Question about full Lorentz transformation

Chenkel said:

If dx^2 + dy^2 + dz^2 - (c*dt)^2 is negative and I can invert it and take the square root? What do you mean by invert?

I mean take off the minus sign. You cannot take the square root of a negative number. So you take the square root of its absolute value instead.

[Yes, we know about imaginary numbers like the square root of minus one. But that is a road that you do not want to go down in this context. It does not end up being helpful].

jbriggs444 said:

I mean take off the minus sign. You cannot take the square root of a negative number. So you take the square root of its absolute value instead.

And if the distance between between events increases because the reference frame is moving then the time between the events needs to increase too to keep the spacetime interval invariant.

Chenkel said:

And if the distance between between events increases because the reference frame is moving then the time between the events needs to increase too to keep the spacetime interval invariant.

Yes. If the coordinate distance increases, the coordinate delta time must increase because the interval is invariant.

I am trying to sprinkle the word "coordinate" in here to emphasize that the "distance" and "time" values here are relative to a coordinate system.

The point is that a massive particle's world line is time-like everywhere, and thus
$$\mathrm{d} \tau =\mathrm{d}t \sqrt{1-\vec{\beta}^2}>0$$
Integrated this means the ##\tau(t)## is a monotonously increasing function. The advantage of using ##\tau## rather than ##t## to formulate the dynamics of point particles in special relativity is that it is a scalar, while time is a component of a four-vector wrt. some arbitrary frame of reference.

The heuristics is as follows:

To "translate" (or rather extrapolate) the equations of motion from Newtonian to special relativsitic equations you can take the Newtonian ones to be valid over short time intervals in the momentaneous inertial rest frame of the particle.

Newton's equation of motion for a particle in some "external field" (e.g., a charged particle in the electromagnetic field as the paradigmatic example) you have
$$\mathrm{d}_t \vec{p}=\vec{F}(t,\vec{x}), \quad \vec{p}=m \mathrm{d}_t \vec{x}.$$
This should be valid in the momentaneous inertial rest frame of the particle. In this frame ##\mathrm{d} t =\mathrm{d} \tau##. So the first step is to define momentum as ##\vec{p}=m \mathrm{d}_{\tau} \vec{x}##. It's very suggestive to rather define four-momentum,
$$p^{\mu} = m \mathrm{d}_{\tau} x^{\mu}.$$
Note that ##m## is defined as a constant, quantifying "inertial" in the momentaneous rest frame of the particle. As such it's a Lorentz scalar. Since ##\mathrm{d} \tau## is a scalar too and ##x^{\mu}## a four-vector, indeed ##p^{\mu}## is a four-vector.

The meaning of the time component becomes clear from (using the (+---) convention for the Minkowski form)
$$\eta_{\mu \nu} p^{\mu} p^{\nu}=m^2 c^2 \; \Rightarrow \; p^0=\sqrt{m^2 c^2+\vec{p}^2}.$$
Now go to the non-relativistic limit, which means ##\vec{p}^2 \ll m^2 c^2##. Then you have
$$p^0 \simeq m c +\frac{1}{2} \frac{\vec{p}^2}{mc} \; \Rightarrow c p^0 \simeq m c^2 + \frac{\vec{p}^2}{2m},$$
i.e., you have
$$c p^0=m c^2+E_{\text{kin}},$$
i.e., adding the "rest mass" to the kinetic energy you have the advantage that ##p^0## combines with ##\vec{p}## to a four-vector. That's why one keeps this additive constant term as part of relativistic energy of a particle.

