Four Velocity Vector: why divide by time according to the particle?

  • #1
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Summary:

Why do we need to divide dX/dT, where dT is time according to particle.

Main Question or Discussion Point

So I understand that time is now part of the four vector, and so dividing delta X by delta t (time according to me), would produce just c as the first dimension of the vector, which gives us no intuition as to how fast time is moving for the observer, so is not useful.

I understand why we divide by dT for the first dimension, but I don't understand why this is useful for the other spacial dimensions. What does dX1/dT, dX2/dT... Really mean intuitively? It means the distance between 2 events in my frame divided by time in the frame of the particle. But what does this enable us to do that couldn't be done by dividing dX1/dt?

So my class is defining the four vector for an event as this
X = (ct, x, y, z) (first is time * c, last 3 are spacial components)
dX/dt = (c, dx/dt, dy/dt, dz/dt)
V = dX/dT = dX/dt * dt/dT = (c/sqrt(1-v^2/c^2), v/sqrt(1 - v^2/c^2))
 
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  • #2
PeterDonis
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time is now part of the four vector
Meaning, one of the four components of the vector is a "time" component? Yes, this is true, provided you are using a chart with one timelike and three spacelike coordinates.

dividing delta X by delta t
What delta X and delta t are you talking about?

which gives us no intuition as to how fast time is moving for the observer
There is no invariant meaning to "how fast time is moving for the observer". The only invariant sense of "how fast time is moving" for any observer is the rate at which they perceive their own clock to tick: one second per second. The fact that the 4-velocity is a unit vector is a reflection of this fact.

I understand why we divide by dT for the first dimension
What dT are you talking about and why are you dividing by it? Taking derivatives, which is how you get the 4-velocity components, is not the same thing as division.

What does dX1/dT, dX2/dT... Really mean intuitively?
I have no way of telling unless I know what dT, dX1, dX2, etc. refer to.

It means the distance between 2 events in my frame divided by time in the frame of the particle
It does? Where are you getting that from? What do you think the 4-velocity components are? What you are describing doesn't look like 4-velocity to me.

A concrete example, or a reference for where you are getting all this from, would help.
 
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  • #3
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Meaning, one of the four components of the vector is a "time" component? Yes, this is true, provided you are using a chart with one timelike and three spacelike coordinates.



What delta X and delta t are you talking about?



There is no invariant meaning to "how fast time is moving for the observer". The only invariant sense of "how fast time is moving" for any observer is the rate at which they perceive their own clock to tick: one second per second. The fact that the 4-velocity is a unit vector is a reflection of this fact.



What dT are you talking about and why are you dividing by it? Taking derivatives, which is how you get the 4-velocity components, is not the same thing as division.



I have no way of telling unless I know what dT, dX1, dX2, etc. refer to.



It does? Where are you getting that from? What do you think the 4-velocity components are? What you are describing doesn't look like 4-velocity to me.

A concrete example, or a reference for where you are getting all this from, would help.
I am so sorry for the unclear question.
So my class is defining the four vector for an event as this
X = (ct, x, y, z) (first is time * c, last 3 are spacial components)
dX/dt = (c, dx/dt, dy/dt, dz/dt)
V = dX/dT = dX/dt * dt/dT = (c/sqrt(1-v^2/c^2), v/sqrt(1 - v^2/c^2))

dt/dT is derived from the space time interval. All observers can agree on the time elapsed according to the particle dT = dt * sqrt(1 - v^2/c^2)

Would it be possible to explain each of the components of the new velocity vector, and their intuitive meaning?
 
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There is no invariant meaning to "how fast time is moving for the observer". The only invariant sense of "how fast time is moving" for any observer is the rate at which they perceive their own clock to tick: one second per second. The fact that the 4-velocity is a unit vector is a reflection of this fact.
Thank you, understood! What I meant by "fast" is the quantity of difference in perceived time
 
  • #5
PeterDonis
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my class is defining the four vector for an event as this
X = (ct, x, y, z) (first is time * c, last 3 are spacial components)
Ok.

dX/dt = (c, dx/dt, dy/dt, dz/dt)
V = dX/dT = dX/dt * dt/dT = (c/sqrt(1-v^2/c^2), v/sqrt(1 - v^2/c^2))
Yes.

