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Is the value of momentum an invariant?,
specificly for instance the momentum value $\mathbf p_{\text{lab}}[~\Lambda^0~]$ of a $\Lambda^0$ baryon (drifting from the (actual) interaction point of a collider experiment towards the beam pipe wall) with respect to suitable(1) constituents of the "lab" (of beam pipe wall, of detectors, of magnets)?

Or does the momentum value $\mathbf p_{\text{lab}}[~\Lambda^0~]$ depend on the assignment of coordinate values to the relevant (unique) events, such as the event of the $\Lambda^0$ baryon under consideration having been produced; or the event of the $\Lambda^0$ baryon under consideration passing the beam pipe wall, or the event of the $\Lambda^0$ baryon under consideration decaying?

Expressing the value of the momentum of the specific $\Lambda^0$ baryon under consideration wrt. the lab constituents as

$$\mathbf p_{\text{lab}}[~\Lambda^0~] := m[~\Lambda^0~] ~ c ~ \frac{\beta_{\text{lab}}[~\Lambda^0~]}{\sqrt{1 - (\beta_{\text{lab}}[~\Lambda^0~])^2}} ~ \mathbf e_{\text{lab}}[~\Lambda^0~], $$

where

  • $m[~\Lambda^0~]$ denotes the invariant mass of the $\Lambda^0$ baryon under consideration,

  • $c$ denotes the signal front speed, and

  • $\mathbf e_{\text{lab}}[~\Lambda^0~]$ denotes the (normalized) direction of motion of the $\Lambda^0$ baryon under consideration wrt. the lab constituents,

is the corresponding real number value $\beta_{\text{lab}}[~\Lambda^0~]$ an invariant, too?
(Or else: How, explicitly, does the value $\beta_{\text{lab}}[~\Lambda^0~]$ depend on the assignment of coordinates?)

(1: Specifily, constituents of beam pipe wall, detectors, magnets which were remaining separate and at rest with respect to each other; i.e. constituting members of an inertial system in the sense of Rindler: "simply an infinite set of point particles sitting still in space relative to each other".)

user12262
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  • Note that you have not defined what $\beta_{\rm lab}\left[\lambda^0\right]$ is, so how can anyone tell you if (a) it is invariant or (b) how it would depend on coordinates. – Kyle Kanos Jul 27 '15 at 17:11
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    If the nature of the quantity is not clear in you personal, look-how-much-of-a-purist-I-am-I-write-move-complicated-notation-than-you-do way of writing things, then perhaps you should reconsider. Seriously, with as much effort as you put into writing things you have managed to obscure whether you mean the 3-vector, the 4-vector, the magnitude of the 3-vector or the magnitude of the 4-vector. Only one of which invariant, and that is properly called the mass. – dmckee --- ex-moderator kitten Jul 27 '15 at 17:12
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    @Kyle Kanos: "you have not defined what $\beta{\text{lab}}[~\lambda^0~]$ is_" -- The definition of this quantity, for the purpose of my question, is implicit in the OP question statement; namely as a (the?) real number value such that the expression $$m[~\lambda^0~] ~ c ~ \frac{\beta_{\text{lab}}[~\lambda^0~]}{\sqrt{1 - (\beta_{\text{lab}}[~\lambda^0~])^2}} ~ \mathbf e_{\text{lab}}[~\lambda^0~] $$ is equal to the momentum value $\mathbf p_{\text{lab}}[~\lambda^0~]$; where all remaining symbols ($m$, $c$, $\mathbf e$) have been named in the OP, too. (Additional definitions may have existed.) – user12262 Jul 27 '15 at 17:22
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    @user12262: well, your formula has explicit $v$'s in it, so how invariant do you expect it to be? – Zo the Relativist Jul 27 '15 at 17:24
  • Well if $\beta=v/c$, then you end up with $p^a=\gamma mv^a$ (where $a$ superscript is for spatial components) which is what everyone else uses and is plainly not invariant. – Kyle Kanos Jul 27 '15 at 17:25
  • @dmckee: "[...] whether you mean the 3-vector, the 4-vector, the magnitude of the 3-vector or the magnitude of the 4-vector." -- Vectors ?? ... – user12262 Jul 27 '15 at 17:28
  • @Jerry Schirmer: "how invariant do you expect it to be?" -- Perfectly independent of any coordinates whatsoever; hence perfectly invariant. If you believe that a momentum value such as the momentum of a specific $\lambda^0$ baryon with respect to (suitable) constituents of a specific lab is not invariant, then please state its specific dependence on coordinates. – user12262 Jul 27 '15 at 17:34
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    @user12262: the frame velocity is 100% not Lorentz covariant. It is explicitly present in the equation. Therefore, a boost will change the frame velocity, and change your momentum. – Zo the Relativist Jul 27 '15 at 17:45
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    Like, the quantity you're talking about is explicitly frame-dependent, as your notation makes manifest. It's not an invariant. – Zo the Relativist Jul 27 '15 at 17:46
  • @Jerry Schirmer: Please consider expanding your extremely terse comments into an answer. Don't omit to make explicit the coordinate dependence of the momentum value (e.g. of the specific $\Lambda^0$ baryon wrt. specific suitable constituents of the beam pipe wall), if you believe that such a dependence exists. – user12262 Jul 27 '15 at 17:53
  • @Kyle Kanos: "$p^a$ [...] what everyone else uses" -- This notation seems to obscure the two arguments required for defining and evaluating "momentum"; namely "of whom" the value is to be obtained (e.g. the specific $\Lambda^0$ baryon), and "by whom and with respect to whom" the value is obtained (e.g. the specific constituents of the beam pipe wall etc.) So, yes, I'm primarily questioning this notation which "everyone else uses" (as far as you seem to know). – user12262 Jul 27 '15 at 17:56
  • This is a trivial question for first year courses on special relativity, completely obscured by bad notation and pedantism. The "usual notation" is clearer and you would do well to learn to use it. – Danu Jul 30 '15 at 11:56
  • @Danu: "This is a trivial question for first year courses [...]" -- Your assessment suggests that you (too) know a plain, unambiguous answer to my question. So which is it: "Yes.", or "No."? If "No." then please consider expanding and submitting this as an answer. (The answer "Yes." has been issued already.) "[...] The "usual notation" is clearer and you would do well to learn to use it." -- What exactly do you consider the "usual notation" for denoting the momentum of one specific particle (such as a $\Lambda^0$) with respect to one specific reference system (such as "the lab"), please? – user12262 Jul 30 '15 at 18:41