Written in terms of the "coordinate velocity" you have
$$(p^{\mu})=m \mathrm{d}_{\tau} t \mathrm{d}_t \begin{pmatrix} c t \\ \vec{x} \end{pmatrix} = m \gamma \begin{pmatrix} c \\ \vec{v} \end{pmatrix}, \quad \gamma=\frac{1}{\sqrt{1-\vec{v}^2/c^2}}=\frac{1}{\sqrt{1-\vec{\beta}^2}}.$$
For the equation of motion itself the four-dimensional covariant guess is
$$\mathrm{d}_{\tau} p^{\mu} = K^{\mu}.$$
Now because of ##p_{\mu} p^{\mu}=m^2 c^2=\text{const}## you get ##p_{\mu} \mathrm{d}_{\tau} p^{\mu}=0##, i.e., the Minkowski force on the right-hand side is subject to the contraint
$$p_{\mu} K^{\mu}=0$$
or, equivalently,
$$K_{\mu} \mathrm{d}_{\tau} x^{\mu}=0.$$
For the motion of a charged particle in an em. field ##(\vec{E},\vec{B})## you have (in Heaviside Lorentz units)
$$\vec{F}=q (\vec{E}+\vec{\beta} \times \vec{B}), \quad \vec{\beta}=\vec{v}/c.$$
Now this means the corresponding Minkowski force in the four-dimensional covariant translation should be linear in the fields and the four-velocity of the particle, ##u^{\mu}=\mathrm{d}_{\tau} x^{\mu}/c## (note that I have made it dimensionless by dividing by ##c##, which is convenient. Of course you have ##u^{\mu}=p^{\mu}/(mc)##. Since the Minkowski force must be a four-vector the only law with these features is
$$K^{\mu}=q F^{\mu \nu} u_{\nu},$$
where ##F^{\mu \nu}## is a Minkowski four-tensor field. The contraint above says
$$u_{\mu} K^{\mu}=0 \; \Rightarrow \; F^{\mu \nu} u_{\mu} u_{\nu}=0.$$
This is automatically fulfilled if ##F^{\mu \nu}## is an antisymmetric tensor, and indeed defining
$$(F^{\mu \nu}) = \begin{pmatrix} 0 & -\vec{E}^{\text{T}} \\ \vec{E} & (-\epsilon^{jkl} B^l) \end{pmatrix},$$
because then you get
$$(K^{\mu}) = (q F^{\mu \nu} u_{\nu}) =q \gamma \begin{pmatrix} \vec{\beta} \cdot \vec{E} \\ \vec{E}+\vec{\beta} \times \vec{B} \end{pmatrix}.$$
Indeed the spatial part gives in the non-relativistic limit the correct Lorentz force in the non-relativistic approximation. So it's pretty convincing to assume that ##K^{\mu}## is the correct relativistic description of this Lorentz force, and indeed experiment agrees with it.

It's also clear that you need to solve only the spatial part of the equation of motion for the particle, i.e.,
$$\mathrm{d}_{\tau} p^{\mu} = q (u^0 \vec{E}+ \vec{u} \times \vec{B})=q \gamma (\vec{E}+\vec{v} \times \vec{B}).$$
Then the time component of the 4D equation of motion is automatically fulfilled, because it says that
$$u_0 K^0=\gamma K^0=\gamma \vec{u} \cdot \vec{E} \stackrel{!}{=}\vec{u} \cdot \vec{K}$$
but indee
$$\vec{u} \cdot \vec{K}=u^0 \vec{E} \cdot \vec{u}=\gamma \vec{E} \cdot \vec{u}.$$
In terms of time derivatives wrt. coordinate time ##t## you have
$$\mathrm{d}_t \vec{p}=\frac{1}{\gamma} \mathrm{d}_{\tau} \vec{p} = \frac{1}{\gamma} \vec{K} = q(\vec{E} +\vec{u} \times \vec{B}/\gamma)=q(\vec{E} + \vec{v} \times \vec{B}).$$
Further
$$\vec{p}=m \gamma \vec{v},$$
i.e.,
$$m \mathrm{d}_{t} (\gamma \mathrm{d}_t \vec{x}) = q (\vec{E}+\vec{v} \times \vec{B}),$$
but that's of course a pretty ugly not manifestly covariant notation for the much nicer covariant formula
$$m \mathrm{d}_{\tau} x^{\mu} = q F^{\mu \nu} u_{\nu} = \frac{q}{c} F^{\mu \nu} \mathrm{d}_{\tau} x_{\nu}.$$

vanhees71 said:

a monotonously increasing function

You mean "monotonically" increasing. "Monotonous" means "boring"--which many people might find it to be, but is not what you meant in context.