dt/dT is derived from the space time interval.
I would say it is derived from the fact that V has to be a unit vector, or more precisely, a "unit" vector in units of ##c##. (In "natural" units, where ##c = 1##, ##V## will be an actual unit vector, with norm ##1##.) If we write ##V = \left( c dt/dT, dx/dT, dy/dT, dz/dT \right) = dt/dT \left( c, v_x, v_y, v_z \right)##, then the norm of V will be

$$
|V| = c = \frac{dt}{dT} \sqrt{ c^2 - v_x^2 - v_y^2 - v_z^2 } = c \frac{dt}{dT} \sqrt{ 1 - \frac{v^2}{c^2} }
$$

from which ##dt / dT = 1 / \sqrt{1 - v^2 / c^2}## follows immediately.

All observers can agree on the time elapsed according to the particle
Yes. The usual term in relativity is "proper time". Geometrically, it's just arc length along the particle's worldline.

Would it be possible to explain each of the components of the new velocity vector, and their intuitive meaning?
Their meaning as velocity components isn't really "intuitive", but if we write ##dt / dT = \gamma##, then we have ##V = \left( \gamma c, \gamma \vec{v} \right)##. If we then multiply by the rest mass ##m## of the object, we have the 4-momentum ##P = m V = \left( m \gamma c, m \gamma \vec{v} \right)##. The components of this are the energy and momentum of the object ("energy" here is in momentum units, so it's the conventional energy ##E## divided by ##c##, but physically the meaning is the same). That's the direct physical meaning.
 
  • #6
PeterDonis
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What I meant by "fast" is the quantity of difference in perceived time
What is "perceived time"? Perceived according to whom?
 
  • #7
PeterDonis
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Btw, @dkhurana, the "LaTeX Guide" link at the bottom left of the box you use to edit posts will take you to a guide to the PF LaTeX feature, which is very helpful for writing equations. I've used it in my posts in this thread.
 
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  • #8
PeroK
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Summary:: Why do we need to divide dX/dT, where dT is time according to particle.
Because the energy-momentum four-vector is a conserved quantity:
$$p = m(\frac{dt}{dT}, \frac{dx}{dT}, \frac{dy}{dT}, \frac{dz}{dT})$$
Whereas, the Newtonian three-momentum is not conserved at relativistic speeds:
$$\vec p = m(\frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt})$$
 
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  • #9
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Ok.



Yes.



I would say it is derived from the fact that V has to be a unit vector, or more precisely, a "unit" vector in units of ##c##. (In "natural" units, where ##c = 1##, ##V## will be an actual unit vector, with norm ##1##.) If we write ##V = \left( c dt/dT, dx/dT, dy/dT, dz/dT \right) = dt/dT \left( c, v_x, v_y, v_z \right)##, then the norm of V will be

$$
|V| = c = \frac{dt}{dT} \sqrt{ c^2 - v_x^2 - v_y^2 - v_z^2 } = c \frac{dt}{dT} \sqrt{ 1 - \frac{v^2}{c^2} }
$$

from which ##dt / dT = 1 / \sqrt{1 - v^2 / c^2}## follows immediately.



Yes. The usual term in relativity is "proper time". Geometrically, it's just arc length along the particle's worldline.



Their meaning as velocity components isn't really "intuitive", but if we write ##dt / dT = \gamma##, then we have ##V = \left( \gamma c, \gamma \vec{v} \right)##. If we then multiply by the rest mass ##m## of the object, we have the 4-momentum ##P = m V = \left( m \gamma c, m \gamma \vec{v} \right)##. The components of this are the energy and momentum of the object ("energy" here is in momentum units, so it's the conventional energy ##E## divided by ##c##, but physically the meaning is the same). That's the direct physical meaning.
What I still don't understand is why is it so important that we have an absolute time T that we use dX/dT instead of dX/dt. I understand that using dX/dt won't transform under the lorentz transformation, but then why is it important that it does? Not all observers agree on the time taken between 2 events, and they also don't agree on the distance between 2 events. So why do we only divide dX by dT and not use some sort of absolute distance measurement dX or something?

