Ah, in that case, I'd agree with you!

We have:

globally conserved locally conserved tensor quantity globally conserved

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By the way, you appear to be using the term "Lorentz invariant" to mean "behaves like a tensor", which is not correct. A quantity is Lorentz invariant if its value at each spacetime point is unchanged by a Lorentz transformation. Scalar fields are Lorentz invariant, but vectors, 1-forms, and general tensors ARE NOT, since their component values change. For example, a vector field pointing in the "+

t" direction no longer points that way after a boost. See

this link for the correct definition.

Total angular momentum

M is most certainly NOT Lorentz invariant!

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While I followed your proof quite well, allow me to present a slightly different version, that might be clearer to some readers:

Let

S(

t) be a "tensor-like quantity" defined as the sum or integral of the tensor

T over all space (that is, over all of

x,

y,

z, with

t held constant), as calculated in a given reference frame F.

By construction,

S is invariant w.r.t. purely spatial translations in F, i.e.:

S(

t,

x,

y,

z)

S(

t)

Now, furthermore suppose that

S is indeed a tensor: That is to say, any Lorentz transformation that takes F to any other reference frame F', when applied to the components of

S at any spacetime point, yields the same result as calculating

S directly at that point in the frame F'.

Since this latter calculation yeilds a tensor field that's invariant w.r.t. purely spatial translations in F', we can conclude that

S (as calculated in frame F) must be invariant w.r.t.

any spacetime translation that is purely spatial in

some reference frame F'.

In particular, the translations

and

in F both represent purely spatial translations in two other (correctly chosen) reference frames, respectively. Individually each translation must leave the value of

S unchanged, and composing them yeilds the translation

which therefore must also. Hence:

S(t) = S() for all

This can only be satisfied when

S is constant for all

t (as calculated in F); i.e.

S is "conserved" and does not change over time.

Such a "conserved"

S is therefore a tensor quantity that is constant across all spacetime, with component values that depend on the frame F in which it is calculated (as usual for a tensor).

Note that in earlier discussion, we described the conservation of

S over time by (equivalently) saying that

T is "

globally conserved".

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Note that we could construct an analogous proof (corresponding to a slightly different definition of "globally conserved") if the sum or integral

S were to be defined over some particular (frame-relative) set of non-intersecting spacelike slices, where a "slice" is defined so as to bisect all of spacetime into +

t and -

t regions, and where for each frame, this set covers all space. (i.e. As an alternative to using the flat surfaces

t=const). All that is required for the proof is that any point in spacetime be reachable from any other, by traversing the union of these slices (as collected from

all reference frames). This works since

S must be constant over any such slice, irrespective of reference frame. (Where we define

S(

t,

x,

y,

z) calculated in reference frame F as the value of the sum or integral of

T across the unique F-relative slice that the point (

t,

x,

y,

z) happens to lie on).

Note that saying that

T is "globally conserved" in this manner is

not as strong a statement as saying that the value of

S in a given frame F will be the same when computed over

any arbitrary spacelike slice. This latter property

is implied by

local conservation of

T, however.

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We have demonstrated that: "

S is a tensor"

"

T is globally conserved"

It remains an open question (

to me) as to whether either of the (successively) stronger implications hold:

"S is a tensor" "the value of S in a given frame F will be the same when computed over any arbitrary spacelike slice"

or

"S is a tensor" "T is locally conserved"

Additionally we could ask whether:

"the value of S in a given frame F will be the same when computed over any arbitrary spacelike slice" "T is locally conserved"

...Anybody have any thoughts on this?