ok: Would vt_2 be equal to -vt? — Gampa Dee
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? :)
ps: Your name tells me that you might be a french speaking person? I am as well...my name is Andre
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
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
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
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
[...]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.
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
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. — Gampa Dee
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.
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 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
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