Understanding the Relative Velocity Paradox

[x0tek]

Hi! I'm an utter novice at physics, so please bear with me if I ask a foolish question.

I just learned that velocities don't purely add: 2m/s + 2m/s does not equal 4m/s, but something like 3.9999999...9 m/s, and so on. Something to do with the upper limit of c.

Logic makes me assume that this is true in reverse - slowing down from 4m/s by 2m/s would result in not 2m/s, but 2.000...1 m/s.

Which means that 1m/s in fact doesn't equal 1 m/s, but something concretely smaller.

So my questions are:
1. Does this mean that the speed of light acts like a universal "infinity" symbol, acting as it does in basic calculus? This feels intuitively right, but seems to have major flaws...

2. Is this the logic behind the proof that "absolute zero" doesn't exist, since we would never be able to fully reach "negative infinity"?

I'm exhausted and going to sleep now, but it was such an intriguing question I had to ask someone haha. Looking forward to learning!

 

[PeterDonis]

Logic makes me assume that this is true in reverse

"Reverse" just means one of the velocities you add is negative. You are correct that the result will be a little larger than simple subtraction (whereas the result of adding two positive velocities results in a velocity a little smaller than simple addition).

Which means that 1m/s in fact doesn't equal 1 m/s, but something concretely smaller.

No, it doesn't. I don't know why you would think that. The relativistic velocity addition formula doesn't change any units or any meanings of numbers.

Does this mean that the speed of light acts like a universal "infinity" symbol, acting as it does in basic calculus?

No.

Is this the logic behind the proof that "absolute zero" doesn't exist

What proof are you talking about? Please give a reference. And what does absolute zero have to do with relativity?

 

[x0tek]

Thanks for the fast response! I'll try to answer you in order, hopefully you can clarify :)

1. Okay, that makes sense. Glad I'm on the right track!

2. 1m/s + 1m/s = 1.999...8m/s. Maybe I'm misunderstanding something fundamental about the case at hand, but I think the arithmeric explains my statement...could you clarify?

3. Okay, thanks!

4. Absolute zero seems the logical antithesis of light speed, hence my positive and negative infinity analogy :) thus, any proof for impossibility of light speed attainment seems proof of impossibility of attaining absolute zero. Or am I missing something here too?
Edit: to answer your question directly, I logically extended my reasoning to absolute zero, since any decrease in velocity would always leave a remainder - making absolute zero impossible

Thank you again so much!

 

[PeterDonis]

1m/s + 1m/s = 1.999...8m/s.

No, it doesn't. You should not use the standard + sign for relativistic velocity "addition" because it is not the same as the addition you learned in elementary school, which is the kind of addition the + sign refers to. 1 m/s + 1 m/s = 2 m/s. But 1 m/s "added" to 1 m/s using the relativistic formula gives 1.999...8 m/s.

I think the arithmeric explains my statement

No, it doesn't. See above.

Absolute zero seems the logical antithesis of light speed

No, it isn't. I strongly advise you to abandon this line of thought. It will only confuse you.

 

[PeterDonis]

any decrease in velocity would always leave a remainder

No, it wouldn't. Try "adding" + 1 m/s and - 2 m/s using the relativistic formula. What do you get?

 

[x0tek]

Thanks for the responses again!

I see, so it's only adding in a loose sense - it's really processing it through a unique formula, with "addition" being the intuitive interpretation rather than the mathematic.

It seems like I'm misunderstanding something about absolute zero too - could you clarify why that -wouldn't- be the antithesis of light speed? If lightspeed is the upper bound of how quickly a given particle can go, then...? Or am I misunderstanding what temperature represents (in my mind, energy through motion of particles)?

I know little about any of this, so I appreciate you helping me see a little more clearly

 

[PeterDonis]

so it's only adding in a loose sense - it's really processing it through a unique formula, with "addition" being the intuitive interpretation rather than the mathematic.

Yes.

could you clarify why that -wouldn't- be the antithesis of light speed?

First, light can move at the speed of light, so it's not correct to say that nothing can move at the speed of light.

Second, while the speed of an object that is not massless, like light, in relativity can't be increased indefinitely--it can only approach the speed of light, never reach it--its energy can. Speeds closer and closer to the speed of light correspond to energies that increase without bound. But absolute zero represents a finite minimum of energy.

 

[Filip Larsen]

To me it sounds like some of the confusion x0tek express is due to not being aware how relativistic velocities add. Exploring the mathematical consequences of an equation for a physical system is qualitatively different than understanding why and to what extend the equation is a valid physical model for that system. Or to put it differently, trying to understand how an equation works without trying to understand the math directly (e.g. by trying to establish a non-mathematical analogy in hope that will make better sense than the math) is almost always going to be misleading, much more complicated, and in the end leave you unable to calculate anything new yourself. Understanding physics is in large parts to understand the math used to model it.

 

[PeroK]

Thanks for the responses again!

I see, so it's only adding in a loose sense - it's really processing it through a unique formula, with "addition" being the intuitive interpretation rather than the mathematic.

If you are measuring velocities in a given frame of reference, then the normal rules of addition apply. For example, if object A is moving at $2m/s$ to the right and object B is moving at $2m/s$ to the left, then the two are separating at $4m/s$. This assumes that all velocities are measured in the same frame of reference.

