frac{ dx^k}{dt}## is the same as the ##x^i## in ##\rho(x^i)##.

##\rho(t, x^1, x^2, x^3)## is the charge density at some arbitrary point ##(x^1, x^2, x^3)## at time ##t##. At that point and time, the charge density has a velocity ##\vec{v} = \left( \frac{dx^1}{dt},\frac{dx^2}{dt},\frac{dx^3}{dt} \right)## which is well-defined. Thus, we can define ##\frac{dx^i}{dt} =\left( c, \vec{v} \right)= \left( c, \frac{dx^1}{dt},\frac{dx^2}{dt},\frac{dx^3}{dt} \right)##.

Then ##j^k = \rho(x^i) \frac{dx^k}{dt}## is well defined. This could also be written as ##j^k(x^i) = \rho(x^i) \frac{dx^k}{dt}(x^i)## in order to show the explicit functional dependence of the factors on the choice of the spacetime point ##x^i = (x^0, x^1, x^2, x^3)##.

In the expression ##j^k(x^i) = \rho(x^i) \frac{dx^k}{dt}(x^i)## , we can imagine varying the spatial point ##(x^1, x^2, x^3)## while keeping ##t## fixed.

Thank you very much!You help me understand more.
I am still confused with the velocity definition ##v^k=\frac{d{x'}^k}{dt}(t,x^1,x^2,x^3)\ \ \ (1)##
{I use ##x'## replace ##x## in this equation,since ##\frac{dx^k}{dt}(t,x^1,x^2,x^3)## may not a single valued function suitable for defining the velocity.Because for the ##dt## combined with different ##dx^i## to form ##(dt,dx^i(1)),(dt,dx^i(2)),...## may sucessfully make the charge density ##\rho(t+dt,x^i+dx^i)## exist.}
Inserting equation (1) into the equation (28.5),we have ##\int\rho{dV}=\int\rho\frac{d{x'}^i}{dt}\frac{dt}{d{x'}^i}dV=\frac{1}{c}\int{j^i}dS_i## which indicates ##\frac{dt}{d{x'}^i}dV=\frac{1}{c}dS_i##,this means ##d{x'}^i## was treated as equivalent as ##dx^i##.But ##\frac{dx^i}{dt}(t,x^i)## is a multivalued function and not suitable for the velocity definition.
 
  • #4
zhouhao said:
{I use ##x'## replace ##x## in this equation,since ##\frac{dx^k}{dt}(t,x^1,x^2,x^3)## may not a single valued function suitable for defining the velocity.Because for the ##dt## combined with different ##dx^i## to form ##(dt,dx^i(1)),(dt,dx^i(2)),...## may sucessfully make the charge density ##\rho(t+dt,x^i+dx^i)## exist.}
I'm having a hard time understanding what you are saying here.

Inserting equation (1) into the equation (28.5),we have ##\int\rho{dV}=\int\rho\frac{d{x'}^i}{dt}\frac{dt}{d{x'}^i}dV=\frac{1}{c}\int{j^i}dS_i## which indicates ##\frac{dt}{d{x'}^i}dV=\frac{1}{c}dS_i##,this means ##d{x'}^i## was treated as equivalent as ##dx^i##.But ##\frac{dx^i}{dt}(t,x^i)## is a multivalued function and not suitable for the velocity definition.

Note that, for example, the ##x^1## component of velocity of the charge density, ##\frac{dx'^1}{dt}##, could equal 0. Then, there is a problem writing its inverse ##\frac{dt}{dx'^1}##.

Also, when you write ##\int\rho\frac{d{x'}^i}{dt}\frac{dt}{d{x'}^i}dV=\frac{1}{c}\int{j^i}dS_i##, the index ##i## is summed in the integral on the right. So, you must be summing over ##i## also in the integral on the left. But, ##\frac{d{x'}^i}{dt}\frac{dt}{d{x'}^i} \neq 1## if you are summing over ##i##. So, it would not be correct to write ##\int\rho{dV}=\int\rho\frac{d{x'}^i}{dt}\frac{dt}{d{x'}^i}dV##.
 
  • #5
TSny said:
I'm having a hard time understanding what you are saying here.
Note that, for example, the ##x^1## component of velocity of the charge density, ##\frac{dx'^1}{dt}##, could equal 0. Then, there is a problem writing its inverse ##\frac{dt}{dx'^1}##.

Also, when you write ##\int\rho\frac{d{x'}^i}{dt}\frac{dt}{d{x'}^i}dV=\frac{1}{c}\int{j^i}dS_i##, the index ##i## is summed in the integral on the right. So, you must be summing over ##i## also in the integral on the left. But, ##\frac{d{x'}^i}{dt}\frac{dt}{d{x'}^i} \neq 1## if you are summing over ##i##. So, it would not be correct to write ##\int\rho{dV}=\int\rho\frac{d{x'}^i}{dt}\frac{dt}{d{x'}^i}dV##.

Thanks.
What I mean is that:
for a trajectory of a point charge the velocity could be defined as ##\frac{dx^i}{dt}##,because the trajectory ##x^i## could be written as ##x^i(t)##.

however, for the movement of density of charge----##\rho(t,x^i)##,I think the velocity ##v(t,x^i)## can not be expressed as ##\frac{dx^i}{dt}##,because at the time ##t##,there will be more than one ##x^i## to make ##\rho(t,x^i)## sense(##x^i## can vary with a fixed time ##t##).
As a result,##x^i## can not be written as ##x^i(t)##.
So I use ##\frac{d{x'}^i}{dt}## to express ##v(t,x^i)##,I can not clarify what ##x'## exactly is...

