Unit of Acceleration: Calculating Speed with Velocity Addition Formula

[Stephanus]

So, basically $X^\mu \rightarrow (\tau,0,0,0)$ is a position not in space but in time axis?

 

[Nugatory]

So, basically $X^\mu \rightarrow (\tau,0,0,0)$ is a position not in space but in time axis?

Neither. It is a position in four-dimensional spacetime (that's why there are four coordinates needed, one for time and three for space) just as (x,t) is a position in the two-dimensional spacetime (two coordinates needed, one for time and one for space) that we illustrate with the two-dimensional spacetime diagrams that we routinely draw and post in this forum.

 

[Stephanus]

Neither. It is a position in four-dimensional spacetime (that's why there are four coordinates needed, one for time and three for space) just as (x,t) is a position in the two-dimensional spacetime (two coordinates needed, one for time and one for space) that we illustrate with the two-dimensional spacetime diagrams that we routinely draw and post in this forum.

Thanks. It confirms my opinion. Altough I didn't express myself clearly (again and again). Perhaps I should say $(\tau,0,0,0)$ is a position directly right in Time Axis.
Btw, what is the differents between $t$ and $\tau$?

Or the other way around. $t$ is in the vertical wl and $\tau$ is in the slanted wordline.
As I recall $\tau = \gamma(t - vx)$ But we can swap $t$ and $\tau$ with this equation.
I just need to know the 'convention' in SR about $\tau$.

Thanks.

 

[Nugatory]

Perhaps I should say $(\tau,0,0,0)$ is a position directly right in Time Axis.

That's right, but of course that point will only be on the time axis if you've chosen a frame in which the time axis goes through that point. Choose a different frame with a different time axis and that point will not be on the time axis and its coordinates will not be of the form $(\tau,0,0,0)$.

Btw, what is the difference between $t$ and $\tau$?

Well, they're both symbols, so their meaning depends on the context in which they're used. When you write something like $(\tau,0,0,0)$ to name a given point in spacetime, $\tau$ is just a number like the zeroes next to it - it's part of the name of that point in a particular frame.

Often, however, $\tau$ is used to represent proper time, which is to say the time that a clock would measure as it moves from one point in spacetime to another. That's how it's being used in the formula $\tau=\sqrt{\Delta{t}^2-\Delta{x}^2}$.

$t$ is most often used to stand for the time coordinate of a point, just as $x$ is used for the space coordinate. That's how both $x$ and $t$ are being used in the formula above. If there is a point in spacetime, and I want to know what its $t$ coordinate is in a given frame, I draw a line through the point and parallel to the x-axis (this trick only works in two-dimensional space-time diagrams, but those are by far the most common) - where that line intersects the t-axis is the value of the t coordinate for that point. It would be agood exercise to try this for yourself on some of the space-time diagrams you've drawn - pick some points that are simultaneous in one frame but not another.

 

[Stephanus]

That's right, but of course that point will only be on the time axis if you've chosen a frame in which the time axis goes through that point. Choose a different frame with a different time axis and that point will not be on the time axis and its coordinates will not be of the form $(\tau,0,0,0)$.

This quote needs some time to understand.
Btw..
https://en.wikipedia.org/wiki/Ladder_paradox#Ladder_paradox_and_transmission_of_force

Axis' don't have to be perpendicular. They can be slanted like the red arrow t' and x'. Is this true?
Just a random question.
Is ladder paradox similar (perhaps not precise the same) as barn paradox?
Do ladder/barn paradox take the concept of the famous train experiment? Where the light from the front and the back of the train hits the center observer at the same time.

Well, they're both symbols, so their meaning depends on the context in which they're used. When you write something like $(\tau,0,0,0)$ to name a given point in spacetime, $\tau$ is just a number like the zeroes next to it - it's part of the name of that point in a particular frame.

Thanks a lot! This helps me greatly in understanding the explanations/articles in SR.

Often, however, $\tau$ is used to represent proper time, which is to say the time that a clock would measure as it moves from one point in spacetime to another. That's how it's being used in the formula $\tau=\sqrt{\Delta{t}^2-\Delta{x}^2}$.

Thanks a lot, this helps me greatly in understanding SR explanations/articles.

$t$ is most often used to stand for the time coordinate of a point, just as $x$ is used for the space coordinate. That's how both $x$ and $t$ are being used in the formula above. If there is a point in spacetime, and I want to know what its $t$ coordinate is in a given frame, I draw a line through the point and parallel to the x-axis

Yep, the most basic cartesian diagram.

(this trick only works in two-dimensional space-time diagrams, but those are by far the most common) - [..]

In 2D (1 time + 1 spatial) spacetime diagram, it would be a horizontal line. In 3D (1 time + 2 spatials) that would still be horizontal line. And in 4D?. Okay don't answer it. I'm trying to understand acceleration (X direction only of course) but I need to know the convention in SR first.

