Several questions in a kind of stream of consciousness format...(adsbygoogle = window.adsbygoogle || []).push({});

I was going over some special relativity notes and I'm not sure why I should buy this. The statement was saying that we can introduce a four-force by differentiating the four-momentum with respect to proper time. That is,

[itex]K^μ \equiv \frac{dp^μ}{dτ}[/itex], where [itex]τ[/itex] is proper time.

I looked in a different reference, and it highlighted that the four-force defined this way is a four-vector because the proper time is an invariant.

I guess I could brute-force check to see if it transforms like a four-vector (it does), but checking this one case doesn't seem like it would prove the general case.

I suppose you could turn everything on its head and also ask the even more "obvious" question "Why is the derivative of a [itex]\mathbb{R}^{3}[/itex] vector with respect to [whatever the 3-space analog to a Lorentz invariant is] a [itex]\mathbb{R}^{3}[/itex] vector?"

What's the general case of a Lorentz invariant? What do you call an invariant for any arbitrary transformation?

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This leads to a second question I thought of while thinking about this. What does it commutation with the Lorentz-transformation operation imply?

The Lorentz-transformation transforms from one space [itex]x[/itex] to another system [itex]\overline{x}[/itex]:

[itex]\overline{x}^{\mu} = \Lambda^{\mu}_{\nu}x^{\nu}[/itex]

or similarly in momentum space...

[itex]\overline{p}^{\mu} = \Lambda^{\mu}_{\nu}p^{\nu}[/itex]

Now we take the derivative on the left...

[itex]\frac{d}{dτ}\overline{p}^{\mu} = \frac{d}{dτ}\Lambda^{\mu}_{\nu}p^{\nu}[/itex]

[itex]\overline{K}^{μ} = \frac{d}{dτ}\Lambda^{\mu}_{\nu}p^{\nu}[/itex]

[itex]\Lambda^{\mu}_{\nu}K^{μ} = \frac{d}{dτ}\Lambda^{\mu}_{\nu}p^{\nu}[/itex]

[itex]\Lambda^{\mu}_{\nu}\frac{dp^{μ}}{dτ} = \frac{d}{dτ}\Lambda^{\mu}_{\nu}p^{\nu}[/itex]

or

[itex]\left[\frac{d}{dτ},\Lambda^{\mu}_{\nu}\right] = 0[/itex]

Is the commutation of an operator with the Lorentz transformation associated with some physical property or observable? Is this the definition of an invariant under a transformation? Something that commutes with the transformation matrix? If so, can [itex]\frac{d}{dτ}[/itex] be inferred to be an invariant because [itex]τ[/itex] is one? This seems strange to say since one is an operator (can an operator be called an invariant?), and the other is a scalar, though I suppose you could represent the proper time as the scalar time the identity matrix.

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# Why is it obvious that the derivative of a 4-vector is also a 4-vector

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