2 Answers2

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If I measure the speed of a particle in the lab and then write down in my notebook the value I measured, the number an observer in a different inertial frame reads from my notebook will be the same (although the numerals may be Doppler shifted, length contracted, etc.). In the same way, the name of my cat is "Mittenz" independent of any choice of coordinates. Perhaps you can give some more of the context that led you to consider this question so that we can give you a more satisfying answer.

d_b
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  • user37496: "If I measure the speed of a particle in the lab and then write down in my notebook the value" ... as real-number multiple of $c$ ... "an observer in a different inertial frame reads [...] the same" -- Correct. Real numbers are presumed unambiguous; they can be copied; +1. "(although [...])" -- Their graph structure (or topology) should remain distinctive enough. "context that led you" -- Foremost this: "speed itself is a coordinate system dependent concept"; then that. – user12262 Jul 27 '15 at 22:15
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    Speed is a coordinate dependent concept. The value of a particle's speed measured by an observer using a particular set of coordinates doesn't change once the measurement is made, but observers using different coordinates may measure different speeds, so there is no such thing as "the speed" of a particle independent of coordinates. – d_b Jul 28 '15 at 00:57
  • user37496: "observers using different coordinates may measure different speeds" -- Suppose that the described specific lab constituents (of beam pipe wall etc.) had measured the speed value of the specific $\Lambda^0$ baryon wrt. those lab constituents (themselves) as $$ \beta_{\text{lab}}[~\Lambda^0~]~c = 0.2~c.$$ Which value do you suppose that "observers using different coordinates" might obtain instead for the speed $\beta_{\text{lab}}[~\Lambda^0~]~c$ of the specific $\Lambda^0$ baryon wrt. those lab constituents?? (Or do you use the word "speed" improperly??) And: ... – user12262 Jul 28 '15 at 05:36
  • user37496: "no such thing as "the speed" of a particle independent of coordinates." -- Why involve any coordinates at all?? Relevant and necessary are (e.g.) values of certain distances and durations; which (btw.) can be expressed as values of intervals $s^2$ "between" certain pairs of the relevant events (e.g. "the passage of the beam pipe wall" and "the passage of the first inner detector plane"). Do you claim "observers using different coordinates may measure different interval values" of this event pair? – user12262 Jul 28 '15 at 05:37
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    How do you measure the spacetime interval without measuring time elapsed and distance traveled? – d_b Jul 28 '15 at 05:44
  • user37496: How do you suppose that the measurements of distance values, or of duration values, might be contingent on coordinates?? – user12262 Jul 28 '15 at 17:02
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    How do you measure a distance without a ruler? How do you measure a time without a clock? – d_b Jul 28 '15 at 17:16
  • user37496: "How do you measure a distance without a ruler? How do you measure a time without a clock?" -- Well, when it comes to off-the-shelf rulers and clocks, the pertinent questions are how to determine whether a given pair of ruler ends were and remained "at rest to each other" (or at least "rigid to each other"); and whether a given clock was "good". That's (of course) addressed by the "Marzke-Wheeler method"; see also MTW §16.4. But returning to my question: What do you suppose this might have to do with coordinates?? – user12262 Jul 28 '15 at 17:51
