B Question about Morin's time dilation explanation

[Chenkel]

The deviation of the angle between the frames, you are talking about, is called aberration.

What you write was only valid in non-relativistic physics (Newton's theory of classical mechanics). Example:

Source:
https://en.wikisource.org/wiki/Tran...on_of_a_Light_Ray_from_its_Rectilinear_Motion

In special relativity, aberation works not only with classical particles, but also with EM waves. The reason is the relativity of simultaneity.

Assume, a horizontal electromagnetic wavefront of the vertical light crosses the horizontal x-axis. In the primed rest frame of the light clock, the left and right side of the wavefront cross the x'-axis simultaneously, that means
$\Delta t' = 0$.
With inverse LT and length contraction follows with reference to the unprimed frame, that the left and right side of the wavefront cross the x-axis with the following time difference:

$\Delta t = \gamma (\Delta t' + {v \over c^2} \Delta x') = \gamma (0 + {v \over c^2} \Delta x') = {v \over c^2}\Delta x$.

For small angles: When the right side of the wavefront crosses the x-axis, the left side has already moved further across the x-axis by

$\Delta d \approx c \Delta t = c {v \over c^2}\Delta x = {v \over c} \Delta x$.

That means the wavefront is tilted by

$\Delta d / \Delta x \approx v/c$.

In both frames, the Poynting vector $\vec S = \vec E \times \vec H$ is oriented perpendicular to the wavefront.

Thanks for the info, I will need to study those equations.

The full Lorentz transform I find a little confusing, I don't fully understand the part where $\frac {v} {c^2} \Delta x'$ the units seem to check out but I'm not sure how that quantity is associated with time.

 

[PeterDonis]

The full Lorentz transform I find a little confusing, I don't fully understand the part where $\frac {v} {c^2} \Delta x'$ the units seem to check out but I'm not sure how that quantity is associated with time.

That term is due to the relativity of simultaneity. That's how it is associated with time.

 

[Orodruin]

Thanks for the info, I will need to study those equations.

The full Lorentz transform I find a little confusing, I don't fully understand the part where $\frac {v} {c^2} \Delta x'$ the units seem to check out but I'm not sure how that quantity is associated with time.

It is simply how the Lorentz transformations work and mix space and time into spacetime and a necessity for keeping the speed of light invariant.

 

[Sagittarius A-Star]

Thanks for the info, I will need to study those equations.

The full Lorentz transform I find a little confusing, I don't fully understand the part where $\frac {v} {c^2} \Delta x'$ the units seem to check out but I'm not sure how that quantity is associated with time.

As others said, this is the (very important to understand) term for "relativity of simultaneity". In Morin's book you find it's derivation in chapter "1.3.1 Loss of simultaneity", paragraph "The 'rear clock ahead' effect".

In the light clock scenario, the left and right side of a wavefront reach a horizontal mirror simultaneously in the light clock's frame and not simultaneously in the lab frame.

If you use the complete Lorentz transformation, then all relevant effects are automatically covered.

• length contraction
• relativity of "same location"
• time dilation
• relativity of "same time"

$\require{color} x' = \color{blue}\gamma \color{black}(x\color{red}-vt\color{black})$
$t'= \color{green} \gamma \color{black}(t \color{orange}-\frac{v}{c^2}x\color{black})$

For comparison the non-relativistic Galilei transformation:
$x'=x-vt$
$t'=t$

 

[Chenkel]

Perhaps I should expand on why I put "make" in scare-quotes in my previous post just now.

You seem to be thinking that the light "wants" to move vertically in the frame in which the light clock is moving horizontally, so something has to "make" it go at an angle instead. That's not a good way to look at it.

The light's path in spacetime is invariant; that path is what is dictated by the physics of the problem. Looking at the light clock in different frames just means looking at the same invariant path in spacetime from different viewpoints. From the viewpoint of the frame in which the light clock is at rest, the spatial path of the light beam is vertical; from the viewpoint of the frame in which the light clock is moving horizontally, the spatial path of the light beam is at an angle. But both are viewpoints of the same path of the light beam in spacetime.

