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Relativistic current density components

  1. Aug 14, 2003 #1

    Relativistic current density components are the vector "J" with the scalar "j*c*rho".

    My questions are:

    Since it's a scalar and not a vector, what is the meaning of "c" times "rho"?

  2. jcsd
  3. Aug 14, 2003 #2


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    the ordinary current density vector J has units "coulombs per square meter per second" C/m2 s

    more correctly:

    Cm-2 s-1

    the amount of charge flowing thru a sq meter of area in one second

    the ordinary spatial charge density ρ has units "coulombs per cubic meter"


    If one multiplies by c = 3E8 m s-1

    one gets cρ also with units of current density Cm-2 s-1

    [[[impressionistically one can think of some charge which simply by existing is "moving at the speed c along the time axis" and so
    could intuitively represent a "current" in the time direction]]]

    the main thing is that now since the charge density has been changed to something the same units as the rest, it can be put
    together with them to make a 4-vector!

    (cρ, j1, j2, j3)

    to be combined into a 4-vector all the quantities must have the same units!

    Be careful about the different formats which people use for writing 4-vectors. Sometimes they use the "imaginary number" i, also sometimes written j, which is sqrt(-1) and they write that with the time component so that the metric can be of the form

    (ict)2 +(x1)2 + (x2)2 + (x3)2

    Other people, perhaps the majority, do not use this "imaginary time" convention and simply introduce a minus sign into the metric

    - (ct)2 +(x1)2 + (x2)2 + (x3)2

    or they may write the ct component as x0 so that the metric is
    - (x0)2 +(x1)2 + (x2)2 + (x3)2

    because there is some variation in the conventions, I was not
    sure I understood your notation. You would need to explain in
    more detail for me to be sure I understand the symbol "j" in your expression jcρ
    Last edited: Aug 14, 2003
  4. Aug 14, 2003 #3
    Merci Marcus,

    The "j" shall be "i".

    I understand that to convert and to fit the right units, "rho" is multiply my "c". That's all? Is the forth part (rho component) a vector or a scalar, or some hidden component?

    Merci encore Marcus
  5. Aug 14, 2003 #4


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    Re: Re: Relativistic current density components

    You say the rho component is the FOURTH part. So we have yet another format! Is this what it looks like?

    (j1, j2, j3, icρ)

    In that case the metric being used requires an "i" in the fourth component of the vector

    (x1)2 + (x2)2 + (x3)2 + (ix4)2

    A more common approach is not to bother with the "i" in the fourth component but simply have a minus sign in the metric:

    (x1)2 + (x2)2 + (x3)2 - (x4)2

    However written, the charge density is assimilated as one component of a 4-vector.

    I would not consider it a scalar, but rather as a PART of a vector.

    these 4-vectors are cleverly constructed so that they TRANSFORM correctly when there is a change of observer.
    what is static charge density (rho) for one person can be flowing and be a current density (j1...etc) for another person, if that
    other person is moving. All these things must be packaged together because they mix around during a change of coordinates. So there is an organic reason one MUST multiply
    the rho by c and incorporate it into a 4-vector.

    If information is packaged poorly some may get lost in "transmission", that is, in the change to a different observer's coordinates.

    Notice, BTW, how conventions differ!
    For example I suppose that some particle's energy-momentum vector which I would write

    (E/c, p1, p2, p3)

    is something that you would write with energy in the FOURTH place, namely

    (p1, p2, p3, iE/c)

    I have put the "i" in because I suspect you like the "imaginary time" formalism---perhaps formality is a better word, the difference is only superficial.

    what matters most is to pack the information into a 4-vector where everything has the same units and so that it will transform correctly into what the other person wants to know

    (the other person may be moving relative to you but may also
    want to know the momentum and energy of the particle---only transformed!)

    I still am not confident that I have identified the formalism you are using. could you describe it? do you write a point in spacetime as

    (x1, x2, x3, x4), where x4 = ct

    or as

    (x1, x2, x3, x4) where x4 = ict

    or as

    (x0, x1, x2, x3) where x0 = ct

    If you tell me which convention you are using I will be able to
    reply more surely.
  6. Aug 15, 2003 #5
    Posing I use this one : (x1, x2, x3, x4) where x4 = ict, I don't think it's relevant to my question. I am not pointing to the imaginary time part here.

