I Question about full Lorentz transformation

[Chenkel]

Given the difficulty you are having understanding this stuff, you might want to consider the possibility that the way you’re imagining what’s going on is part of the problem….

“An observer at rest with respect to an IRF recording events” can only record the events that happen at one place, where the observer is: these are the events with x-coordinate equal to zero, forming a vertical line through the origin.

It is possible to think in terms of observers recording events, but you need more observers. Taylor and Wheeler (“Spacetime Physics”) take this approach: we imagine that the entire universe is full of observers stationed one meter apart in a grid, at rest relative to one another, and carrying synchronized clocks. Whenever something happens, the one at that spot makes a note on a slip of paper, something like “I am the observer at x=10,y=0,z=0. This spaceship passed right where I am when my clock read 12:55 PM”. After our thought experiment is done we gather all these slips of paper and use them to reconstruct what happened according to the frame in which they were all at rest. We’ll call this the unprimed frame.

Now suppose that we have another such flock of observers, also all at rest relative to one another and carrying synchronized clocks but all moving relative to our first group. We’ll call the frame in which they are at rest the primed frame.

The Lorentz transformations let me calculate, from all the notes written by the unprimed frame observers, what the various primed-frame observers will have written down. The inverse transformations go the other way, they let me calculate the unprimed-frame results from the primed-frame ones.

So if the frames are moving with a relative velocity of 86.6 percent the speed of light (gamma factor of 2) then an event at x=1 y=0, and z=0, and t=20 seconds will be recorded in the primed frame at t' = 2*(20 -.866*1) = 38.268 and x'= 2*(1-.866*20) = -32.64 light seconds

The event of recording an event at x'=-32.64, y' = 0, z' = 0, and t' = 38.268 seconds in the primed frame will happen in the unprimed frame at t=2(38.268 + .866*(-32.64)) = 20.00352
And x=2(-32.64 + .866*38.268) = 1.000176

We got roughly the same value for t and x as when we started so this makes some sense to me.

 

[Nugatory]

We got roughly the same value for t and x as when we started so this makes some sense to me.

yep, that’s how it should work (I haven’t checked your calculations though).

When you’re setting up examples, try choosing $v=\frac{3}{5}c$ or $\frac{4}{5}c$ so $\gamma=\frac{5}{4}$ or $\frac{5}{3}$ - the arithmetic will be a lot easier and you’ll find that the inverse comes out exact.

 

[Chenkel]

yep, that’s how it should work (I haven’t checked your calculations though).

When you’re setting up examples, try choosing $v=\frac{3}{5}c$ or $\frac{4}{5}c$ so $\gamma=\frac{5}{4}$ or $\frac{5}{3}$ - the arithmetic will be a lot easier and you’ll find that the inverse comes out exact.

I just plugged the Lorentz transformation into the inverse Lorentz transformation algebraicly and found that after simplifying x = x and t = t so the Lorentz transformation and the inverse Lorentz transformation must be true for an event in the primed frame and a corresponding event in the unprimed frame for both Lorentz transformation and inverse Lorentz transformation.

 

[Chenkel]

I was testing my intuition about space time intervals.

So if you have an event where a juggler is juggling balls at 1 meter round trip intervals and a ball returns to his hand every a second then the spacetime interval is 1 meter minus 1 second times the speed of light.

If the reference frame the juggler is juggling in is moving very fast relative to earth then the distance between balls leaving the hand and coming back will increase but the spacetime interval will be invariant across inertial reference frames, so the ball will travel a longer distance but the time between the ball leaving the hand and coming back will also increase when viewed relative to someone on earth to maintain invariance of space time intervals.

What do you guys think about this?

 

[Nugatory]

I just plugged the Lorentz transformation into the inverse Lorentz transformation algebraicly and found that after simplifying x = x and t = t so the Lorentz transformation and the inverse Lorentz transformation must be true for an event in the primed frame and a corresponding event in the unprimed frame for both Lorentz transformation and inverse Lorentz transformation.

You got it!

Now, if you want your next challenge, try deriving the time dilation formula $\Delta T' = \Delta T/\gamma$ for when a spaceship leaves earth travelling at speed $v$ - when one year has passed on earth, how much time has passed on the spaceship?
There are two relevant events:
1) Spaceship passes earth, x=0, t=0, x'=0, t'=0.
2) The spaceship is where it is after $\Delta T$ years have passed on earth: $x=v\Delta T$, $t=\Delta T$. This will require calculating the x' and t' coordinates of that event.