Also is the energy and momentum of a particle conserved throughout different reference frames?
 
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  • #10
vanhees71
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It's just much simpler to use scalars, vectors, and generally tensors in Minkowski space than some complicated non-invariant objects to describe reality by the special theory of relativity. Since ##(ct,\vec{x})## are components of a four-vector, it's thus much simpler to use the four-velocity ##u=\gamma(1,\vec{v}/c)##, which is a four-vector rather than ##\vec{v}## to describe "velocity". For the same reason we nowadays prefer the invariant mass ##m## over older concepts of relativistic mass (though there are some also new textbooks, which stick to old-fashioned ideas, but there's no reason to read them ;-))).
 
  • #11
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It's just much simpler to use scalars, vectors, and generally tensors in Minkowski space than some complicated non-invariant objects to describe reality by the special theory of relativity. Since ##(ct,\vec{x})## are components of a four-vector, it's thus much simpler to use the four-velocity ##u=\gamma(1,\vec{v}/c)##, which is a four-vector rather than ##\vec{v}## to describe "velocity". For the same reason we nowadays prefer the invariant mass ##m## over older concepts of relativistic mass (though there are some also new textbooks, which stick to old-fashioned ideas, but there's no reason to read them ;-))).
Ok so just to confirm, the whole reason we transform use 4 velocity is just to make it easier to transform velocity from one reference frame to another?
 
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  • #12
PeroK
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Also is the energy and momentum of a particle conserved throughout different reference frames?
There are two different concepts here:

A quantity is conserved if it does not change over time. Energy-momentum, for example, is conserved in collisions and particle decay etc.

The conservation of energy-momentum in SR replaces the separate conservation of mass, energy and three-momentum in Newtonian physics. That is something of a unification.

A quantity is invariant if it is the same in all (inertial) reference frames. The magnitude of a four-vector, for example, is the same in all (inertial) reference frames. For energy-momentum, this means that:
$$E^2 - (p_x^2 + p_y^2 + p_z^2)c^2$$
is the same in all (inertial) reference frames.

Question: what is this invariant quantity in simple terms?
 
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  • #13
Ibix
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I understand that using dX/dt won't transform under the lorentz transformation, but then why is it important that it does?
Ok so just to confirm, the whole reason we transform use 4 velocity is just to make it easier to transform velocity from one reference frame to another?
The principle of relativity says that the laws of physics take the same form in all inertial frames. If that's the case, it must be possible to write the laws of physics in terms of things that transform covariantly, or else different frames aren't using the same thing.

You can certainly write down ##(\frac{d(ct)}{dt},\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt})##. But if you were using another frame you would want to write ##(\frac{d(ct')}{dt'},\frac{dx'}{dt'},\frac{dy'}{dt'},\frac{dz'}{dt'})##, and this is not the same thing (consider the case of an object stationary in one frame - what's the magnitude of those two "vectors"?). So physical laws formulated in terms of this quantity, but using different frames' coordinates, are talking about different things or need to have a different form.

However, ##(\frac{d(ct)}{d\tau},\frac{dx}{d\tau},\frac{dy}{d\tau},\frac{dz}{d\tau})## turns out to be the same vector as ##(\frac{d(ct')}{d\tau},\frac{dx'}{d\tau},\frac{dy'}{d\tau},\frac{dz'}{d\tau})##, albeit described in different terms, whatever frame you use. So it's the natural object to use in the definition of physical laws that take the same form in all frames (that's what @vanhees71 means by "simpler", I think).
 
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  • #14
PeterDonis
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why is it so important that we have an absolute time T
T is not an "absolute time". It is the proper time along the object's worldline. There is no "absolute time" in relativity.

why is it important that it does?
Because experimentally we have found that the laws of physics are Lorentz invariant, so we want our mathematical representations of those laws to be Lorentz invariant as well.

why do we only divide dX by dT
We don't "divide" anything by anything. As I have already said: taking derivatives is not the same as division. The 4-velocity of an object is obtained by taking derivatives of the four coordinates with respect to the proper time along the object's worldline.