That means it is not possible for the light's path in the frame in which the light clock is moving horizontally to be anything other than what it is. Nothing needs to "make" it the way it is. It has to be the way it is, because that's what happens when you take the light's path in spacetime and project it into that particular frame. There is simply no room for it to be any other way. It's as if you looked at the same object from two different angles and then asked what "makes" it look different, as if changing your viewpoint meant something had to happen to the object. Nothing happens to the object when you change your viewpoint. And nothing happens to the light clock when you change frames.

So a light clock will still work in any inertial frame, but the path may look different depending on which inertial frame you're looking at it from.

 

[Ibix]

So a light clock will still work in any inertial frame, but the path may look different depending on which inertial frame you're looking at it from.

A light clock will work if it is moving inertially. Or, to say the same thing in different words, it will work if it is at rest in an inertial frame.

The path will be different depending on what frame (inertial or otherwise) you use to describe it.

 

[PeterDonis]

the path may look different depending on which inertial frame you're looking at it from.

The spatial path may look different. The spacetime path will be the same.

 

[Herman Trivilino]

So what is giving the light the horizontal movement in the lab?

The observer's motion relative to the lab.

If you're cruising in a jetliner and toss a ball upward so that it falls back down and lands in your hand, the ball moves up and down along a straight line. But imagine a distant observer on mountain top looking at you through a telescope. He sees the ball's trajectory to be a parabola. So what is the path of the ball? A straight line or a parabola? The answer is, "yes".

 

[Sagittarius A-Star]

The full Lorentz transform I find a little confusing, I don't fully understand the part where $\frac {v} {c^2} \Delta x'$ the units seem to check out but I'm not sure how that quantity is associated with time.

1. From the light clock scenario, derive time dilation factor $1/\gamma$ and calculate $\gamma=\frac{1}{\sqrt{1-v^2/c^2}}$ with the Pythagorean theorem.
2. Use this result and the standard twin paradox scenario to argue, that the length contraction factor must be $1/\gamma$.
3. Derive the Lorentz transformation from length contraction by using the following diagram.

Solve the shown equation for $x'$:
$$x'=\gamma(x-vt) \ \ \ \ \ \ \ \ \ \ (1)$$With a symmetry argument, you get the inverse transformation:
$$x=\gamma(x'+vt')\ \ \ \ \ \ \ \ \ \ (2)$$Eliminate $x'$ between the two previous equations and then solve for $t'$ to get the time transformation:
$$t'=\gamma(t-\frac{v}{c^2}x)\ \ \ \ \ \ \ \ \ \ (3)$$With a symmetry argument, you get the inverse transformation:
$$t=\gamma(t'+\frac{v}{c^2}x')\ \ \ \ \ \ \ \ \ \ (4)$$

Effective learning is possible by not only reading, but if you also solve problems by yourself.
It is good, that you started using LaTeX for writing formulas in posting #1 of this thread.

I propose, that you try to derive step-by-step the above equation (3) from equations (1) and (2) while using $\gamma=\frac{1}{\sqrt{1-v^2/c^2}}$.

 

[Chenkel]

1. From the light clock scenario, derive time dilation factor $1/\gamma$ and calculate $\gamma=\frac{1}{\sqrt{1-v^2/c^2}}$ with the Pythagorean theorem.
2. Use this result and the standard twin paradox scenario to argue, that the length contraction factor must be $1/\gamma$.
3. Derive the Lorentz transformation from length contraction by using the following diagram.

View attachment 337548​

Solve the shown equation for $x'$:
$$x'=\gamma(x-vt) \ \ \ \ \ \ \ \ \ \ (1)$$With a symmetry argument, you get the inverse transformation:
$$x=\gamma(x'+vt')\ \ \ \ \ \ \ \ \ \ (2)$$Eliminate $x'$ between the two previous equations and then solve for $t'$ to get the time transformation:
$$t'=\gamma(t-\frac{v}{c^2}x)\ \ \ \ \ \ \ \ \ \ (3)$$With a symmetry argument, you get the inverse transformation:
$$t=\gamma(t'+\frac{v}{c^2}x')\ \ \ \ \ \ \ \ \ \ (4)$$

Effective learning is possible by not only reading, but if you also solve problems by yourself.
It is good, that you started using LaTeX for writing formulas in posting #1 of this thread.

