Question about Morin's time dilation explanation

Hello everyone,

I'm reading Morins book which I like, and I feel I kind of understand the part on time dilation, however I'm a little confused by the geometry of the Pythagorian theorem when applied to velocities.

On the moving clock he shows the velocity of light on the diagonal it traces out is c, so far I understood this.

Then he says the tangential velocity is v and it makes a right triangle with c as the hypoteneus and v as the adjacent and where the vertical velocity component (opposite side of the triangle) is ##\sqrt {c^2 - v^2}##

The way I think about it is I imagine a 1 second interval, and the adjacent will trace (v * 1 second) meters, the hypoteneus will trace (c * 1 second) meters and the opposite will trace (##\sqrt {c^2 - v^2}## * 1 second) meters.

I'm wondering if my understanding is rigorous enough or maybe I need to get more rigorous?

Please let me know what you think, thank you!

I'd be wary of specifying a 1s interval - better to use some arbitrary ##\delta t##. Then you can say that in time ##\delta t## the light travels ##c\delta t##, the clock moves laterally ##v\delta t##, and if this is the time it takes the light to reach the far mirror in this frame then the length of the clock, ##L##, must satisfy ##(v\delta t)^2+L^2=(c\delta t)^2##. That seems to be what you've written, only with the possibly dangerous assumption that the clock ticks once per second in this frame replaced with the statement that it ticks once every ##\delta t##.

The ##\LaTeX## instruction for ##\times## is \times, by the way, which is a lot better to use than *, which can have other meanings in non-programming contexts.

Using the "one-second" trick is fair enough in this case. That shows that in one second in the ground frame, the light (being informal with units) has attained a height of ##\sqrt{c^2 - v^2} = c\sqrt{1 - v^2/c^2}##. Which is the height attained in the moving frame after ##\sqrt{1 - v^2/c^2}## seconds.

And that is, essentially, time dilation in a nutshell. The ground frame measured that it took the light one second to reach a height that represents only ##\sqrt{1 - v^2/c^2}## seconds in the moving frame.

PeroK said:

Using the "one-second" trick is fair enough in this case. That shows that in one second in the ground frame, the light (being informal with units) has attained a height of ##\sqrt{c^2 - v^2} = c\sqrt{1 - v^2/c^2}##. Which is the height attained in the moving frame after ##\sqrt{1 - v^2/c^2}## seconds.

And that is, essentially, time dilation in a nutshell. The ground frame measured that it took the light one second to reach a height that represents only ##\sqrt{1 - v^2/c^2}## seconds in the moving frame.

Sometimes I get a little confused by the term "moving frame" is the moving frame the frame that is moving or the frame where something is moving?

Chenkel said:

Sometimes I get a little confused by the term "moving frame" is the moving frame the frame that is moving or the frame where something is moving?

The moving frame is the frame moving with the light clock - the frame in which the light clock is at rest! That said, the term "moving frame" only makes sense where you have established a "ground frame". Generally, things move in all frames.

PeroK said:

The moving frame is the frame moving with the light clock - the frame in which the light clock is at rest! That said, the term "moving frame" only makes sense where you have established a "ground frame". Generally, things move in all frames.

So the ground frame is considered stationary where the clock moves within it?

Chenkel said:

Sometimes I get a little confused by the term "moving frame" is the moving frame the frame that is moving or the frame where something is moving?

Both. It is a frame moving relative to the one first specified and in the moving frame things that were at rest in the original frame are moving.

Then it is a pretty bad terminology. Since there is no absolute rest, all frames will be moving relative to some (most!) other frame.

Edit: A better terminology would be to use ”the clock’s rest frame” and ”frame where the clock moves at v”. One can introduce letters such as ##S## and ##S’## to denote the frames if one does not want to write out the above every time.

Orodruin said:

Both. It is a frame moving relative to the one first specified and in the moving frame things that were at rest in the original frame are moving.

Then it is a pretty bad terminology. Since there is no absolute rest, all frames will be moving relative to some (most!) other frame.

Edit: A better terminology would be to use ”the clock’s rest frame” and ”frame where the clock moves at v”. One can introduce letters such as ##S## and ##S’## to denote the frames if one does not want to write out the above every time.

So most frames are moving relative to a chosen rest frame

Chenkel said:

So most frames are moving relative to a chosen rest frame

If you limit yourself to frames related by boosts (ie, distrgard simple rotations of space), then all other frames are moving relative to any rest frame you choose.