    Take an other quadri-vector, relativistic momentum: P = (p,E/c), with imaginary part or not. P and p are vectors, E/c is scalar.

    May I consider the partial derivative of (E/c) over time as scalar pointing toward the positive direction of time and, as per, an implicit vector?

    Or should I consider P, quadri-vector, being expressed as tri-vector plus scalar? IMHO, that would means special interpretation about quadri-vector? Right?

    Note: If I remember correctly, quadri-vector was introduced consequently to Lorentz transformation where, depending on the observer's relative velocity, the "scalar component" becomes the "tri-vector" component (and vice-versa) when conservation law applied.
  7. Aug 15, 2003 #6


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    Imagine, where you are involved I do not like to say "No!" or to occupy a post of authority, so I think now would be a good time for one of the others to answer.

    In the middle ages, if you would suggest that one can make a vector by putting a scalar in the fourth position, to make it "point" in the "time direction", then your house would be visited by the officers of the Inquisition.
    And if you persisted in this opinion and told them that one of the components of a quadrivector is a "scalar" you would be burnt at the stake.
    My child, you are in great danger of heresy!

    A "scalar" quantity is one which is not changed by passage to a different system of coordinates

    but each of the components of a quadrivector may change (it may also not change depending on which transformation is used but it MAY change) and so it is not a scalar.

    I must defer to a PF mentor or one of several other competent authorities who can instruct you

  8. Aug 15, 2003 #7
    Sorry Marcus to have been heretical. My be I losted something during translation. I'll check that.

    Returning to the relativistic current density question, the quadri-vector is composed of a non-relativistic vector (tri-vector) and "ic*rho", which is not by itself a vector.

    Shall I consider "c*rho" to be at current density at rest as per "E/c" is at momentum at rest?
  9. Aug 15, 2003 #8


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    Imagine, I was trying to be funny about the issue of
    heresy. It wasnt very intelligent of me---now I see it
    was not very amusing after all.

    We must find someone else to help----a mentor, or perhaps
    "Pete" who signs himself "pmb". There are plenty of people
    here at PF who can give you a little relativity lecture.

    I actually enjoy talking to you, but not explaining to you! I like your
    tone of voice, style, whatever je-ne-sais-quoi.
    Pardon me for not being useful!
  10. Aug 17, 2003 #9
    Bonjour Marcus,

    I wouldn't like to discourage your explanation, I always appreciate your discussions. I used to be funny in french but that's another story in english. It's seams that I have two language-oriented brain's instances, a french emotional and an english rational. IMHO, being heretically stamped was very french funny and a compliment for my english mind-side.

    Shall I reformulate my mind and interrogation to have precious helps and aimed answers from PF members?

    For what I know (FWIK), physics have conservation laws for Energy, Electric charge, Momentum and Angular momentum. (Did I missed something?) FWIK, these physical properties are fundamentally major due to these laws. Also, I just red about quadri-vector (4-V) current density and 4-V momentum. I can understand that, by definition and consequences, 4-V have a 3-V and a non-vector as components (I didn't said scalar . Thanks Marcus! And didn't talk about vectorized-time component. Wouldn't like to be burnt at the stake:wink:)

    To help me further understandings:

    1) What are the physical properties where 4-V applies? Electric current, Momentum, Flows, ...? All properties that are 3-V shall become 4-V when relativistic is considered?

    2) For 4-V considerations, is it better to handle the property's density instead of the property itself?

    3) When I looked at the 4-V current density, I saw charge density times "c". When I looked at the 4-V momentum, I saw "E/c" which is mass times "c". I was surprised since I don't have, in my memories, any thing that looks like "E=qc2" and don't have mass conservation law. That's why I try to understand the fourth component of 4-V.

    Everyone is welcome to help me orient or clarify my mind. I can understand that my interrogations could be considered junior stuff for full-time physicists but for me these are extremely important.