I propose, that you try to derive step-by-step the above equation (3) from equations (1) and (2) while using $\gamma=\frac{1}{\sqrt{1-v^2/c^2}}$.

I tried to get equation 3 from 1 and 2:

$x' = \gamma (x - vt)$

$x = \gamma (x' + vt')$

$x = \gamma ( \gamma (x - vt) + vt')$

$\frac x \gamma = \gamma (x - vt) + vt'$

$vt' = \frac x \gamma - \gamma (x - vt)$

Something tells me I might be on the wrong track with my equations but I told you I would try to solve the problem you proposed to me.

 

[PeterDonis]

I tried to get equation 3 from 1 and 2

Why did you stop in the middle? Keep going.

 

[Chenkel]

Why did you stop in the middle? Keep going.

I'll try again

 

[Chenkel]

I tried to get equation 3 from 1 and 2:

$x' = \gamma (x - vt)$

$x = \gamma (x' + vt')$

$x = \gamma ( \gamma (x - vt) + vt')$

$\frac x \gamma = \gamma (x - vt) + vt'$

$vt' = \frac x \gamma - \gamma (x - vt)$

Something tells me I might be on the wrong track with my equations but I told you I would try to solve the problem you proposed to me.

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

$vt' = x \sqrt{1 - \frac {v^2}{c^2} } - \frac {(x - vt)} {\sqrt{1 - \frac {v^2}{c^2} }}$

$vt' = \frac {x(1 - \frac {v^2}{c^2})} {\sqrt{1 - \frac {v^2}{c^2}}} - \frac {(x - vt)} {\sqrt{1 - \frac {v^2}{c^2} }}$

$vt' = \frac {vt - x\frac {v^2}{c^2}} {\sqrt{1 - \frac {v^2}{c^2}}}$

$vt' = \gamma ({vt - x\frac {v^2}{c^2}})$

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

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

 

[Chenkel]

1. From the light clock scenario, derive time dilation factor $1/\gamma$ and calculate $\gamma=\frac{1}{\sqrt{1-v^2/c^2}}$ with the Pythagorean theorem.
2. Use this result and the standard twin paradox scenario to argue, that the length contraction factor must be $1/\gamma$.
3. Derive the Lorentz transformation from length contraction by using the following diagram.

View attachment 337548​

Solve the shown equation for $x'$:
$$x'=\gamma(x-vt) \ \ \ \ \ \ \ \ \ \ (1)$$With a symmetry argument, you get the inverse transformation:
$$x=\gamma(x'+vt')\ \ \ \ \ \ \ \ \ \ (2)$$Eliminate $x'$ between the two previous equations and then solve for $t'$ to get the time transformation:
$$t'=\gamma(t-\frac{v}{c^2}x)\ \ \ \ \ \ \ \ \ \ (3)$$With a symmetry argument, you get the inverse transformation:
$$t=\gamma(t'+\frac{v}{c^2}x')\ \ \ \ \ \ \ \ \ \ (4)$$

Effective learning is possible by not only reading, but if you also solve problems by yourself.
It is good, that you started using LaTeX for writing formulas in posting #1 of this thread.

I propose, that you try to derive step-by-step the above equation (3) from equations (1) and (2) while using $\gamma=\frac{1}{\sqrt{1-v^2/c^2}}$.

I'm trying to understand the symmetry argument to obtain $x=\gamma(x'+vt')$

 

[PAllen]

I'm trying to understand the symmetry argument to obtain $x=\gamma(x'+vt')$

If the primed frame is moving v with respect to the unprimed frame, then the unprimed frame is moving -v with respect to the primed frame. If all inertial frames are equivalent, the transform from prime to unprimed must the the same as the other way, with -v as v.