I agree with @Orodruin - do not use "moving frame" and "stationary frame" if at all possible. It just leads to talking about things that are stationary in the moving frame or moving in the stationary frame, and this stuff is tricky enough without terminology that's asking for you to mix it up. (Things that are stationary in the moving frame are moving in the stationary frame, but not all things that are moving in the stationary frame are necessarily stationary in the moving frame - clear?!)

Lab frame is useful. Say something like "the clock moves at speed ##v## relative to the lab" and then refer to the "clock frame" (meaning the rest frame of the clock) and "lab frame" (meaning the rest frame of the lab) would be my suggestion.

Ibix said:

I agree with @Orodruin - do not use "moving frame" and "stationary frame" if at all possible. It just leads to talking about things that are stationary in the moving frame or moving in the stationary frame, and this stuff is tricky enough without terminology that's asking for you to mix it up. (Things that are stationary in the moving frame are moving in the stationary frame, but not all things that are moving in the stationary frame are necessarily stationary in the moving frame - clear?!)

Lab frame is useful. Say something like "the clock moves at speed ##v## relative to the lab" and then refer to the "clock frame" (meaning the rest frame of the clock) and "lab frame" (meaning the rest frame of the lab) would be my suggestion.

I'm going to try to label my frames from now on and describe them by what's at rest within them, or what is moving within them.

So in the book Morin says the light clock could be constructed in its rest frame with a laser.

I'm just curious how the diagonal speed of light is c considering the vertical laser setup.

I'm thinking that the slope of the line is something like ##\frac {\sqrt{c^2 - v^2}} {v}## but I'm thinking my analysis is a little naive and I'm not sure why.

If the clock is carried horizontally at a speed v then the light will travel up and horizontally so it will have a diagonal path and the trajectory of this path should travel at speed c according to special relativities postulate.

I'm thinking that I'm maybe jumping to some conclusions here and my analysis is maybe not rigorous enough.

Chenkel said:

I'm just curious how the diagonal speed of light is c considering the vertical laser setup.

That one is easy: It doesn’t. And it doesn’t need to.

With the laser pointing vertically in the clock rest frame and moving at velocity v in the lab frame, the light will come out diagonally. The laser does not need to be tilted.

Orodruin said:

That one is easy: It doesn’t. And it doesn’t need to.

With the laser pointing vertically in the clock rest frame and moving at velocity v in the lab frame, the light will come out diagonally. The laser does not need to be tilted.

So if light has a trajectory the speed of light in the trajectory is always c?

Yes, that is one of Einstein’s postulates for special relativity.

Orodruin said:

That one is easy: It doesn’t. And it doesn’t need to.

I'm just getting a little o.c.d here about my interpretation of what you're saying.

What "doesn’t" do what and why does it not need to do the thing?

The light moves diagonally in the lab frame but vertically in the clock rest frame, am I understanding things correctly?

Chenkel said:

The light moves diagonally in the lab frame but vertically in the clock rest frame, am I understanding things correctly?

Yes.

Chenkel said:

If the clock is carried horizontally at a speed v then the light will travel up and horizontally so it will have a diagonal path and the trajectory of this path should travel at speed c according to special relativities postulate.

Yes. And since it is traveling at ##c##, but covers a longer distance in the lab frame, the light clock will be time dilated in the lab frame--one "tick" of the light clock will correspond to more than one "tick" of time in the lab frame.

PeterDonis said:

Yes. And since it is traveling at ##c##, but covers a longer distance in the lab frame, the light clock will be time dilated in the lab frame--one "tick" of the light clock will correspond to more than one "tick" of time in the lab frame.

So if we only use clocks that tick once per second then in the rest frame the clock is ticking at 1 tick per second and gamma is 2 then in the lab frame two ticks will happen in the lab for every one tick that happens in the clock's rest frame.

Chenkel said:

So if we only use clocks that tick once per second then in the rest frame the clock is ticking at 1 tick per second and gamma is 2 then in the lab frame two ticks will happen in the lab for every one tick that happens in the clock's rest frame.

Yes.

Chenkel said:

The light moves diagonally in the lab frame but vertically in the clock rest frame, am I understanding things correctly?

Yes, and that’s not strange at all, it happens with ordinary objects moving at ordinary speeds with no relativity involved at all. Consider an airliner cruising at altitude, and a passenger sitting in their seat drops their pen. Using the frame in which the aircraft is at rest the pen falls vertically and its path makes a 90 degree angle with the floor. Using the frame in which the earth is at rest the pen follows a slanting path to hit the same spot on the floor, so its path isn’t a 90-degree angle with the floor.