Get past that challenge and we'll have another one for you involving the symmetry of time dilation.

 

[Chenkel]

You got it!

Now, if you want your next challenge, try deriving the time dilation formula $\Delta T' = \Delta T/\gamma$ for when a spaceship leaves earth travelling at speed $v$ - when one year has passed on earth, how much time has passed on the spaceship?
There are two relevant events:
1) Spaceship passes earth, x=0, t=0, x'=0, t'=0.
2) The spaceship is where it is after $\Delta T$ years have passed on earth: $x=v\Delta T$, $t=\Delta T$. This will require calculating the x' and t' coordinates of that event.

Get past that challenge and we'll have another one for you involving the symmetry of time dilation.

I'll work on this one, thanks!

 

[robphy]

Now, if you want your next challenge

Bonus points: next, draw a spacetime diagram of the situation!

 

[Chenkel]

I was testing my intuition about space time intervals.

So if you have an event where a juggler is juggling balls at 1 meter round trip intervals and a ball returns to his hand every a second then the spacetime interval is 1 meter minus 1 second times the speed of light.

If the reference frame the juggler is juggling in is moving very fast relative to earth then the distance between balls leaving the hand and coming back will increase but the spacetime interval will be invariant across inertial reference frames, so the ball will travel a longer distance but the time between the ball leaving the hand and coming back will decrease when viewed relative to someone on earth.

What do you guys think about this?

Maybe I should have said the time interval between the juggling events relative to someone on earth increases but the light travel time between the events also increases to keep the spacetime interval invariant.

Because the distance between the ball leaving the hand and coming back increases the time interval must also increase because the spacetime interval is event distance minus light travel time in meters.

 

[Chenkel]

You got it!

Now, if you want your next challenge, try deriving the time dilation formula $\Delta T' = \Delta T/\gamma$ for when a spaceship leaves earth travelling at speed $v$ - when one year has passed on earth, how much time has passed on the spaceship?
There are two relevant events:
1) Spaceship passes earth, x=0, t=0, x'=0, t'=0.
2) The spaceship is where it is after $\Delta T$ years have passed on earth: $x=v\Delta T$, $t=\Delta T$. This will require calculating the x' and t' coordinates of that event.

Get past that challenge and we'll have another one for you involving the symmetry of time dilation.

These are my equations:

$t' = \gamma(t - vx/c^2)$

$x=vt$

So

$t' = \gamma{(t - \frac {v^2{t}} {c^2})}$

$t' = \gamma{t(1 - \frac {v^2} {c^2})}$

$t' = t\sqrt{(1 - \frac {v^2} {c^2})}$

$t' = \frac t \gamma$

 

[jbriggs444]

So if you have an event where a juggler is juggling balls at 1 meter round trip intervals and a ball returns to his hand every a second then the spacetime interval is 1 meter minus 1 second times the speed of light.

You have lost me. We are using an inertial frame where the juggler is moving at 1 meter per second and the ball takes 1 second to rise and fall back down?

So in our chosen frame, the displacement from toss event to catch event is 1 meter in space and 1 second in time?

Then the squared separation is ${(\Delta x)}^2 - {(c \Delta t)}^2$ which is approximately minus one squared light-second. Because it is negative, it is a time-like interval. Within measurement accuracy, we might simply say that the [unsquared] interval is "one second".

The key point being that you square before doing the subtraction.

If one repeated the calculation using a different frame, the full Lorentz transform could be used to obtain new coordinates and new coordinate deltas. The computed interval would again be approximately one second regardless of the chosen frame.

 

[Chenkel]

You have lost me. We are using an inertial frame where the juggler is moving at 1 meter per second and the ball takes 1 second to rise and fall back down?

So in our chosen frame, the displacement from toss event to catch event is 1 meter in space and 1 second in time?

Then the squared separation is ${(\Delta x)}^2 - {(c \Delta t)}^2$ which is approximately minus one squared light-second. Because it is negative, it is a time-like interval. Within measurement accuracy, we might simply say that the [unsquared] interval is "one second".

The key point being that you square before doing the subtraction.

If one repeated the calculation using a different frame, the full Lorentz transform could be used to obtain new coordinates and new coordinate deltas. The computed interval would again be approximately one second regardless of the chosen frame.