In the book,the equation ##(28.5)## is ##\int\rho{dV}=\frac{1}{c}\int{j^0}dV=\frac{1}{c}\int{j^i}{dS_i}##,I think the index ##i## is not summed and equation ##(28.5)## is derived from equation ##(28.2)##----##j^i=\rho{\frac{dx^i}{dt}}##----which means inverse ##\frac{dt}{dx^i}## is used and we have problem when ##dx^i## equal ##0##.

I feel it is hard to read this book coutinuously if confusion was not solved.Any suggestion for learning?
 
  • #6
zhouhao said:
Thanks.
What I mean is that:
for a trajectory of a point charge the velocity could be defined as ##\frac{dx^i}{dt}##,because the trajectory ##x^i## could be written as ##x^i(t)##.

however, for the movement of density of charge----##\rho(t,x^i)##,I think the velocity ##v(t,x^i)## can not be expressed as ##\frac{dx^i}{dt}##,because at the time ##t##,there will be more than one ##x^i## to make ##\rho(t,x^i)## sense(##x^i## can vary with a fixed time ##t##).
As a result,##x^i## can not be written as ##x^i(t)##.
So I use ##\frac{d{x'}^i}{dt}## to express ##v(t,x^i)##,I can not clarify what ##x'## exactly is...
Your description seems OK to me. The continuous distribution of charge is like a fluid. If you arbitrarily pick a time ##t_0## and a point in space ##\vec{x}_0##, then the charge density at that time and place is ##\rho(t_0, \vec{x}_0)##. The fluid at that time and place will generally be in motion with some velocity ##\vec{v}(t_0, \vec{x}_0)##. This velocity could be written as ##\frac{\vec{dx}}{dt}## where ##dt## is an infinitesimal time interval from ##t = t_0## to ##t = t_0 + dt##. ##\vec{dx}## is the corresponding displacement of the fluid from ##\vec{x}_0## to ##\vec{x}_0 + \vec{v} dt##.

In the book,the equation ##(28.5)## is ##\int\rho{dV}=\frac{1}{c}\int{j^0}dV=\frac{1}{c}\int{j^i}{dS_i}##,I think the index ##i## is not summed and equation ##(28.5)## is derived from equation ##(28.2)##----##j^i=\rho{\frac{dx^i}{dt}}##----which means inverse ##\frac{dt}{dx^i}## is used and we have problem when ##dx^i## equal ##0##.
Consider ##\int\rho{dV}=\frac{1}{c}\int{j^0}dV##. The integration is done in a particular inertial reference frame and the integral is over all of space at one fixed time ##t_0## in this frame of reference. All of space at one particular time is an example of a "hypersurface" in spacetime, as discussed in the text starting at the bottom of page 19. An element of volume ##dV## at fixed time ##t_0## can be considered as the zeroth component, ##dS^0##, of a four-vector ##dS^i##, as explained on page 20. Since ##t_0## is considered fixed, the components ##dS^1, dS^2## and ##dS^3## are all zero in this inertial frame. Thus, it is then true that we can trivially write ##j^0 dV = j^0 dS_0 = j^0 dS_0 + j^1 dS_1 + j^2 dS_2 + j^3 dS_3 = j^i dS_i##, where we sum over ##i## in the last expression. This expresses the integrand in an invariant form.

I feel it is hard to read this book coutinuously if confusion was not solved.Any suggestion for learning?
I find the Landau Lifshitz books to be quite challenging. They are awesome but somewhat intimidating (for me). The writing is elegant, but terse. There is very little elaboration or repetition. Sometimes I need to spend quite a bit of time trying to decipher and understand a paragraph or equation. I often find that my difficulty was due to overlooking one particular phrase in one particular sentence. But the time spent is often well worth it.
 
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FAQ: Meaning of trajectory in four-electricity in Landau's book

What is the meaning of trajectory in four-electricity in Landau's book?

In Landau's book, trajectory refers to the path that a charged particle takes in a four-electricity field. This field is described by four components: the electric field, the magnetic field, and two additional fields that arise from relativity. The trajectory of a charged particle is affected by these fields, causing it to curve or bend as it moves.

How does Landau define four-electricity in his book?

In Landau's book, four-electricity is a theoretical concept that combines the traditional three-dimensional electric and magnetic fields with two additional fields that arise from the theory of relativity. These four components together make up the four-electricity field, which describes the behavior of charged particles in a relativistic context.

Why is the concept of four-electricity important in Landau's book?

The concept of four-electricity is important in Landau's book because it allows for a more complete and accurate understanding of the behavior of charged particles in a relativistic context. By incorporating the additional fields from relativity, the four-electricity field can better describe the complex motion of particles in high-speed or high-energy situations.

What are the equations used to describe four-electricity in Landau's book?

The equations used to describe four-electricity in Landau's book are the Maxwell's equations, which describe the behavior of electric and magnetic fields, and the Lorentz force law, which describes the force on a charged particle in an electric and magnetic field. These equations are extended to include the additional fields from relativity, resulting in a set of four equations that describe the four-electricity field.

How does understanding four-electricity contribute to our understanding of physics?

Understanding four-electricity contributes to our understanding of physics by allowing us to accurately describe and predict the behavior of charged particles in a relativistic context. This knowledge is essential in fields such as particle physics, astrophysics, and cosmology, where high-speed or high-energy phenomena are common. Additionally, it provides a deeper understanding of the fundamental principles of electromagnetism and relativity.

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