Geometrically, proper time is just the arc length along the worldline, so all we are doing is taking derivatives of the coordinates with respect to arc length, which is the only invariant we have for the object's worldline.

and not use some sort of absolute distance measurement dX or something?
Arc length along the worldline is the closest thing we have to the "absolute distance measurement" you describe.

is the energy and momentum of a particle conserved throughout different reference frames?
Energy and momentum by themselves are frame-dependent, but the norm of the 4-momentum, the rest mass of the object, is invariant; it's the same in all frames.
 
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  • #15
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I'm apologize for all my confusion, and I really appreciate everyone helping.

However, ##(\frac{d(ct)}{d\tau},\frac{dx}{d\tau},\frac{dy}{d\tau},\frac{dz}{d\tau})## turns out to be the same vector as ##(\frac{d(ct')}{d\tau},\frac{dx'}{d\tau},\frac{dy'}{d\tau},\frac{dz'}{d\tau})##, albeit described in different terms, whatever frame you use. So it's the natural object to use in the definition of physical laws that take the same form in all frames (that's what @vanhees71 means by "simpler", I think).
I think I'm getting where you're coming at but I don't really understand it when you say "same thing" or the "same vector" but different terms. Would it be possible elaborate a little more on that?

I think where I am confused is that if you try to find the the quantity dX/dTau in nature, you won't be able to, although you can derive it. Each observer in some sense only really sees dX/dt. It just seems to me like a mathematical construct that somehow lead to the uncovering of the correction to energy and velocity.

Because experimentally we have found that the laws of physics are Lorentz invariant, so we want our mathematical representations of those laws to be Lorentz invariant as well.
Is it possible to transform four velocity without using a lorentz transformation? Does the usefulness of it only come from the convenience?


So this vector ##(\frac{d(ct)}{dt},\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt})## accurately describes the velocity of time and space in my frame (time being 1 second per second), and this accurately describes time and space in another frame of reference ##(\frac{d(ct')}{d\tau},\frac{dx'}{d\tau},\frac{dy'}{d\tau},\frac{dz'}{d\tau})## (time is one second per second here). We can use the relative velocity transformation for the spacial coordinates and just leave time the same. Why can't I use this to transform between the 2 equations? Is it because this encodes no information about the relative time between reference frames?
 
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  • #16
PeterDonis
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I don't really understand it when you say "same thing" or the "same vector" but different terms.
A vector is a geometric object, independent of any choice of coordinates. But once you choose coordinates, you can write the vector in terms of components in those coordinates; and the values of the components will depend on the coordinates you choose.

Is it possible to transform four velocity without using a lorentz transformation?
Not between inertial frames; the transformation between inertial frames is the Lorentz transformation.

If you want to use non-inertial coordinates, then there will be other kinds of transformations involved.

So this vector ##(\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt})## accurately describes the velocity of time and space in my frame
No, it describes the motion of an object, not "time and space". Time and space don't move. Objects do.

We can use the relative velocity transformation for the spacial coordinates and just leave time the same.
No. You can't just make up transformations however you want. We're doing physics here; that means the transformations have to accurately represent the physics.
 
  • #17
Ibix
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I think I'm getting where you're coming at but I don't really understand it when you say "same thing" or the "same vector" but different terms. Would it be possible elaborate a little more on that?
Draw an arrow of unit length on a piece of paper and put the paper on a table. Turn the paper so that the arrow is pointing straight away from you. You'd now describe it as the vector (0,1) in Cartesian coordinates, right? Now walk around the side of the table. The arrow is pointing to the left or right, depending which way you went. You'd now describe it as (1,0) in Cartesian coordinates (or (-1,0), depending). But it's the same arrow - the same vector. You're just describing it in terms of a different coordinate system.