So in the lab the photons are moving horizontally at v but at the same time they are moving vertically at ##\sqrt {c^2 - v^2}##

Chenkel said:

So in the lab the photons are moving horizontally at v but at the same time they are moving vertically at ##\sqrt {c^2 - v^2}##

If you want to separate out components of their velocity, yes.

By the phytageoren theorem $$c^{2}=c_{{\rm horiz}}^{2}+c_{{\rm vert}}^{2}$$ thus $$(1)\quad c_{{\rm vert}}=\sqrt{c^{2}-c_{{\rm horiz}}^{2}}$$
In the frame relative to which the light clock is at rest we have ##c_{{\rm horiz}}=0## and it follows from (1): $$c_{{\rm vert}}=c$$
In the frame relative to which the light clock is moving we have ##c_{{\rm horiz}}=v## and it follows from (1): $$c_{{\rm vert}}=\sqrt{c^{2}-v^{2}}$$

Just so I understand better the laser could be in a vertical tube where if light touches the tube a paint bomb can explode that covers the clock in paint.

During the entire experiment everyone can agree the clock is not covered in paint.

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

Is it that the photons are being emitted from a moving emitter and this makes a trajectory of photons on a diagonal because the photons move vertically and the emitter moves horizontally?

Chenkel said:

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

What gives you movement in a frame of reference in which you are moving?

PeroK said:

What gives you movement in a frame of reference in which you are moving?

I would say energy and force.

Chenkel said:

I would say energy and force.

If you provide force you would be accelerating.

Chenkel said:

I would say energy and force.

You move because the coordinates of your position change with time in that particular reference frame. Motion is not absolute.

PeroK said:

You move because the coordinates of your position change with time in that particular reference frame. Motion is not absolute.

So positioned changing with time = motion within the frame where your position is measured

Chenkel said:

So positioned changing with time = motion within the frame where your position is measured

Yes. The light's velocity has a horizontal component because we deliberately chose a reference frame to give it that horizontal component.

PeroK said:

Yes. The light's velocity has a horizontal component because we deliberately chose a reference frame to give it that horizontal component.

The way we gave it a horizontal component was by having a moving emitter.

It seems that we are saying that the horizontal component of the velocity of light is v based on the velocity of the emitter.

The emitter isn't doing anything to the photons it's just emitting them upward and because the emitter has a horizontal velocity the upward emitting turns into a diagonal trajectory of photons.

I guess with this info and I can kind of "get it" but it seems a little like the trajectory of light is incidental in the theory and that the trajectory doesn't seem like a continuous stream of light but some photons that happened to get in the right place at the right time.

Chenkel said:

It seems that we are saying that the horizontal component of the velocity of light is v based on the velocity of the emitter.

That will be true of anything, not just light. Read post #20 by @Nugatory and think about it carefully.

Chenkel said:

The way we gave it a horizontal component was by having a moving emitter.

It seems that we are saying that the horizontal component of the velocity of light is v based on the velocity of the emitter.

The emitter isn't doing anything to the photons it's just emitting them upward and because the emitter has a horizontal velocity the upward emitting turns into a diagonal trajectory of photons.

I guess with this info and I can kind of "get it" but it seems a little like the trajectory of light is incidental in the theory and that the trajectory doesn't seem like a continuous stream of light but some photons that happened to get in the right place at the right time.

In my experience, many students struggle with the idea of more than one reference frame. In this case, you have a source of light, emitting light vertically (in the rest frame of the source). And that's it. The physics must be done in that reference frame - all other frames are invalid in some way. That seems to be the logic. Or, perhaps, even if you allow an alternative reference frame, the source remains at rest at emits light vertically in that frame.

Being able to consider a scenario in more than one reference frame is useful in classical physics. In SR it not only becomes vital, but it's what the theory is all about!!

If you don't see these things, then I guess it's tough.

Why should all trajectories of light travel at c when we just formed the trajectory with some random photons?

The emitter emits photons with a vertical component of velocity, but their horizontal component of velocity doesn't seem intrinsic to the light beam or the photons but extrinsic to the emitter and how it places them on the trajectory based on the emitters movement.

Chenkel said:

Why should all trajectories of light travel at c when we just formed the trajectory with some random photons?

You should not be thinking of light as photons. Photons are a QM concept and this is the relativity forum, and you should not be trying to mix the two. For this discussion you should think of the light in the light clock as a short pulse of radiation that travels at ##c##.

Chenkel said:

their horizontal component of velocity doesn't seem intrinsic to the light beam

That's because it isn't. It's an artifact of your choice of reference frame. As @Nugatory explained in post #20, this is not limited to light; it's true of anything.