Momentum Measurement Confusion

In summary, Dr. Polyzou told me that the equation for a particle's momentum,\eta_{\alpha\beta}p^\alpha p^\beta =(mc)^2holds true for individual measurements, but that after each measurement, the probability distributions lose some of the values.
  • #1
jfy4
649
3
Hi everyone,

I was thinking about what it means to talk about the "momentum of a particle" and I tried to come up with a little scheme to help me, but as of now it has only made more confusion... This is how my thoughts went.

I know from quantum mechanics can one can relate classical quantities, like momentum, to expectation values. I was also under the impression that a particle never really has a value for momentum until it is measured, just a probability distribution; that the value of a particle's momentum isn't already existent, it comes about after measurement. With these in mind I was thinking about what
[tex]
\eta_{\alpha\beta}p^\alpha p^\beta =(mc)^2
[/tex]
means quantum mechanically. Does it mean
[tex]
\langle p^0\rangle^2-\langle p^1\rangle^2-\langle p^2\rangle^2-\langle p^3\rangle^2=(mc)^2
[/tex]
It seems to me this might be true, but that this isn't the whole story, and can't be the last word. This only says the average has to be equal to [itex](mc)^2[/itex], not the individual measurements, but I feel like if I took the square sum of the four measured values of an arbitrary particle that they would equal [itex](mc)^2[/itex] every time. So then I thought that perhaps the above formula applies to the individual measurements of a particle's momentum. Then [itex]p^\alpha =\langle p^0,\vec{p}\rangle[/itex] are measured values. But I also know something else about a particle's momentum: measurements of different components of a particle's momentum leave the other components alone, I believe [itex][\hat{p}_\alpha , \hat{p}_\beta ]=0[/itex] tells us this.

Now here I take a step and assume that the equation [itex]\eta_{\alpha\beta}p^\alpha p^\beta =(mc)^2[/itex] holds true for individual measurements of a particles momentum. I also assume that a measurement of a particles momentum in the [itex]\alpha[/itex] direction does not affect a measurement of the particle's momentum in the [itex]\beta[/itex] direction. Then let's say I make measurements of a particles momentum for [itex]p_0, p_1, p_2[/itex], then using [itex]p_\alpha p^\alpha =(mc)^2[/itex] I can find [itex]p_3[/itex] without having to measure it. To me this seems to be saying one of two things: either

1) After I measured the three components of the particle's momentum, the [itex]p_3[/itex] component was decided, in which case the previous measurements of the various components of the particle's momentum did affect the measurements of other components.

or

2) the particle had [itex]p_3[/itex] to begin with.

Neither of these seem correct to me. Where along this did my thinking go askew.

Thanks,
 
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  • #2
Of the four components of pμ only three are independent. You can specify a state by specifying the three-momentum, in which case p0 = E, is determined, since it is a function of the other three. The relation ημνpμpν is not just an expectation value, it's an operator relation, holding whether the pμ have been measured or not.
 
  • #3
Thanks Bill,

That is just what Dr. Polyzou told me also. There seem to be some cool consequences for this then. Consider that if we measure two different spatial components, [itex]p_1, p_2 [/itex], then the relativistic equation reads
[tex]
p_{3}^2 - p_{0}^2 = p_{5}^2
[/tex]
where [itex]p_{5}^2=-(mc)^2-p_{2}^2 - p_{1}^2 [/itex] and is a negative number. The this is the equation for a hyperbola, with [itex]p_{0}[/itex] on the x-axis . The cool part is that the values for [itex]-p_{5}< p_0 < p_5[/itex] are no longer accessible which means that the probability for these values in the probability distribution should be zero. So after each measurement, the probability distributions, lose in a sense, some of the values. For me this was a learning experience about the power of relativity even over probability.
 

What is momentum measurement confusion?

Momentum measurement confusion refers to the difficulty in accurately measuring the momentum of a moving object.

Why is it important to accurately measure momentum?

Momentum is a fundamental concept in physics and is necessary for understanding the motion of objects. Accurate measurement of momentum allows for precise calculations and predictions of an object's motion.

What factors contribute to momentum measurement confusion?

Several factors can contribute to momentum measurement confusion, including the speed and direction of the object, the precision of the measuring instruments, and external forces acting on the object.

How can momentum measurement confusion be minimized?

To minimize momentum measurement confusion, it is important to use precise and accurate measuring instruments, take multiple measurements, and account for external factors such as air resistance or friction. Additionally, using mathematical equations and concepts such as impulse and conservation of momentum can help to reduce errors in measurement.

What are some real-world applications of accurate momentum measurement?

Accurate momentum measurement is crucial in various fields such as engineering, sports, and transportation. For example, engineers use momentum measurements to design safer vehicles, while athletes and coaches use it to enhance performance and improve techniques. In transportation, momentum measurement is essential for designing efficient and safe transportation systems.

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