If that arrow represents something real (e.g. the velocity of a particle) then the velocity doesn't really change in any sense just because you walked round the side of the table. We want the quantities in our mathematical model to be like that - we might say that the velocity is (1,0) or (0,1) depending on our choice of reference frame, but we always want to be talking about the same vector.
I think where I am confused is that if you try to find the the quantity dX/dTau in nature, you won't be able to, although you can derive it. Each observer in some sense only really sees dX/dt.
Well, both ##dx/d\tau## and ##dx/dt## are interpretations of data but, actually, ##dx/dτ## is the simpler concept. If I want to measure the proper time you experience, I just look at your watch. If I want to measure the coordinate time, I need to set up and synchronise clocks at the start and end of your route - so instead of just one clock I need two clocks and a synchronisation convention.
It just seems to me like a mathematical construct that somehow lead to the uncovering of the correction to energy and velocity.
Remember that relativity talks about spacetime. Your path through spacetime is a line, usually called your worldline. At any event on your worldline, your worldline is pointing in a particular direction. A unit vector (or a vector of length ##c##, definitions vary) in that direction is your four-velocity.
Is it possible to transform four velocity without using a lorentz transformation?
You can make up any transformation you want. But the only one that matches reality is the Lorentz transform.
We can use the relative velocity transformation (v + u/(1 + uv/c^2)) to get the transformation for the spacial coordinates, and we can also use the time dilation dt' = dt/sqrt(...). Why can't I use this to transform between the 2 equations?
Assuming you meant "spatial components", you can do (although some care is needed with the time transformation - it depends on ##v_x## too in general). But those velocity transformations are derived from the Lorentz transforms, so you are just using the Lorentz transforms in a disguised way.
 
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  • #18
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Assuming you meant "spatial components", you can do (although some care is needed with the time transformation - it depends on ##v_x## too in general). But those velocity transformations are derived from the Lorentz transforms, so you are just using the Lorentz transforms in a disguised way.
Why is it so important that time doesn't transform?

Ok so I am now realizing that there is a problem in defining the four velocity my way, assuming I make all the right corrections. The spacial components transform just fine with the relative velocity formula. But the first component by my definition will remain c because c * dt' / dt' is just c. This accurately describes that each observer sees time as moving 1 second per second. So my four velocity defines these components exactly like the observer sees them. If you could confirm with me this that I am right with diagnosing the problem here:

This gives us no description as per what my 1 second translates to in your reference frame. They both describe time as being the same in all reference frames, which is not true.
 
  • #19
Ibix
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Why is it so important that time doesn't transform?
The time component transforms too, just not quite the way you wrote in general. I was saying that you wrote coordinates and I think you meant components.

Apologies - have to rush off. I'll reapond to the rest later.
 
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  • #20
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The time component transforms too, just not quite the way you wrote in general. I was saying that you wrote coordinates and I think you meant components.

Apologies - have to rush off. I'll reapond to the rest later.
Hi! No problem at all. That question was more for myself and I only accidently included it. I was mainly wondering why it is important that we take the proper time derivative of X that doesn't transform from reference frame to reference frame. But I think I get it kind of. If we have some sort of universal time, then we can explain any change in the velocity vector of a particle by a different relative velocity in another r.f. But still something about spacetime just fundamentally confuses me. I think I'm too caught up on newtonian mechanics where it is ok for a velocity to have different magnitudes based on the observer. So I guess I don't really get why there's this limit, c, in spacetime that all vector magnitudes have to obey to.
 
  • #21
PeroK
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Hi! No problem at all. That question was more for myself and I only accidently included it. I was mainly wondering why it is important that we take the proper time derivative of X that doesn't transform from reference frame to reference frame. But I think I get it kind of. If we have some sort of universal time, then we can explain any change in the velocity vector of a particle by a different relative velocity in another r.f. But still something about spacetime just fundamentally confuses me. I think I'm too caught up on newtonian mechanics where it is ok for a velocity to have different magnitudes based on the observer. So I guess I don't really get why there's this limit, c, in spacetime that all vector magnitudes have to obey to.
First, the limit of ##c## applies to relative velocity of particles and reference frames. That emerges from the Lorentz Transformation and/or the velocity transformation formula:
$$u' = \frac{u + v}{1 + uv/c^2}$$
Uisng that formula, you cannot get higher than ##c##.

Having discovered SR what you could say is that now there is now no such thing as conservation of energy or conservation of momentum, if you insist on defining energy and momentum in Newtonian terms:
$$KE = \frac 1 2 mv^2, \ \ \text{and} \ \ p = mv$$
You stay with these definitions. You use only coordinate time ##t##. You have no laws of physics now. No conservation of energy-momentum and physics is dead. Time to take up basket-weaving instead!