PeterDonis said:

"Monotonous" means "boring"

This is the name of a company that builds tunnels, founded by Elon Musk

Source:
https://de.wikipedia.org/wiki/The_Boring_Company

vanhees71 said:

$$p^{\mu} = m \mathrm{d}_{\tau} x^{\mu}.$$...$$\mathrm{d}_{\tau} p^{\mu} = K^{\mu}.$$

It follows: ##K^{\mu} = \mathrm{d}_{\tau} p^{\mu} = m \mathrm{d}_{\tau}(\mathrm{d}_{\tau} x^{\mu}) + (\mathrm{d}_{\tau}x^{\mu})({d}_{\tau}m)##

vanhees71 said:

Now because of ##p_{\mu} p^{\mu}=m^2 c^2=\text{const}## you get ##p_{\mu} \mathrm{d}_{\tau} p^{\mu}=0##, i.e., the Minkowski force on the right-hand side is subject to the contraint
$$p_{\mu} K^{\mu}=0$$or, equivalently,$$K_{\mu} \mathrm{d}_{\tau} x^{\mu}=0.$$

An additional assumption is here, that the force is mass-preserving.

Source, before equation (39):
http://www.scholarpedia.org/article/Special_relativity:_electromagnetism

NO!!! ##m=\text{const}## is the invariant mass. Today we work with manifestly covariant equations!

vanhees71 said:

NO!!! ##m=\text{const}## is the invariant mass. Today we work with manifestly covariant equations!

An EM-force on a classical resistor model i.e. can create heating: ##{d}_{\tau}m \ne 0##.

I talked about charged particles in a electromagnetic field. Of course, the notion of classical relativistic point particles is a quite problematic concept, and you come not very far with it within classical theory.

Chenkel said:

If dx^2 + dy^2 + dz^2 - (c*dt)^2 is negative and I can invert it and take the square root? What do you mean by invert?

For a timelike interval, you can write:
##{\Delta s}^2 = {(c \Delta t)}^2 - {\Delta x}^2 - {\Delta y}^2 - {\Delta z}^2 \ \ \ \ \ (1)##

The inverse LT for coordinate differences is:
##c\Delta t = \gamma (c\Delta t' + {v \over c}\Delta x')##
##\Delta x = \gamma (\Delta x' + {v \over c}c\Delta t')##
##\Delta y = \Delta y'##
##\Delta z = \Delta z'##

I plug the inverse LT into equation (1):

##{\Delta s}^2 = {\gamma^2 (c\Delta t' + {v \over c}\Delta x')}^2 - {\gamma^2 (\Delta x' + {v \over c}c\Delta t')}^2 - {\Delta y'}^2 - {\Delta z'}^2##

##{\Delta s}^2 = \gamma^2 [(c\Delta t' + {v \over c}\Delta x')^2 - (\Delta x' + {v \over c}c\Delta t')^2] - {\Delta y'}^2 - {\Delta z'}^2##

##{\Delta s}^2 = \gamma^2 (1-v^2/c^2)[(c\Delta t')^2 - \Delta x'^2] - {\Delta y'}^2 - {\Delta z'}^2## ##{\Delta s}^2 = (c\Delta t')^2 - \Delta x'^2 - {\Delta y'}^2 - {\Delta z'}^2 \ \ \ \ \ (2)##

Set equation (1) equal equation (2) to show the invariance of the squared spacetime interval:

$$ (c\Delta t')^2 - \Delta x'^2 - {\Delta y'}^2 - {\Delta z'}^2 = {(c \Delta t)}^2 - {\Delta x}^2 - {\Delta y}^2 - {\Delta z}^2$$
Assume a clock at rest in the primed frame: ##\Delta x' = \Delta y' = \Delta z' = 0##.
It follows:

##(c\Delta \tau)^2 = {(c \Delta t)}^2 - {\Delta x}^2 - {\Delta y}^2 - {\Delta z}^2##

##\Delta \tau^2 / \Delta t^2 = 1 - {\Delta x^2 + \Delta y^2 + \Delta z^2 \over c^2 \Delta t^2}##

Time dilation with respect to the unprimed frame of a clock at rest in the primed frame:
##\Delta \tau / \Delta t = \sqrt{1 - {{v}^2 \over c^2 }}##

Sagittarius A-Star said:

For a timelike interval, you can write:
##{\Delta s}^2 = {(c \Delta t)}^2 - {\Delta x}^2 - {\Delta y}^2 - {\Delta z}^2 \ \ \ \ \ (1)##

The inverse LT for coordinate differences is:
##c\Delta t = \gamma (c\Delta t' + {v \over c}\Delta x')##
##\Delta x = \gamma (\Delta x' + {v \over c}c\Delta t')##
##\Delta y = \Delta y'##
##\Delta z = \Delta z'##

I plug the inverse LT into equation (1):

##{\Delta s}^2 = {\gamma^2 (c\Delta t' + {v \over c}\Delta x')}^2 - {\gamma^2 (\Delta x' + {v \over c}c\Delta t')}^2 - {\Delta y'}^2 - {\Delta z'}^2##

##{\Delta s}^2 = \gamma^2 [(c\Delta t' + {v \over c}\Delta x')^2 - (\Delta x' + {v \over c}c\Delta t')^2] - {\Delta y'}^2 - {\Delta z'}^2##

##{\Delta s}^2 = \gamma^2 (1-v^2/c^2)[(c\Delta t')^2 - \Delta x'^2] - {\Delta y'}^2 - {\Delta z'}^2## ##{\Delta s}^2 = (c\Delta t')^2 - \Delta x'^2 - {\Delta y'}^2 - {\Delta z'}^2 \ \ \ \ \ (2)##

Set equation (1) equal equation (2) to show the invariance of the squared spacetime interval:

$$ (c\Delta t')^2 - \Delta x'^2 - {\Delta y'}^2 - {\Delta z'}^2 = {(c \Delta t)}^2 - {\Delta x}^2 - {\Delta y}^2 - {\Delta z}^2$$
Assume a clock at rest in the primed frame: ##\Delta x' = \Delta y' = \Delta z' = 0##.
It follows:

##(c\Delta \tau)^2 = {(c \Delta t)}^2 - {\Delta x}^2 - {\Delta y}^2 - {\Delta z}^2##

##\Delta \tau^2 / \Delta t^2 = 1 - {\Delta x^2 + \Delta y^2 + \Delta z^2 \over c^2 \Delta t^2}##

Time dilation with respect to the unprimed frame of a clock at rest in the primed frame:
##\Delta \tau / \Delta t = \sqrt{1 - {{v}^2 \over c^2 }}##

I followed the math except for the last part how does the following equation make sense?##\frac {\Delta x^2 + \Delta y^2 + \Delta z^2} {c^2 \Delta t^2} = \frac {v^2} {c^2}##

Chenkel said:

I followed the math except for the last part how does the following equation make sense?##\frac {\Delta x^2 + \Delta y^2 + \Delta z^2} {c^2 \Delta t^2} = \frac {v^2} {c^2}##

According to Pythagoras, a squared distance is ##\Delta d^2 = \Delta x^2 + \Delta y^2 + \Delta z^2##.
Velocity is ##v= \Delta d / \Delta t##. Velocity squared is ##v^2= \Delta d^2 / \Delta t^2##.

In this special scenario, ##\Delta y = \Delta z = 0## anyway, because ##\Delta y' = \Delta z' = 0##.

