Relativistic Velocity Addition: Is c Ever Exceeded?

[mahoneywingo]

I was watching this video by Brian Green about relativistic velocity addition when he said something at the end that I wondered about.

https://www.britannica.com/video/222286/Your-Daily-Equation-07-Relativistic-Velocity-Combination

It was along the lines of the formula will always result in the relative velocities of two frames of reference being less than c "as long as the velocities of each frame is less than c".

Is there any circumstance where one (or even both) would more than c? I was always under the impression that wasn't possible, but why would he point out a possible exception, if that was the case?

 

[Ibix]

Without watching the video I'm not sure of the context, but the relative velocity addition formula takes any two velocities less than $c$ and "adds" them together giving you another velocity less than $c$. On the other hand, if one of the velocities is $c$ then the result is always $c$, whatever the other velocity is. You are correct that a frame cannot have a velocity of $c$ (or greater) relative to anything, but you might be interested in transforming the velocity of a light pulse, which is $c$ and transforms to $c$.

So the exception he's mentioning is probably just that if one of the velocities is $c$ then the transformed velocity is also $c$. Or he may not be speaking entirely accurately. Physicists often say things that are slightly misleading when trying to say things to the public, because it's hard to be precise and write something people will read. Greene has a bit of a reputation for that hereabouts, I'm afraid.

I'm not watching a 13 minute video to try to catch one sentence he said, but if you can give a time stamp I can take a look.

 

[mahoneywingo]

It starts at 10:17.

 

[Ibix]

Yeah, he does say "as long as you put in speeds at or below the speed of light".

You can get speeds above the speed of light out of the addition formula if you put speeds above the speed of light in - it makes no physical sense to put such speeds in because nothing can have those speeds, but the maths won't object if you try. So I think he's just qualifying his comment that the formula will always give you a speed below $c$.

It might have been better to make an aside there, something like "it'll always give you a speed at or below $c$. Of course, you can plug in numbers above $c$ if you like and that will sometimes give you a result above $c$, but that's physically meaningless because you can't get speeds above $c$."

 

[PeroK]

It was along the lines of the formula will always result in the relative velocities of two frames of reference being less than c "as long as the velocities of each frame is less than c".

The formula is a mathematical equation. The hypothesis is that both velocities are less than $c$ and the conclusion is that the resulting velocity is less than $c$.

Is there any circumstance where one (or even both) would more than c? I was always under the impression that wasn't possible, but why would he point out a possible exception, if that was the case?

It's less an exception than a (mathematical) hypothesis, which itself is justified by the physical theory.

Note that, to be precise, if $u$ is the velocity of an object in one reference frame (unprimed, say). And that reference frame has a velocity of $v$ relative to a second (primed) reference frame, then we have the velocity of the object in the primed frame:
$$u < c, \ v < c \ \Rightarrow \ u' = \frac{u + v}{1 + uv/c^2} < c$$$$u = c, v < c \ \Rightarrow \ u' = \frac{u + v}{1 + uv/c^2} = c$$The second case represents the invariance of the speed of light, $c$, across all inertial reference frames. Where the "object" is a light wave.

 

[PeterDonis]

You can plug in numbers above $c$ if you like and that will sometimes give you a result above $c$,

More specifically, if neither number is $c$ itself, then if exactly one of the numbers is greater than $c$, the result will be as well. But if both are greater than $c$, the result will be less than $c$.

The proof is straightforward if you use units in which $c = 1$. There are two cases:

(1) Exactly one number greater than $1$. Then we write $u = 1 - e$ and $v = 1 + f$, with $e > 0$ and $f > 0$, and we have

$$
u' = \frac{2 - e + f}{2 - e + f - ef} > 1
$$

(2) Both numbers greater than $1$. Then we write $u = 1 + e$ and $v = 1 + f$, with $e > 0$ and $f > 0$, and we have

$$
u' = \frac{2 + e + f}{2 + e + f + ef} < 1
$$

 

[FactChecker]

A "frame" does not need to have any real object stationary in it, so it is free to move theoretically at any speed. No object can travel faster than c, but a frame with no real objects stationary in it can be moving theoretically faster than c. That being said, I don't know if there would be any use for it.
CORRECTION: This is true for a "coordinate chart" but not for a "frame", which has implications about relativistic time and distance.

 

[PeterDonis]

a frame with no real objects stationary in it can be moving theoretically faster than c.

No, it can't, at least not with the standard meaning of "frame". You can have a coordinate chart in which the $x^0$ coordinate is spacelike instead of timelike, but you cannot have a frame "moving faster than c", because the definition of a frame includes measuring rods and clocks (or, in more technical language, orthonormal tetrads), and a clock can't move on a spacelike worldline.

 

[FactChecker]

No, it can't, at least not with the standard meaning of "frame". You can have a coordinate chart in which the $x^0$ coordinate is spacelike instead of timelike, but you cannot have a frame "moving faster than c", because the definition of a frame includes measuring rods and clocks (or, in more technical language, orthonormal tetrads), and a clock can't move on a spacelike worldline.

I'll buy that. I stand corrected. Thanks.

 

[mahoneywingo]

Thanks, everyone for taking the time to respond. The comment was just something that made me go ....hmmm.

