The magnitude of a four velocity

[Arman777]

I want to clarfiy some ideas, so let us assume an inital observer measures a 4 velocity of an object. $\vec{u} = (c\gamma_u, \gamma_uu^x, \gamma_uu^y, \gamma_uu^z)$. When we calculate the magnitude of the 4-velocity we get,

$$\vec{u} \cdot \vec{u} = -c^2\gamma_u^2 + \gamma_u^2U^2$$ where $U^2 = (u^x)^2 + (u^y)^2 + (u^z)^2$

and after some calculation, we see that the magnitude of the velocity vector is $c$. So in space-time, we all are moving with a velocity $c$. Is this a true statement? More interestingly the same calculation for photons gives 0. So the magnitude of the velocity of the photons in ST is 0. It seems interesting...

Also in SR the dot product of the two vectors are frame invariant. I think the main reason for that is the fact that $ds^2$is frame invariant such that
$$l = g_{\mu \nu}A^{\nu}B^{\mu} = g_{\mu \nu}A'^{\nu}B'^{\mu}$$.

Is this means that for a coordinate transformation such that, $A^{\nu}(y) = \frac{\partial y^{\nu}}{\partial y^{\alpha}}A'^{\alpha}(x)$ and $B^{\mu}(y) = \frac{\partial y^{\mu}}{\partial y^{\alpha}}B'^{\alpha}(x)$ , $\frac{\partial y^{\nu}}{\partial y^{\alpha}}\frac{\partial y^{\mu}}{\partial y^{\alpha}}=1$ . Here x and y represents two different coordinates. I also used the contravarient vector transformation. But I guess I should have just use the lorentz transformation (?)

How we can prove that the dot product is frame invariant by using the coordinate transformation? Can we say that scalers are frame invariant? So under any coordinate transformation if the metric does not change the dot product will always give the same result.

 

[Ibix]

So in space-time we all are moving with a velocity cc. Is this a true statement ?

Not really. Your maths is correct, but what you've done is show that the normalised tangent vector to your worldline is normalised - which is kind of trivial (edit: this has not stopped hundreds of pop-sci publications making hay with this). The point is that $\tau$ ticks off intervals of 1s (or whatever time unit you use) along your worldline. Thus the rate of change of your "location" in spacetime with respect to $\tau$ is a constant because you've defined it as such.

As you note, the "four velocity" of a null worldline is null. In fact, this is an invalid construction because attempting to define $\tau$ along such a worldline doesn't work - the interval along the worldline is zero. So the derivative with respect to $\tau$ is ill-defined. You need to define a different parameter that does vary along the worldline, or just use a different normalisation of the tangent vector, such as the four momentum.

How we can prove that the dot product is frame invarient by using the coordinate transformation ? an we say that scalers are frame invarient quantities right. So under any coordinate transformation if the metric does not change the dot product will always give the same result.

You forgot to transform $g$ (I think - your notation seems slightly confused).
$$g_{\mu\nu}A^\mu B^\nu=g'_{\alpha\beta}A'^\alpha B'^\beta$$
If you note that the partial derivatives in the transformation for upper indices are the other way up for those for lower indices, the result should drop out.

 

[PeterDonis]

in space-time, we all are moving with a velocity $c$. Is this a true statement? More interestingly the same calculation for photons gives 0. So the magnitude of the velocity of the photons in ST is 0. It seems interesting...

As @Ibix says, many pop science sources have tried to make something out of this, but there is nothing there: it's just a restatement of the trivial fact that you can always find a vector tangent to a timelike worldline that has unit norm, whereas the tangent vector to a null worldline must have zero norm.

Also in SR the dot product of the two vectors are frame invariant. I think the main reason for that is the fact that $ds^2$ is frame invariant

No, this is not correct. The line element $ds^2$ is not the same as the metric. The line element $ds^2$ is associated with one vector, not two. The dot product of two vectors involves the metric $g_{\mu \nu}$, but not $ds^2$. The fact that the dot product formula, when written out, happens to look a lot like the formula for $ds^2$ does not mean they're the same thing.

 

[PAllen]

The key thing about a 4 velocity as a 4 vector tangent is that it represents a spacetime direction that is timelike. As such, the only purpose of normalization is to remove its magnitude from being significant. There are authors who normalize it to 1 even when c is not set to 1. You could even normalize 4 velocity to 42 without any change to the physics (to please hitchhikers).

 

[Arman777]

No, this is not correct. The line element ds2ds2ds^2 is not the same as the metric. The line element ds2ds2ds^2 is associated with one vector, not two. The dot product of two vectors involves the metric gμνgμνg_{\mu \nu}, but not ds2ds2ds^2. The fact that the dot product formula, when written out, happens to look a lot like the formula for ds2ds2ds^2 does not mean they're the same thing.