The juggler is in a spaceship and is at rest with respect to the spaceship and a ball leaves the hand and comes back to the hand in 1 meter roundtrip distance and 1 second of travel time so 1 second of light travel time relative to the rest frame of the spaceship. This means 1 - 300000000 * 1 second = -299999999 meters is the spacetime interval in meters. (I rounded the speed of light to 300000000 meters per second for simplicity.)

If the spaceship is moving really fast then the balls will go a longer distance relative to a rest frame on earth, but the spacetime interval will be the same (-299999999), so if we let D be the distance traveled by the ball relative to earth then $D - \Delta t * c = -299999999$ if D is 225000000 (the distance an object travels at 75 percent the speed of light in one second) then we have $225000000 - \Delta t * 300000000 = -299999999$ so $-300000000\Delta t = -524999999$ so $\Delta t = 524999999/300000000 = 1.75 seconds$

So an event on the spaceship of a ball leaving the hand and coming back will be 1 second on the spaceship but it will be 1.75 seconds for the ball to leave the hand and come back relative to someone on earth if the spaceship is going 75 percent the speed of light.

 

[jbriggs444]

The juggler is in a spaceship and is at rest with respect to the spaceship and a ball leaves the hand and comes back to the hand in 1 meter roundtrip distance and 1 second of travel time so 1 second of light travel time relative to the rest frame of the spaceship. This means 1 - 300000000 * 1 second = -299999999 meters is the spacetime interval in meters. (I rounded the speed of light to 300000000 meters per second for simplicity.)

Assuming that we are using the flat space of special relativity, the ball is traversing a parabolic free fall trajectory. A curved trajectory. Not a straight line trajectory. Not a pair of straight line trajectories.

The invariant length of this time-like trajectory will be the proper time recorded by a wrist-watch attached to the ball. One could obtain that time by integrating the space-time interval over small segments of the curved path.

Subtracting 1 meter from 300000000 meters strikes me as a nonsense calculation for this purpose. If we had a straight-line trajectory covering 1 meter of coordinate distance over 1 second of coordinate time then we could compute interval $t$ as:$$t = \frac {\sqrt{300000000^2-1^2}} {300000000} = 0.99999999999999999444444444444444 \text{ sec}$$

 

[Chenkel]

Assuming that we are using the flat space of special relativity, the ball is traversing a parabolic free fall trajectory. A curved trajectory. Not a straight line trajectory. Not a pair of straight line trajectories.

The invariant length of this time-like trajectory will be the proper time recorded by a wrist-watch attached to the ball. One could obtain that time by integrating the space-time interval over small segments of the curved path.

That's an interesting idea to attach a clock to the ball.

In my calculations I made some simplification, I also included some gravity on the spaceship (how can you juggle without gravity?)

 

[Ibix]

The juggler is in a spaceship and is at rest with respect to the spaceship and a ball leaves the hand and comes back to the hand in 1 meter roundtrip distance and 1 second of travel time so 1 second of light travel time relative to the rest frame of the spaceship. This means 1 - 300000000 * 1 second = -299999999 meters is the spacetime interval in meters. (I rounded the speed of light to 300000000 meters per second for simplicity.)

No, this isn't right. The ball started and ended in the same place, so $\Delta x=0$. Thus the spacetime interval between throw and catch is $\Delta s^2=\Delta x^2-c^2\Delta t^2=-c^2×1\ \mathrm{s^2}$. So you forgot the squares, and you have misapplied the formula anyway.

If you want to calculate the spacetime interval along the path followed by the ball (rather than the "straight line" distance calculated above) you can do so, but there's an integral involved.

 

[jbriggs444]

That's an interesting idea to attach a clock to the ball.

In my calculations I made some simplification, I also included some gravity on the spaceship (how can you juggle without gravity?)

I assumed a gravitational force and, hence, a parabolic trajectory.

 

[Chenkel]

No, this isn't right. The ball started and ended in the same place, so $\Delta x=0$. Thus the spacetime interval between throw and catch is $\Delta s^2=\Delta x^2-c^2\Delta t^2=-c^2×1\ \mathrm{s^2}$. So you forgot the squares, and you have misapplied the formula anyway.

If you want to calculate the spacetime interval along the path followed by the ball (rather than the "straight line" distance calculated above) you can do so, but there's an integral involved.

Thanks for pointing that out, I did forget the square.

Also interesting I would need an integral, I was trying to simplify the problem by approximating the path with a straight line.

 

[Ibix]

Thanks for pointing that out, I did forget the square.

Also interesting I would need an integral, I was trying to simplify the problem by approximating the path with a straight line.