    Answers always appreciated and therefore thanks everyone in advance.
  11. Aug 18, 2003 #10
    Hi Imagine,

    I'll throw in my two cents if it helps. I believe that it would help you to consider what is actually meant by the terms "vector" and "scalar". A scalar, for example, it not just some real number, and a vector is not simply a collection of 3 (or 4) numbers.

    A scalar is, as Marcus pointed out, a quantity that is invariant under coordinate transformations. In Special Relativity, the coordinate transformations are Lorentz transformations in 4d spacetime. So the only quantities that can be properly called "scalar" are those which have the same value regardless of the reference frame in which they are measured. Charge, for example, is a scalar; the charge of a moving electron is the same as one at rest. Charge DENSITY, on the other hand is most definitely not a scalar, since volumes change (due to length contraction) depending on the frame of reference.

    In the case of vectors, consider 3-vectors for a moment where our coordinate transformations are simply rotations of the axes. An arbitrary 3-vector has components (x1,x2,x3). However, it is possible to change to a coordinate system in which one or even two of these components are zero. In particular, I can take my new 3-axis to point in the direction of the vector. The new components are then (0,0,x3') where x3' is equal in magnitude to the length of the vector. From this example you can see that none of the components can be called a scalar since none of them was invariant under the coordinate transformation. What DIDN'T change, however, is the (squared) length of the vector, given by L^2 = x1^2 + x2^2 + x3^2 which implies that L^2 = (x3')^2. So, while none of the components of the vector is a scalar, its magnitude IS a scalar. A set of 3 numbers must have this property before it can be called a vector.

    The same is true of relativistic 4-vectors. Note, however, that we have some room to define what is meant by "length of a vector". In the above example, I used the Pythagorean Theorem to define length.
  12. Aug 18, 2003 #11
    cont...(sorry for being long-winded)

    Consider a 4-d displacement in spacetime. Its components can be given as (ct, x1, x2, x3), where these components measure displacement from the origin. But I need some notion of their length, which furthermore is a scalar. It turns out that in SR, a quantity called the spacetime interval, defined (for displacements from the origin) as:
    s^2 = -(ct)^2 + (x1)^2 + (x2)^2 + (x3)^2 IS invariant under Lorentz transformations!! If for an arbitrary 4-vector (v0,v1,v2,v3), I define length to be L^2 = -(v0)^2 + (v1)^2 + (v2)^2 + (v3)^2, then my above displacement can properly be called a vector, with the magnitude given by the spacetime interval. It is important to realize that without the invariant spacetime interval, there would be no good notion of length, and therefore no displacement 4-vector.

    The same is true of all SR 4-vectors. Their length, defined in the above way, MUST form a spacetime invariant. In your example, we want to turn the current density into a 4-vector. It so happens that an invariant IS formed by J^2 - (c rho)^2 = (c rho_0)^2, where J is the set of 3 spatial components of the current density, rho is the charge density and (rho_0) is the rest density (which is then a scalar). You can verify this by applying the appropriate transformations. This means that (c rho, J1,J2,J3) forms a proper 4-vector. Furthermore, c rho is the ONLY choice for the fourth component, since any other choice will not lead to invariant length.

    So you can see that 4-vectors do not arise arbitrarily by tacking on any old extra component with the right units. Rather, the existence of a 4-vector reflects the existence of a corresponding spacetime invariant. There are also other ways of defining vectors, one of which Marcus mentioned, but they turn out to be equivalent.

    Hope this helps,
    Last edited: Aug 18, 2003
  13. Aug 18, 2003 #12
    Merci beaucoup dhris !

    Now, I really understand 4-V and scalar. Particularly when I red the zero in "J^2 - (c rho)^2 = (c rho_0)^2". This help me to recover some losted neurons.

    You wrote that electric charge is a scalar and charge density is not a scalar (due to length contraction), that's right. The length of the 4-V current density is invariant, in Lorentz transformation, but not its components. Ok!

    Here, I have electric charge and current-density's length that are invariant but the first has a position (4-D) and the second has also a direction (4-D).

