Is rapidity a measure of acceleration?

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9 hours ago, Mordred said:

Acceleration is easily handled in both SR and GR. It's simply a type of boost called rapidity.

I've seen this posted on this site many times over the years and I think it's wrong but never saw a correction or explanation. It's repeated often in posts labelled "expert" but I don't understand what it means. As a Lorentz transformation doesn't a boost imply constant velocity? How can a measure of velocity be called an acceleration? How is a measure of velocity a type of Lorentz transformation?

Is there some sensible meaning to what I quoted that I'm just not comprehending?

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Posted (edited)
2 hours ago, md65536 said:

I've seen this posted on this site many times over the years and I think it's wrong but never saw a correction or explanation. It's repeated often in posts labelled "expert" but I don't understand what it means. As a Lorentz transformation doesn't a boost imply constant velocity? How can a measure of velocity be called an acceleration? How is a measure of velocity a type of Lorentz transformation?

Is there some sensible meaning to what I quoted that I'm just not comprehending?

Well, I had no idea rapidity had a technical meaning in the context of relativity, so I've learned something already! But having looked it up here: https://en.wikipedia.org/wiki/Rapidity,  I'm definitely going to leave it to you chaps in long trousers.......

Edited by exchemist
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Posted (edited)
18 hours ago, md65536 said:

I've seen this posted on this site many times over the years and I think it's wrong but never saw a correction or explanation. It's repeated often in posts labelled "expert" but I don't understand what it means. As a Lorentz transformation doesn't a boost imply constant velocity? How can a measure of velocity be called an acceleration? How is a measure of velocity a type of Lorentz transformation?

Is there some sensible meaning to what I quoted that I'm just not comprehending?

Both velocity and accelerations are boosts in the the Lorentz transforms. I know you and I had tried discussing this in the past. Later on when I'm not at work I will try to get you far better detail on the difference of a boost due to velocity as opposed to acceleration. Part of the confusion  is that both velocity and accelerations are also described by rapidity. However the transforms for each slightly  differ .

Edited by Mordred
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13 hours ago, Mordred said:

Both velocity and accelerations are boosts in the the Lorentz transforms. Rapidity is just a particular type of boost. I know you and I had tried discussing this in the past. Later on when I'm not at work I will try to get you far better detail on the difference of a boost due to velocity as opposed to acceleration. Part of the confusion  is that both velocity and accelerations are also described by rapidity. However the transforms for each slightly  different .

Well I look forward to the details because nothing you wrote here makes sense to me. You are an expert on this? I was hoping someone else would chime in because the only confusing parts of it are things you wrote.

A boost is a Lorentz transformation, do you agree? Rapidity is a measure for relativistic velocity, do you agree?

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Posted (edited)
19 hours ago, md65536 said:

I've seen this posted on this site many times over the years and I think it's wrong but never saw a correction or explanation. It's repeated often in posts labelled "expert" but I don't understand what it means. As a Lorentz transformation doesn't a boost imply constant velocity? How can a measure of velocity be called an acceleration? How is a measure of velocity a type of Lorentz transformation?

Is there some sensible meaning to what I quoted that I'm just not comprehending?

I'm actually going to state that I agree the manner its often presented requires a more detailed presentation. I know I'm one of those  parties as velocity can also be treated via rapidity its actually more accurate to state acceleration is boosting or rotating the rapidity. As we have two forms of acceleration change in direction and change in velocity.

lets take a problem set. Lets have a constant acceleration for however many years.  A couple of key notes there is more than one type of boost. the boost for velocity depends on $$\beta$$ the velocity parameter. Now I'm sure you agree using velocity addition for a constant accelerating observer can get clunky.  So this is where rapidity steps in.

this describes the Lorentz boost equations ( for other readers I know you know these details)

$\acute{x}=(x-vt)$

$\acute{t}=\gamma(t-\frac{vx}{c^2})$

$\gamma=\frac{1}{\sqrt{1-b^2}}$

$\gamma=v/c$

so the constant velocity observer will have the above Lorentz boost.. however constant acceleration its more useful to use  the boost parameter not the speed parameter. This is is the rapidity given by the tanh function.

$\varsigma=\tanh{-1}\beta$

from which $$cosh\varsigma=\gamma$$ and $$\sinh\varsigma=\beta\gamma$$. Now that's the Lorentz boosts in terms of rapidity. So rapidity can be used instead of velocity.

however we need acceleration so lets  have the x,y,t planes. for simplicity. now a rotation in the x,y plane describes the change in angle

$\begin{pmatrix}x_a\\y_a\end{pmatrix}=\begin{pmatrix}\cos\varphi&-sin\varphi\\sin\varphi&cos\varphi\end{pmatrix}\begin{pmatrix}x_b\\v_b\end{pmatrix}$

$\frac{dy}{dx}=tan(\theta)$

for changes in {x,t} we are boosting the velocity or alternately the rapidity . As rapidity can also describe velocity.

