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The four-velocity(world-velocty) is defined by : $u^μ=\frac{dx^μ}{dτ}$ ,where $τ$ is the proper time of the object.

I don't understand why it's defined with respect to the proper time but not the time of reference frame of the observer.

I understand that the proper time is a Lorentz invariant, but if that's the purpose of the definition which is making the four-velocity independent of the reference frame, the numerator of the derivative ($dx^\mu$)is still not a Lorentz invariant, so the four-velocity is not a Lorentz invariant after all.

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    $\begingroup$ The magnitude of the four-velocity is invariant. $\endgroup$
    – m4r35n357
    Nov 14, 2023 at 11:52

7 Answers 7

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$dx^\mu$ is covariant and $d\tau$ is invariant. So $dx^\mu/d\tau$ is manifestly covariant while $dx^\mu/dt$ is not.

Covariant quantities are of interest because the laws of physics are covariant. So they should be able to be written exclusively in terms of manifestly covariant quantities. So we know from this that we should be able to write our laws of physics without any $dx^\mu/dt$ quantities appearing.

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Any well-defined quantity in special relativity must behave covariantly under Lorentz transformations.

Covariant behavior includes

  1. being invariant: in this case the object is called a Lorentz scalar, like $\tau$ in your question.
  2. obeying $V^{\mu'}=M^{\mu'}_{\nu}V^{\nu}$: in this case the object $V^{\mu}$ is called a 4-vector, like $x^{\mu}$ and $\frac{dx^{\mu}}{d\tau}$ in your question. ($M^{\mu'}_{\nu}$ consists of boosts and rotations)
  3. obeying $T^{\mu'\nu'}=M^{\mu'}_{\alpha}M^{\nu'}_{\beta}T^{\alpha\beta}$: in this case the object $T^{\mu\nu}$ is called a rank-2 tensor, like the energy momentum tensor: $T{\mu\nu}$.
  4. You can continue to construct out covariant objects with more number of indices.
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Dale and Navid gave really good answers but I want to write an answer to say $\textbf{why}$ we care that the velocity transforms covariantly.

When we teach Newtonian Mechanics, we do not dwell much on a frame of reference. We acknowledge that the laws of physics is the same in all inertial frame of reference but then go on to compute relevant quantities of interest. However, even though it is not apparent, this "symmetry" is ingrained in the equations of Classical Mechanics. Most prominently, Newton's $2^{nd}$ law is invariant under a Galilean Transformation. Another example comes to mind when one solves the box on an inclined plane problem. Regardless of what coordinate system you choose, you should get the same physics. The two coordinate system you might choose here (mainly one where the $x$ axis is parallel to the ground and the other where the $x$ axis is parallel to the inclined plane) is related by a rotation. Therefore, you expect the equations of motion to be same under a passive rotation. Since

$$ \vec{x}' = R\cdot\vec{x} \tag{1}$$

under a passive rotation, you want the force to change as $\vec{F}'= R\cdot\vec{F}$ so that Newton's Law doesn't change.

In Special Relativity or even in GR, you want to keep this "coordinate independence" in your equations of motion. Therefore, you need to write down equations of motion for objects that transform in a similar way to equation (1). To achieve this, you need to consider the proper velocity and not something else.

Hope this helps.

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    $\begingroup$ I would also like to add another reason. In SR and GR, time is now another coordinate: it cannot be separated from space. Therefore, defining $u^{\mu} = \frac{d x^{\mu}}{dt}$ is weird as you are taking the derivative of a coordinate with respect to another coordinate. $\endgroup$
    – emir sezik
    Nov 14, 2023 at 14:28
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    $\begingroup$ I think your comment is one of the best answers. In particular, a general coordinate system cannot necessarily be decomposed into "spatial" and "time" coordinates (e.g. $u = 2t-x$ and $v = 2t+x$ in $(1+1)$-D Minkowski space), so $\frac{dx^\mu}{dt}$ cannot be made sense of in general. $\endgroup$
    – jawheele
    Nov 15, 2023 at 16:24
  • $\begingroup$ @jawheele I definitely agree with you and thank you! $\endgroup$
    – emir sezik
    Nov 15, 2023 at 22:16
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By definition, the four-vector is:$$X=(ct,\vec{r})$$ a four-velocity by $$V=\frac{dX}{d\tau}=\left(c\frac{dt}{d\tau},\frac{d\vec{r}}{d\tau}\right)=\gamma(c,\vec{v})$$ The Minkowskian scalar product gives: $$V^{2}=\gamma^{2}(c^{2} -\vec{v}^{2})$$

