Why Is the Speed of Light Essential in 4-Velocity Invariance?

[omidj]

TL;DR

4-velocity vector dot itself will not be equal to C^2, if we don't multiply time by coefficient C (speed of light).

If we don't multiply time by the speed of light C, in Lorentz transformation equatins, the invariant tensor of U.U will not be equal to c^2. So it seems the coefficient of C is an obligation for parameter t.
I thought it's just for visual convenience in graphs, besides C is not dimensionless.

 

[Ibix]

$c$ is effectively a unit conversion factor in this context, and appears so we are using the same units in all four dimensions. It's analogous to distances over water measured in nautical miles and distance below the surface measured in fathoms. Attempting to write a three vector in those units won't give you an invariant object either, just a collection of three numbers. You would need to introduce a fathom-per-nautical mile conversion factor on the z dimension to adapt your naive choice of convenient units to the mathematical framework.

 

[omidj]

$c$ is effectively a unit conversion factor in this context, and appears so we are using the same units in all four dimensions. It's analogous to distances over water measured in nautical miles and distance below the surface measured in fathoms. Attempting to write a three vector in those units won't give you an invariant object either, just a collection of three numbers. You would need to introduce a fathom-per-nautical mile conversion factor on the z dimension to adapt your naive choice of convenient units to the mathematical framework.

So, if we don't make all four dimensions same, then it shouldn't have any effect on invariancy of products, but plz try it, calculate the inner product of 4-velocity to itself (concluding C^2 as the result is impossible).

 

[Ibix]

So, if we don't make all four dimensions same, then it shouldn't have any effect on invariancy of products,

Yes it will have an effect (unless you correct for it). A trivial 2d example would be choosing to measure distances in one dimension in meters and the other one in centimeters. Then consider the vector (1,0) and the vector (0,100). Naively taking the inner products with themselves for those two vectors gives you 1 and 10,000 respectively, but we know they are the same length (because 1m=100cm) so their inner products with themselves should be the same.

The problem is naively taking an inner product as the sum of the products of components. You can correct this either using a metric tensor of $\mathrm{diag}(100^2,1)$ or manually inserting that you need to multiply the first component by 100cm/m to get consistent results.

That's exactly the same as what $c$ is doing in the four-velocity. It corrects for the naive choice of different units in different directions.

calculate the inner product of 4-velocity to itself (concluding C^2 as the result is impossible).

If you do this naively as $\left(\dfrac{dt}{d\tau}\right)^2-\left(\dfrac{dx}{d\tau}\right)^2$ with $t$ in seconds and $x$ in meters you will find that the result is not $c^2$ in general, true. But that would be the same mistake as complaining that $1^2+0^2$ and $0^2+100^2$ are not equal in my example above. You aren't actually calculating a modulus.

 

[div_grad]

So, if we don't make all four dimensions same, then it shouldn't have any effect on invariancy of products, but plz try it, calculate the inner product of 4-velocity to itself (concluding C^2 as the result is impossible).

The vacuum-speed-of-light $c$ is a scalar invariant under Lorentz transformations, while the 4-velocity $v^\mu$ is a 4-vector and is therefore not invariant (although the scalar product $v_\mu v^\mu$ of two 4-vectors is invariant).

If you define 4-velocity as a derivative $v^\mu = \frac{\mathrm{d} x^\mu}{\mathrm{d} s}$ of the position vector $x^\mu = (x^0, \mathbf{x})$ with respect to the interval $s$ (where $\mathrm{d}s = c\sqrt{1-\frac{v^2}{c^2}}$), then you obtain the 4-velocity of the form
$$
v^\mu = \left(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}, \frac{\frac{v}{c}}{\sqrt{1-\frac{v^2}{c^2}}}\right) \rm{,}
$$
which gives you the product $v_\mu v^\mu = 1$ (which is constant, but not equal $c^2$).