$\begin{pmatrix}t_a\\y_a\end{pmatrix}=\begin{pmatrix}\cosh\varsigma&-sinh\varsigma\\sinh\varsigma&cosh\varsigma\end{pmatrix}\begin{pmatrix}x_b\\v_b\end{pmatrix}$

$\frac{dx}{cdt}=tanh(\varsigma)$

so yes we need to be more clear I agree or rather I need to be more clear. you certainly use rapidity for both types types of acceleration but you can also use rapidity for velocity. As mentioned I should state acceleration via change in velocity is a boost in the rapidity, while a change in direction is a rotation of rapidity. Does acceleration require rapidity no and you and I have agreed on this in the past if you recall. However as shown it certainly does apply to acceleration.

2 hours ago, md65536 said:

Well I look forward to the details because nothing you wrote here makes sense to me. You are an expert on this? I was hoping someone else would chime in because the only confusing parts of it are things you wrote.

A boost is a Lorentz transformation, do you agree? Rapidity is a measure for relativistic velocity, do you agree?

yes of course I agree on that but it also applies to acceleration as shown you can also boost the rapidity or rotate it. If you understood the relation  of rapidity to relativistic velocity I really have a hard time understanding why you would think it wouldn't apply to change in velocity at relativistic speeds or changes in direction at relativistic velocity ?

Edited by Mordred
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1 hour ago, Mordred said:

Now that's the Lorentz boosts in terms of rapidity.

So that's the same Lorentz boost, just represented using a different expression for the constant velocity.

1 hour ago, Mordred said:

for changes in {x,t} we are boosting the velocity or alternately the rapidity .

Where's the acceleration? Everything you wrote seems to be for a constant velocity ie. a constant rapidity. If one changes, the other changes. I don't see that expressed anywhere.

1 hour ago, Mordred said:

you can also boost the rapidity

Are you still using "boost" to refer to a Lorentz transformation or are switching between meanings of the word "boost" here? If it's "a type of boost called rapidity" you're saying you can boost the boost? What does that mean?

It looks to me like you're just showing the rapidity form of the (constant velocity) Lorentz boost, which still describes constant rapidity. However since it's easier to use changes in rapidity to describe relativistic acceleration because rapidity is additive, you're assuming that rapidity describes the change in velocity itself (ie. change in rapidity) rather than just representing the velocity alone.

Your problem example was, "Lets have a constant acceleration for however many years." What's the result?

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Posted (edited)
3 hours ago, md65536 said:

So that's the same Lorentz boost, just represented using a different expression for the constant velocity.

Where's the acceleration? Everything you wrote seems to be for a constant velocity ie. a constant rapidity. If one changes, the other changes. I don't see that expressed anywhere.

sigh you do realize that rapidity is a hyperbolic function. it isn't the velocity itself . I gave you those functions above. However here it is again.

$\varsigma=\tanh{-1}\beta$

from which $$cosh\varsigma=\gamma$$ and $$\sinh\varsigma=\beta\gamma$$

if your not familiar with hyperbolic functions.

the hyperbolic function describes the region a/2 in that link.

the hyperbolic tangent is on that link tanH

now look at the image in this link.

the area of the hyperbolic sector is the yellow in the second graph.

I also provided the hyperbolic rotations for observer a to observer b in the above the underscript A is Alice.  under b for Bobs whose x,y inertial frame changes according to the rotation group above.

as per the second link for proper acceleration.

"Proper acceleration (the acceleration 'felt' by the object being accelerated) is the rate of change of rapidity with respect to proper time (time as measured by the object undergoing acceleration itself). Therefore, the rapidity of an object in a given frame can be viewed simply as the velocity of that object as would be calculated non-relativistically by an inertial guidance system on board the object itself if it accelerated from rest in that frame to its given speed.

The product of β and γ appears frequently, and is from the above arguments"

quoting the second link...note rate of change of rapidity I provided you the transformation matrix involved in both case of acceleration.

