ok: Would vt_2 be equal to -vt? — Gampa Dee

No, since vt is the distance travelled by the mirror during the forward path while vt_2 is the distance travelled by the source during the return path, and the return path takes a shorter time. t_2 < t.

This is where it gets foggy for me; you seem to divide by a relative velocity (c+v) instead of adding another distance vt....I’m not saying that you’re wrong, only that I can’t visualize it . If v causes a foreshortening of L, then why not say T = L / (c – v) + L / ( c+v) ? but here, we would be in Newtonian mechanics wouldn’t we? :)

I followed the method you recommended and then applied an algebraic simplification to make the gamma factor explicit.

Using your method we get

1) d1 = L/gamma + vt = ct
2) d2 = L/gamma - vt_2 = ct_2

There was a typo in my previous post where I had placed a "+" sign rather than a "-" sign. Apologies for this. The rest of the calculation was correct but here it is again, more explicitly:

From the first equation, t = L/(gamma(c - v))
From the second equation, t_2 = L/(gamma(c + v))

Therefore, the total time

T = t+t_2 = L/(gamma(c - v)) + L/(gamma(c + v))

To simplify this, we can multiply the first term by (c+v)/(c+v) and the second term by (c-v)/(c-v), effectively multiplying each term by 1.

T = L(c + v)/(gamma(c - v)(c + v)) + L(c - v)/(gamma(c - v)(c + v))
= L((c + v) + (c - v))/gamma(c2 - v2) = 2Lc/gamma(c2 - v2)

Next, we can express gamma explicitly as 1/sqrt(1 - v^2/c^2), and divide both the numerator and denominator by c^2 to get

T = (2L/c)sqrt(1 - v^2/c^2)/(1 - v^2/c^2) = 2Lgamma/c

As long as all velocity terms are expressed relative to the same inertial frame of reference, the relative velocities between two material objects (or between a material object and a photon) can be expressed as simple sums or differences (e.g., c + v or c - v). This doesn't involve any Newtonian assumptions. When the light pulse travels at c and the clock at v, their relative velocity is c + v (or c - v). This is because it's the rate at which their separation changes, as measured in the stationary reference frame.

ps: Your name tells me that you might be a french speaking person? I am as well...my name is Andre

Yes, from the Montreal area, in Quebec, Canada. Nice meeting you, André.

ok: Would vt_2 be equal to -vt?
— Gampa Dee

No, since vt is the distance travelled by the mirror during the forward path while vt_2 is the distance travelled by the source during the return path, and the return path takes a shorter time. t_2 < t. — Pierre-Normand

Ok...Maybe this is where I’m off the rail.I was visualizing the light going to the mirror as traveling a distance vt further away from the length of the apparatus (distance between source and mirror)and a distance vt less than the apparatus when coming back towards the source being the reason why it took a shorter time.Maybe, somehow I will focus on this part.

From the first equation, t = L/(gamma(c - v))
From the second equation, t_2 = L/(gamma(c + v))

Therefore, the total time

T = t+t_2 = L/(gamma(c - v)) + L/(gamma(c + v))

To simplify this, we can multiply the first term by (c+v)/(c+v) and the second term by (c-v)/(c-v), effectively multiplying each term by 1.

T = L(c + v)/(gamma(c - v)(c + v)) + L(c - v)/(gamma(c - v)(c + v))
= L((c + v) + (c - v))/gamma(c2 - v2) = 2Lc/gamma(c2 - v2)

Next, we can express gamma explicitly as 1/sqrt(1 - v^2/c^2), and divide both the numerator and denominator by c^2 to get

T = (2L/c)sqrt(1 - v^2/c^2)/(1 - v^2/c^2) = 2Lgamma/c

As long as all velocity terms are expressed relative to the same inertial frame of reference, the relative velocities between two material objects (or between a material object and a photon) can be expressed as simple sums or differences (e.g., c + v or c - v). This doesn't involve any Newtonian assumptions. When the light pulse travels at c and the clock at v, their relative velocity is c + v (or c - v). This is because it's the rate at which their separation changes, as measured in the stationary reference frame. — Pierre-Normand

What I will do for now, Pierre-Normand, is try to understand on my own your derivation using (c+v)(c-v)method since it does indeed seem to make sense. Right now, the way I’m visualizing this, I am calculating the distances travelled by the moving apparatus vt and then claiming that the light is travelling with the velocity of "c" in between those distances (L+vt). The other thing I will focus on is my visualizing the return path as being vt shorter than the apparatus. If you think of some way to clarify any of this, I will
certainly look at it; if not, it’s ok; for this is simply a pastime of mine, I don’t feel the “need” to understand any of this.

ps: Your name tells me that you might be a french speaking person? I am as well...my name is Andre