    In regards to "mass" domain instead of electric domain, my newbie following-quest question is: What "properties" are invariant? Mass density? Momentum length? Energy density?
  14. Aug 18, 2003 #13
    Well, in that case, it's the rest mass (or rest energy). You have:

    m^2 c^4 = -E^2 + c^2 p^2

    meaning that we can form a 4-vector out of energy and momentum. You can verify this relation by applying the Lorentz transformations for energy and momentum.

    Last edited: Aug 18, 2003
  15. Aug 18, 2003 #14
    You means rest energy or rest momentum like:

    (m0 c2)2 = - (E)2 + (c p)2
    (m0 c)2 = - (E/c)2 + (p)2
  16. Aug 18, 2003 #15
    Well, those are the same thing (they are both invariant), they just differ by a factor of c^2. I suppose, though, that the momentum 4-vector is (E/c,p1,p2,p3) so the magnitude of this particular vector is given by your second expression. Multiplying the components by c does yield a different 4-vector with magnitude given by the first expression, but this difference is trivial. In fact, in many theoretical applications units are scaled such that c=1.
  17. Aug 20, 2003 #16
    Do you have an equation expressing the four-vector current similar to four-vector current density?
  18. Aug 21, 2003 #17
    The real question is: Do you know an invariant four-current expression?
  19. Aug 22, 2003 #18
    I will guest !
    I2 - (c q)2 = (c q0)2
    I4 = ( I3 , icq )
    Is it right or the volume changed somethings?
  20. Aug 22, 2003 #19


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    Hello Imagine,
    dhris was providing some very welcome explanations
    (aussi-bien lucides que gentiles) but he seems temporarily absent
    his last post was of the 18th and now it is 22nd!

    the units of conventional electric current are "Coulombs per second"

    that is the same as "Amperes"-----measuring the passage of charge-per-unit-time.

    Is what you mean by the symbol "I" an ordinary current with units coulomb/second?

    The units of every component in a 4-vector must be the same.

    but the units of qc are coulomb-meter per second!!!!
    this is not the same as coulomb-per-second!!!

    edit: two afterthoughts
    hats off to you for guessing, I admire your taking initiative

    the idea of current density is elegant, I urge you to be satisfied with it (Bach chorale BWV 511 "Gib dich zufrieden.")
    it is defined at every point in space-----one does not need to describe ACCESSORIES like wires and particle beams and other annoying hardware, in order to talk about the current density
    at every point in space there is a current density in the x-direction and in the y, and in the z, direction----so 3 current densities----the idea is very economical: once one has said those three the thing is completely described

    (oh, plus the charge density to make it a quadrivector so that it
    will transform to moving coordinates, but being a sedentary person I ignore moving coordinates most of the time and only think of trivectors, as you call them)
    Last edited: Aug 22, 2003
  21. Aug 23, 2003 #20
    Merci Marcus,
    I revisited my EM books and discovered that, you're right, Ampere is Coulomb/sec. I was "thinking" that Ampere was Coulomb times speed. I waked up disgracefully.

    Now, suppose that O4 and O3 are energy density vectors, and Rp is momentum density value.
    Could I write:
    O4 = ( O3, icRp )
    and therefore
    (O4)2 = (O3)2 - (cRp)2
    where (O4)2 is invariant under Lorentz transformation.

    The point here, from what I learned from your posts, is to write something like:
    div4(O) = div3(O) + dRp/dt (partial)

    Rp has (kg-m/s)/(m3)
    O4 and O3 have (kg-m/s)/(m2-s)

    Rem: Replacing Coulomb by Momentum in Current density quadri vector.

    I would physically interpret like this:
    Considering a small volume, T.
    The rate at which the momentum T*Rp, enclosed in the volume T increase with time is T*dRp/dt (partial)
    The rate at which the enclosed momentum decrease with time is T*div3(O) since the energy density, O, is the momentum flowing out per unit time and per unit area.

    P.S.1: I red the Howto: Making math symbols post but I wasn't able to preview these code. Some help will be appreciated.

    P.S.2: Est-ce que (Bach chorale BWV 511 "Gib dich zufrieden.") signifie "À quoi cela sert-il de souffrir"?
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