The alternative is to look for the upgraded laws of physics in this new framework of SR. That means not just looking for but accepting new ideas. One new idea is proper time of a particle. Can we find some laws of physics using that?

Another new idea is that of a four-vector. Perhaps instead of sticking with three-vectors we should explore this new idea of four-vectors and their properties? Maybe that's interesting and useful?
 
  • #22
PeterDonis
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why it is important that we take the proper time derivative of X that doesn't transform from reference frame to reference frame.
I'm not sure what you think "doesn't transform". The ##t## component of 4-velocity changes when you change frames just as the ##x##, ##y##, and ##z## components do.

I strongly suggest trying out some explicit computations, instead of talking purely in generalities. For example, suppose a particular object has 4-velocity ##(1, 0, 0, 0)## in one frame (where I have given the ##t## component first and then the ##x, y, z## components). What will the components of this 4-velocity be in a frame that is moving, relative to the first one, with speed ##v## in the ##x## direction?

If we have some sort of universal time
There is no "universal time" in relativity.
 
  • #23
PeterDonis
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I think I'm too caught up on newtonian mechanics where it is ok for a velocity to have different magnitudes based on the observer.
This is also true in relativity for anything that moves slower than light.

guess I don't really get why there's this limit, c, in spacetime that all vector magnitudes have to obey to.
Not "all vector magnitudes". Just all velocities.

Also, it might help to think of it, not as a limit of c on all velocities, but as a limit on the Lorentz transformation: you can only do a Lorentz transformation with a relative velocity ##v## whose magnitude is less than ##c##. The transformation is singular (i.e., mathematically invalid, hence not allowed) for ##v = c##.
 
  • #24
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I'm not sure what you think "doesn't transform". The ##t## component of 4-velocity changes when you change frames just as the ##x##, ##y##, and ##z## components do.

I strongly suggest trying out some explicit computations, instead of talking purely in generalities. For example, suppose a particular object has 4-velocity ##(1, 0, 0, 0)## in one frame (where I have given the ##t## component first and then the ##x, y, z## components). What will the components of this 4-velocity be in a frame that is moving, relative to the first one, with speed ##v## in the ##x## direction?



There is no "universal time" in relativity.
What I meant is dTau is the same the same for all observers. I understand V0 transforms, but dTau doesn't transform from one observer to the other. They can all agree on what is dTau. What I meant by universal time is a time frame transform to that everyone agrees. From the transform equations it's (1/sqrt(1 - v^2/c^2), v/sqrt(1 - v^2/c^2), ..), and I understand the derivation pretty well.

I want to understand what dX1/dTau really means. I understand it mathematically well. It's the infinitesimally small distance between points of the particle in my reference over infinitesimally small time difference in the particles' reference frame. I understand it can be transformed through the lorentz transformation. But I want to really understand it's significance, and it's difference to just plain old velocity. Why is it important in general to use proper time and not rely on variable time by an observer? Why is it a wrong to take the derivative of a vector over one of it's components like dX/dt to describe the velocity vector (also side question: is dX/dt a vector or a scalar?): a mathematical explanation is preferred?
 
  • #25
PeterDonis
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What I meant is dTau is the same the same for all observers.
dTau for a particular object is the same for all observers. But dTau for different objects with different worldlines is different.

What I meant by universal time is a time frame transform to that everyone agrees.
The Lorentz transformation, which is what you seem to be referring to here, is not a "time frame transform". It's a transformation between reference frames, which are more than just "time". The fact that everyone agrees on the transform does not mean there is a "universal time".

I want to understand what dX1/dTau really means.
See the last part of my post #5.

Why is it a wrong to take the derivative of a vector over one of it's components like dX/dt to describe the velocity vector
Because a vector is not the same as one of its components.

is dX/dt a vector or a scalar
I don't know exactly what dX/dt you are referring to, since you have shifted notation several times in this thread so I don't know what you mean by X.

If X means one of the space coordinates (x, y, or z), then dX/dt is neither a vector nor a scalar. It's a component of a frame-dependent 3-vector.

If X means the spatial 3-vector (x, y, z), then dX/dt is a frame-dependent 3-vector.
 

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