Yes, from the Montreal area, in Quebec, Canada. Nice meeting you, André. — Pierre-Normand

Hello Pierre-Normand, I’m your next door neighbor from northern Ontario.
Take care....we’ll probably chat again sometime.
Grampa Dee
Andre

Ok...Maybe this is where I’m off the rail.I was visualizing the light going to the mirror as traveling a distance vt further away from the length of the apparatus (distance between source and mirror)and a distance vt less than the apparatus when coming back towards the source being the reason why it took a shorter time.Maybe, somehow I will focus on this part. — Gampa Dee

Your thinking was correct, but since both the forward and the return path are travelled by the light pulse at the same speed c (in any referential frame), and since the return path is shorter, the time to travel it is also shorter. This is why you must setup two distinct equations:

d1 = L/gamma + vt_1 = ct_1
d2 = L/gamma - vt_2 = ct_2

The first equation states that the distance the mirror has travelled away from the source (with a L/gamma head start) by the time the light pulse reaches it is the same as the distance travelled by the light pulse at that time; and similarly for the second equation. Each equation allows to solve for t_1 and t_2. The rest only involves algebraic manipulation.

[...]The other thing I will focus on is my visualizing the return path as being vt shorter than the apparatus. If you think of some way to clarify any of this, I will certainly look at it; if not, it’s ok; for this is simply a pastime of mine, I don’t feel the “need” to understand any of this.

Think of two people walking an unleashed dog at a steady pace v thereby keeping a constant distance L between them. Picture the dog running back and forth between them at constant speed v_2 > v. The forward path for the dog is longer than L since the person ahead keeps moving the 'goalpost' further away until the dog reaches them. Conversely, the person behind moves the 'goalpost' closer during the time when the dog returns.

I hope those further explanations help, Andre. Feel free to reach out if you have any further questions. Happy thinking!

Hello Pierre-Normand, nice to hear form you again

Your thinking was correct, but since both the forward and the return path are travelled by the light pulse at the same speed c (in any referential frame), and since the return path is shorter, the time to travel it is also shorter. This is why you must setup two distinct equations:

d1 = L/gamma + vt_1 = ct_1
d2 = L/gamma - vt_2 = ct_2 — Pierre-Normand

I don't think I have any problem with upper statement.

Think of two people walking an unleashed dog at a steady pace v thereby keeping a constant distance L between them. Picture the dog running back and forth between them at constant speed v_2 > v. The forward path for the dog is longer than L since the person ahead keeps moving the 'goalpost' further away until the dog reaches them. Conversely, the person behind moves the 'goalpost' closer during the time when the dog returns. — Pierre-Normand

I was able to visualize the situation using (c+v) (c-v) instead of the increase/decrease of length to the light path thanks to your example of the two people and the dog.However, I must have something missing here as well because I still arrive at the same conclusion. Maybe you will be able to spot my error.

If we have a moving platform with a velocity of v relative to us (moving from left to right),and let’s say that a camera is recording the event and what we see is the platform on a monitor as though it isn’t moving.The source is at the left and the mirror is located to the right of the platform.

Now we will observe a ray of light travelling from the source to the mirror as having a velocity of (c-v) towards the mirror. Afterwards, the light ray will travel in the opposite direction, back towards the source with a velocity of (c+v).
Here is where I might be wrong (if I’m not wrong already :) ), the average velocity the light ray will have is going to be c. However, we are also observing the platform as having been contracted by gamma^-1. This, in my mind, identifies a time contraction of gamma^1 instead of a time dilation of gamma, it seems.

I hope you don’t think that I’m merely trolling here. I sincerely have problems understanding s.r.
The example that you gave (people and dog) was very good and easy to understand, however, my problem stems directly in the invariant speed of light.

Thanks again for your time and patience..
Andre

I was able to visualize the situation using (c+v) (c-v) instead of the increase/decrease of length to the light path thanks to your example of the two people and the dog.However, I must have something missing here as well because I still arrive at the same conclusion. Maybe you will be able to spot my error.

If we have a moving platform with a velocity of v relative to us (moving from left to right),and let’s say that a camera is recording the event and what we see is the platform on a monitor as though it isn’t moving. The source is at the left and the mirror is located to the right of the platform. — Gampa Dee

In this case, you're using the camera to register the events as they occur relative to the moving reference frame (i.e., the inertial frame in which the light clock is stationary).

Now we will observe a ray of light travelling from the source to the mirror as having a velocity of (c-v) towards the mirror. Afterwards, the light ray will travel in the opposite direction, back towards the source with a velocity of (c+v).

This, along with the Lorentz contractions of the platform, are what you register in the inertial referential frame in which you are at rest, in which the platform is travelling at v, and in which the time registered by the light clock is (consequently) dilated.