However, if you decided to define the 4-velocity as a derivative $u^\mu = \frac{\mathrm{d} x^\mu}{\mathrm{d} \tau}$ of $x^\mu$ with respect to the proper time $\tau$ (in order to make the expression more similar to the 3-dimensional case $\mathbf{v} = \frac{\mathrm{d} \mathbf{x}}{\mathrm{d} t}$) and use the fact that $\mathrm{d} \tau = \mathrm{d} s / c$, then you will obtain that $u^\mu = c v^\mu$ where $v^\mu$ was defined above. Then since $v_\mu v^\mu = 1$ now you get that $u_\mu u^\mu = c^2$ like you wanted.

The difference is that you define the 4-velocity either by taking the derivative with respect to the space-time interval or with respect to the proper time.

 

[Dale]

If we don't multiply time by the speed of light C, in Lorentz transformation equatins

If you don’t multiply time by $c$ in the Lorentz transforms then the resulting transformation no longer accurately predicts the results of experiments.

So it seems the coefficient of C is an obligation for parameter t

There are options. You can multiply the time coordinate by $c$ and use the metric $$\left( \begin{array}{cccc} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right)$$ or you can not multiply the coordinate by $c$ and use the metric $$\left( \begin{array}{cccc} -c^2 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right)$$ I usually take the second approach.

 

[Ibix]

while the 4-velocity is a 4-vector and is therefore not invariant

Language differs on this. Vectors themselves can be discussed entirely without reference to any coordinate system, so IMO it is reasonable to call them invariant. However their component representations are, naturally, coordinate dependent with components that vary in contravariant way. So some people will call the vector invariant and its components contravariant, and some people just call the whole shebang contravariant.

The difference is that you define the 4-velocity either by taking the derivative with respect to the space-time interval or with respect to the proper time.

I think the issue is that the OP is not doing either of the things you propose (which I agree are valid), but rather combining your $v^0$ with your $u^i$ to get in a mess.

 

[Dale]

So some people will call the vector invariant and its components contravariant, and some people just call the whole shebang contravariant.

I still always struggle with this.

 

[div_grad]

Vectors themselves can be discussed entirely without reference to any coordinate system, so IMO it is reasonable to call them invariant.

In that case, what would be the difference between a scalar, a vector and a symmetric traceless tensor? Why would there be a need for distinguishing these objects from one another? Even without referring to particular coordinates, a transformation such as rotation changes a (3D) vector because it makes it point in a different direction from the original "free" vector; what is invariant under this rotation is the length of a vector but its direction changes.

 

[Ibix]

In that case, what would be the difference between a scalar, a vector and a symmetric traceless tensor?

I don't think saying they are all invariant objects removes the distinction between them. A vector has a magnitude and direction whether you describe it as invariant or not.

Even without referring to particular coordinates, a transformation such as rotation changes a (3D) vector because it makes it point in a different direction from the original "free" vector;

What do you mean by a rotation here? If you mean a passive transform then the vector is the same in both cases but you must have chosen a coordinate system. If you mean an active transform then you are talking about two different vectors and invariance wouldn't seem to me to be a relevant concept.

 

[Orodruin]

In that case, what would be the difference between a scalar, a vector and a symmetric traceless tensor? Why would there be a need for distinguishing these objects from one another? Even without referring to particular coordinates, a transformation such as rotation changes a (3D) vector because it makes it point in a different direction from the original "free" vector; what is invariant under this rotation is the length of a vector but its direction changes.

A scalar is a function on the base manifold. A vector is a member of the tangent space of spacetime. A traceless symmetric tensor is a linear map $T$ from the tangent space to itself such that, for any orthonormal basis $\{\vec E_\mu\}$, it holds that
$$ \sum_\mu g(\vec E_\mu, T(\vec E_\mu)) = 0 $$

 

[cianfa72]

I still always struggle with this.

Yes, I have no idea why so many confusing definitions are used