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

[wotanub]

Several questions in a kind of stream of consciousness format...

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,

$K^μ \equiv \frac{dp^μ}{dτ}$, where $τ$ 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 $\mathbb{R}^{3}$ vector with respect to [whatever the 3-space analog to a Lorentz invariant is] a $\mathbb{R}^{3}$ 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 $x$ to another system $\overline{x}$:

$\overline{x}^{\mu} = \Lambda^{\mu}_{\nu}x^{\nu}$

or similarly in momentum space...

$\overline{p}^{\mu} = \Lambda^{\mu}_{\nu}p^{\nu}$

Now we take the derivative on the left...
$\frac{d}{dτ}\overline{p}^{\mu} = \frac{d}{dτ}\Lambda^{\mu}_{\nu}p^{\nu}$

$\overline{K}^{μ} = \frac{d}{dτ}\Lambda^{\mu}_{\nu}p^{\nu}$

$\Lambda^{\mu}_{\nu}K^{μ} = \frac{d}{dτ}\Lambda^{\mu}_{\nu}p^{\nu}$

$\Lambda^{\mu}_{\nu}\frac{dp^{μ}}{dτ} = \frac{d}{dτ}\Lambda^{\mu}_{\nu}p^{\nu}$

or

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

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 $\frac{d}{dτ}$ be inferred to be an invariant because $τ$ 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.

 

[dextercioby]

In 3D space, the correspondent of tau is t, the normal (absolute) time parameter. By differentiating wrt time, the vectorial characteristic of any measurable is kept: position vector differentiated wrt time equals the velocity vector, then further the acceleration vector.

 

[Nugatory]

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,

$K^μ \equiv \frac{dp^μ}{dτ}$, where $τ$ 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.

The expression above isn't a definition of the four-force, it's a recipe for calculating the components of the four-force in a given coordinate system. If you express the definition of the four-force without coordinates (the difference between two invariant four-vectors, divided by an invariant scalar) it's a lot easier to buy that it is itself an invariant four-vector.

Of course this isn't a proof; it's just a hand-wave that may persuade you to buy the proposition without the proof.

 

[GRGuy]

...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,

$K^μ \equiv \frac{dp^μ}{dτ}$, where $τ$ is proper time...
...What's the general case of a Lorentz invariant? What do you call an invariant for any arbitrary transformation?,,,

Actually the way you are defining it in your first equation, its not generally a 4-vector. That only works out to be the 4-force according to a rectilinear inertial frame that way. Generally the derivative you are taking using element notation isn't that, but is a "covariant derivative" operation. It includes Christoffel symbol terms. Those symbols are defined just that the covariant derivative operation maps tensors to tensors so defining a 4-force as a covariant derivative of 4-momentum with respect to proper time ensures that the 4-force is a 4-vector. Now it turns out that when you use a rectilinear inertial frame as you do in special relativistic physics that the Christoffel symbol terms are zero leaving you with 4-force just being the derivative of 4-momentum with respect to proper time.
Call something that's invariant just invariant. We don't need the term Lorentz because GR in local free fall reduces to SR anyway. The "length" so to speak of any tensor is invariant to frame transformation. When people use the term Lorentz scalar, they are just trying to be clear to you that this is a scalar that is invariant to Lorentz transformation.

 

[arkajad]

In any vector space V if you have a vector v(s) depending on an external parameter s, the derivative with respect to s is again a vector. Its components in any basis are just derivatives of the components of v(s).

I physics however, sometimes, we are taking derivatives with respect to one parameter in one coordinate system (say, inertial frame) and with respect to another parameter in a different coordinate system. In such a case you must be careful. This happens in Relativity if you differentiate with respect to, say, "t". But this is not your case, so a general rule of differentiating vector-valued functions applies.