Here is where I might be wrong (if I’m not wrong already :) ), the average velocity the light ray will have is going to be c. However, we are also observing the platform as having been contracted by gamma^-1. This, in my mind, identifies a time contraction of gamma^1 instead of a time dilation of gamma, it seems.

The velocity of all light rays always is c as measured from any referential frame. This is a postulate of the Special Theory of Relativity. The Lorentz contraction of the platform only is being registered in the 'stationary' referential frame. There is no Lorentz contraction of the platform observed in the reference frame in which the platform is at rest.

I hope you don’t think that I’m merely trolling here. I sincerely have problems understanding s.r.
The example that you gave (people and dog) was very good and easy to understand, however, my problem stems directly in the invariant speed of light.

Yes, it's precisely because the speed of light in a vacuum, c, is invariant, that the relative velocity between an object (such as a light clock) and a light ray is frame-dependent. When you measure events in a frame where the object is at rest, its velocity relative to all light rays is c. When you measure events in a frame where the object moves inertially at v, its relative velocity is c - v (or c + v) relative to a light ray moving in the same (or opposite) direction. And that's simply because after some time t has elapsed, the light ray has travelled a distance ct while the object has travelled a lesser distance vt. The distance between the object and the light ray therefore varies at the rate d/t = (ct - vt)/t = c - v (or (ct + vt)/t = c + v). This is what I call their relative velocity in a given referential frame.

I assure you, I never had the impression that you were trolling. Some of these issues are indeed subtle, and it's understandable to have questions. Keep asking - no worries at all!

Hello Pierre Normand; how are you? Thank you again for this interaction.


Pierre Normand wrote ....within our first exchange....I think :)

The time required for the light ray to reach the receding mirror is therefore t1 = d/v1, where d is the distance between the source and the mirror (d = L/gamma) and v1 is the relative velocity between the source and the mirror (v1 = c - v). Similarly, the time required for the light ray to return to the source is therefore t2 = d/v2 where v2 = c + v.

The total time elapsed is therefore t1 + t2 = L/(gamma(c - v)) + L/(gamma(c+v)).

I agree

Pierre Normand wrote:
To simplify, you can multiply the numerator and denominator of the first term by c + v and the numerator and denominator of the second term by c - v.

As you’re multiplying by 1, I fully agree.


Pierre Normand wrote:
You get
t = L(c + v)/(gamma(c^2-v^2))+L(c - v)/(gamma(c^2-v^2)) = 2Lc/(gamma(c^2-v^2))

ok...good

Pierre Normand wrote:
Since gamma = 1/sqrt(1-v^2/c^2), 2Lc/(gamma(c^2-v^2)) simplifies to 2L*gamma/c which is the same time registered by the vertical light clock.

Shouldn’t it be 2Lc*gamma instead?

Grampa Dee (Andre)

Pierre Normand wrote:
Since gamma = 1/sqrt(1-v^2/c^2), 2Lc/(gamma(c^2-v^2)) simplifies to 2L*gamma/c which is the same time registered by the vertical light clock.

Shouldn’t it be 2Lc*gamma instead? — Gampa Dee

That wouldn't work since this has the dimensions of squared meters per seconds and we want something that has the dimensions of seconds.

To arrive at the simplification, note that gamma can be rewritten as $\frac{1}{\sqrt{\frac{c^2}{c^2}-\frac{v^2}{c^2}}} = \frac{1}{\sqrt{\frac{(c^2-v^2)}{c^2}}}$ and therefore $\gamma^2$ is $\frac{c^2}{c^2-v^2}$

Our expression $\frac{2Lc}{\gamma(c^2-v^2)}$ can therefore be rewritten as $\frac{2L \cdot \gamma^2}{c \cdot \gamma}$, which is $\frac{2L\gamma}{c}$. As gamma is dimensionless, this expression indeed has the dimensions of seconds.

(Thanks to GPT-4 for help with the formatting of the mathematical expressions.)

That wouldn't work since this has the dimensions of squared meters per seconds and we want something that has the dimensions of seconds.

To arrive at the simplification, note that gamma can be rewritten as 1c2c2−v2c2√=1(c2−v2)c2√1c2c2−v2c2=1(c2−v2)c2 and therefore γ2γ2 is c2c2−v2c2c2−v2

Our expression 2Lcγ(c2−v2)2Lcγ(c2−v2) can therefore be rewritten as 2L⋅γ2c⋅γ2L⋅γ2c⋅γ, which is 2Lγc2Lγc. As gamma is dimensionless, this expression indeed has the dimensions of seconds. — Pierre-Normand

Thank you Pierre - Normand; it's now clear as a bell :)
I was impressed with you're mathematical juggling
and I understood everything you did.