we knew that $\gamma^{2}=\frac{1}{1-\frac{v^{2}}{c^{2}}}$ so :$$V^{2}=c^{2}$$ By construction of the theory of special relativity, the speed of light is invariant (postulate).

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Your question addresses a fundamental aspect of relativistic kinematics. The definition of four-velocity, $$ \mathbf{u}^\mu = \frac{dx^\mu}{d\tau} $$ indeed uses proper time ($ \tau $) in the denominator, and this choice is pivotal for maintaining consistency with the principles of special relativity.

Proper Time as a Lorentz Invariant: The reason for using proper time ($ \tau $) instead of the time in the observer's reference frame is indeed its Lorentz invariance. Proper time is the time interval measured by a clock moving with the particle, and it remains the same in all inertial frames. This is a crucial property because it ensures that the description of motion is consistent across different frames of reference.

Role of the Four-Velocity: The four-velocity $ \mathbf{u}^\mu $ is not a Lorentz invariant quantity; you are correct in this observation. However, its transformation properties under Lorentz transformations are well-defined. The transformation of the four-velocity components from one inertial frame to another is governed by Lorentz transformation equations. This predictable behavior under transformations is what makes the concept valuable.

Spacetime Interval: The four-velocity is defined with respect to the proper time because the spacetime interval (or the line element), $ ds^2 $, which is invariant under Lorentz transformations, is related to the proper time by $ ds^2 = -c^2 d\tau^2 $ for a particle moving through spacetime. Thus, the denominator in the four-velocity is directly tied to this fundamental invariant of spacetime.

Physical Interpretation: The four-velocity's magnitude is always the speed of light, which reflects the constancy of the speed of light in all inertial frames, a cornerstone of special relativity. This wouldn't hold if the time of the observer's frame were used.

Differentiation Concern: Your concern regarding the non-invariance of $ dx^\mu $ is valid. However, the key is that while each component of four-velocity transforms under Lorentz transformations, the way it transforms ensures that the physics described by these quantities is frame-independent. That is, the laws of physics, as described through these quantities, take the same form in every inertial frame, which is a fundamental requirement in relativity.


In summary, the use of proper time in defining the four-velocity ensures that the description of motion adheres to the principles of relativity and is consistent across different inertial frames, even though the four-velocity itself is not Lorentz invariant. The transformation properties of the four-velocity under Lorentz transformations preserve the form of physical laws, reflecting the underlying symmetry of spacetime in special relativity.

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One thing that the other answers haven't mentioned is the Hamiltonian formulation of special relativity.

The vector transformation property of the four velocity can be obtained by differentiation wrt any arbitrarily chosen scalar parameter $\lambda$. This does not necessitate proper time.

In the Hamiltonian formulation of special relativity, we have the Hamiltonian $\sqrt{p_i p^i+m^2}$. The $p_i$ is the three momentum given by differentiation wrt proper time : $m\frac{dx^i}{d\tau}$. This Hamiltonian produces the time evolution:

$$\frac{dp^i}{dt}=0$$

$$\frac{dx^i}{dt}= \frac{p^i}{\sqrt {p^2+m^2}}$$

Notice how the second equation can never exceed the speed of light (We have chosen units where $c=1$)

You can also start with the above equation and derive that the parameter must be proper time. Substitute $p^i=m\frac{dx^i}{d\lambda}$ to get:

$$(\frac{dx^i}{dt})^2=\frac{(\frac{dx^i}{d\lambda})^2}{(\frac{dx^i}{d\lambda})^2+1}$$

or

$$\frac{dx^i}{dt}=\frac{1}{\gamma} \frac{dx^i}{d\lambda}$$

Implying $dt= \gamma d\lambda$, which means $\lambda$ must be proper time.