Where the relativistic velocity addition formula comes into play is where you want to add velocities that are measured in different frames of reference. For example:

Suppose object A is moving to the right at $2m/s$ in your frame of reference (i.e. as measured by you) and object B is moving to the right at $2m/s$ in A's frame of reference (i.e. as measured by A), then the velocity of object B in your frame of reference is not $4 m/s$. It is in fact $\frac{4}{1 + 4/c^2}m/s$.

To see why this is not directly any sort of contradiction to the laws of arithmetic, you could use the follow notation:

$v_A$ is the velocity of object A in your frame of reference and $v_B$ is the velocity of object B in your frame of reference. And, $v_{BA}$ is the velocity of object B in A's frame of reference, then we have:
$$v_B = \frac{v_A + v_{BA}}{1 + v_A v_{BA}/c^2}$$

This is a relationship between measured physical quantities. In principle that relationship could have any mathematical formula. It doesn't have to be simply of the form $A + B$. Note that you still use the normal rules of algebra and arithemtic to evaluate these expressions. That formula in fact relies on the normal rules of algebra and arithmetic in order for you to be able to calculate those quantities properly.

 

[Dale]

2. 1m/s + 1m/s = 1.999...8m/s. Maybe I'm misunderstanding something fundamental about the case at hand, but I think the arithmeric explains my statement...could you clarify?

The mathematical operation of addition is unchanged. So 1 m/s + 1 m/s = 2 m/s.

What is different is that the formula for velocity composition is no longer addition. If an object in A’s reference frame is moving at velocity v, and A itself is moving at u in B’s reference frame then the velocity of the object in B’s reference frame is given by an operation called composition. Usually velocity composition is confusingly called velocity addition, but you should understand that the operation is technically a composition.

The formula for relativistic velocity composition is $$\frac{u+v}{1+\frac{uv}{c^2}}$$

 

[x0tek]

Thanks everyone ^^ I'm grateful for the depth and thoroughness of your explanations. You all are awesome

 

[robphy]

There is something that is still additive... the rapidity (the minkowski-angle) between the two timelike-vectors:
(the spacelike Minkowski arc-length along the unit-Minkowski circle [the hyperbola])
$V_{AC}=\tanh(\theta_{AC})= \tanh(\theta_{AB}+\theta_{BC}) =\displaystyle \frac{\tanh(\theta_{AB})+\tanh(\theta_{BC})}{1+\tanh(\theta_{AB})\tanh(\theta_{BC})} =\frac{V_{AB}+V_{BC}}{1+V_{AB}V_{BC}}$
then velocity is the hyperbolic-tangent of the rapidity.

The Euclidean analogue is the "composition-of-slopes".
To find it, we add angles.
$m_{AC}=\tan(\phi_{AC})= \tanh(\phi_{AB}+\phi_{BC}) =\displaystyle \frac{\tan(\phi_{AB})+\tan(\phi_{BC})}{1-\tan(\phi_{AB})\tan(\phi_{BC})} =\frac{m_{AB}+m_{BC}}{1-m_{AB}m_{BC}}$
then slope is the tangent of the angle.

There is a Galilean analogue that uses a funny but consistent function (named by the mathematician IM Yaglom)
the Galilean tangent $\mbox{tang}(\beta)=\beta$, where the argument of $\mbox{tang}$, $\beta$, is the Galilean rapidity (spacelike Galilean arc-length along the unit-Galilean circle).
Then to compose Galilean velocities, we first add Galilean rapidities.
$v_{AC}=\mbox{tang}(\beta_{AC})= \mbox{tang}(\beta_{AB}+\beta_{BC}) =\displaystyle \frac{\mbox{tang}(\beta_{AB})+\mbox{tang}(\beta_{BC})}{1+0\mbox{tang}(\beta_{AB})\mbox{tang}(\beta_{BC})} =\frac{v_{AB}+v_{BC}}{1+0v_{AB}v_{BC}}=v_{AB}+v_{BC}$
then velocity is the Galilean-tangent of the Galilean-rapidity.
Our "common sense" got so used to "adding velocities" (rather than "adding angles") that we expect anything else to be weird...
but, among these three cases, the Galilean approach is the weird one.

(And we don't seem to have any paradoxical issues with the Euclidean analogue.
Maybe we need to make one up!)

 

[Grinkle]

It seems like I'm misunderstanding something about absolute zero too - could you clarify why that -wouldn't- be the antithesis of light speed? If lightspeed is the upper bound of how quickly a given particle can go, then...?

The speed of light is the same in for all reference frames. This is the only speed that is the same for all reference frames. There is no other speed that is same for all reference frames, including a zero speed. Something at rest in one frame will not be at rest in other frames. Something (light) moving at c in one frame will be moving at c in all other frames as well, c is unique in this respect.

Abandon the reasoning that leads you to think there must or should be some symmetric-to-maximum-speed minimum speed that is frame independent; there is not.

 

[Mister T]

I see, so it's only adding in a loose sense - it's really processing it through a unique formula, with "addition" being the intuitive interpretation rather than the mathematic.

You are combining speeds. One way to combine them is to add them, but as it turns out, that only works as a low-speed approximation.

It seems like I'm misunderstanding something about absolute zero too - could you clarify why that -wouldn't- be the antithesis of light speed? If lightspeed is the upper bound of how quickly a given particle can go, then...? Or am I misunderstanding what temperature represents (in my mind, energy through motion of particles)?

Speed is a property of a singe particle. Temperature is a property of a collection of particles. Note that you could, for example, observe a speeding bullet to be at rest, but that does not mean it has a temperature of absolute zero.