Back
Forum Replies Created
-
The numbers seem to be not accurate. Gamma = 1/sqrt(1-0.994^2) = 9.14243. Dividing 5.9 km by gamma = 0.645343 km. To obtain the speed of the muon in km/s, we divide 0.645343 km by (2.2 * 10^-6 s) = 293,338 km/s. Speed of light = 299,792 km/s. Dividing these two numbers = 293,338 / 299,792 = 0.978472. This not equal to 0.994 that we started with.
The numbers seem to be not accurate. Gamma = 1/sqrt(1-0.994^2) = 9.14243. Dividing 5.9 km by gamma = 0.645343 km. To obtain the speed of the muon in km/s, we divide 0.645343 km by (2.2 * 10^-6 s) = 293,338 km/s. Speed of light = 299,792 km/s. Dividing these two numbers = 293,338 / 299,792 = 0.978472. This not equal to 0.994 that we start with.
The numbers seem to be not accurate. Gamma = 1/sqrt(1-0.994^2) = 9.14243. Dividing 5.9 km by gamma = 0.645343 km. To obtain the speed of the muon in km/s, we divide 0.645343 km by (2.2 * 10^-6 s) = 293,338 km/s. Speed of light = 299,792 km/s. Dividing these two numbers = 293,338 / 299,792 = 0.978472. This not equal to 0.994 that Prof. Green mentions.
The numbers seem to be not accurate. Gamma = 1/sqrt(1-0.994^2) = 9.14243. Dividing 5.9 km by gamma = 0.645343 km. Now, dividing 0.645343 km by (2.2 * 10^-6 s) = 293,338 km/s, which is the speed of the muons. Speed of light = 299,792 km/s. Dividing these two numbers = 293,338 / 299,792 = 0.978472. This not equal to the speed of 0.994 that Prof. Green mentions.
The numbers seem to be not accurate. Gamma = 1/sqrt(1-0.994^2) = 9.14243. Dividing 5.9 km by gamma = 0.645343 km. Now, dividing 0.645343 km by (2.2 * 10^-6 s) = 293,338 km/s. Speed of light = 299,792 km/s. Dividing these two numbers = 293,338 / 299,792 = 0.978472. This not equal to 0.994 that Prof. Green mentions.
The numbers seem to be not accurate. Gamma = 1/sqrt(1-0.994^2) = 9.14243. Dividing 5.9 km by gamma = 0.645343 km. Now, dividing 0.645343 km by (2.2 * 10^6 s) = 293,338 km/s. Speed of light = 299,792 km/s. Dividing these two numbers = 293,338 / 299,792 = 0.978472. This not equal to 0.994 that Prof. Green mentions.
We will not agree. While I see an isosceles triangle, George will see a triangle titled forward but with the same hight. The reason is as follows. We know that the triangle will shrink in the direction of motion (horizontally) but not in the vertical direction. Each of the two equal sides of the triangle consists of two components: vertical and horizontal. The horizontal component will shrink, but the vertical one won’t. Moreover, for the front side, the horizontal component will shrink more than the horizontal component of the rear side of the triangle. The base of the triangle will shrink horizontally. Hence, the result is that the triangle will be tilted forward with the same hight: the front side will be shorter and its angle with the base will be larger, while the rear side will be longer and its angle with the base will be smaller, and the base will be shorter.
If we define being “real” to be the ability to establish physical communications, then the answer is no. The reason is simple. There is no possibility of the alien to influence the past or the future of the concerned person. There is no way to establish physical communications. Hence, the supposed “now” is only “virtual” or hypothetical, not real, by definition.
There are actually two right-triangles in this lecture: (1) One in which the hypoteuse = D and the opposite = L (but the adjacent is not specified). (2) One in which the hypotenuse = c, the adjacent = c.cos(theta), and the opposite = c.sin(theta). Prof. Greene shifts from the first triangle to the second then back to the first without making these shifts clear. Not only this is confusing, it also involves additional steps of derivation that could have been avoided if there were only one triangle. Other writers show the derivation of gamma from only one triangle using only Pythagoras theorem.