It would be agood exercise to try this for yourself on some of the space-time diagrams you've drawn - pick some points that are simultaneous in one frame but not another.

Are these what you mean? I have had before. Thanks.

Event C and A.

Event E1 to E6 or E3.
Event E2 to TM
----------------------------------------------
I have a question:
Can Axis' be slanted, not perpendicular?

Many thanks for your help.

 

[Nugatory]

This quote needs some time to understand.

Spend that time. It will be time well spent.

Is ladder paradox similar (perhaps not precise the same) as barn paradox?

Well, seeing as how the very first sentence of that wikipedia article say "The ladder paradox (or barn-pole paradox)..." i'd have to say yes.

Axes don't have to be perpendicular. They can be slanted like the red arrow t' and x'. Is this true?

Yes (and I seem to recall some previous posts saying this). In fact, when you're drawing space-time diagrams, if the axes of one inertial frame are perpendicular, then the axes of all other inertial frames will not be. A corollary is that whenever you draw a spacetime diagram, you get to choose which frame's axes are the perpendicular ones... Just remember that this angle has no physical significance whatsoever, and no matter which frame's axes are drawn perpendicular, the spacetime diagram is showing the same events and the same physics. Which frame's axes are perpendicular is about as important as the color of the ink you use to draw them.

Does the ladder/barn paradox take the concept of the famous train experiment? Where the light from the front and the back of the train hits the center observer at the same time.

No. They are related but not the same. The train experiment shows why there is relativity of simultaneity and the pole-barn paradox shows why you'll get in trouble if you forget it.

 

[Stephanus]

Well, seeing as how the very first sentence of that wikipedia article say "The ladder paradox (or barn-pole paradox)..." i'd have to say yes.

What?? My eyes must have been so far-sighted to miss that sentence.

Axis' don't have to be perpendicular. They can be slanted like the red arrow t' and x'. Is this true?

Yes (and I seem to recall some previous posts saying this).[..]

Thanks. This helps me much in understanding SR.
"Previous posts saying this?" Perhaps they implied this, but I just didn't realize.

No. They are related but not the same. The train experiment shows why there is relativity of simultaneity and the pole-barn paradox shows why you'll get in trouble if you forget it.

Ok

 

[Stephanus]

Dear PF Forum,
In http://www.lecture-notes.co.uk/susskind/special-relativity/lecture-6/relativistic-kinematics/

When we talk about the particle frame, we shall really mean this ‘momentarily co-moving reference frame’, and use primed coordinates to refer to it. In this frame, we have
Velocity $U^{\mu} = \frac{dX^{\mu}}{d\tau} \rightarrow (1,0,0,0)$

$(1,0,0,0)$ or let me use $(1,0)$ is understandable. If we transform $(1,0)$ by 0.6c we'll get $(1.25,0.75)$ and we'll get the velocity $\frac{0.75}{1.25} = 0.6c$ To make it easier velocity is always $(1,0)$
But why $U^{\mu} = \frac{dX^{\mu}}{d\tau}$? Why not just $U = \frac{X}{\tau}$? Does the velocity change?

 

[Stephanus]

But why $U^{\mu} = \frac{dX^{\mu}}{d\tau}$? Why not just $U = \frac{X}{\tau}$? Does the velocity change?

DON'T ANSWER THAT! I want to delete this post. But I'm afraid like previous case. My deleted post got answer any way.

If we suppose that the velocity of a particle is changing over time - that is, it can accelerate and decelarate - then the particle's rest frame is not necessarily an inertial reference frame. However, at any given time (either proper time or coordinate time), we can consider a frame that, at that instant, is moving at the same velocity as the particle and has the particle at its origin.

When we talk about the particle frame, we shall really mean this ‘momentarily co-moving reference frame’, and use primed coordinates to refer to it. In this frame, we have... $U^{\mu} = \frac{dX^{\mu}}{d\tau}$

That is in one paragraph above.

 

[Vanadium 50]

Oh sorry, I haven't clicked it.

Why ask questions if you won't read the answer? This is highly inefficient.

Wild guess:

Why guess when you can read the answer?

 

[Stephanus]

Why ask questions if you won't read the answer? This is highly inefficient.
Why guess when you can read the answer?

Come on Vanadium 50. There are many post before Pwz answer came. And I haven't finished reading them. And then came another post.

If you go through the link I gave in my previous post, you will understand where the hyperbolic function comes from.

So I go back to PWiz post and gave it priority.

Wild guess?
I typed that when I was in PF Forum "reply window". Then right away I click the lick.

You even corrected your answer before I had a chance to read it.
https://www.physicsforums.com/threads/unit-of-acceleration.825542/#post-5184050