Sagittarius A-Star said:

According to Pythagoras, a squared distance is ##\Delta r^2 = \Delta x^2 + \Delta y^2 + \Delta z^2##.
Velocity is ##v= \Delta r / \Delta t##. Velocity squared is ##v^2= \Delta r^2 / \Delta t^2##.

In this special scenario, ##\Delta y = \Delta z = 0## anyway, because ##\Delta y' = \Delta z' = 0##.

I thought ##\Delta x, \Delta y, \Delta z## is the location of an event in the unprimed frame and v is the velocity of the primed frame in the positive x direction of the unprimed axis.

Chenkel said:

I thought ##\Delta x, \Delta y, \Delta z## is the location of an event in the unprimed frame and v is the velocity of the primed frame in the positive x direction of the unprimed axis.

The clock is at rest in the primed frame. Two tick-events have the same ##x'##-coordinate. The delta then is ##x_2' - x_1' = \Delta x' = 0##.

Then with the LT ...
##\Delta x' = \gamma (\Delta x - {v \over c}c\Delta t)##
... follows:
##\Delta x = v \Delta t##

##v=\Delta x / \Delta t##.

Here ##\Delta x## is the distance between the two tick-events in ##x##-direction and ##\Delta t## the coordinate-time interval between both events, both with reference to the unprimed frame.

Sagittarius A-Star said:

The clock is at rest in the primed frame. Two tick-events have the same ##x'##-coordinate. The delta then is ##x_2' - x_1' = \Delta x' = 0##.

Then with the LT ...
##\Delta x' = \gamma (\Delta x - {v \over c}c\Delta t)##
... follows:
##\Delta x = v \Delta t##

##v=\Delta x / \Delta t##.

Here ##\Delta x## is the distance between the two tick-events in ##x##-direction and ##\Delta t## the coordinate-time interval between both events, both with reference to the unprimed frame.

It's starting to make some sense to me, thanks!

Chenkel said:

It's starting to make some sense to me, thanks!

The next thing you may calculate by yourself with the inverse LT:

Assume a train moves with velocity ##v## with respect to the ground. In the train, a passenger walks with velocity ##u'## with respect to the train from the back end of the train to the front of the train. What is the velocity ##u## of the passenger with respect to the ground?

Call the rest-frame of the train the primed frame ##S'##.
Call the rest-frame of the ground the unprimed frame ##S##.
The train moves in positive ##x##-direction.

Inverse LT:
##\Delta x = \gamma (\Delta x' + v \Delta t')##
##\Delta t = \gamma (\Delta t' + {v \over c^2}\Delta x')##

Plug into the inverse LT:
##\Delta x' := u' \Delta t'##.

Calculate ##\Delta x / \Delta t## and simplify the term on the right side of the equation.

Compare ##u = \Delta x / \Delta t## as function of ##u'##, ##v## and ##c## with the Wikipedia formula for relativistic velocity composition for collinear motion.

Sagittarius A-Star said:

The next thing you may calculate by yourself with the inverse LT:

Assume a train moves with velocity ##v## with respect to the ground. In the train, a passenger walks with velocity ##u'## with respect to the train from the back end of the train to the front of the train. What is the velocity ##u## of the passenger with respect to the ground?

Call the rest-frame of the train the primed frame ##S'##.
Call the rest-frame of the ground the unprimed frame ##S##.
The train moves in positive ##x##-direction.

Inverse LT:
##\Delta x = \gamma (\Delta x' + v \Delta t')##
##\Delta t = \gamma (\Delta t' + {v \over c^2}\Delta x')##

Plug into the inverse LT:
##\Delta x' := u' \Delta t'##.

Calculate ##\Delta x / \Delta t## and simplify the term on the right side of the equation.

Compare ##u = \Delta x / \Delta t## as function of ##u'##, ##v## and ##c## with the Wikipedia formula for relativistic velocity composition for collinear motion.