$\begin{pmatrix}x_a\\y_a\end{pmatrix}=\begin{pmatrix}\cos\varphi&-sin\varphi\\sin\varphi&cos\varphi\end{pmatrix}\begin{pmatrix}x_b\\v_b\end{pmatrix}$

$\frac{dy}{dx}=tan(\theta)$

this expression describes that change in direction How is that not showing acceleration? the first matrix on the RHS is the rotation matrix.... The second matrix on the right hand side is Bob's 2d reference plane

$\begin{pmatrix}t_a\\y_a\end{pmatrix}=\begin{pmatrix}\cosh\varsigma&-sinh\varsigma\\sinh\varsigma&cosh\varsigma\end{pmatrix}\begin{pmatrix}x_b\\v_b\end{pmatrix}$

$\frac{dx}{cdt}=tanh(\varsigma)$

this expression is your transformations due to change in velocity via a boost of the rapidity which is a also a change in velocity

Do you understand this much ? Its obvious proper acceleration uses rapidity in the quoted section of link 2. it clearly shows you in that link.

these expressions are the hyperbolic tangents

The acceleration is there for both cases. Can you agree on that  with the links provided  ?

I also gave the curves in each case by the way.

Edited by Mordred
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On 4/19/2024 at 5:15 AM, md65536 said:

How can a measure of velocity be called an acceleration? How is a measure of velocity a type of Lorentz transformation?

Is there some sensible meaning to what I quoted that I'm just not comprehending?

The way I've heard 'rapidity' used is as a proper velocity.  It is proper acceleration integrated over proper time, and thus it adds the normal way.

Two obvious examples:

I accelerate my ship away at 1g in the same direction for 3 years ship time. That gets me up to a rapidity of a bit over 3c, meaning I travel 3 light years (Earth inertial frame) for every year I age. At that rapidity, the ship has a velocity relative to Earth's inertial frame of about 0.995c

Second example is recession rates of galaxies.  Gnz-11 (no longer the most distant object seen) is increasing its distance from us at a rate of about 2.3c.  That's a rapidity.   Rapidities add the normal way. If galaxy X recedes from us at 1.2c and galaxy Y (further away from us on the same line) recedes from X at 1.3c, then Y recedes from us at 2.5c.  If an inertial frame was valid over such distances (such as it would be in say a zero energy Milne metric), then that rapidity of GNz-11 would correspond to a velocity of 0.98c.  This underscores the fact that recession rates are not expressed relative to any inertial frame.

Anyway, since the usage comes from Mordred, he may be using the term in a different way than I tend to see it.

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Posted (edited)

I believe part of the confusion may have to do with the fact that there is numerous ways to express proper velocity.  Further adding to the mix different hyperbolic functions are used different between coordinate speed and proper speed

This link provides several different methods. Rapidity being one of them.

The graph in the link shows the distinctions between each. It also describes rapidity in more detail

"Proper speed divided by lightspeed c is the hyperbolic sine of rapidity η, just as the Lorentz factor γ is rapidity's hyperbolic cosine, and coordinate speed v over lightspeed is rapidity's hyperbolic tangent"

so you can see that the hyperbolic sine of rapidity is the proper speed whereas the coordinate speed uses the hyperbolic tangent...

the link provides details as other key and useful proper velocity methods and relations.

This is also one of the reasons I always present other references when concepts such as this being discussed. Sometimes it gets rather tricky finding a mathematical method that a given person can relate to.

A good example is components of a vector such as where and when you use the inner/outer or cross product of those components they each have their own purpose and unfortunately v being a vector is subjective to those components in velocity addition rules. I already provided a link above showing the vector addition rules for rapidity. The link included a good listing of other related velocity additions.

The other problem is that although rapidity describes the proper velocity it also describes coordinate time depending on the trig function used.

Lmao it also doesn't help that in particle physics we use rapidity in reference to a bar. Short hand for a probability current that's describes the path integrals with a number density of state's describing an ensemble for a specific particle. For particle resonance it's often used in the lab frame via Breit Weigner. (Used for cross sections).

I mentioned as if one isn't careful and was trying to self study rapidity he could very well google up a reference particular to particle treatment as opposed to how it's used with Lorentz/Minkiwskii. Typically those papers will mention in reference to a bar.

I think maybe an example of rapidity in the twin paradox may help. Later on I will type in the Lewis Ryder treatment

Mainly because he further extrapolated an interesting function with regards to rapidity. As well as other useful relations.

Edited by Mordred
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5 hours ago, Halc said:

The way I've heard 'rapidity' used is as a proper velocity.  It is proper acceleration integrated over proper time, and thus it adds the normal way.

You're referring to "celerity", which is different. https://en.wikipedia.org/wiki/Proper_velocity

https://en.wikipedia.org/wiki/Rapidity "For one-dimensional motion, rapidities are additive."

6 hours ago, Halc said:

Anyway, since the usage comes from Mordred, he may be using the term in a different way than I tend to see it.

I'm sorry to hear that, but yes he's not using the terms correctly.

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Posted (edited)
31 minutes ago, md65536 said:

You're referring to "celerity", which is different. https://en.wikipedia.org/wiki/Proper_velocity

https://en.wikipedia.org/wiki/Rapidity "For one-dimensional motion, rapidities are additive."