Yes I should have written something like,

$$ds^2 = g_{\mu \nu}dx^{\nu}dx^{\mu}$$

So okay let me fix my notation. We have two vectors that are meaured in $x$ frame $A(x)$ and $B(x)$and then we have another frame $y$measures the same vectors $A(y)$, $B(y)$.

In this case we make a dot product in the frame of $x$, $l = g_{\mu \nu}(x)A(x)^{\nu}B(x)^{\mu}$

In y frame $l' = g_{\alpha \beta}(y)A(y)^{\alpha}B(y)^{\beta}$

Here we can make transformations such that,

$A^{\alpha}(y) = \frac{\partial y^{\alpha}}{\partial x^{\nu}}A^{\nu}(x)$ and $B^{\beta}(y) = \frac{\partial y^{\beta}}{\partial x^{\mu}}B^{\mu}(y)$ also

$g_{\alpha \beta}(y) = \frac{\partial x^{\nu}}{\partial y^{\alpha}}\frac{\partial x^{\mu}}{\partial y^{\beta}} g_{\mu \nu}(x)$

So from here we see that $l=l'$. So we proved that the dot product of the two vectors are frame invarient quantity.

As you note, the "four velocity" of a null worldline is null. In fact, this is an invalid construction because attempting to define ττ\tau along such a worldline doesn't work - the interval along the worldline is zero. So the derivative with respect to ττ\tau is ill-defined. You need to define a different parameter that does vary along the worldline, or just use a different normalisation of the tangent vector, such as the four momentum.

Do you mean that for light $d\tau^2 = 0$ so that $dx^{\alpha}/d\tau$ does not make sense ?

The key thing about a 4 velocity as a 4 vector tangent is that it represents a spacetime direction that is timelike. As such, the only purpose of normalization is to remove its magnitude from being significant.

What do you mean by using different normalisation ?

 

[PAllen]

@Arman777 , given vectors such that one is a scalar multiple of the other, their direction is the same. If only the direction is significant, the you can remove magnitude a spurious parameter by choosing one arbitrary magnitude for all direction vectors. This process of choosing a standard magnitude for direction vectors is called normalization. In the case of a tangent vector (which is what a 4 velocity is), there cannot be any significance to the magnitude. Thus, you are free to normalize to whatever you want, as a matter of convenience or even aesthetics.

 

[Ibix]

Do you mean that for light $d\tau^2=0$ so that $dx^{\alpha}/d\tau$ does not make sense ?

Yes.

 

[Arman777]

@Arman777 , given vectors such one is a scalar multiple of the other, their direction is the same. If only the direction is significant, the you can remove magnitude a spurious parameter by choosing one arbitrary magnitude for all direction vectors. This process of choosing a standard magnitude for direction vectors is called normalization. In the case of a tangent vector (which is what a 4 velocity is), there cannot be any significance to the magnitude. Thus, you are free to normalize to whatever you want, as a matter of convenience or even aesthetics.

Hmm I understand it. So it does not actually matter the magnitude of our 4 velocity ?

 

[Ibix]

Hmm I understand it. So it does not actually matter the magnitude of our 4 velocity ?

In a physical sense, no. But you need to be consistent with whatever source you are working with, or else you'll get erroneous factors of $c$ (or 42, as @PAllen suggests) popping up in your work.

 

[PAllen]

Just to add a point: ratio of magnitudes of 4 momenta are significant as the ratio of invariant masses. The value of each magnitude reflects only unit choice. Any 4 velocity normalization choice (even a silly value like 42) makes no change to ratios of 4 momenta magnitudes.

 

[vanhees71]

As @Ibix says, many pop science sources have tried to make something out of this, but there is nothing there: it's just a restatement of the trivial fact that you can always find a vector tangent to a timelike worldline that has unit norm, whereas the tangent vector to a null worldline must have zero norm.

It's also unfortunate to call it "norm". It is not a norm. The Minkowski product is not a scalar product, because it's not positive definite. It's a fundamental form of the affine pseudo-Euclidean space with signature (1,3) (or (3,1), depending on the convention used). A lot of confusion arises from this sloppy use of mathematical language!

 

[PeterDonis]

It's also unfortunate to call it "norm". It is not a norm.

Is there a better term?

 

[vanhees71]

It's the fundamental form of an affine space. You can also say it's a "pseudo scalar product". The same holds of course for GR, where you have a pseudo-Riemannian manifold with a "pseudo metric".