##\Delta x = \gamma(\Delta x' + v \Delta t')##
##\Delta t = \gamma(\Delta t' + \frac v {c^2} \Delta x')##

##\Delta x = \gamma(u' \Delta t'+ v \Delta t')##
##\Delta t = \gamma(\Delta t' + \frac v {c^2} u'\Delta t')##.

##\frac {\Delta x}{\Delta t} = \frac {u'+v}{1 + \frac {v}{c^2}u'}##

Are these equations correct?

Chenkel said:

##\Delta x = \gamma(\Delta x' + v \Delta t')##
##\Delta t = \gamma(\Delta t' + \frac v {c^2} \Delta x')##

##\Delta x = \gamma(u' \Delta t'+ v \Delta t')##
##\Delta t = \gamma(\Delta t' + \frac v {c^2} u'\Delta t')##.

##\frac {\Delta x}{\Delta t} = \frac {u'+v}{1 + \frac {v}{c^2}u'}##

Are these equations correct?

So if the train is going .9 light seconds per second and the person is heading to the front of the train at .9 light seconds per second relative to the train(I know, super hero speed) then the speed of the person with respect to the ground is .9+.9/(1 + .9*.9) = .994 or 99.4 percent the speed of light with respect to the ground.

Chenkel said:

So if the train is going .9 light seconds per second and the person is heading to the front of the train at .9 light seconds per second relative to the train(I know, super hero speed) then the speed of the person with respect to the ground is .9+.9/(1 + .9*.9) = .994 or 99.4 percent the speed of light with respect to the ground.

Could the train be going so fast that moving to the front of the train could be impossible for a human because the amount of kinetic energy one would have to exert to move 1 meter toward the front of the train increases to the point where forward movement becomes impossible?

Chenkel said:

##\frac {\Delta x}{\Delta t} = \frac {u'+v}{1 + \frac {v}{c^2}u'}##

Are these equations correct?

Yes, that's correct. In Morin's book you can find this formula in the chapter "Longitudinal velocity addition", see pages XI-24 and XI-25 under the following link:
https://scholar.harvard.edu/files/david-morin/files/cmchap11.pdf
via:
https://scholar.harvard.edu/david-morin/classical-mechanics

Now you could plug into your last formula
##u' := c## and then simplify again the right side of the equation.

Then you will see, how the speed of light transforms from the primed frame to the unprimed frame.

Chenkel said:

Could the train be going so fast that moving to the front of the train could be impossible for a human because the amount of kinetic energy one would have to exert to move 1 meter toward the front of the train increases to the point where forward movement becomes impossible?

No. The passenger has the right to regard the train to be at rest. That's according to the principle of relativity. Then why should the very high velocity of the ground in ##(-x')##-direction change anything for him/her?

Sagittarius A-Star said:

No. The passenger has the right to regard the train to be at rest. That's according to the principle of relativity. Then why should the very high velocity of the ground in ##(-x')##-direction change anything for him/her?

But doesn't the mass of an object increase the faster it's traveling with respect to the ground because the relativistic mass equation?

Chenkel said:

I thought ##\Delta x, \Delta y, \Delta z## is the location of an event in the unprimed frame and v is the velocity of the primed frame in the positive x direction of the unprimed axis.

x, y, z are the spatial coordinates, using the unprimed frame, of an event (and likewise x', y', z' are the spatial coordinate of an event using the primed frame). That is a location.

##\Delta x##, ##\Delta y##, ##\Delta z## are the differences between the x, y, z coordinates of two events: for example, ##\Delta x=x_1-x_2## is the difference betweem the x coordinates of two events, one at ##x_1,y_1,z_1,t_1## and the other at ##x_2,y_2,_z2,t_2##
That is not a location, it is the difference between two locations.

Nugatory said:

x, y, z are the spatial coordinates, using the unprimed frame, of an event (and likewise x', y', z' are the spatial coordinate of an event using the primed frame). That is a location.

##\Delta x##, ##\Delta y##, ##\Delta z## are the differences between the x, y, z coordinates of two events: for example, ##\Delta x=x_1-x_2## is the difference betweem the x coordinates of two events, one at ##x_1,y_1,z_1,t_1## and the other at ##x_2,y_2,_z2,t_2##
That is not a location, it is the difference between two locations.