I'm sorry to hear that, but yes he's not using the terms correctly.

pray tell how am I not using it correctly in your understanding Yes rapidity's are additive that's literally one of its conveniences. I provided the link showing the vector additions for rapidity.

Edited by Mordred
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10 minutes ago, Mordred said:

pray tell how am I not using it correctly in your uderstanding

Rapidity is not equal to proper velocity. The chart at the top of https://en.wikipedia.org/wiki/Proper_velocity shows how they (and velocity in natural units) relate.

Also,

On 4/18/2024 at 5:16 PM, Mordred said:

It's simply a type of boost called rapidity.

Rapidity is not a type of boost.

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yes I'm fully aware rapidity is not equal to proper velocity.

why do you think I went through the equations for the hyperbolic relations ? If  I didn't know that?

6 minutes ago, md65536 said:

Rapidity is not a type of boost.

did you miss my earlier comment in regards to this ?

14 hours ago, Mordred said:

as shown you can also boost the rapidity or rotate it.

14 hours ago, Mordred said:

I'm actually going to state that I agree the manner its often presented requires a more detailed presentation. I know I'm one of those  parties as velocity can also be treated via rapidity its actually more accurate to state acceleration is boosting or rotating the rapidity. As we have two forms of acceleration change in direction and change in velocity.

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52 minutes ago, md65536 said:

You're referring to "celerity", which is different.

So it seems. Imagine how many posts I've made that are then wrong. My definition of proper velocity seems right, but that's different than rapidity.

I must admit I need to digest that page better to figure out what the rapidity actually is, how it is useful.

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Posted (edited)
58 minutes ago, Halc said:

So it seems. Imagine how many posts I've made that are then wrong. My definition of proper velocity seems right, but that's different than rapidity.

I must admit I need to digest that page better to figure out what the rapidity actually is, how it is useful.

yes its different than proper velocity. This might help first rapidity is a dimensionless parameter that alone is a significant difference from velocity however rapidity is the hyperbolic function relating (v/c)

$TanH\alpha=(v/c)$

it adds as $$\alpha=\alpha_1+\alpha_2$$

with the following

$tanh\alpha=\frac{tanh\alpha_1+tanh\alpha_2}{1+tanh\alpha_1tanh\alpha_2}=v=\frac{v_1+v_2}{1+v_1v_2/c^2}$

Edited by Mordred
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Posted (edited)
4 hours ago, Halc said:

I must admit I need to digest that page better to figure out what the rapidity actually is, how it is useful.

One of its useful relations is how it increases. Rapidity increases linearly when describing the four momentum and four force of the object. So if you have some object/craft/particle undergoing constant acceleration. You need some constant force as well. So as we all know the particle velocity will never reach it will always approach but reach c. However the rapidity of the the object under constant acceleration grows linearly so it can grow arbitrarily large. Velocity has c as a limiting factor but that isn't true for rapidity. That is very useful when you want to use the four force to maintain the constant acceleration of the spacecraft.

Edited by Mordred
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On 4/20/2024 at 11:39 AM, Halc said:

So it seems. Imagine how many posts I've made that are then wrong. My definition of proper velocity seems right, but that's different than rapidity.

I think we'd both have been helped a lot if it was more common on this site for people to discuss and accept corrections, but it always feels like a fight to try. Because of that I think this site is harmful for learning at least about relativity.

On 4/20/2024 at 3:57 PM, Mordred said:

One of its useful relations is how it increases. Rapidity increases linearly when describing the four momentum and four force of the object. So if you have some object/craft/particle undergoing constant acceleration. You need some constant force as well. So as we all know the particle velocity will never reach it will always approach but reach c. However the rapidity of the the object under constant acceleration grows linearly so it can grow arbitrarily large. Velocity has c as a limiting factor but that isn't true for rapidity. That is very useful when you want to use the four force to maintain the constant acceleration of the spacecraft.

All of that makes sense now. The constant acceleration implies constant proper acceleration here, I just point that out because I've confused it with constant coordinate acceleration in a single reference frame before. Am I wrong to assume that someone who understands what you wrote here would see that previous statements from you (going back years) don't make sense?

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Posted (edited)

No you were never wrong to question it not in the slightest. It's also why I chose to work with you to straighten it out. It's all good glad we could work it out. After all its how we all learn

I will try to be more accurate and clear in further mentions of rapidity.

Edited by Mordred
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One other way to view rapidity it to consider three observers, A, B, and C in relative collinear motion. The Doppler ratio between observers A and C is the product of the Doppler ratios between A and B, and B and C. Taking logarithms of the Doppler ratios, the resulting quantity combines additively.

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