Worth Mentioning : The Lorentz Force Law

The Lorentz force law takes the simple form in terms of proper time:

$$m\frac{d^2x^{\mu}}{d\tau^2}=q F^{\mu \nu}v_{\mu}$$

In summary

The dynamics of special relativity favor the proper time parameter, even though tangent vectors can be defined wrt any scalar parameter.

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Space

To describe the direction of a path through space, you use a forward unit tangent vector: a unit vector which is tangent to the path, and points forward rather than backward.

If you have a path $\mathbf{x}^c$, parameterized by some variable $s$ that increases as you go forward, you can find the unit tangent vector $\mathbf{u}^c$ at each point using the formula $$\mathbf{u}^c = \left. \frac{d\mathbf{x}^c}{ds} \right/ \sqrt{g_{ab} \frac{d\mathbf{x}^a}{ds} \frac{d\mathbf{x}^b}{ds}}.$$ Here, $g_{ab}$ is the metric of space. By calculating $g_{cd} \mathbf{u}^c \mathbf{u}^d$ from the formula, you can confirm that $\mathbf{u}^c$ is a unit vector.

You can simplify this formula by using an arc length parameterization. That means choosing a path parameter $\sigma$ for which $$g_{ab} \frac{d\mathbf{x}^a}{d\sigma} \frac{d\mathbf{x}^b}{d\sigma} = 1.$$ In terms of an arc length parameter $\sigma$, the forward unit tangent vector has the elegant expression $$\mathbf{u}^c = \frac{d\mathbf{x}^c}{d\sigma}.$$

Spacetime

To describe the direction of a timelike path through spacetime, you use a forward unit tangent vector (image source): a unit vector which is tangent to the path, and points forward rather than backward. (We typically orient timelike paths so that going forward along the path also means going forward in time.)

If you have a timelike path $\mathbf{x}^c$, parameterized by some variable $t$ that increases as you go forward, you can find the unit tangent vector $\mathbf{u}^c$ at each point using the formula $$\mathbf{u}^c = \left. \frac{d\mathbf{x}^c}{dt} \right/ \sqrt{g_{ab} \frac{d\mathbf{x}^a}{dt} \frac{d\mathbf{x}^b}{dt}}.$$ Here, $g_{ab}$ is the metric of spacetime. Since the path is timelike, the number under the square root is always positive.

You can simplify this formula by using a proper time parameterization. That means choosing a path parameter $\tau$ for which $$g_{ab} \frac{d\mathbf{x}^a}{d\tau} \frac{d\mathbf{x}^b}{d\tau} = 1.$$ In terms of a proper time parameter $\tau$, the forward unit tangent vector has the elegant expression $$\mathbf{u}^c = \frac{d\mathbf{x}^c}{d\tau}.$$

World-velocity and spatial velocity

As you may have guessed by now, the world-velocity is another name for the forward unit tangent vector of a worldline. To see how the world-velocity is related to the familiar spatial velocity from Galilean mechanics, suppose spacetime is flat, so we can choose coordinates $t$, $x$, $y$, $z$ with $$g_{ab} = dt_a\,dt_b - \left(dx_a\,dx_b + dy_a\,dy_b + dy_a\,dz_b\right).$$ Writing the world-velocity in terms of these coordinates as $$\mathbf{u}^c = \left[\begin{array}{c}u_t \\ u_x \\ u_y \\ u_z\end{array}\right],$$ we can express the spatial velocity as a slope: $$\frac{1}{u_t} \left[\begin{array}{c} u_x \\ u_y \\ u_z \end{array}\right].$$

To get a more explicit expression for the world-velocity, let's use the time coordinate $t$ as our path parameter, giving $$\mathbf{u}^c = \frac{1}{\sqrt{(dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2}}\left[\begin{array}{c} 1 \\ dx/dt \\ dy/dt \\ dz/dt \end{array}\right].$$ When we take the slope, that square-root normalization factor cancels out, and we see that the spatial velocity is $$\frac{1}{1} \left[\begin{array}{c} dx/dt \\ dy/dt \\ dz/dt \end{array}\right],$$ just like we'd expect.

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