  • Experimentalists measure real numbers. If you want to measure a distance or time in a lab, you need a way to get from spacetime events to real numbers. – d_b Jul 28 '15 at 18:24
  • user37496: "[...] you need a way to get from spacetime events to real numbers." -- Correct. The "canonical way" to achieve this is to count successive pings (a.k.a. "signal roundtrips") between participants; as shown e.g. in MTW Box 16.4 (part of which happens to be publicly visible in this link). These counts are of course integer numbers; their ratios provide rational values; their limits in turn reals. So: do you suppose that counting pings is in any way contingent on coordinates (which may or may not be assigned to the events under consideration)? – user12262 Jul 28 '15 at 18:58
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Yes, any (measured) momentum value (of someone or something specific, with respect to a specific system of suitable participants) is invariant, i.e. unchanged by any and all coordinate transformations, such as Lorentz transformations; and so is any (measured) value of velocity, or of speed (of someone or something specific, with respect to a specific system of suitable participants).

user12262
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    This is in no way related to the meaning of the word "invariant" used by relativists. – dmckee --- ex-moderator kitten Jul 27 '15 at 18:06
  • @dmckee: "This is in no way related to the meaning of the word "invariant" used by relativists." -- Please note the Edit of the answer: "invariant, i.e. unchanged by any and all coordinate transformations, such as Lorentz transformations"; referring specificly to the meaning of the word "invariant" used by relativists. – user12262 Jul 27 '15 at 18:35
  • Can you explicitly show how you're getting that $p_\nu={\Lambda_\nu}^\mu p_\mu$ is invariant? – Kyle Kanos Jul 27 '15 at 19:08
  • @Kyle Kanos: "Can you explicitly show how [...]" -- I can illustrate that coordinates are irrelevant: Given two specific coordinate assignments $$\mathbf p :=\mathbf b^{\mu}~p_{\mu}$$ and $$\mathbf p :=\mathbf b^{\nu}~p_{\nu},$$ such that the transformation $\Lambda_{\nu}^{\mu}$ between these two assignments is invertible (one-to-one), and where $\mathbf b^{\mu}$ and $\mathbf b^{\nu}$ are suitable "bases" (spanning the range of momentum values). Then $$p_{\nu}=\Lambda_{\nu}^{\mu}~ p_{\mu}, \quad \mathbf b^{\nu} = (\Lambda_{\nu}^{\mu})^{(-1)} \mathbf b^{\mu},$$ leaving $\mathbf p$ invariant. – user12262 Jul 27 '15 at 20:03
  • Note that $p_\mu$ is a list of components while $\mathbf p$ is that list projected onto a basis; they are not the same thing. The transformation of $p_\mu$ does not contain the basis vector transformation; further $\mathbf b^\nu$ and $\mathbf b^\mu$ here are completely different vectors; one is not the transformation of the other into a different reference frame. – Kyle Kanos Jul 27 '15 at 20:47
  • @Kyle Kanos: "[...] transformation [...] into a different reference frame" -- I wonder whether that's even a proper notion. As far as it is meaningful at all, remember that such "transformation" should be applied both to the specific $\Lambda^0$ baryon "of whom" the momentum value is to be obtained as well as to the constituents of the specific lab equipment by whom and with respect to whom the value is obtained. p.s. Admittedly, since "bases" had been brought up, the notion of "vectors" cannot be denied either. – user12262 Jul 27 '15 at 22:00