I've read things that have said ,theoretically, if you had an energy source that could supply infinite energy for an infinite amount of time, c could be exceeded. I don't know if that is true or not, but I thought maybe that was what he was referring to.

 

[Ibix]

I've read things that have said ,theoretically, if you had an energy source that could supply infinite energy for an infinite amount of time, c could be exceeded. I don't know if that is true or not, but I thought maybe that was what he was referring to.

I'm afraid that's the kind of not-quite-true thing that gets repeated a lot. Fundamentally, an object with mass travelling at the speed of light turns out to be a self-contradiction in relativity. It cannot happen. The various infinities that come up when you try to talk about it, therefore, don't mean "if you had infinite energy/time/whatever you could reach the speed of light". Rather, they mean "however much energy/time/whatever you have, you can't get to the speed of light".

 

[PeterDonis]

I've read things

This is not a valid basis for PF discussion. You should give a specific reference, and it should be something like a textbook or a paper on the topic.

 

[PeroK]

I've read things that have said ,theoretically, if you had an energy source that could supply infinite energy for an infinite amount of time, c could be exceeded. I don't know if that is true or not, but I thought maybe that was what he was referring to.

That statement is what I would describe as vacuously false! It's both meaningless and wrong at the same time.

 

[vanhees71]

This cannot be true by construction of relativity. It's most easily seen on an example. Take a charged particle in a homogeneous electric field. The relativistic equation of motion for the motion in direction of the field (the $1$-direction) reads (with $c=1$)
$$m \mathrm{d}_{\tau}^2 \begin{pmatrix}t \\ x \end{pmatrix} = q E \mathrm{d}_{\tau} \begin{pmatrix} x \\ t \end{pmatrix}.$$
Using $u=\mathrm{d}_{\tau} (t,x)$ we can write this as a first-order equation
$$\mathrm{d}_{\tau} \begin{pmatrix}u^0\\u^1 \end{pmatrix} = \frac{q E}{m} \begin{pmatrix}u^1 \\ u^0 \end{pmatrix}.$$
With the initial condition that the particle starts from rest and the constraint $\eta_{\mu \nu} u^{\mu} u^{\nu} = (u^0)^2-(u^1)^2=1$ you find the solution
$$\begin{pmatrix}u^0 \\ u^1 \end{pmatrix} = \begin{pmatrix} \cosh(q E \tau/m) \\ \sinh(q E \tau/m) \end{pmatrix}.$$
This shows that the "three-velocity"
$$v=\frac{u^1}{u^0}=\tanh \left (\frac{q E}{m} t \right)$$
goes to 1 for $\tau \rightarrow \infty$, i.e., even in a homogeneous electric field the speed of the particle never gets greater than $c$. The kinetic energy the particle has is
$$E_{\text{kin}}=m (u^0-1)=m \left [\cosh \left(\frac{q E}{m} \tau \right) -1\right]$$
which goes to $\infty$ for $\tau \rightarrow \infty$, i.e., you already need an infinite amount of energy to accelerate the particle to the speed of light. Since you always have only a finite amount of energy (i.e., this example is physically oversimplified, because you cannot create a homogeneous electric field all over space), you can never reach the speed of light, let alone exceed it.

 

[jbriggs444]

it makes no physical sense to put such speeds in because nothing can have those speeds

It can make physical sense. Suppose that one has a super-luminal trajectory such as a laser dot being swept across the face of the moon. Or the intersection point of a pair of very long scissors. Of course, no physical object is following the trajectory. It is a space-like path after all. Nonetheless, having chosen an inertial frame, one can [edit: usually^(*)] calculate the ordinary 3-velocity for the trajectory at various events along the path.

Now suppose that one wishes to predict what velocity would be calculated for the 3-velocity using some other frame of reference. Easy. Take the original super-luminal velocity and the relative velocity between the two frames. Put them into the velocity addition formula and out pops the super-luminal velocity relative to the new frame.

Note that if you choose the relative velocity between the frames right, what is a super-luminal velocity in one frame may become infinite in the other frame. Or the traversal direction for the trajectory in terms of the chosen coordinate time may flip.

Edit: (*) If the trajectory (or its tangent) is purely spatial in a particular frame, the calculation for 3-velocity will involve division by zero elapsed coordinate time, leaving the velocity technically undefined in that frame. Though one might choose to call it "infinite" instead.

 

[Mister T]

It was along the lines of the formula will always result in the relative velocities of two frames of reference being less than c "as long as the velocities of each frame is less than c".

I've read things that have said ,theoretically, if you had an energy source that could supply infinite energy for an infinite amount of time, c could be exceeded. I don't know if that is true or not, but I thought maybe that was what he was referring to.

These are statements made about the mathematical formulas. The authors are just blindly looking at the formula and talking about how one variable would change if another variable were changed.

They imply nothing about faster than light speeds because the formulas are derived from the principle that the speed of light is invariant, which implies that the speed of light is the maximum speed for a frame of reference or a particle.

 

[Sagittarius A-Star]

Suppose that one has a super-luminal trajectory such as a laser dot being swept across the face of the moon.

Another example for transforming a super-luminal velocity is regarding the phase velocity of de Broglie matter waves.