So the Lorentz transformation also works on Delta x Delta t Delta z and Delta time between two events.

Chenkel said:

But doesn't the mass of an object increase the faster it's traveling with respect to the ground because the relativistic mass equation?

Relativistic mass is not a very useful concept. Click here for an insight that explains why.

Special relativity is built upon an axiom that the laws of physics work the same regardless of what frame of reference you adopt. Relativistic mass (to the extent that it is a viable concept at all) rises from the math of special relativity. It does not contradict the postulates of the theory.

Regardless of how fast the train is moving relative to some inertial frame zipping to the rear at high speed, the train can be regarded as being at rest.

If you want to argue the opposite, you need to do the math. How much chemical energy is present in the muscles of the human who is about to run forward in the moving train? Please assess this using a frame of reference where the train is moving at 0.999 c. How much kinetic energy is present in a stationary human in the 0.999 c train? How much kinetic energy in a running human in the same train? Are those figures consistent with energy conservation?

Feel free to replace the human with ideal batteries, motors and toy cars if that will make things easier.

Chenkel said:

So the Lorentz transformation also works on Delta x Delta t Delta z and Delta time between two events.

You should be able to figure that out for yourself by writing things out explicitly.

We are almost 100 messages in.

@Chenkel , I see you often respond almost instantaneously to messages. You need to think about them, not just react.

I also see you are posting new threads on this same subject - do you think that will be helpful? That your problem is that you aren't working on enough things at once?

Finally, I see a lot of good advice that you appear not to be taking.

I don't think you are on a path to success. Perhaps you should take on a different strategy, like rereading what people have already written more slowly and carefully and see if that helps.

Chenkel said:

But doesn't the mass of an object increase the faster it's traveling with respect to the ground because the relativistic mass equation?

There were some discussions about relativistic mass in the past. Today most physicists agree to no longer use this terminology, but instead ##{E \over c^2}##, which is the same.

Maybe, the following insights article, which was also linked by @jbriggs444, answers also your question, especially the last paragraph:
https://www.physicsforums.com/insights/what-is-relativistic-mass-and-why-it-is-not-used-much/

Chenkel said:

So the Lorentz transformation also works on Delta x Delta t Delta z and Delta time between two events.

Yes. In addition to the good suggestion of @PeterDonis in posting #97 the following remark:

Also the LT without the Deltas can be interpreted as transforming coordinate-deltas, because of
##x = x-0##.

Then one of the two events is the common origin of the primed and unprimed frame: ##t=x=y=z=t'=x'=y'=z'=0##.

Source:
http://www.scholarpedia.org/article...nematics#Galilean_and_Lorentz_transformations

For the LT without Deltas it is required to use the "standard configuration" for the two 4D-coordinate systems with a common origin ##(0, 0, 0, 0)##.

This common origin is not required for a LT with explicit ##\Delta##'s in it, because i.e. different offsets on the ##x##-axis and ##x'##-axis would not change coordinate-differences.

Vanadium 50 said:

We are almost 100 messages in.

Just to note that @Sagittarius A-Star must have contributed almost as much material to this thread as Morin has in the whole of Chapter 11!

This is why textbooks are written and why students should study them. Our job should then be to elucidate anything that isn't clear. Rather than construct a textbook ad hoc post by post.

PeroK said:

Just to note that @Sagittarius A-Star must have contributed almost as much material to this thread as Morin has in the whole of Chapter 11!

This is why textbooks are written and why students should study them. Our job should then be to elucidate anything that isn't clear. Rather than construct a textbook ad hoc post by post.

I am indebted to all of the members of physics forums and I love the posts, equations and diagrams that @Sagittarius A-Star
contributed, I feel I'm understanding things much better and I will try to use text books more when I'm feeling curious instead of depending on the physics forums for understanding things.

Thank you all.