But it's two straight lines, one upwards and one downwards. If you actually simplify it to a no gravity situation where you bounce a ball off the ceiling then (idealising the collision as instantaneous) then you don't need an integral - you can compute $\Delta s$ for each leg and add them up (the integral is just the limit of this process for curved paths, with infinitesimal straight lines segments). Note that you need to take a square root because you need to add the two $\Delta s$ values, not the $\Delta s^2$ values, so in this case you are better off working with the $\Delta s^2 =c^2\Delta t^2-\Delta x^2$ convention - otherwise you have to compute $\sqrt{|\Delta s^2|}$.

 

[Chenkel]

But it's two straight lines, one upwards and one downwards. If you actually simplify it to a no gravity situation where you bounce a ball off the ceiling then (idealising the collision as instantaneous) then you don't need an integral - you can compute $\Delta s$ for each leg and add them up (the integral is just the limit of this process for curved paths, with infinitesimal straight lines segments). Note that you need to take a square root because you need to add the two $\Delta s$ values, not the $\Delta s^2$ values, so in this case you are better off working with the $\Delta s^2 =c^2\Delta t^2-\Delta x^2$ convention - otherwise you have to compute $\sqrt{|\Delta s^2|}$.

I thought to get the spacetime interval you need to find the square of the distance between two events minus the square of distance light travels between the two events occurring.

 

[jbriggs444]

I thought to get the spacetime interval you need to find the square of the distance between two events minus the square of distance light travels between the two events occurring.

The "distance light travels" is not exactly relevant. The $c$ enters into the formula as a conversion factor between the chosen units for time and space. Light is not travelling anywhere.

You start with two events. You express them each in coordinate form using your chosen inertial frame. E.g. as $(x_1, y_1, z_1, t_1)$ and $(x_2, y_2, z_2, t_2)$.

You subtract the two coordinate tuples yielding $(\Delta x, \Delta y, \Delta z, \Delta t)$

You compute $${\Delta x}^2 + {\Delta y}^2 + {\Delta z}^2 - {(c \Delta t)}^2$$That is the "squared separation".

If the result is positive, you have a spacelike interval. Take the square root and you have the length of the interval in your chosen units of distance.

If the result is negative, you have a timelike interval. Invert it and take the square root. You have $c$ times the duration of the interval. Divide by $c$ and you can extract a value for time in your chosen units of time.

If the result is zero, you have a lightlike or "null" interval.

This assumes a normal coordinate system on a flat space-time with a (-+++) metric signature. You can save yourself some paper and ink if you use units where $c = 1$.

 

[Chenkel]

If the result is negative, you have a timelike interval. Invert it and take the square root. You have c times the duration of the interval. Divide by c and you can extract a value for time in your chosen units of time.

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?

 

[jbriggs444]

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].

 

[Chenkel]

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.

 

[jbriggs444]

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.

 

[vanhees71]

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
$$m \mathrm{d}_{\tau} x^{\mu} = q F^{\mu \nu} u_{\nu} = \frac{q}{c} F^{\mu \nu} \mathrm{d}_{\tau} x_{\nu}.$$

 

[PeterDonis]

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.

 

[Sagittarius A-Star]

"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

 

[Sagittarius A-Star]

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

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

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

 

[vanhees71]

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

 

[Sagittarius A-Star]

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: ${d}_{\tau}m \ne 0$.

 

[vanhees71]

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.

 

[Sagittarius A-Star]

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:
$\Delta \tau / \Delta t = \sqrt{1 - {{v}^2 \over c^2 }}$

 

[Chenkel]

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?$\frac {\Delta x^2 + \Delta y^2 + \Delta z^2} {c^2 \Delta t^2} = \frac {v^2} {c^2}$

 

[Sagittarius A-Star]

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 $\Delta y' = \Delta z' = 0$.

 

[Chenkel]

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.

 

[Sagittarius A-Star]

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.

 

[Chenkel]

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!

 

[Sagittarius A-Star]

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.

 

[Chenkel]

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]

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

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

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

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]

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?

 

[Sagittarius A-Star]

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

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.

 

[Sagittarius A-Star]

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?

 

[Chenkel]

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?

 

[Nugatory]

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.

 

[Chenkel]

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.

 

[jbriggs444]

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.

 

[PeterDonis]

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.

 

[Vanadium 50]

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.

 

[Sagittarius A-Star]

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/

 

[Sagittarius A-Star]

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.