 

[Sagittarius A-Star]

I'm trying to understand the symmetry argument to obtain $x=\gamma(x'+vt')$

See the very good explanation of @PAllen in posting #65.

For a visualization, I modified the scenario from posting #59. Now, a red rod is at rest with respect to the "moving" frame $S$ (and length-contracted with respect to frame $S'$).

If you solve the shown equation for $x$, then you get the equation you ask for.

 

[Chenkel]

See the very good explanation of @PAllen in posting #65.

For a visualization, I modified the scenario from posting #59. Now, a red rod is at rest with respect to the "moving" frame $S$ (and length-contracted with respect to frame $S'$).

If you solve the shown equation for $x$, then you get the equation you ask for.

View attachment 338416​

I noticed in the first picture the origin of the primed frame is vt relative to the unprimed frame, this made some sense to me.

In the second image you linked you show the origin of the unprimed frame relative to the primed frame is -vt' (minus v t primed)

If two frames are moving relative to each other then one frame will measure the origin of the "other" as vt, and if you switch which frames you are using you will measure the origin of the "other" as -vt, bur you have -vt' (minus v t primed)

Why would you use the unprimed time over the primed time or the primed time over the unprimed time?

If a primed frame is in relative motion to an unprimed frame do you use the primed time to describe the motion or the unprimed time?

Thanks in advance for any input.

 

[Sagittarius A-Star]

Why would you use the unprimed time over the primed time or the primed time over the unprimed time?

• If I define the unprimed frame as reference frame (i.e. posting #59), then I calculate the scenario with unprimed coordinates.
• If I define the primed frame as reference frame (i.e. posting #66), then I calculate the scenario with primed coordinates.

If a primed frame is in relative motion to an unprimed frame do you use the primed time to describe the motion or the unprimed time?

It depends on, which frame I defined as reference frame. See answer above.

 

[Ibix]

If a primed frame is in relative motion to an unprimed frame do you use the primed time to describe the motion or the unprimed time?

You use the quantities from the frame you chose to use. Typically, you will start with the frame where you and your clocks and rulers are at rest and transform to a frame where somebody else is at rest in order to deduce that other person's measurements.

Frames are always a matter of choice. You can use any you like for anything. There are often ones that are a smart choice (for example, if you are interested in the measurements of a particular clock or ruler then transforming to its frame will let you read its measurements from that frame's coordinates), but you can always choose to use a different frame if you prefer. Or if your professor tells you to.

 

[A.T.]

If a primed frame is in relative motion to an unprimed frame do you use the primed time to describe the motion or the unprimed time?

If a primed frame is in relative motion to an unprimed frame, then the unprimed frame is also in relative motion to the primed frame. It's symmetrical, so there is no way to pick the frame "to use" based on that alone.

 

[robphy]

If a primed frame is in relative motion to an unprimed frame do you use the primed time to describe the motion or the unprimed time?

Quoting myself from another thread you started,

I suggest first looking at the Euclidean analogue: https://en.wikipedia.org/wiki/Rotation_matrix

Many questions and confusions in special relativity
can be addressed by first asking what happens in
hopefully simpler and more familiar analogues in Euclidean geometry or Galilean physics.

IMHO, https://www.merriam-webster.com/dictionary/walk before one can run

 

[Herman Trivilino]

If a primed frame is in relative motion to an unprimed frame do you use the primed time to describe the motion or the unprimed time?

The entire point of creating the primed and unprimed frames is that they are in relative motion. You use both of them! Otherwise there would be no reason to create both of them.

 

[Sagittarius A-Star]

The full Lorentz transform I find a little confusing, I don't fully understand the part where $\frac {v} {c^2} \Delta x'$ the units seem to check out but I'm not sure how that quantity is associated with time.

It's the term for "relativity of simultaneity". Such a term appeared also in your calculation:

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

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

 

[Chenkel]

It's the term for "relativity of simultaneity". Such a term appeared also in your calculation:

I appreciate that you helped me get that result hopefully I see the big picture and things start making sense.

Thank you for the help.